INDEX
S. No
Topic
Week 1
Page No.
1
Magnetic Circuit and Transformer
1
2
Magnetising Current from B-H Curve
11
3
Ideal Transformer, Dot Convention and Phasor Diagram
20
4
Operation of Ideal Operation with Load Connected
30
5
39
6
Equivalent Circuit of Ideal Transformer
Rating of Single Phase Transformer Rated Current and Rated Voltage
with Example
7
Transformer with Multiple Coils
56
47
Week 2
8
Modelling of Practical Transformer - I
66
9
Modelling of Practical Transformer - II
79
10
Modelling of Practical Transformer - III
88
11
Core Loss - Eddy Current Loss
96
12
Factors on Eddy Current Loss Depends
106
13
Hysteresis Loss
117
14
Exact Equivalent Circuit
125
15
Approximate Equivalent Circuit
135
16
Determination of Equivalent Circuit Parameters - No Load Test
143
Week 3
17
Short Circuit Test
156
18
Choosing Sides to Carry Out O.C / S.C Test
165
19
Efficiency of Transformer - Losses
176
20
Efficiency (Contd.)
187
21
196
22
Condition for Maximum Efficiency When Load Power Factor Constant
Family of Efficiency Curve at Various Power Factor and Energy
Efficiency
23
Load Description and Energy Efficiency
216
24
Regulation its Expression
224
25
Regulation its Expression (Contd.)
232
205
Week 4
26
Auto Transformer - Introduction
244
27
AutoTransformer versus Two Winding Transformer
255
28
AutoTransformer versus Two Winding Transformer (Contd.)
264
29
Numerical Problems on Ideal Auto Transformer
274
30
Two Winding Transformer Connected as Auto Transformer
285
31
Practical Auto Transformer
294
32
Equivalent Circuit of an Auto Transformer
305
33
Polarity Test and Sumpner Test
314
Week 5
34
3 Phase Transformer Using 3 Single Phase Transformer
324
35
Various Connections of 3-Phase Transformer - I
333
36
Various Connections of 3-Phase Transformer - II
344
37
Vector Group of 3-Phase Transformer
354
38
Vector Group (Contd.)
362
39
Open Delta Connection
372
40
3-Phase Cone Type and Shell Type Transformer
382
41
Zig Zag Connection
393
42
Effect 3rd Harmonic Exciting Current and Flux
402
43
Choosing Transformer Connection
411
44
Choosing Transformer Connection (Contd.)
423
Week 6
45
Phase Conversion using Transformer Scott Connection
432
46
Scott Connection (Contd.)
443
47
3 Phase to 6 Phase Conversion O.C / S.C Test on 3 Phase Transformer
453
48
Parallel Operation of Transformers - I
464
49
Parallel Operation of Transformers - II
475
50
Parallel Operation of Transformers - III
484
51
Specific Magnetic and Electric Loadings
492
52
Cooling of Transformer and Fillings of Transformer
502
53
Output Equation of 3- Phase Transformer
514
Week 7
54
Introduction to D.C Machine
524
55
Single Conductor D.C Generator / Motor Operation
533
56
Homopolar D.C Generator
547
57
Homopolar D.C Motor
554
58
Introduction to Rotating D.C Machines
564
59
Armature Winding of D.C Machine - I
575
60
Armature Winding of D.C Machine - II
588
61
Armature Winding of D.C Machine - III
596
Week 8
62
Generated Voltage Across the Armature
604
63
Electromagnetic Troque in D.C Machine
615
64
Generator and Motor Operation - Basics
624
65
O.C.C and Load Characteristic of Separately Excited Generators
635
66
Voltage Build Up in Shunt Generator
643
67
Load Characteristic of Shunt Generator
651
68
Qualitative Discussion on Armature Reaction
661
69
Ill Effects of Armature Reaction
671
70
Compensating and Interpoles
680
Week 9
71
689
72
Armature Reaction (Contd.)
Field Flux Density, Armature Flux Density and Resultant Field
Distribution
73
Field Patterns for Both Motor and Generators
707
74
Demagnetising and Cross Magnetising mmf for Brush Shifted Machine
717
75
Calculation of Compensating, Interpole and Series Field Turns
726
76
Estimating Armature and Field Resistance from its Rating
734
77
Power Flow Diagram, Rotational Loss
742
78
Shunt Motor Basic Equation
752
698
Week 10
79
Starting of D.C Motor - 3- Point Starter
761
80
Speed Control of Shunt Motor - I
772
81
Speed Control of Shunt Motor - II
782
82
Speed Control of Shunt Motor - III
792
83
Field Control (Contd.)
801
84
D.C Motor Braking
811
85
Introduction to Series Motor
820
Week 11
86
Series Motor Characteristics
832
87
Series Motor Speed Control
842
88
Universal Motor
852
89
Swinburne Test
861
90
Hopkinson Test
870
Week 12
91
Efficiency Calculation
879
92
Field Test on D.C Series Motor
889
93
Simplex Wave winding
899
94
Wave Winding (Contd.)
909
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 01
Magnetic Circuit and Transformer
Welcome to this course on Electrical Machines – I, where we will discuss about
Transformers and DC machines. These are the two major topics. And we will begin with
transformers.
Transformers as we will see will generally consist of a magnetic core material made of
soft iron, over which there will be at least two coils wound. And one of the coil will be
energized with AC source of known frequency and in the other coil will get voltage of
same frequency, but at different levels. So, that is primarily the job of a transformer is, that
is suppose you have a 200 volt 50 Hertz supply, you require 400 volt 50 Hertz supply, then
I will use a transformer to change the level of the voltage from 200 to 400 volt keeping the
frequency constant.
It can be similarly a step down situation where you have a say 400 volt you want to step it
down to 100 volt level then use a transformer. Transformer is a static device its efficiency
is very large. In case of power transformer efficiency could be as large as 99 percent unlike
a rotating machine efficiency which may be 80 percent, 85 percent. Very good efficiency
because there is no rotating parts in it.
There will be of course, losses which accounts for that 1 percent or 2 percent loss in power,
that is there because of conductors will carry current it will have a 𝑖 2 𝑟 loss. And also core
we will see when it is having a time varying flux then also there will be heat loss inside
the core of the material. Anyway, those things we will discuss in detail.
1
(Refer Slide Time: 02:58)
But it looks like then we have to deal with this situations, that is suppose you have a
magnetic core of this kind ok. And it has got a thickness, like this in 3 dimension I am
trying to draw ok, and it will be like this. So, this is called it is a made of soft iron and I
have drawn it right now as a solid iron block. We will see what is to be done because solid
irons are not used.
But this is the structure of the iron and over which there will be coils wound like this ok.
When there is a single coil and there is a core like this, core material then if this coil carries
current it will produce flux inside the core. So, we start with magnetic circuit ok. We just
review that because that will be essential in understanding the.
Student: (Refer Time: 04:51).
Magnetic circuit because there is a magnetic material, there will be coils wound over it
and the coils are supposed to carry current therefore, they are going to produce flux in the
core. And we know that if this is the direction of the current, suppose DC current first then
the direction of the current will be like this in the coil, I. Suppose we have connected a DC
source this is the current. Then what happens, in the core there will be a flux produced 𝜑.
Direction of the flux will be as you know you wrap your fingers around the coils and thumb
will give you the direction of the flux produced.
2
Of course, if it is a constant DC current whatever flux will be produced inside the core that
too will have constant values; and how to estimate that flux? You know we apply the
ampere circuital law to find out the flux produced in the core. For example, you see if I
draw the sectional view of the core, like this, then if you draw the sectional view the
conductors can be shown to be like this, this is the coil sections. How many coils sections
you will see? As is the number of turns of this one. And the direction of the current as
shown can be shown by cross and dot like this. This will be cross current, cross and these
will be dot, is it not?
Then the flux inside the core what do we do is we take a mean path of the flux which I am
showing by dotted lines. This is the flux and this red one is the length of the mean path,
length of mean path of the flux that length let me call 𝑙. If these are 𝑁 then ampere circuital
laws says that
⃗⃗⃗ = 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑
⃗ . 𝑑𝑙
∮𝐻
⃗ and 𝑙 are same.
is it. But in this case the direction of 𝐻
⃗ . ⃗⃗⃗
⃗ and
Therefore, what will come out to be 𝐻
𝑑𝑙 that dot product because the direction of 𝐻
⃗ , direction of 𝐵
⃗ , direction of 𝜑 they are all same
𝑙 as I move they are same, direction of 𝐻
⃗⃗⃗ means 𝐻
⃗ . 𝑑𝑙
⃗ into 𝑙 should be equal to the current enclosed.
and also length is like that, 𝐻
How much current is enclosed by this path? 𝑁𝐼. We know this, so I will not spend much
of this time. So
𝐻=
𝑁𝐼
𝑙
⃗ then I will calculate
Its unit is ampere turns per meter this will be the case. Once I know 𝐻
⃗.
𝐵
⃗ = 𝜇0 𝜇 𝑟 𝐻
⃗ = 𝜇0 𝜇𝑟
𝐵
𝑁𝐼
𝑙
⃗
where 𝜇𝑟 is called the relative permeability of the core and if the relationship between 𝐵
⃗ is linear, then 𝜇𝑟 is a constant value and so this will be equal to 𝜇0 𝜇𝑟 and for 𝐻
⃗ I
and 𝐻
3
𝑁𝐼
can write 𝑙 . Then the then I go here, then the flux produced which is in weber will be
equal to:
𝜑 = 𝐵 × 𝐴 = 𝜇0 𝜇𝑟
𝑁𝐼
𝐴
𝑙
What is this 𝐴? 𝐴 is this cross sectional area that is this, this, this, this, through which flux
will be flowing, so, this is 𝐴 ok.
So, this cross sectional area is perpendicular to the flux at any sections therefore, flux in
weber will be 𝐵 × 𝐴 and 𝐵 already I have calculated, so I substitute that 𝜇0 𝜇𝑟
𝑁𝐼
𝑙
, this is
𝐻 into area. Now, if you see this can be written as 𝑁𝐼 I am sorry, if you see this it will be
1
𝑁𝐼 and bring all the other things in the denominator which will become 𝜇 𝜇
𝑙
0 𝑟𝐴
𝜑 = 𝐵 × 𝐴 = 𝜇0 𝜇𝑟
𝑁𝐼
𝐴=
𝑙
.
𝑁𝐼
1 𝑙
𝜇0 𝜇𝑟 𝐴
Why I have written in this fashion is this, that this is the cause I can identify this, this is
the cause and 𝜑 is the effect, is not? I have supplied with mmf and this magnetic circuit
returns you a flux 𝜑. And how to calculate that flux? Mmf, this is called mmf and you this
is called reluctance. And that is why we name magnetic circuit.
In case of electrical circuit,
𝐼=
𝐸𝑀𝐹 𝐸𝑀𝐹
=
𝜌𝑙
𝑅
𝐴
𝜌𝑙
and if you recall 𝐸𝑀𝐹 is 𝐸𝑀𝐹 voltage resistance is 𝐴 . So, there is a striking similarity and
that is why it is called reluctance. Reluctance limits the value of the flux when you apply
an mmf. Although, the mmf and the flux in electrical circuit as you know the circuit is like
𝐸
this resistance and this is your EMF say 𝑅. And this is the current, but in magnetic circuit
this mmf and this flux wherever it is flowing they are totally decoupled, I mean it is not
that flux is in the conductor.
4
But nonetheless this equation prompts us to simplify the matter and say that as so far as
calculation of mmf is concerned you do it like this 𝑁𝐼 and connect it by a reluctance which
is shown by a curly letter ℜ and here you show the flux. Although, in this diagram I show
they are as if connected, but they are not in practice, but only prompted by this relationship
corresponding in the existing in the electrical circuit, it is better, that is why it is called
magnetic circuit.
So, if mmf is known. I will first calculate, what is the reluctance? Reluctance of the
magnetic circuit path depends upon its geometry, what is the length, what is the cross
sectional area, cross sectional area is this one of this, this is the cross sectional area then
𝑁𝐼
you calculate 𝐻 = 𝑙 and then multiply it with 𝜇0 𝜇𝑟 to get 𝐵, then multiply with 𝐴 and get
that one. Therefore, in a simple excitation with DC current this value of the flux whatever
you will get, if you divide the mmf with reluctance that too will be constant and its
direction will be also fixed in this case it is clockwise ok.
The reverse problem is also very simple. In that case I will say that I want to create a
certain amount of flux what should be the current necessary. So, I will calculate reluctance,
reluctance into flux will give me 𝑁𝐼 and if I know the number of turns I will simply divide
that 𝑁𝐼 with 𝑁 to get the current necessary. Anyway, these are known stuff and we have a
magnetic circuit like this ok. And I will presume that you know about it, so no question of
further telling about it. Now, we will go to next page, next page.
Student: L that keyboard is (Refer Time: 16:21) useful.
Let us go to next page.
5
(Refer Slide Time: 16:42)
Now, I will tell you that what happens if the same magnetic circuit that is this one, this
point you listen very carefully and this is the most interesting thing, sorry, I am sorry. Let
me try to draw another thing here, that is the core I am drawing you must understand. Why
it is getting circle, anyway.
So, suppose this is the core of the material let me draw slowly ok. And suppose you have
the let us consider a single coil same magnetic circuit with 𝑁 turns, but this time what I
will do is this I will connect it to an AC source this is AC source of known voltage.
For example,
𝑣1 (𝑡) = √2𝑉1 cos 𝜔𝑡
𝑉1 is equal to √2 I am sorry will write it like this 𝑣1 (𝑡) is equal to √2𝑉1, some say sin 𝜔𝑡
or cos 𝜔𝑡 I will write it applied voltage to this coil. Now, if you pass a sinusoidal voltage
across the coil, in case of DC circuit magnetic circuit what I was telling when the current
is constant then 𝑁𝐼 you calculate mmf divide it by reluctance get the flux. But in AC circuit
AC magnetic circuit the coil is connected across a AC supply voltage of known frequency,
rms value 𝑉1 and supply frequency is 𝑓.
Then what I am telling, first I will tell this statement the flux in the core gets fixed I mean
no question of once the supply voltage and rms value is known, I mean we do not start
6
telling that I will first calculate current, then calculate mmf, then divide that mmf with
reluctance to get the flux, not like that; it will be the moment you apply a known rms
voltage at certain frequency 𝑓 where 𝜔 = 2𝜋𝑓 and AC voltage across the coil.
The level of flux which will be produced which will be also time varying it is expected
flux gets fixed. Now, what is the reason for that? Reason for that is it is expected whatever
current it draws that will be also sinusoidally varying, because after all it is some sort of
coil or inductance we have connected across an AC supply, some alternating current it will
draw and since current value and magnitude is changing with time the this 𝜑(𝑡) too will
be time dependent, sometimes it will be flowing to clockwise sometimes in the
anticlockwise direction, sometimes it will be 0 when the instantaneous value of the current
will be 0 and so on. But the moment an alternating flux is created inside the core of the
coil, inside the core of this magnetic circuit between these two points there appears an AC
induced voltage.
Suppose the number of turns of this coil is 𝑁 or say 𝑁1 single coil 𝑁1 . Then according to
Faraday if there is a coil if there is a time varying flux in the coil then this coil itself become
a seat of EMF. The value of instantaneous value of which is some −𝑁1
𝑑𝜑
𝑑𝑡
about that sign
it is not necessarily so important, but what I am telling this is the induced voltage the flux
linking the coil is changing with time and therefore, there will be induced voltage across
this coil.
In this case this flux is created by the current carried by the coil itself, but it does not
matter, Faraday says, if there is a change of flux rate of change of flux exists linking a coil
it is time varying then between these two points it will become a seat of EMF. Negative
sign is the Lenz’s law it tells that the polarity of this induced voltage will be such that it
will try to oppose the very cause for which it is due.
For example, this coil can be modeled as here is your AC supply 𝑣1 (𝑡) I have applied.
Then what I am telling across this coil if I neglect the resistance of the coil there appears
another source here and that induced voltage is 𝐸1 and the polarity of this voltage will be
𝑑𝜑
such that if this side is 𝑑𝑡 is positive then it will try to oppose the cause, so its polarity will
be like this ok. Or in other words what I am telling, so these thing can be now be modeled
like these applied voltage and there is another induced EMF.
7
In case of DC magnetic circuit DC 1 will not be there, there is applied voltage which is
𝑉
constant. Of course, current will be limited by the resistance of the circuit 𝑅 that is why
current gets fixed 𝐼. But here what I am telling the induced voltage in the coil will be same
as the applied voltage, only thing about the polarity its polarity will be such that it will try
to oppose the cause very cause that is the flux it will try to oppose it. What was the reason
for flux existing and increasing? This current it was increasing in the positive direction.
So, it will try to limit reduce that value of the current that was the reason.
Anyway, after this I can say that in this loop KVL equation is to be satisfied. Therefore,
applied rms voltage must be equal to the rms voltage of 𝐸1 what else, is not? Now, you see
this flux
𝜑 = 𝜑𝑚𝑎𝑥 sin 𝜔𝑡
will be some 𝜑𝑚𝑎𝑥 , say let us forget about this cos 𝜔𝑡 sin suppose I start with these one
𝜑𝑚𝑎𝑥 equal to sin 𝜔𝑡. Why I am assuming this? Because the circuit cannot, but draw
sinusoidally varying current let call 𝜑 = 𝜑𝑚𝑎𝑥 sin 𝜔𝑡. If that be the case then the rms value
of the then the induced voltage in the coil
𝑒1 = 𝑁1
𝑑𝜑1
= 𝑁1 𝜑𝑚𝑎𝑥 𝜔 cos 𝜔𝑡
𝑑𝑡
ok.
If you differentiate these this will become equal to 𝑁1 𝜑𝑚𝑎𝑥 𝜔 cos 𝜔𝑡. This will the value
of the induced voltage, is not? So, what will be the rms value of that voltage? It will be
𝐸1 =
𝑁1 𝜑𝑚𝑎𝑥 𝜔
√2
= √2𝜋𝑓𝜑𝑚𝑎𝑥 𝑁1
For 𝜔 if you substitute 2𝜋𝑓, so it will become √2𝜋𝑓𝜑𝑚𝑎𝑥 𝑁1 . See I am not so much
bothered about this −1, I should not struggle, my intention here is to calculate what is the
rms value of the induced voltage ok. Differentiate it, peak value you get divide it by √2
and that will give you rms voltage.
8
What I am telling is in the circuit KVL equation must be satisfied 𝑅 is vanishingly small.
So, applied rms voltage must be equal to the induced rms voltage it cannot be other than
that. Therefore, if the applied rms voltage is 𝑉1 it must be equal to
𝑉1 = √2𝜋𝑓𝜑𝑚𝑎𝑥 𝑁1
it has to be nothing other than that.
So, what I have told you that in case of AC same magnetic circuit which I considered with
DC excitation, if you connect it to an external voltage source which is alternating in nature
sinusoidally and the rms value of that applied voltage is known current whatever flows has
to be sinusoidal therefore, 𝜑 created inside the core of this magnetic material too will be
time dependent and vary sinusoidally. And if this flux vary sinusoidally then Faraday tells
us that across this two points; this two points there will be induced voltage.
What is the magnitude of this induced voltage rms value? That magnitude of this rms
voltage is √2𝜋𝑓𝜑𝑚𝑎𝑥 𝑁1 . Applied rms voltage is known 𝑉1, induced rms voltage is known
capital 𝐸1 and these two must be same because KVL is to be satisfied in the primary. If
that be the case in this equation what are the things I know, 𝑉1 is known, supply frequency
is known, 𝑁1 is known number of turns and your
𝜑𝑚𝑎𝑥 =
𝑉1
√2𝜋𝑓𝑁1
This is the crucial thing.
As I told you see mind you in case of alternating magnetic circuit if you apply an AC
voltage the level of flux 𝜑𝑚𝑎𝑥 is decided, decided by whom? By the supply voltage rms
value by the supply frequency and number of turns; so, is it not something different from
DC circuit; DC circuit you apply some known current, DC current in the circuit calculate
mmf divide by reluctance, get the flux. But here it is somewhat interesting.
The moment you connect an AC voltage 𝑉1 and 𝑓 I can tell you the flux is sinusoidally
varying. What is the maximum value of that flux? It is fixed it is this one. Therefore, this
𝜑𝑚𝑎𝑥 how much will be produced inside the core is decided. I do not have any control
over it. In fact, I will go a step ahead I will tell you. If this core material it has got a
9
permeability of 𝜇𝑟1 you replace this core material with another magnetic material having
relative permeability 𝜇𝑟2 then also 𝜑𝑚𝑎𝑥 is 𝑉1 by these one.
So, no matter what is the kind of magnetic material it is good magnetic material, back bad
magnetic material, the moment supply rms voltage and supply frequency are known 𝜑𝑚𝑎𝑥
in the core is decided. A very crucial point to go ahead with the concept of transformer.
We will continue with that.
Thank you.
10
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 02
Magnetizing Current from B-H Curve
Welcome to the second lecture on the Electrical Machine - I course, where we will be
primarily discussing about transformers and DC machines. So, we started in our first
lecture about transformers.
Transformer essentially is nothing but at least two coupled coils, wound on a common
magnetic circuit and one coil will be excited with alternating source and will get voltage
in the other coil. Details will discuss. But before that we told you in the last class that a
magnetic circuit essentially comprises of a magnetic core which is a ferro magnetic
material and it is wound by coil, may be by several coils. One of the coil is suppose
carrying current we discussed in our last lecture what will be the flux and what will be its
direction.
If you pass DC current, in case of DC magnetic circuit then we say it is a DC magnetic
circuit, you excite the coil with some constant current and what will be generated is a
constant amount of flux in the core. The amount of flux can be calculated from the equation
𝜑=
𝑁𝐼
ℜ (𝑅𝑒𝑙𝑢𝑐𝑡𝑎𝑛𝑐𝑒)
if it is a linear magnetic circuit. And also, the direction of the flux whether it will be
clockwise or anticlockwise can be figured out by knowing the direction of the current in
the coil.
So, in case of DC magnetic circuit you excite it with a DC current and there will be flux
produced and to calculate the flux there are two options you can use that formula that is
𝑁𝐼
ℜ (𝑅𝑒𝑙𝑢𝑐𝑡𝑎𝑛𝑐𝑒)
or alternatively if the B-H curve is available which also encompasses the
non-linearity case. Then you calculate first 𝑁𝐼, then calculate
𝑁𝐼
𝑙
that is 𝐻 from the
geometry of the magnetic circuit, you get the mean flux length and from the B-H curve
11
corresponding to that 𝐻 read what is 𝐵 produced and 𝐵 multiplied by the cross sectional
area of the magnetic circuit will give you the flux, that was in DC.
Also in DC if you want to create a definite amount of flux you can go in the reverse way
that is if the flux to be generated is this much then calculate first 𝐵 then from the B-H
𝑁𝐼
curve you get the value of 𝐻 corresponding to that 𝐵 and from that 𝐻 is nothing, but 𝑙 . I
will be able to calculate the mmf required.
(Refer Slide Time: 03:52)
In case of AC magnetic circuit that is what we were talking about the interesting thing is
you will excite the coil with a suppose this is the core of the magnetic circuit, this is the
core material and over which there is winding ok. Only one coil let us first understand.
And what I am telling is this coil I will excite it from a sinusoidally time varying voltage
say 𝑣1 (𝑡) which has got an rms value of 𝑉1 , say 𝑉1, 𝑉1 is the rms voltage. And suppose the
number of turns of the coil is 𝑁1 and this is the core of the material. See it is time varying
current and also, we neglect winding resistance which is suppose vanishingly small
winding resistance.
And also let us assume that all the flux; all the flux created is confined to the core only, is
confined to the core that is in other words that is no leakage flux. So, all the flux that will
be produced will be confined to the core. And there will be several lines I have just drawn.
I am so sorry, I forgot to do that.
12
Student: (Refer Time: 06:19).
So, thank you. So, this is the flux path and suppose let us also assume the instantaneous
current is 𝑖(𝑡) ok. So, with winding resistance is neglected without going thinking too
much about it I can only say this much look there is a coil, having no resistance and
therefore, this coil will perhaps act as an ideal inductor and therefore, it will draw some
current 90 degree lagging ok. The value of the current will be decided by the inductance
of the coil. That way one can think, but I will tell the same thing in slightly different way.
Suppose, it draws current and this current will be also sinusoidal. Why not it will be
sinusoidal because your supply voltage is sinusoidal there cannot be anything else. Now,
the moment this sinusoidal voltage is applied, the flux developed direction of the flux will
be also time varying.
When 𝐼 is 𝐼𝑚𝑎𝑥 , 𝜑 will be 𝜑𝑚𝑎𝑥 , when 𝐼 is 0 𝜑 will be 0, because current is expected to
be sinusoidal why not and therefore, it will be like this. But one should also remember in
this circuit this flux is time varying therefore, this coil will become, this coil means coil
with 𝑁1 turns will become a seat of EMF that is 𝑁1
𝑑𝜑
𝑑𝑡
because of Faraday’s law. And the
polarity of this induced EMF will be such that it will try to oppose the very cause for which
it is (Refer Time: 08:56).
Let us also tell about one thing about 𝑖(𝑡), let us consider this to be the instantaneous
𝑑𝑖
𝑑𝜑
direction of the current and not only that 𝑑𝑡 is positive that is 𝑑𝑡 is positive, it is increasing,
𝑑𝑖
ok. So, 𝑖(𝑡) is positive and also 𝑑𝑡 is positive. That is current is increasing and also having
𝑑𝜑
a positive slope ok. Therefore, 𝑑𝑡 in the direction shown is also positive that is what we
𝑑𝜑
are attributed to time ok; 𝑑𝑡 may be negative, but these will be taken care of by the
equation. But this is the two things I have attributed to 𝑖(𝑡) 𝑖(𝑡) is positive that is what I
𝑑𝑖
have assumed and not only that 𝑑𝑡 is also positive and that is done.
𝑑𝜑
Therefore, 𝑑𝑡 is also positive; that means, at this instant flux linkage with turns 𝑁1 is
𝑑𝜑
positive number it is increasing, 𝑑𝑡 is positive. And what is the cause of this induced
voltage in this coil 𝑁1 ? Because 𝑖(𝑡) is increasing, therefore, polarity of this induced EMF
will be such it will try to oppose this increase in current 𝑖(𝑡). That is why it will oppose
13
the supply voltage, mind you I have assumed supply voltage like this. So, this is the thing
a single coil excited by AC will have some induced voltage here 𝐸1 and this without
writing any equations I can tell instantaneous value of this induced voltage ok.
Therefore, applied voltage and induced voltage. And as I told you it is a magnetic circuit
only thing coil is excited by a voltage source AC voltage source. Now, in this case what
happens second argument was that ok, there is expected to be current is varying
𝑑𝜑
sinusoidally therefore, flux will also vary sinsusoidally, 𝑑𝑡 fellow will be also sinusoidally
it will be changing therefore, induced voltage 𝐸1 too has to be sinusoidal.
What will be the nature of this induced voltage in relation to 𝑉1? This too will be in same
time phase because KVL is to be satisfied only thing this sense in which 𝐸1 is acting is
opposite to this sense of 𝑉1 applied to this circuit that is if 𝐸1 was allowed to drive current
alone it would have driven current from right to left.
So, this is the Lenz’s law that is the induced voltage will oppose the supply voltage
essentially. But nonetheless this two voltages will have same with respect to time I if you
draw the phasor diagram I will draw this is 𝑉1 and this is also equal to 𝐸1 has to be, that is
both 𝑉1 and 𝐸1 are in phase. In books many of the books they will show 𝑉1 is like this, 𝐸1
is in opposition to indicate that 𝐸1 acts in opposition with 𝑉1, but I will not do that, because
I know time phasor means what with respect to time how 𝑉1 is changing and how 𝐸1 is
changing. When 𝑉1 will be maximum 𝐸1 too will be maximum and so on. So, they are in
time phase.
So, I will go by this way of looking at the things, because time phasor I know if there are
two quantities 𝑖1 (𝑡) and 𝑖2 (𝑡), some currents in two different branch and if there in time
phase there is when 𝑖1 attains maximum 𝑖2 also attains maximum they are in phase. So,
draw along same line why not. So, mind you this is the thing.
In this loop in this coil loop KVL is to be satisfied therefore, the rms value of the applied
voltage at rms value of the induced voltage they are to be same and has to be we have seen
that. Therefore, what is known to me? 𝑉1 is the rms value of the applied voltage. In my
last class I show the rms value of this induced voltage is nothing, but
𝐸1 = √2𝜋𝑓𝜑𝑚𝑎𝑥 𝑁1
14
What is 𝜑𝑚𝑎𝑥 ? 𝜑𝑚𝑎𝑥 is the sinusoidally varying flux peak value of that. So, this is the
equation Therefore, your this equation is true.
𝑉1 = √2𝜋𝑓𝜑𝑚𝑎𝑥 𝑁1
This 𝜑(𝑡) can be expressed in terms of say
𝜑(𝑡) = 𝜑𝑚𝑎𝑥 cos 𝜔𝑡
differentiate that last time we did that. So, the rms value of the applied voltage is equal to
𝑑𝜑
rms value of the induced voltage which I got from 𝑁1 𝑑𝑡 this thing is your 𝑒1 rms value
of that. How did you get? Put 𝜑(𝑡) = 𝜑𝑚𝑎𝑥 cos 𝜔𝑡 differentiate it and peak value of that
divided by √2. So, this is the thing. This equation implies that this is this a most important
statement that phi max is equal to
𝜑𝑚𝑎𝑥 =
𝑉1
√2𝜋𝑓𝑁1
In other words, what I am trying to tell, in case of DC circuit excite with DC current 𝑁𝐼 is
known then calculate flux, but in case of the same coil which is excited from an rms voltage
sinusoidally varying 𝑉1 the value of the peak value of the flux is fixed
𝑉1
√2𝜋𝑓𝑁1
. That is no
question of trying to calculate current then calculate mmf, then flux unlike DC circuit.
𝜑𝑚𝑎𝑥 gets decided, decided by whom? Decided by this two numbers, rms value of the
supply voltage and applied frequency. Of course, this is the peak value of the flux. How
flux is changing? Sinusoidally.
If you want to describe it as 𝜑𝑚𝑎𝑥 as a sine function then write it 𝜑𝑚𝑎𝑥 sin 𝜔𝑡 or
𝜑𝑚𝑎𝑥 cos 𝜔𝑡 whichever is convenient to you. And this flux is also sinusoidally varying
with the same supply frequency 𝑓. Therefore, to describe a sinusoidal function completely
what you need is its peak value and the frequency, so 𝜑𝑚𝑎𝑥 . So, we say that if a coil is
excited from a sinusoidal voltage source the value of 𝜑𝑚𝑎𝑥 gets fixed means the value of
flux gets fixed in the core of the machine ok. So, this is the thing one should remember
very carefully ok.
Now, let us go to a new page after learning these. So, in this a magnetic circuit once again
this is the thing, so we now know this is what is going to happen ok. Now, I will ask you
15
I will refer to back these once again re-drawing is not necessary. Now, what I am telling
is suppose I want to know and what is the another important implication of this equation,
that is if this core material is replaced by another core material other things remaining
same, that is same voltage you apply with same number of turns only thing replace these
magnetic material with another magnetic material, then also 𝜑𝑚𝑎𝑥 will remain same.
In other words, if 𝜇𝑟 changes change it by a different magnetic material 𝜑𝑚𝑎𝑥 you cannot
alter, it is solely decided by this 𝑉1 and the supply frequency. They say the last word about
the strength of the flux which will be created inside the core, nobody else definitely not
the relative permeability of the core material. Supply voltage and frequency is fixed 𝜑𝑚𝑎𝑥
is decided.
Then how to apply the magnetic circuit problem; then rather relevant question will be if
that be the case, I would like to know what will be the current drawn, suppose you connect
an ammeter in series with this line. And I ask you that, it is a magnetic material which has
got a B-H characteristics and I want to know what is 𝜑𝑚𝑎𝑥 . So, let us not make this page
too dirty. Let us go to new page.
(Refer Slide Time: 20:26)
So, if necessary let me draw once again. So, suppose once again the same coil with 𝑁1
turns. Now, I ask I will apply an AC voltage with rms value 𝑉1 and frequency 𝑓 to this two
points with this side plus this side minus and I will connect an ammeter here. I am asking
you what will be the reading of this ammeter? That is the question I am asking.
16
The answer to that question will be you have applied 𝑉1 𝑓1 then I will say this will be time
varying flux and the peak value of the flux is 𝜑𝑚𝑎𝑥 which is equal to
𝑉1
√2𝜋𝑓𝑁1
, this I know.
Then what I will do is these I will calculate
𝐵𝑚𝑎𝑥 =
𝜑𝑚𝑎𝑥
𝐴 (𝐶𝑟𝑜𝑠𝑠 𝑆𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝐶𝑜𝑟𝑒)
Suppose it is uniform constant, this is A, this is the cross sectional area. Then I will I have
to calculate the value of the current, I will ask what is the B-H characteristics of this
material, B-H characteristics of the material as you know it remains linear then it goes like
this. This is the B H curve of the material.
Then what I will do. So, what is 𝐼? That is the question asked. So, I will calculate 𝜑𝑚𝑎𝑥 ,
then 𝐵𝑚𝑎𝑥 and B-H characteristics is given to me, given. Then what I will do this 𝐵𝑚𝑎𝑥 I
will come here whatever it is 𝐵𝑚𝑎𝑥 and corresponding to this 𝐵𝑚𝑎𝑥 I will note down how
much 𝐻 is necessary that I will call the 𝐻𝑚𝑎𝑥 and this is nothing but
𝐻𝑚𝑎𝑥 =
𝑁𝐼𝑚𝑎𝑥
𝑙
What is 𝑙? 𝑙 is the mean length of the flux path which I am not drawing I have shown in
the previous. And from this I will say that
𝐼𝑚𝑎𝑥 =
𝑙𝐻𝑚𝑎𝑥
𝑁1
So, maximum value of the rms current is current drawn is known and then ammeter reading
if it is MI meter I will say
𝐼=
𝐼𝑚𝑎𝑥
√2
This is the thing.
So, this the now tells me that if a single coil is excited from an alternating voltage source
of frequency 𝑓 and this coil is wound on a magnetic material whose B-H characteristics is
like this. This is known to you; I hope from magnetic circuit analysis. This portion is
saturation zone it is initially linear things like that. But the moment you apply a known
17
voltage and frequency flux peak value of the flux gets fixed and I am asking you what is
the current drawn?
So, what I will do is this I will 𝜑𝑚𝑎𝑥 I will first calculate then I will calculate 𝐵𝑚𝑎𝑥 , if
𝐵𝑚𝑎𝑥 is known then I will calculate 𝐻𝑚𝑎𝑥 and from that I will be able to calculate 𝐼𝑚𝑎𝑥
that divided by √2 will be the rms value of the current. This you must understand.
Therefore, in AC magnetic circuit everything is sinusoidally varying time varying thing,
𝜑 is time varying, instantaneous quantity is 𝑖(𝑡) is time varying, only thing is 𝑖(𝑡) and
𝜑(𝑡) will be in same phase, when 𝑖 is maximum, 𝜑 is maximum and so on. But the rms
voltage induced in this coil 𝐸1 they will be in phase with this supply voltage. When 𝑉1 is
maximum 𝐸1 too will be maximum and so on ok.
Now, suppose this part if you understand then we really can discuss much more about an
ideal transformer, although I have not used the term ideal yet at least in today’s lecture.
Suppose, I say that this magnetic material and its B-H characteristics is this one. So, this
is suppose material 1, material 1 for which B-H characteristics is known.
Suppose, I say this material I will replace it by a newer magnetic material which is whose
B-H curve is like this. So, material 2, B-H curve is this. Then I apply same voltage and
frequency to this coil, then how much current this coil is going to draw. Then if this is the
material 2 now 𝜑𝑚𝑎𝑥 is fixed no matter whether it is material 1 or material 2.
So, same 𝐵𝑚𝑎𝑥 you read, but now you have to read the value of 𝐻𝑚𝑎𝑥 this is for material
1, this is for this is material 1. If it is material 2, I will say it will take it will draw current
this much only is not corresponding to these that is 𝐻𝑚𝑎𝑥 material 2, this green color is
𝐻𝑚𝑎𝑥 corresponding to material 2. And 𝐻𝑚𝑎𝑥 is lower therefore, current drawn will be
lesser. Therefore, you see this current which establishes the flux in the core I will use now
another term is called magnetizing current, usually denoted by instantaneous value by 𝑖𝑚
or rms value by 𝐼𝑚 , no maximum it is magnetizing current.
We see that if you go on replacing this core material by better and better material current
drawn ammeter reading will then become smaller and smaller, because another material if
you choose suppose somebody says, ok I will replace these by another material whose BH curve is like these material 3. So, this axis is 𝐻 or also 𝐼, let us forget about this constant
18
𝑁
𝑙
, this axis also represents the current drawn. Therefore, current drawn because 𝜑𝑚𝑎𝑥
remains same, so current drawn can be further reduced.
In other words, to establish a given flux if you use better and better material current needed
will be smaller and smaller. So, more efficient way of creating flux provided you use better
and better material. Suppose there is no restriction plenty of better materials are available,
suppose some material is like this very steep material 4, then current needed will be only
this much. Further, you can reduce the magnetizing current. What is this better material?
𝑑𝐵
Better material means this slope of 𝑑𝐻 is 𝜇𝑟 , it is a measure of 𝜇𝑟 . So, we are increasing
relative permeabilities for this different curves essentially I mean.
Therefore, suppose 𝜇𝑟 , is very high tending to infinity what does that mean, it means that
B-H curve will be something like this very high value of. Therefore, to create flux you
require very little current. Therefore, the magnetizing current can be reduced to any extent
provided you have got very good magnetic material. So, we will continue with this in our
next lecture.
Thank you.
19
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 03
Ideal Transformer, dot Convention and Phasor Diagram
Welcome to lecture 3. So, in my last class, I was talking about the magnetizing current
necessary to create a flux.
(Refer Slide Time: 00:31)
The flux is decided by the supply voltage and I will be writing many a times this. So, that
you also become accustomed with this one. So, this is the thing.
𝜑𝑚𝑎𝑥 =
𝑉1
√2𝜋𝑓𝑁1
So, 𝜑𝑚𝑎𝑥 gets fixed and then from this I will be able to calculate 𝐵𝑚𝑎𝑥 and from 𝐵𝑚𝑎𝑥 then
I will say, where is your B-H curve? What is the core material? B-H curve is like this
suppose then corresponding to 𝐵𝑚𝑎𝑥 , you read 𝐻𝑚𝑎𝑥 . This axis can be for a given value of
number of turns and magnetic length 𝑙, this can be also treated as 𝑖. 𝐻𝑚𝑎𝑥 or 𝐼𝑚𝑎𝑥 . Is not?
𝑁
I can always do that because 𝑙 come as a constant.
20
So, corresponding to this 𝐵𝑚𝑎𝑥 , you get 𝐻𝑚𝑎𝑥 , hence 𝐼𝑚𝑎𝑥 hence the current drawn from
this supply and the another important thing I told if you have better and better magnetic
material another material, then the current needed will be only this much not this much
have another better material current will be needed this much. So, essentially this 𝜇𝑟 is
increasing ok. Now, I will be telling you about the ideal transformer ok. Therefore, if the
magnetic material is very highly permeable, if 𝜇𝑟 tends to infinity very large that is this
curve will become almost like a vertical line, then to create a flux current needed will be
vanishingly small to establish the flux ok.
(Refer Slide Time: 02:50)
Therefore, we can say that, if this material 𝜇𝑟 tends to infinity what will be the ammeter
reading? You think a bit. Ammeter reading will be vanishingly small ok some finite current
is necessary, but that current you can make it as small as possible provided you do not
have any restriction on using better and better magnetic material.
Important thing is flux is finite like this, 𝜇𝑟 tending to infinity means that this ammeter
reading 𝑖𝑚 magnetizing current will tend to 0 that is all, it only means that. No point in
telling why 0 current how it can create flux ok, it takes a current no matter how good your
magnetic material is, it will definitely take a current, but that current is very small and will
see that, it will be very small compared to the rated current of a transformer.
21
So, like that therefore, in a ideal situation; very ideal situation will say that as 𝜇𝑟 attains to
infinity the current drawn that is 𝐼 and this is 𝐵. This magnetizing current drawn can be
made as small as possible. So, we are now in a position to talk about ideal transformer ok.
(Refer Slide Time: 04:37)
Now, for the first time I am writing this ideal transformer and if the concept of ideal
transformer is clear, you can deal with any situation you can deal with a practical
transformer very nicely, because the concepts are so interesting.
So, ideal transformer let us say that the first two conditions remains same winding or coil
resistance are 0, 2 no leakage flux, that is all the flux is confined to the core or all the flux
or all the flux is mutual flux of course, this mutual flux what it is I will tell write now. And
number 3 is magnetic core material is extremely good 𝜇𝑟 → ∞ I put a this mark also here
means vanishingly small current extremely good.
So, that magnetizing current necessary to create flux is practically 0. These are the 3
properties. Now, it is I will now draw the core of the transformer from my previous
diagram, which I have already drawn that is I come here and copy it and then please bear
with me till that time, because and is the best; this is the thing. So, this was the thing. Now
you see in this it is not a transformer a single coil. So, far I was discussing.
Now what I will do is this; I will draw another coil suppose here is another coil. So, on the
magnetic circuit, now 2 coils are connected and this two terminals of this second coil I
22
have not connected anything. Therefore, whatever I discussed in my previous lectures a
single coil excited with a voltage 𝑉1 from frequency 𝑓 remains intact I mean, because this
coil whether it is present or not no one bothers, because nothing have connected no current
in this coil.
Therefore, you and suppose this has got a 𝑁1 number of turns 𝑁2 is that clear? Therefore,
you have created these then flux is created mind you although you require vanishingly
small current, but 𝜑𝑚𝑎𝑥 is finite and its strength is this one. That is what? But current
needed magnetizing current, it will draw magnetizing current and this current is practically
0. If I assume the magnetic material is of very high quality having very large permeability
and all the flux here is confined to the core.
So, what is mutual flux that is what I have to tell you mutual flux is the flux which is
common to both 𝑁1 and 𝑁2 turns. So, mutual flux will be the flux, which is confined to the
core ok. Therefore, same flux will be linking both the primary coil, this I will call now
primary and in this coil no source is connected, this flux is also changing with respect to
time therefore, with respect to I mean applying Faraday’s law I will then also conclude
that these too will become a seat of emf.
So,
𝑒1 = 𝑁1
𝑑𝜑
𝑑𝑡
𝑒2 = 𝑁2
𝑑𝜑
𝑑𝑡
same flux. Rms value of the induced voltage, see polarity of that voltage I have found it
out without I mean thinking so, much about that negative sign. I have applied physical
reasoning and told that if supply voltage at any time is increasing, then flux is increasing
this terminal has to become plus and minus so far as 𝐸1 is concerned.
Similarly, and
𝐸1 = √2𝜋𝑓𝜑𝑚𝑎𝑥 𝑁1
23
which of course, happens to be equal to 𝑉1, because KVL is to be satisfied. Similarly, in
the second one; if I say induced voltage in the second coil 𝐸2 in which way it will be
different same flux only 𝑁2 comes in therefore, it will be
𝐸2 = √2𝜋𝑓𝜑𝑚𝑎𝑥 𝑁2
That is all. Is not I am not deriving once again differentiating and trying to write. This is
going to be what else. So, rms value of the induced voltage in the primary, which happens
to be equal to 𝑉1, because there is no resistance, no leakage flux, 𝑉1 = 𝐸1.
On the secondary side this will be this one. Now I should not of course, jump to the
conclusion that, this is plus, this is minus will see now that one. So, at a particular instant,
if this is plus this is minus, what will be the polarity of this induced voltage 𝐸2 here;
whether, this is plus this is minus or this is minus this is plus, that I have to decide ok. The
polarity of the induced voltage will be such that, it will try to oppose the very cause for
which it is due, I am repeating the same statement. Means what this flux about this flux,
what I told the flux is positive and it is increasing ok.
Now, there are only two possibilities, either this is plus this is minus or this is minus this
is plus, either of one of this is true correct, that way if you think. Suppose I say that, I am
not sure, suppose this has happened, when this is plus this is minus this has become plus
this is minus; suppose, let us assume it is like that. Now let us verify whether this
assumption is correct or not it has become a seat of emf.
Now you imagine that, this 𝐸2 I will allow it to act; 𝐸2 as it is open circuited nothing
happens no current etc, only thing it has become a seat of EMF and somebody says this is
plus this is minus. Now I am telling you that, if it 𝐸2 is allowed to act on a circuit, then it
will deliver current at that instant and what will be the direction of this current? If you
imagine that you have connected some loads; some resistance here I will connect it some
resistance here.
Then at that instant current supplied by the coil will be like this is not it will go this is the
source like a battery current will go like this and the direction of the current in the windings
will be like this. Now this is what in accordance with Lenz’s law the answer is no why?
24
Because the cause of the voltage induced is that 𝜑 was increasing is not? 𝜑 was positive
𝑑𝜑
and increasing 𝑑𝑡 was positive that was our assumption.
Now I find at that instant if current flows the flux in the core produced by this current will
be; flux produced by core in this coil will be also in this same direction; are you getting?
This was the flux created by 𝐼𝑚 , it was going up in this direction now also flux. So, the
flux is strengthened, the cause for which it is due is strengthened, but that is not in
accordance with Lenz’s law.
It will try to oppose the very cause for which it is due. What is the very cause? Phi from
top to bottom was increasing. So, the polarity of this induced voltage will be such that, if
this 𝐸2 is allowed to act, it should pass current through this winding in such a direction,
that it will try to oppose this flux that is what has to be. That is therefore, it looks like this
is not correct no. So, this cannot be like this, what is the other possibility? Other
possibilities is this has become plus, this has become minus.
(Refer Slide Time: 18:12)
Let us see, whether this will be consistent mind you this is the continuous thing I have
been only corrected no.
Student: Ok.
25
What is the other possibility? This is plus this is minus is it consistent? Let us see, if this
emf is allowed to act on its own to an external circuit like 𝑟 you have connected, then at
that instant it will try to send current in this way. Because this is battery it is the source,
current comes current goes current goes current goes like this current goes. So, this coil,
when it delivers current it will create flux in a direction opposite to these 𝜑(𝑡) which has
been created by this one.
So, it is trying to oppose the very cause for which it is due. Therefore, at a given instant of
time, if this terminal is plus this terminal is also plus. Let me repeat this point in a nicer
diagram; it is like this.
(Refer Slide Time: 19:50)
So, let me sketch it that will be faster. So, I sketch because let us spend some time on this;
sorry, I have not selected I will cut it out. See so, let us do it like this, suppose this is your.
Now I had got 2 coils let me repeat this point emphasize this point. This is suppose one
coil having 𝑁1 turns, this is suppose another coil with 𝑁2 turns, you have applied I will not
just simply say that you have applied 𝑉1 rms voltage of frequency 𝑓 with this is plus this
is minus.
𝐸1 has appeared here, it will have polarity like this here also I applied the Lenz’s law and
the flux in the core is 𝜑(𝑡). Then this two terminals also become a seat of emf. The value
of this voltage 𝑉2 is nothing, but 𝐸2 only 𝑉2 is the terminal voltage there is no distinction
between 𝐸2 and 𝑉2 right now.
26
So,
𝑉2 = 𝐸2 = √2𝜋𝑓𝜑𝑚𝑎𝑥 𝑁2
and also I know
𝑉1 = 𝐸1 = √2𝜋𝑓𝜑𝑚𝑎𝑥 𝑁1
and it is easy to show that
𝑉1 𝐸1 𝑁1
=
=
𝑉2 𝐸2 𝑁2
This is one famous formula ok. So, that is the thing, but I was discussing about the polarity
of this induced voltage instantaneous polarity.
If at any instant of time this is plus this terminal has to become plus no way; the other
possibility that at that instant this is plus this is minus will violate Lenz’s law and this ratio
𝑁
of this voltage is this one and from this many things can be told it is merely the ratio of 𝑁1
2
𝑁1
manipulating this ratio 𝑁 you can step up the voltage this is called secondary coil or step
2
down the voltage depending upon, whether 𝑁2 is greater than 𝑁1 or 𝑁1 is greater than 𝑁2
depending on that.
But anyway, this is the thing. Now people use you know some dot marking to communicate
this particular important aspect of a transformer by using known as dot convention. It
merely tells you that any given instant of time, if the polarity of this induced voltage is this
is plus, induced voltage in the other coil instantaneous polarity will be also this is plus and
this is done by dot convention.
Now I must also point out see, I have drawn the core material and drew the coils primary
coils and the secondary coils rather carefully. What do I mean by carefully; that you can
draw suppose you have a magnetic circuit like this as a you draw the coils like these. It is
for better than this, what I mean to say that this coils I have drawn. So, that you know what
is the sense of the winding, you take the coil this way then turn it like that, but from this
sense you cannot figure out, how this coils went somebody draws like this.
27
Then of course, you cannot point out if this is dot which one is dot this side. No out of
question you cannot do that. Perhaps by doing some experiments you will be able to do,
but as such on pen and paper you cannot predict ok, this is plus this will be plus that you
can only do provided you know the sense of the winding; sense of winding the coil. If that
is meticulously shown, then you can figure out. Before I tell some more interesting thing;
also you note that if these two are dots after some time; you know because it is after all
AC supply, it will be better if we say dot terminals are those terminals which have like
polarities of the induced voltage.
For example if you say this is plus this will be plus, you rub this off for the same
transformer I am talking this point you listen, suppose this dot I remove if somebody says
no this is dot. Then also he is correct are you getting like terminals, if it is minus that will
be minus whenever it will become plus because polarity reverses this will be also plus.
In other words, what I am telling no point in showing so many dotteds it is understood
that, wherever you have shown dots; good enough other two terminals are also like
terminals you can put some square brackets to indicate that. They are like terminals they
are like, whenever this is plus this will be plus, whenever this is plus this will be plus,
whenever this is minus this will be minus and so on like that.
So, we have we had discussing about ideal transformer and the ideal transformer is that
transformer, who which requires vanishingly small magnetizing current to create a finite
flux of strength
𝜑𝑚𝑎𝑥 =
𝑉1
√2𝜋𝑓𝑁1
current necessary, that is magnetizing current is vanishingly small 0.
Since, 𝜇𝑟 is infinitely large, that is the idea. Apart from the fact that all flux is confined to
the core there is no winding resistance, which allows me to write 𝑉1 = 𝐸1 = √2𝜋𝑓𝜑𝑚𝑎𝑥 𝑁1
and the last thing I will tell in this class, then if I draw the phasor diagram, what should I
draw this is the applied voltage is not?
Let me just draw it give you some idea, then the current drawn; magnetizing current drawn
although vanishingly small it will be lagging 90°. So, magnetizing current will be along
this line, because after all pure inductance it has got only inductance; inductance value is
28
very large if 𝜇𝑟 is tending to infinity we have shown wrote some expression for inductance,
if 𝜇𝑟 goes to infinity inductance goes to infinity, you can now interpret the things from
different angle inductance point of view very large 𝜇𝑟 , but the current drawn from the
supply will be 90° lagging and this length is pretty small 0, so, this is 𝑉1.
So, primary current is this where is 𝐸1 ? 𝐸1 will be also like this same as these one and
𝑑𝜑
where will be 𝐸2 same, because same 𝑁 𝑑𝑡 . So, 𝐸2 will be also like this, all voltages
induced voltages, applied voltages this is 𝑉1 they will be all in time phase and this is the
direction of 𝜑. Why 𝜑? Because 𝜑 is proportional to 𝐼𝑚 although 𝐼𝑚 is vanishingly small,
but it creates a finite flux. So, flux phasor will be along this line. We will continue with
this in the next lecture.
Thank you.
29
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 04
Operation of Ideal Operation with Load Connected
Welcome to this next lecture that is lecture 4 on Electrical Machines I and we have started
our discussion on transformer and we discussed several important concepts, which is
essential to understand the working principle of transformer.
(Refer Slide Time: 00:43)
For example now, in our last class I told several important things, one is that suppose there
are now 2 coils and we are discussing about ideal transformer. Ideal transformer that is
these winding resistances are 0, both the coils. There is no leakage flux and permeability
of the code is infinitely large, which means that if we energize this primary coil with a
known voltage at a frequency sinusoidal voltage; then, the current drawn is vanishingly
small. And if that with the case then, there will be 𝜑(𝑡), peak value of this 𝜑(𝑡) is 𝜑𝑚𝑎𝑥
and this 𝜑 being time varying also links this coil and you have induced voltage.
Then, I told you about the dot marking; these are called dot marking., how to find out the
dot markings? By applying Lenz’s law. So that everything is now, I mean in place, one
should not be thinking this way that way apply a given voltage at frequency RMS voltage
of the induced voltage and its polarity with respect to this plus minus all things are now
30
drawn polarity; instantaneous polarity plus minus. Then, the secondary instantaneous
polarity is also this one and 𝑉1 and 𝐸1 will be same, there is nothing in between. Similarly,
𝐸
𝑉2 and 𝐸2 will be same and also we noted that 𝐸2 rms value of then this voltage is:
1
𝐸2 𝑉2 𝑁2
=
=
𝐸1 𝑉1 𝑁1
This is the essential thing of a transformer; therefore, you can change the level of voltage.
Simply by manipulating the number of trans ratio ok.
Now, this is the thing and then, in my last class I told you how to draw phasor diagram.
You see, magnetizing current with the assumption that there is no leakage flux and mutual
flux only. So, this coil to these AC supply, this coil will appear as an inductance what else
𝑑𝜑
𝑑𝑖
because no resistance. See, essentially 𝑁1 𝑑𝑡 is nothing, but 𝐿 𝑑𝑡 from your circuit analysis.
You know, if time permits we will dwell upon that, but the point is so, if I want to draw
the phasor diagram I will do it like this; this is my applied voltage. Carefully see, the
current drawn by the circuit will be 0 will it be 0; not really vanishingly small current.
So that current will be lagging this by 90°, a small current, but that current produces the
finite flux 𝜑 getting. These are the phasors. This flux will induced voltage in the primary
coil and 𝐸1 has to be equal to 𝑉1; therefore, primary coil induced voltage will be like this
also equal to 𝐸1 . I told you in some books find 𝐸1 in the 90°, but I know the whole story
now, 𝐸1 only it takes in opposition with supply voltage, but with respect to time they are
in phase when, 𝑉1 attends maximum; 𝐸1 attends maximum. So, I will draw like that.
Similarly, 𝐸2 will be along the same lines its length may be different depending upon if
𝑁2 is greater than 𝑁1 . It will above and this angle will be 90°; 𝐼𝑚 magnetizing current is
vanishingly small. Now, in today’s lecture we will further go that is so far the secondary
circuit nothing is connected except while it deciding about dot convention I told something
you connect very casually I told ok; connect then, the current direction whether you allow
this e m f to act like that. But, now today we will see much more deeply what is going to
happen if you connect something here.
So, I will, what I am trying to tell? You imagine that there is some load which is still now
open circuited that is a switch. Our discussion till now with S opened whatever we
31
discussed with S opened. Now, the question is what happens if I close S; that is very
interesting; mind you with S opened let me write with S opened, open condition S open
condition; what is going to happen? Apply voltage frequency flux is created 𝐸1 𝐸2 , if you
connect a voltmeter you can measure all the voltages and these current if you connect an
ammeter is going to be very close to 0.
Since, it is an ideal transformer and so on. With S is opened, it is a magnetic circuit; single
coil and mmf acting in this magnetic circuit with S opened; mmf acting is equal to 𝑁1 into
this 𝐼𝑚 magnetizing current, which is very close to 0; no doubt, isn’t? That was the net
mmf which was acting and this mmf created a flux 𝜑. This created, this flux 𝜑 inside the
𝑁 𝐼
1 𝑚
core 𝑅𝑒𝑙𝑢𝑐𝑡𝑎𝑛𝑐𝑒
, isn’t? 𝑁1 𝐼𝑚 created the flux 𝜑 in the core. So, 𝑁1 into this flux; so, amount
of mmf necessary to create the flux phi is known to me.
𝑀𝑀𝐹 = 𝑁1 𝐼𝑚
Now, listen carefully what I am telling. You close the switch; imagine, you have closed
the switch. The moment you close the switch, this coil 2 will carry current; isn’t? This coil
2 because there is a source of emf we have connected an impedance 𝑍2 here. So, the
𝐸
magnitude of the current will be 𝑍2 and so on. Therefore, to find out, what is the flux in the
2
core of the transformer?
It looks like that this mmf; this mmf I have to take and then, net mmf I have to calculate
divide it by the reluctance that will decide the flux ok. In fact, that is what is going; we
have to do, but before that I tell you one thing that when, the S was open, what was the
flux? How to tell flux? You tell 𝜑𝑚𝑎𝑥 and frequency ok; 𝜑𝑚𝑎𝑥 was fixed. I am writing it
many a times
𝜑𝑚𝑎𝑥 =
𝑉1
√2𝜋𝑓𝑁1
that was 𝜑𝑚𝑎𝑥 ; it was fixed and this 𝜑𝑚𝑎𝑥 has to be there in order that KVL will be satisfied
on the primary loop.
Now, what I am telling? Whether the switch is opened or closed the flux in the core cannot
change that is what I am telling; that 𝜑𝑚𝑎𝑥 and its frequency of course, will remain same
flux must prevail; why? Because primary has to satisfy the KVL equation that is
32
𝑉1 = √2𝜋𝑓𝜑𝑚𝑎𝑥 𝑁1 = 4.44𝑓𝜑𝑚𝑎𝑥 𝑁1 = 𝐸1
𝑉1 is fixed. So, on the primary winding the KVL is to be satisfied.
Therefore, 𝜑𝑚𝑎𝑥 cannot change; no matter whether you have connected something on the
secondary coil or not. In other words, what I am telling? Even, if you have connected some
load on the secondary side. So that secondary coil is carrying current. Then, flux; this flux
cannot change. What it was? It will remain. Therefore, I will write it whether S is opened
or closed 𝜑 that is core flux will not change; cannot change. If earlier, it was 𝜑𝑚𝑎𝑥 sin 𝜔𝑡
it will be still 𝜑𝑚𝑎𝑥 sin 𝜔𝑡. S is opened or closed, it does not matter, but then, we are
slightly perplexed.
Second coil is carrying current. Primary was initially carrying a very small current. now,
now I will tell you that way you better you think. If flux remains same and these 2 coils
are carrying current now. So, what will be the net mmf necessary to create that original
flux? You must understand this 𝑁1 𝐼𝑚 when, S was opened created this flux in the core.
What I am telling?
With S closed when, both coils carry current; when, both coils when both coil carry current
with s closed net mmf must be once again 𝑁1 𝐼𝑚 because the 𝑁1 𝐼𝑚 created this original flux
and I am telling with S so closed or opened flux remains same; it cannot change. Why it
cannot change? Because, this flux will make the KVL equation valid on the primary side;
what was the mmf; which created this flux 𝜑, 𝑁1 𝐼𝑚 ? Where from we concluded that?
When S was opened?
So, with S closed; I am once again telling that flux remain same, but both the coils perhaps
will carry current, but I am sure the net mmf once again has to be 𝑁1 𝐼𝑚 , it cannot be other
than that because net mmf divided by reluctance; reluctance remain same. We will decide
the flux and flux remain same. So, let us see, what happens? What I mean by these. This
is the most interesting part of it.
33
(Refer Slide Time: 15:47)
So, this is the thing. So, here what I am telling. We are now discussing this a this thing 𝑍2 ;
this is my S and instead of telling that 𝐼𝑚 is 0; I will tell let magnetizing current, A is very
small finite. So, initially 𝑁1 𝐼𝑚 is the magnetizing current magnetizing mmf with S opened,
net mmf with S closed must be also 𝑁1 𝐼𝑚 when, both the coils carry current. Now, how
such a thing can happen? Let us see, suppose you now connect this switch, it has become
a seat of e m f. Therefore, it will deliver some current 𝑖2 ; RMS value of that current is
suppose 𝐼2 , instantaneous current.
Therefore, secondary coil will create and mmf 𝑁2 𝑖2 , instantaneous values or 𝑁2 𝐼2 RMS
value of the current. Primary prior to closing of the switch was carrying a current of 𝐼𝑚
magnetizing current. Now, when the secondary will carry a current 𝐼2 , it will try to create
flux; you will see this one. It will try to create flux in the opposite direction; isn’t? Your
thumb rule it try to creates flux in the opposite direction.
Therefore, when you close the secondary switch like this your primary cannot be a idle
spectator may a spectator to this event that you are doing this. What it will do?
Immediately, the moment you try to draw some current through the dot of the secondary
of the transformer; it will draw additional current say 𝑖2′ of such magnitudes that
𝑁1 𝑖2′ = 𝑁2 𝑖2
34
This 𝑖2′ is called reflected current. Therefore, I want to draw current from the secondary
of the transformer, I know. Then, it has produced 𝑁2 𝐼2 , which is acting in the anti
clockwise direction; it tries to create flux, but I am telling the moment you do that primary
draws extra current. So, what is the primary mmf
𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝑚𝑚𝑓 = 𝑁1 (𝑖𝑚 + 𝑖2′ ) = 𝑁1 𝑖𝑚 + 𝑁1 𝑖2;
this will be the total that is equal to 𝑁1 𝑖𝑚 which acts to create flux in this direction plus
𝑁1 𝑖2; from the dot current is coming out it also create flux in this direction clockwise
direction flux.
And, on this magnetic circuit apart from this 2 mmf say third fellow occurs, which is 𝑁2 𝑖2
or I will write plus create this way.
𝑀𝑀𝐹 = 𝑁1 𝑖𝑚 + 𝑁1 𝑖2; + 𝑁2 𝑖2
Therefore, in the earlier case when S was opened net mmf acting in the magnetic circuit
was 𝑁1 𝑖𝑚 ; when you close the switch, what is the net mmf acting in the circuit?
Let there be 𝑁1 𝑖𝑚 and this two will cancel out because 𝑁2 𝑖2 creates flux in this one and
𝑁1 𝑖2; create flux in this one and if this two are equal they will cancel out.
𝑁1 𝑖2; = 𝑁2 𝑖2
So, net mmf how much it is acting now? Once again, 𝑁1 𝑖𝑚 ; this is the most crucial point.
And this, the transformer has to do for all the time. That is why; it is true for instantaneous
values also.
So, what is the thing going on? Initially, the net mmf in the transformer was 𝑁1 𝑖𝑚 ; I am
telling with S closed. Once again, the net mmf has to be 𝑁1 𝑖𝑚 . Therefore, if secondary
delivers a current of mmf 𝑁2 𝑖2 ; if it delivers a current 𝑖2 hits mmf. Therefore, this mmf it
will be compensated by the primary by drawing additional current 𝑖2; over an above 𝑖𝑚
such that not any value of 𝑖2; of this value 𝑁1 𝑖2; = 𝑁2 𝑖2
So that this two mmf s once again balance and net mmf will be once again 𝑁1 𝑖𝑚 . That is
why people say the moment you want to draw current or power out of the secondary
terminals of a transformer. Primary cannot be a may are spectator to this event. The
35
moment you do this primary will react. It will draw additional current such that 𝑁1 through
the dot terminals mind you through the dot if you show the current going out; then, only
this two fluxes will mmf.
Therefore, net mmf once again will be 𝑁1 𝑖𝑚 ; these two will cancel out because from
physical directions I know. So, this will be the thing. Later, I will a tell you about this
whole lot of physic physical way of understand in the thing; there is easier way of
understanding that a look here if you draw power output from the secondary of the
transformer. Primary current must increase; how it can? Because this is the only two points
where I am pumping power into the system and you are taking power out of the system
between these two.
Therefore, if you consume power here; ultimately, power has to come from this place;
from this place here. Therefore, this power must balance. We will discuss about those
things later, but I hope you have understood. So, what I am trying to tell because this is so
important a point in understanding the transformer.
(Refer Slide Time: 24:51)
So, what I told? Let me review this because this is so important. There I connected a switch
and a load; isn’t? I am trying to understand what is going to happen when the switch is
closed because after all you cannot just step up or step down the voltage without
connecting not connecting anything across this secondary; we have to deliver this power
to some load.
36
So, this is the thing. So, initially with S opened; S opened I am concluding that S opened.
Net mmf acting is 𝑁1 𝑖𝑚 . With S closed; once again, I am telling. In this case, this was also
net mmf; secondary current was 0 net mmf was this. With S closed net mmf 𝑁1 𝑖𝑚 must be
net mmf.
Why? Because flux level cannot change, but mmf is contributed by this coil current and it
is number of turns; this coil current and this number of turns. So, net mmf of the primary
coil is
𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝑚𝑚𝑓 = 𝑁1 (𝑖𝑚 + 𝑖2′ ) = 𝑁1 𝑖𝑚 + 𝑁1 𝑖2;
and this is in the clockwise direction flux it creates. Net mmf of the I mean mmf
contributed by secondary, if u show this is the current; it is 𝑁2 𝑖2 and it creates flux in the
opposite direction. Therefore, this two mmf s
𝑁1 𝑖2; = 𝑁2 𝑖2
if this two things are equal for all the time mind you 𝑖2 is wearing sinusoidally.
Then, everything is fine; this two will cancel out. So, if I; if I draw the magnetic circuit
like this suppose magnetic equivalent circuit magnetic circuit. It will be; there are two
mmfs plus minus 𝑁1 𝑖𝑚 and here also plus minus 𝑁1 𝑖2; . Here is the reluctance of the
magnetic circuit and here is another mmf acting in the opposite direction; what is that?
𝑁2 𝑖2 and this is 𝑖2; .
What is the net mmf acting?
𝑁𝑒𝑡 𝑀𝑀𝐹 = 𝑁1 𝑖𝑚 + 𝑁1 𝑖2; − 𝑁2 𝑖2
and this has to be equal to 𝑁1 𝑖𝑚 . How this can happen? If this were equal; this must vanish
to 0, that is why the moment you want to draw some current 𝑖2 primary will immediately
react and draw additional current 𝑖2; on top of 𝑖𝑚 which was vanishingly small that is fine.
So, it will draw a current of 𝑖2; .
So, please go through these particular topic discus among yourselves with your friends;
the arguments put here to conclude that no matter whether the switch is opened or not flux
level in the core of the transformer is remains fixed and it is decided by only 𝑉1 and 𝑓 no
37
matter whether this is closed this is opened. Because, the moment S is closed it will draw
current, but flux level in the core has to remain same because it has to satisfy. It has a duty
to satisfy the KVL equation on the primary loop. You cannot have circuit KVL is not
satisfied.
Therefore, it must be 𝐸1 and 𝐸1 = √2𝜋𝑓𝜑𝑚𝑎𝑥 𝑁1. So, 𝜑𝑚𝑎𝑥 cannot change. The moment
you close; then, what happens? Initial mmf was 𝑁1 𝑖𝑚 with S opened. Final net mmf has to
be 𝑁1 𝑖𝑚 ; if these two are equal; these and these are equal, these two mmfs will cancel each
other and flux remains
𝑁1 𝑖𝑚
𝑅𝑒𝑙𝑢𝑐𝑡𝑎𝑛𝑐𝑒
that was the idea. Please, go through it; hope you have
understood.
Thank you.
38
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 05
Equivalent Circuit of Ideal Transformer
Welcome to lecture 5 on Electrical Machines - I, where we were discussing about
transformers. And in fact, we were discussing about Ideal Transformer, ok, so ideal
transformer.
(Refer Slide Time: 00:27)
So, ideal transformer is that transformer where magnetizing current required is vanishingly
small, there is no winding resistance, there is no leakage flux. And another thing I did not
tell, but right now I can also add that there is no core loss, there is no core loss in the
magnetic material in the core.
We will see what core loss is later which comprises of eddy current and hysteresis loss.
And those losses are neglected, winding resistance neglected and magnetizing current
required is very small because the permeability of the core going to change to going to be
very high. So, to establish the flux you practically do not require any magnetizing current.
In my last class this was the thing I did this is the ideal transformer, ideal transformer and
therefore, under no load condition the phasor diagram I drew it was like this, this is the
39
applied voltage 𝑉1 and this is the induced voltage in the primary which is also of same
length as that of 𝑉1 and that is 𝐸1 . Secondary induced voltage 𝐸2 having 𝑁2 turns and this
is 𝑁1 turns secondary induced voltage will be also in phase with this. Of course, its length
will be different maybe more than 𝑉1, maybe less than 𝑉1 depending upon whether 𝑁2 is
greater than 𝑁1 or not. So, this is the applied voltage.
And the magnetizing current will be lagging 90°, but in this case it is 0, there is no
magnetizing current. And in my last class, so this is the open circuit phasor diagram, OC
phasor diagram. Of course, magnetizing current is 0, but flux is here, is not. Why? Because
after all purely inductive circuit whatever little magnetizing current drawn that will be 90°
lagging the applied voltage and therefore, 𝜑 will be here, 𝜑 is finite nothing like that
because you are using very best material having newer tending to infinity. So, this is the
open circuit phasor diagram.
Then I told you, so to establish this flux 𝜑 mmf necessary was equal to 𝑁1 𝐼𝑚 which is
vanishingly small, ok. Let it be there, but that mmf was necessary to establish this flux.
Then when you close the switch here is a EMF available and there is an impedance here
𝑍2 therefore, there will be current in the circuit, instantaneous current is 𝑖2 or phasor
representation is capital 𝐼2 . This current will be delivered to the load.
But the moment some current is drawn from the secondary primary cannot remain silent
cannot see it idle. The moment it does, so primary will also draw an extra current whose
𝑁
value will be 𝐼2 𝑁2 . This current it will draw and the direction of the current are like this.
1
Therefore, two extra a mmfs come in to play in this magnetic circuit. What is the mmf of
this coil? 𝑁2 𝑖2 . In which direction? Anti-clockwise direction. And what is the mmf
produced by this coil? It is also 𝑁2 𝑖2 , but in this direction. Therefore, these two mmfs
cannot result into flux and primary original that vanishingly magnetizing current will
produce that necessary flux, is that clear.
Therefore, secondary if you load primary will draw extra current and this is called reflected
current drawn from the supply. Mind you, all the EMFs this dot conventions also we
discussed like polarity we fact in instant this is plus this would be plus. So, through the
potential of this point with respect to this; I will define potential of this point with respect
to this. And secondary current I will think it is leaving the dot that is the convention I have
adopted, and if the current leaves through the dot it must invite current from the source
40
into the dot on the primary side, so that these two mmfs balance off. Therefore, this was
the open circuit phasor diagram with S opened.
Now, with S closed the phasor diagram will be the currents I have to draw. So, current will
be, first I draw the secondary current. Secondary current will be lagging this 𝑉2 voltage
𝑉2 = 𝐸2 mind you, by the power factor of the load current this will be your 𝐼2 , ok. So, this
𝐼2 will lag 𝑉2 by the power factor angle of the impedance which is 𝜃𝑍 . So, this will be 𝐼2 .
And what will be the primary current? Primary current will be let me use other colour, it
will be in same phase with that of 𝐼2 , but its length will be different because it is multiplied
𝑁
with the ratio 𝑁2.
1
So, if a 𝑁2 is greater than 𝑁1 , it will be like this will be your 𝐼2 ′ , but with respect to time
they are co-phasor. The mmf produced by because it has to be, because that all times it has
to balance of these two mmfs. So, that the flux does not change. Therefore, whether the
switch is opened or closed flux in the core remain same, same flux. And this flux to create
this flux you require a vanishingly small magnetizing current into 𝑁1 that is the mmf
necessary. So, everything is now in place.
So, this is the phasor diagram with this red things with S opened, S closed or we say phasor
diagram of the ideal transformer phasor diagram with load, with load on secondary side.
Load only can be connected on the secondary side that is the thing. So, so this in simple
terms is the phasor diagram of an ideal transformer under no load and loaded condition.
Now, after we have learned these what I will do I will tell you one very interesting thing.
Therefore, in this ideal transformer when S will be opened what will be the ammeter
reading suppose somebody has connected an ammeter? With S opened what will be the
ammeter reading? 0, because ideal transformer no magnetizing current is necessary. When
you close the switch what will be the ammeter reading? It will be 𝐼2 ′ that is decided by the
𝑁
secondary current 𝐼2 . Now, also we define one thing 𝑁1 henceforth I will call that is called
2
the turns ratio I will denote it by this small letter 𝑎, ok. Therefore, this 𝐼2 ′ is nothing, but
𝐼2
𝑎
, ok. So, this is the thing.
𝑁1
=𝑎
𝑁2
41
𝐼2 ′ =
𝐼2
𝑎
Now, I will tell you about one very important thing that is what is the equivalent circuit of
this ideal transformer. So, equivalent circuit I will now tell. Now, equivalent circuit
generally we say with respect to the referred to the primary side one is referred to the
primary side primary or supply side. And secondly, also equivalent circuit referred to the
secondary side secondary or load side. I will tell you what does all this means.
See, here is a supply and then you have connected a transformer then on the secondary of
the transformer there is 𝑍2 and this two coils are not electrically connected. They are
magnetically coupled, so energy is transferred from the source to the load side via this
flux, flux communicates with the second coil time varying flux and you get voltage and
power in this secondary side. Now, if you look from the supply end supply side supply
does not know whether you have connected a transformer and then across this secondary
side you have connected a load.
What it only knows is that here is a supply and I am supplying a current 𝐼2 ′ when S is
closed and this 𝐼2 ′ I will call it 𝐼1 in this case, primary current. If 𝐼2 is 0 with S opened 𝐼1
also vanishes. So, supply will always, supply is totally unaware of the fact that there is a
transformer and the load connected across the secondary. It will interpret that as if across
the supply 𝑉1 you have connected a load and it is delivering a current 𝐼1 that is all.
𝐼1 =
𝐼2
𝑎
Therefore, from the supply side that is this one I will I can draw it like this, here is a 𝑉1
AC rms voltage it is supplying a current 𝐼1 when S is closed. Therefore, it will interpret
that you have connected an impedance, impedance seen by the supply, impedance seen by
𝑉
the supply is equal to 𝐼1 . He will say oh, somebody has connected across me a an
1
impedance and that is why I am delivering a current 𝐼1 , that is all.
Now, what I can do is this I know that let me be on this page only let it be a bit dirty, but
we are following logically. So, it will be easier for me to establish this important
𝑉
relationship 𝐼1 . But I know that
1
42
𝑉1 𝑁1
=
=𝑎
𝑉2 𝑁2
this I know. Therefore, I can say that
𝑉1 = 𝑎𝑉2
𝐼
Similarly, I can express 𝐼1 in terms of 𝐼2 because 𝐼1 = 𝑎2
𝐼
𝑉
𝑉
2
2
So, I will write it as a 𝐼1 = 𝑎2 . And this then becomes equal to 𝑎2 𝐼2 . But it is 𝐼2 = 𝑍2 , is
𝑉
not, this 𝑍2 = 𝐼2. So, this is very interesting. Therefore, in this transformer if you connect
2
a load across the secondary of magnitude 𝑍2 that load will appear to be different impedance
across the source 𝑉1.
𝑉1 𝑎𝑉2
𝑉2
=
= 𝑎2 = 𝑎2 𝑍2
𝐼2
𝐼1
𝐼2
𝑎
What is the value of that impedance? 𝑎2 𝑍2 , suppose we have connected 10Ω impedance,
trans ratio is say 10, small 𝑎 is 10 then 𝑎2 means 100, so 100 into 10 it will appear to be
an 1kΩ resistance across the supply.
So, the equivalent circuit referred to the supply side it means that across the supply what
is the effective impedance that has been connected 𝑎2 𝑍2 . In other words, what I am telling
is that, this supply you make a circuit like this your supply will not be able to distinguish
between this simple circuit and this whole circuit. So, for as supply side is concerned,
because here also 𝑉1 applied voltage current supplied by it is 𝐼1 then transformer then load
𝑉2, then 𝑍2 etcetera, but in simple terms it is as if you have connected an impedance across
𝐼
the source whose impedance value is 𝑎2 𝑍2 and this 𝐼1 = 𝑎2 .
So, this is called the equivalent circuit of the ideal transformer your actual circuit is these,
but you can to simplify matter. So, coupling circuit drawing this that to find out the currents
in the circuit. You draw the equivalent circuit referred to the source side primary side, if
you have connected an impedance 𝑍2 here, solve this simple circuit where it is 𝑎2 𝑍2 , get
the current 𝐼1 and we have solved the problem. You may only ask that, but my actual circuit
was this, so I have solved for 𝐼1 . But you have solved for 𝐼1 you can easily calculate 𝐼2
43
simply by multiplying these 𝐼1 with this small 𝑎, that is the duty of this particular method
of analyzing the circuit.
So, to analyze this circuit you can leave with the actual circuit and go by this physical
𝐼
argument 𝑎2 mmf balance solve this things you do get the current here, get the current there,
and power also you can calculate if you like all the things you can do. But here is a nicer
way of solving the circuit. What is that? Draw the equivalent circuit referred to primary if
𝑁
there is 𝑍2 any impedance on the secondary side multiply with 𝑎2 . What is 𝑎? 𝑎 = 𝑁1 turns
2
ratio.
𝑉
And then solve this simple circuit 𝑎21𝑍 , you will get 𝐼1 . Then you ask yourself, I have to
2
solve these circuit currents in every parts of this circuit then what you do. 𝐼1 is known you
can easily say what this 𝐼2 will be because there is a definite ratio between 𝐼1 and 𝐼2 . So, it
is useful to draw the equivalent circuit and solve big networks problems which are using
transformers, ok.
After you have done this then you may also think. So, this is the equivalent circuit referred
to the supply side. Similarly, one can see from the load point of view if you sit on this
impedance you know you have been supplied with a voltage 𝑉2 before that there is a
transformer it is not going to change the current in the circuit.
(Refer Slide Time: 21:45)
44
Therefore in the next page if I go, come to the next page. So, I will now tell you what is
the equivalent circuit referred to the load side. So, this was my 𝑍2 , this was this switch is
𝐼
closed, this is 𝐼2 and recall this current is 𝐼1 = 𝑎2 , this we have done right now and this is
the situation. So, with respect to the supply the equivalent circuit we have just got 𝑉1, and
an impedance 𝑍2 , and this is your 𝐼2 . So, this is referred to primary side.
Similarly, I can think that impedance from the load side here is an impedance 𝑍2 and it is
supplied with AC voltage who bothers about transformer before that I am not concerned.
My current here is decided by this voltage 𝑉2. Then solve this circuit, but how to solve this
𝑁
circuit? Z 2 is known, but 𝑉2 I know this 𝑉2. What is known to me? 𝑉1; so, 𝑉1 𝑁2 = 𝑉2, is
1
𝑉1
it not. So, 𝑎 I will do and get 𝑉2 and then I will solve this simple circuit. Because mind
you 𝑉1 𝑉2 𝐸1 𝐸2 they are all in time phase. Therefore, I will there is suppose take the
voltage on reference phasor. So, all are magnitude and angle. So, this divided by 𝑍2 you
do, you will get 𝐼2 then he will ask oh that is fine, but I have to solve for currents this is
𝐼
the actual thing. So, we have solved for 𝐼2 . The moment we have solved for 𝐼2 your 𝐼1 = 𝑎2
that is all, are you getting.
So, you can either draw the, so this is equivalent circuit referred to secondary side,
secondary side or to load side, load side and this is the equivalent circuit referred to primary
side or source side, source side.
So, rule is very clear. Any impedance you bring from the secondary side to the source side,
you multiply it with 𝑎2 𝑍2 , it should not be 𝑍2 , 𝑎2 𝑍2 . And this current is 𝐼1 . And how it is
𝐼
related with this 𝑎2? And any current actual current on this secondary side should be divided
by 𝑎. And any impedance from the secondary to this side is 𝑎2 .
𝑉1 is already here. So, no transformation is necessary for 𝑉1. On the load side when I draw
the equivalent circuit, I am in the secondary circuit, so I should not disturb 𝑍2 , 𝑍2 is 𝑍2 ,
𝑉
here is 𝐼2 , here is 𝑉2, but the primary applied voltage 𝑉1 gets modified transformed to 𝑎1 ,
and your this is 𝐼2 , and 𝐼2 = 𝑎𝐼1 . Therefore, a transformer when you connect it will make
some given impedance to appear as impedance of different values. Of course, phase angle
of the impedance does not change that is there. Therefore, this is the important thing and
phasor diagram also I have seen already.
45
Now, one another important thing I will derive from this two is that 𝑉1 𝐼1, suppose you
multiply 𝑉1 𝐼1 and express 𝑉1 and 𝐼1 in terms of 𝐼2 what this product becomes. So, 𝑉1 =
𝐼
𝑎𝑉2 . And what is 𝐼1 ? 𝐼1 = 𝑎2 is equal to 𝑉2 𝐼2 . This is one very interesting results. Therefore,
volt ampere that is KVA or volt ampere product of voltage and current on the primary side
is same as product of volt ampere on the secondary side. So, so this is called volt ampere
or KVA remain same on both sides, on both sides. And current and voltages are related by
𝐼
this that 𝑉1 = 𝑎𝑉2 and 𝐼1 = 2 . And what is 𝑎?
𝑎
𝑁
Do not forget, 𝑎 = 𝑁1 . What is 𝑁1 ? Number of trans on the source side and number of trans
2
on the load side. So, KVA of the transformer remain same, and current and voltages are
related of the both sides like this and because of this relations exist the impedance whatever
you have connected across this secondary appears somewhat differently on the primary
side.
𝑉1 𝐼1 = 𝑎𝑉2 ×
𝐼2
= 𝑉2 𝐼2
𝑎
A lot of simple problems to illustrate this idea can be solved now. I will do in the next
class. But try to understand the meaning of equivalent circuit. It means that ok, it is a
complicated I mean apparently so many things we have to draw in the actual circuit, it is
true you have to find out current, voltages in all the parts of the circuit.
But after knowing this relations exist you can either choose this equivalent circuit from
which you can only get 𝐼1 , and if you wish you can always find out 𝐼2 , then go back to the
actual circuit and say this is 𝐼1 , this is 𝐼2 and so on. Or you can consider this simple circuit
which is much simpler than this apparently complicated circuit here just a source and then
impedance solve for the current 𝐼2 then you come back to 𝐼1 etcetera by noting the
transformation ratio. And lastly, I told you that KVA remain same on the primary and the
secondary side. So, we will continue with this in the next class.
Thank you.
46
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 06
Rating of Single Phase Transformer: Rated Current & Rated Voltage with
Example
(Refer Slide Time: 00:22)
Welcome to 6th lecture. And we were discussing about the Equivalent Circuit of
Transformers, Equivalent Circuit. And equivalent circuit can be drawn equivalent circuit
is a simple model of this whole complicated things you live in one side and get the results,
ok. We discussed with that.
In the last class I told you that product of Voltage and current on the primary and the
secondary side they remain constant, ok. Let us suppose say that, now I will tell you that
it is because of this fact that V A product remain same. Suppose, the rating of a transformer
let me tell you rather in this way rating, how rating should be specified, rating of
transformer. If you look at the rating of a transformer see, what will be the ratings of the
transformer? The Voltage rating of the two sides must be specified how much Voltage
should you apply, how much Voltage should you apply to the secondary side and so on.
And also, what is the current carrying capacity of these coils on both the sides
47
So, you can specify the Voltage rating of the primary side and current, Voltage rating of
the secondary side and the current. For example, somebody says that a transformer, he
writes it like this the Voltage ratings are 200V/100V. Instead of telling that primary side
is V, secondary side is 100V which you we will also use this terminology no doubt, but
more scientific way of telling that this is the Voltage rating this coil I will call HV coils
suppose this is HV.
And this is called low Voltage coil, whichever side has got low Voltage we can also
identify the coils like that because you must be now feeling that no point either of them
can be used as primary.
But, only thing when LV is used as primary you should apply 100V and then you will get
200V there and vice versa. Therefore, it is a sometimes a more scientific to tell in terms of
HV LV to describe a transformer rather than primary secondary. User will decide which
one will become your primary, which one will become your secondary. Suppose, 100V is
available you have this transformer you required 200V, then there is no other way how it
energize the LV side with 100V get 200V to supply your load or vice versa. So, these are
called the Voltage rating.
Then I told you the current ratings of the coils should also be specified. Instead of that
because we know the V A product remain same what is specified current ratings are
specified indirectly. What is specified is a KVA rating, suppose the KVA rating is 1 KVA
and also frequency will be specified 50 Hz. And also, it will be told about what kind of
transformer, it is single phase transformer, got the point. So, a transformer is specified in
this way.
From these numerical values, what we conclude. I will now draw it instead of drawing this
elaborate diagram, I will simply draw two coils like this and core material may be shown
by this vertical line a simplified way of telling all this things, instead of drawing this core
along with coils I will draw it like this two are dots that is important I will specify, and this
is suppose HV winding and this is LV winding. Now, from this I can calculate what is the
rated current of the HV side, it will be this
𝐼𝐻𝑉 (𝑅𝑎𝑡𝑒𝑑) =
48
1000
𝐴 = 5𝐴
200
Similarly, I will calculate
𝐼𝐿𝑉 (𝑅𝑎𝑡𝑒𝑑) =
1000
𝐴 = 10𝐴
100
So, the moment of these ratings is known I can draw this diagram and say that, HV side
maximum Voltage you can apply 200V 50 Hz, and you will get 100 V 50 Hz here. And
you will you are going to connect some load, but do not connect any load you connect
such a load such that the current while it will deliver to the load does not exceed these 10
A, it should not exceed. Your load current must be less than equal to 10 A.
So, any impedance you do not connect. And when it will deliver this rated current
𝐼𝐿𝑉 (𝑅𝑎𝑡𝑒𝑑) , let us assure you do not have to do anything your primary at that time will carry
5 A that is what we have learned. If 10 A goes, what is the value of 𝑎?
𝑎=
𝑁1 𝑉1
= =2
𝑁2 𝑉2
that is all therefore, the ratings current ratings of the HV and LV windings are indirectly
given in terms of KVA and Voltages. You have to do a little bit of calculations here to
know exactly what is the rated current.
Let me do something more with this. Suppose, this is the transformer I have applied here
200 V and I have connected an impedance here we will get 100 V immediately if you
apply 200 V 100 V. Suppose, you have somebody connects an impedance of 5 Ω, what is
100
going to happen? 200 V 100 V 5 = 20𝐴 you are trying to draw and you will say oh rated
current is 10 A. No, this is not allowed. So, do not connect 5 Ω then. Are you getting? You
must understand this.
100
Therefore, connect what is the minimum impedance you can connect? 10 = 10Ω. Do not
connect impedance. If you connect 10 Ω fine, 10 A here, if it is 10 A you be rest assured
it has it will be automatically 5 A, ok. If you are these things I will do quickly, so that
things go to your mind
Suppose somebody connects 20 Ω here impedance then this current will be 5 A and it is
well within the red below the rated current it is allowed, but if it is 5 A this current will
become 2.5 A because it will be always half. So, several simple, but important things; this
49
should be in your mind always playing, ok. You have connected 20 Ω, this current is 5 A,
𝐼
HV side current is always 𝐼1 = 𝑎2 . And here what is 𝑎?
𝑎=
𝑁1 200
=
=2
𝑁2 100
So, current will be reduced by factor of 2.5 and so on.
With respect to suppose, so I have used HV as my source side, source I have connected to
HV. I could also connect source to the LV side and you can do that you try that. So,
henceforth I will draw this circuit instead of drawing all the time this and getting the
results. So, this will be the thing. Another interesting thing comes out that this ratio of
current whatever impedance you have connected, only you have to see that secondary
impedance you have connected does not make the current in the LV side greater than the
rated current that is only should be your concept.
But, the moment this current is known this current is fixed. What will be the equivalent
circuit of the transformer? It will be a source 200 V very quickly I will do, and across it
you have connected an impedance for this if you have connected 20 Ω I will say you have
connected an impedance of,a is 2, 80 Ω And 200 by 80, so this current is 2.5. So, I will
draw the equivalent circuit oh impedance has.
𝑎2 20 = 80Ω
200
𝑎 = 2, 80 Ω And 80 , so this current is 2.5. So, I will draw the equivalent circuit oh
impedance has.
So, sometimes it transfer of impedance people say transformer means it changes the value
of the impedance source you are physically connecting an impedance of 20 Ω, but to the
source it appears to be 80 Ω. You can manipulate the impedance, ok. So, this is the thing.
Then after calculating this I will say I will come back here and say, this current will be
then 2.5 × 2 = 5𝐴.
Now, after going through this, corresponding to this rated current here I am under utilizing
the transformer, the impedance have it can carry deliver a current of 10 A, but I have
50
connected 20 this I can always do. Under full load condition rated condition I should
connect an impedance which is equal to 10. 10.
At rated condition if you want to apply I will connect here 10 Ω and then this current will
be 10 A rated current, and this current will be 5 A and I will say the transformer is operating
at full load condition when the currents are rated. So, at rated condition at rated currents
sorry, at rated currents transformer operates at full load condition. We will discuss much
more with the loading etc full, we have not at full load condition, got the point.
And another thing we immediately come to the conclusion that on the high Voltage side
Voltage is high current will be less compared to the LV side, where Voltage is less current
is high because the product is to remain same. So, if Voltage has reduced current must
increase by the same proportion that is why KVA remain same V A. Kilo V A is a very
practical unit. But what I want to say if you want to see after all while making this coil you
have to use piece of wire, wind it like this. The cross section of this wire gage of the wire
is decided by the magnitude of the current. Therefore, on HV side thicker conductor will
be used, HV side conductor cross section will be thicker I am just drawing symbolically.
LV side conductor cross section will be less, LV side it will be high correct sorry because
LV side conductor, ok.
Thinner wire, means section, thinner section. And the section of the LV side because it has
to carry rated current. When LV side carries rated current HV side has to carry rated current
therefore, if a transformer is given if you look if the windings are available to you, you can
see you can always immediately see the section of this wire this wire will be LV side will
be much higher than the HV side. So, what we have learned till now? We are still
considering mind you ideal transformer it does not matter. This is what happens and based
on that I told you about the ratings of the transformer importance of KVA rating and
Voltage ratings.
On the HV side of course, if you apply instead of 200 V if you apply 100 V what is going
to happen? On this side you will be getting 50 V that you are allowed to do because you
should not exceed the rated Voltage of the winding. So, you should not apply on the HV
side here a Voltage greater than 200 V, less than 200 V you can apply no problem. And
then current ratings I know.
51
If I ask you this question that, this is the actual rating of the transformer suppose on the
HV side I have applied 100 V, how much current you will allow to flow? What will be the
KVA rating then? Current you can go up to rated values, no problem. So, transformer will
then be de rated, that is there. We will see several problems, we will solve. So, this is the
thing.
(Refer Slide Time: 18:32)
Next thing in this lecture, so that it creates interest in you, I now say that we are continuing
with ideal transformer, mind you, ideal transformer. Suppose, somebody has connected a
transformer here, and the turns ratio of this transformer is suppose 2 and yes connected
another transformer here. And, it is turns ratio is suppose 0.5 suppose somebody has
connected like this, and understood. Suppose, and he has connected an impedance here on
this side 10 Ω, and suppose you have connected a 200 V 50 Hz source here, 200 V AC
rms. Tell me what is the Voltage here? I i want to find out the currents.
So, if you have applied 200 V, if the ratio is this what will be this Voltage? 100 V because
𝑉1
200
1
1
will come here; so, this will be 𝑎 = 100𝑉. So, this 100 V is applied to this transformer.
𝑎
1
And you have suppose let us consider this to be 4 turns ratio, I will this transformer is
1
100
100
2
4
having a turns ratio of 4. What will be the Voltage here? It will be 𝑎 = 1⁄ = 400𝑉. If I
make any mistake you please point me out through your mails or whatever it is.
52
But what I am trying to tell that you can see the nicety of analyzing of circuits involving
transformer, it is so easy I mean in the first place actual transformer there will be equivalent
circuit which will be slightly complicated things like that. But if you even if you assume
that to be ideal you will be getting results very close to the correct values that is why this
exercise, and also it tells you how the ratio of the current, Voltages these are decided.
1
100
Now, now tell me this is 𝑎2 = 4. So, it supplied Voltage 100, 𝑎 = 400𝑉. If you know
2
this is 400 V this is 10. So, this current will be 40 A. If this is 40 A this and mind you I
𝐼
should have shown you the dots, ok. If this is 40 A this current will be 𝑎2 this secondary
current is 40. So, this current will be
160
40
𝑎2
= 160𝐴. If it is 160 A this secondary what will be
160
this current? It will be 𝑎 = 2 = 80𝐴, got the point.
1
Now, what I am telling is the equivalent circuit of this whole system with respect to this is
the only source supply. So, this is 200 V equivalent circuit referred to source will be, if
you have solved these I am telling you this impedance will be simply this Voltage by this
current
200
80
= 2.5Ω. How much it is?
Student: (Refer Time: 23:55).
2.5 Ω, 2.5 Ω, it is coming 2.5 Ω. You see I will now do what, this 10 Ω to this source I
will I will try to get back this same result This is the original 𝑍 load, discrete value, 𝑍 load
1 2
is this. So, to this here, this impedance will appear to be 10 × 𝑎2 = 10 × (4) this
impedance. This impedance is connected across these and this
1 2
10 × ( ) × 4 = 2.5Ω
4
Therefore, this is the thing therefore, this is the equivalent circuit referred to this source.
Similarly, you can say the equivalent circuit referred to the load is nothing but this 10 Ω
supplied from a 400 V source. So, this simple calculation will make you understand better
what is happening. If you know this current you should be able to calculate this current.
See whether I will just give you a tips. See, 𝑎 I have defined for a transformer, I have
53
𝑁
defined 𝑎 = 𝑁1 , 𝑎 is this it may be greater than or less than 1, depending upon whether
2
stepping up stepping down Voltage etcetera.
Therefore if you, current should be divided by 𝑎, Voltage should be multiplied by 𝑎,
students get sometimes confused while solve solving numerical problems. But what I am
telling you calculate this a turns ratio, you should always remember current on the LV side
will be higher Voltage on the LV side will be lower.
For example, take this transformer turns ratio is 2, 200 V LV side Voltage will be lower
and 𝑎 is greater than 1. So, you have to divide it by 2. And if you know this current this
side current this side current which is LV side, I know which one is LV, which one is LV,
HV and I am telling you LV side current is always higher therefore, this current 80 A
should be multiplied by 2 then, to get a 160 or if you know 160, it should be divided by
this 𝑎 to get 80. So, I have a lot of practice, so that you can transfer the impedance.
See, these one just to conclude this I am telling you it this whole system can be drawn like
this also. First transformer I will keep, if I go step by step this is the transformer, first
transformer. I am drawing this whole circuit here this is 200 V fine. And here was a
transformer. Treat once transformer at a time, eliminate this transformer first. Then I will
say across this transformer and impedance is connected whose value is 𝑎2 × 10, here this
transformer this transformer I will replace. So, this impedance will appear to be 𝑎2 × 10
1
1
𝑎 = 4. So, it will be 16 × 10, so much Ω if any mistake point out. So, I can do it like that
also.
10
4
And then this 160 and this transformer turns ratio is 2. So, it will be 16. So, then the next
10
step is sub supply and this is 200 V, and this was 16 × 4 = 2 and there are several ways
you find out your own convenient way of interpreting the results this time.
Therefore, I so far what I have done is this one that a rating of a transformer this is very
important, KVA remain same on both primary and secondary. And then LV side current
will be very large therefore, cross sectional area of the wire with which you make this LV
winding will be larger compared to the cross sectional area of the wire of the HV side.
Of course, the Voltage ratings I have just taken, so that calculations become easier, 200
V/100 V, but in practice Voltage ratings could be one side of kilo Volts another side may
54
be 220 V or 440 V, ok. So, current ratings decides the cross sectional area of the wires and
Voltage ratings is going to decide the insulation level.
You must have seen transformer from where terminals are coming out with sort of
insulations, bushings, they call it. You have seen distribution transformer on the road side,
3-phase transformer. So, these are bushings through which, so this is the these are
insulation, ok. So, insulation level will be decided by the Voltage rating of the transformer
and cross sectional area of the copper wire with which you make the coils they will be
decided by the rated currents of the transformer. So, have a thorough knowledge about the
topics which I have discussed here which will make our live much simpler in future. We
will continue next class.
Thank you.
55
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 07
Transformer with Multiple Coils
A welcome to this Machine - I course. We are discussing about an ideal transformer, and
in my last lecture I told you how to solve a circuit symbol being several ideal transformers
in the circuit because ideal transformer is so simple. But the problems are interesting.
Actually, what impedance you connect, at the end of the across the secondary coil that
values get transferred and we know how to do it.
(Refer Slide Time: 01:12)
Another interesting problem, I will discuss so that you become really conversant with ideal
transformer. For example, consider this is the core first I will draw, this is necessary to
explain what is happening. Suppose you have this transformer with a, so far two coils I
have considered, ok. What we will do is I connect another third coil, another third coil and
perhaps here also I will connect a one switch and load one.
Another switch this is S1 say S2, this is a S3, there is no S1, S1 may be there. So, this is
suppose 𝑍2 and this is suppose 𝑍3 . See, what I am telling I am slightly greedy, that is I will
energize this with a Voltage 𝑉1 frequency 𝑓 and I want to generate a two levels of Voltage
56
because same flux will be linking this coil to therefore, the induced Voltage here will be
𝑉
𝑉
1
1
also 𝑁1 𝑁3 here the Voltage will be, here the Voltage will be 𝑁1 𝑁2 .
One thing I will tell you in the primary and secondary Voltages are different, currents are
different KVA are same, that is all fine. But also try to understand this one that is suppose
you have a transformer, of turns 𝑁1 and 𝑁2 it is very useful while solving problems and
𝑉
𝑁
𝑉1
1
other things. What happens is this 2 = 𝑁2 that is good, but it is also true that
𝑉1 𝑉2
=
𝑁1 𝑁2
This equation is very useful, at least I find it is tells you that in a several coupled coils what
𝑉
𝑉
𝑉
1
1
2
remains constant is Voltage per turns. What is 𝑁1 ? Voltage per turn is constant 𝑁1 = 𝑁2 it
has to be.
Therefore, what you do to calculate Voltage induced here, you simply calculate Voltage
per turn then multiply with respective turns to get the Voltages here and there. So, you can
have more than one secondary coils. For example, here so, what is Voltage per turn here
𝑉
𝑉
1
1
if it is 𝑁1 ? 𝑁1 that will remain constant. What will be the Voltage available here? 𝑁1 𝑁3 ,
multiply with actual number of turns.
Here also this Voltage will be 𝑉3 say, 𝑉3 will be equal to Voltage per turn that what where
from I get from the supplied Voltage and its number of turns into 𝑁3 ; Voltage per turn
here, what will be this Voltage? It will be
𝑉1
𝑁
𝑁1 2
Therefore, you can have two sources available to you of different Voltage level and you
can supply two different loads that is what I am trying to tell. Let us take a numerical
numbers, so that things become much more easier.
57
(Refer Slide Time: 05:58)
And this sort of thing I can then draw it like this. Suppose, you have a transformer, I will
draw now simplified diagrams, and you have two separate secondary coils and there is a
common magnetic circuit. This is how I can represent. And suppose the dots are here, this
is dot, this is dot, this is dot. Suppose, I say this number of turns 𝑁1 is equal to say 200 and
here you apply an AC Voltage of 400 V then what I am telling, immediately you know
𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑝𝑒𝑟 𝑡𝑢𝑟𝑛 =
400
=2
200
2 V per turn.
Now, suppose I say this one is 𝑁2 is equal to 50 turns and this one I say 𝑁3 is equal to say
25 turns then I will immediate say this Voltage will be Voltage per turn 2 × 50 you will
get here 100 V. And 2 × 25 you will get here 50 V. Are you with me? Then this is the
how. So, Voltage per turn is a very useful concept in transformers that remains same in
several coupled coils.
Therefore, suppose the problem is like this is 200 V, 𝑁1 400 V is the applied Voltage 50
Hz, these are the dots and if the secondary terminals are open circuited. So, I can now
supply two loads, one load demanding 50 V I will supply, another load demanding 100 V
I will supply. The question is that, when you supply loads how the primary currents how
do I calculate? Ok. It is slightly not that easy problem, ok.
58
Let me a put me in this way suppose you have a switch here and you have connected a
resistive load here of say 10 Ω, and also I have a switch here and here I will connect a
capacitive load−𝑗5Ω. I have planned to connect these two loads.
So, my problem is find out currents in all the windings and this switch is suppose S1 this
which is S2. Calculate I will indicate the currents as 𝐼2 , this is as 𝐼3 , and this is as 𝐼1 .
Calculate 𝐼1 , 𝐼2 and 𝐼3 . Part 1, when only S1 is closed S2 opened. Part 2, only S2 closed
and S1 opened. And finally, both S1 and S2 are closed. I to have this find out. And let this
transformer be ideal. What does it mean? If you close the switch and with both the switches
are open no current is drawn magnetizing current necessary is 0. So, that is the problem.
First part of the problem is pretty simple; this one S1 is close. So, it is like primary
secondary. This fellow although you will get Voltage here, but no current no
complications; so, first part of the problem will be Voltage per turn is this. So, here you
will get
𝑉2 = 100𝑉
𝐼2 =
100
= 10𝐴
10
𝐼3 of course will be 0, S2 is open. And what will be 𝐼1 ?
𝐼1 =
𝐼2
𝐼2
100
=
=
= 2.5𝐴
𝑎 200⁄
4
50
And let us also draw the phasor diagram under this ideal transformer. What is the phasor
diagram? Phasor diagram will be like this. This is your 𝑉1, , 400 V, this will be your 𝑉2
100 V. All the induced Voltage will be in phase. 𝑉2 = 100𝑉 and to a slightly practical I
will this is one-forth, so it will be much higher this length. This is 𝑉1 = 400𝑉
All Voltages are in phase 𝑉1, 𝑉2, 𝐸1 , 𝐸2 , there is no distinction between 𝐸1 , 𝐸2 , 𝐸3 , 𝐸4 .
And what will be 𝐸3 will be there, 𝐸3 is how much? 50 V so, 𝐸3 𝑉3 will be there. 𝑉3 =
50𝑉 these will be the Voltages. Where is your flux? Flux is along this line. Magnetizing
current is also along this line, but that is 0 that is the thing.
59
Now, if S1 is close there will be 𝐼2 , and this 𝐼2 will be in phase with that of a because it is
resistive load. So, the current 𝐼2 will be I have calculated it 10 A it will be in some another
scale, 𝐼2 will be like this. This 𝐼2 will try to upset them in a balance therefore, primary has
𝐼
to draw extra current. And that value is 𝐼2 ′ = 𝑎2
So, what will be the primary current? And this is the H V side. So, H V side current will
400
be less, ratio is 100 = 4,so it should be divided by 4. So, 𝐼2 = 10𝐴. So, primary current
𝐼1 = 2.5 𝐴. 𝐼3 nothing is there, your problem is over. If you wish you can calculate the
equivalent impedance in by the source and you can verify it is equal to 𝑎2 × 10. This part
is very simple.
Second part; in the second part this current 𝐼3 , S1 is opened only S2 is closed. So,
everything remains same. So, I will better not to re-draw here only thing what I will do
part 2, I am solving. So, Voltages will be as it is induced, but the only thing is now 𝐼2 is
not there although 𝑉2 will be there open circuited and 𝐼3 is present. How much is 𝐼3 ?
𝐼3 =
50
10𝐴
5
but this current will lead 𝑉3 this is your 𝑉3 ; is not.
(Refer Slide Time: 18:00)
It will lead 𝑉3 by. 90°.
60
And what is the magnitude of this current? So, second part 𝐼3 Voltages remain same it will
be suppose I take Voltage on reference
𝐼3 =
50∠0°
= 10∠90°
5∠ − 90°
Therefore, 𝐼3 will be here The moment 𝐼3 flows then will be 𝐼3 ′ and that will be equal to
your 𝐼1 and this is LV side this is HV side, so current will be less. What is the ratio of turns
200
here? 25 = 8
So, your 𝐼1 and they will be in phase why because they have to balance of the mmfs.
𝐼1 = 𝐼3′ =
10
∠90°
8
Please try to follow me.
So, far I was in the previous examples just telling impedance values to tell you about the
rated current. But now I am in a position to go slightly in advance stage that, impedance
may be complex and how to calculate the currents. Had it been RL circuit the phase angle
of the impedance, I could also take into account. But where ever is your secondary current
the reflected current because of that must be there. So, this is the thing.
Now, the last part which is slightly tricky for this what I will do is this. I will copy this, so
that it is easier to talk.
61
(Refer Slide Time: 20:55)
So, now copy go to the next page, paste it. And this one may be deleted. This part you just
try to see how it is to be done. I am sorry for taking some time, they will be edit it.
Now, listen carefully what I am telling. So, this is the thing now. Now, both the coils
switches are closed what do I do, that is the part 3. Both S2 and S3 are closed. Therefore,
what is going to happen? Thing is that the clue to this problem, once again is that
fundamental thing. What is that fundamental thing? No, matter what is the positions of the
switch either closed or open flux in the core of the transformer cannot change. It has to
remain same. Why? Because KVL is to be satisfied on the primary side either S2 is closed,
S3 is closed or both of them are closed it does not matter this is S3 actually and this is or
S1 S2 I told it whatever it is, both the switches are closed.
Then once again what I will do is this first I will calculating Voltage per turn, Voltage per
turn which is equal 2 which I have already calculated and this Voltage 100 open circuit
Voltage has got. And then I will draw the phasor diagram. Phasor diagram is this one. This
is the 400 V, 𝑉1 400 V; then your 100 V and 50 V. So, this is suppose 100 V 𝑉2. Scale you
forget and this is your 50 V, 𝑉3 = 50𝑉. All are in phase. And suppose this I take as
reference phasor all Voltages are magnitude angle 0 °, 0 °, 0 ° and so on.
Now, you see that this is resistive now both 𝐼2 and 𝐼3 exist.
𝐼2 =
100∠0°
= 10𝐴
10∠0°
62
𝐼3 =
50∠0°
= 10∠90°𝐴
5∠ − 90°
both 𝐼2 and 𝐼3 are present, ok. First, I will draw 𝐼2 and 𝐼3 . So, 𝐼2 will be in phase with 𝑉2
because it is resistive load. Suppose this is 𝐼2 phasor and 𝐼3 will be leading 𝑉2 by 90 °, so
of same magnitude mind you I have taken the numbers such that they become, so it will
be 90°.
Now, the big question is what is 𝐼1 . Two ways we can, I will tell you much simpler method,
but I will go by the basic rule. Basic rule is because these two coils carry current extra
mmf is introduced into the transformer that is namely 𝑁2 𝐼2 in this direction 𝑁3 𝐼3 .
Therefore, these extra current drawn from the supply should be such that we it will nullify
both these mmfs. It has to; because applied Voltage is fixed here KVL is to be satisfied.
Therefore, what I will do? I will calculate 𝐼2 ′ , because of this there will be an 𝐼2 ′ , because
of this there will be an 𝐼3 ′ and then these 𝐼2 ′ and 𝐼3 ′ I will add vectorially phasor by phasor
sum.
So, what is this is 𝐼2 , 𝐼2 is how much?
𝐼2 = 10∠0°
𝐼3 = 10∠90°
On the H V side, I am calculating current. So, 𝐼2 ′ will be lesser current It will be how
𝐼
much? 𝑎2 therefore,
Student: (Refer Time: 27:51).
𝐼2 ′ =
10
∠0° = 2.5∠0°A
4
Similarly,
𝐼3 ′ =
10∠90°
= 1.25∠90°A
8
So, this is 𝐼2 ′ and this is𝐼3 ′ And then I will say
𝐼1 = 𝐼2 ′ + 𝐼3 ′
63
This is one complete this problem you must go through very carefully what are the steps
are because there may be situations you from the same source you want to get on the same
core, you connect a several coils like this 2 coils, may be 3 coils another coil is connected,
get different level of Voltages. Of course, in a two winding transformer only two windings
will be present. But if you do this exercise it will further strengthen your understanding of
reflected current. For this there will be reflected current here. This current divided by this
turns ratio. For this current there will be reflected current here. What will be the reflected
current? These actual current divided by turns ratio between these two and so on.
Now, this I will not do, but I will request you to think, ok. These you have found out, 𝐼1
you will find out. Once you find out 𝐼1 therefore, mathematically you will say equivalent
circuit of all these things across the supply will be this 400 V supply and it is delivering
this 𝐼1 therefore, impedance seen by the transformer 𝑍 ′ will be simply this Voltage by this
current, are you getting. 𝐼1 you have to calculate it will have some magnitude and some
angle.
This Voltage divided by this current will then give you the impedance seen by the
transformer. I will request you to think about it that, this is suppose in general 𝑍2 , this is
in general 𝑍3 , are you getting. That ok, supply sees what impedance what should be
equivalent circuit, I have to draw with respect to the supply side. How this R will appear
here, how this Z will appear here that you think in the next class. Of course, I will discuss
about that that is also very interesting.
So, I hope you have understood that if it is ideal transformer magnetizing current is 0 and
everything is so simple. If any coil secondary coil carriers current, primary coil will
immediately draw reflected current corresponding to that winding current what will be that
reflected current. If it is 𝐼2 on the secondary coil it will be 𝐼2 ′ . And we have also seen that
𝐼
𝑁
how to take any general impedance into account. So, 𝐼2 ′ = 𝑎2. What is 𝑎? 𝑎 = 𝑁1 .
2
Similarly, if there is another coil supplying another load wound on the same magnetic core,
if you know 𝐼3 there will be 𝐼3 ′ here, such that the flux remains same. And how to calculate
𝐼3 ′ ? It will be 𝐼3 by turns ratio between the primary coil and this new another coil that is
𝑁
whatever it is 𝑁1 now. So, these things if you take in to account and coupled with this
3
phasor diagram; the comfort zone is all the Voltages are in safe phase.
64
So, you have you can take them on reference. And then first calculate the secondary coil
currents, and then calculate reflected currents for each of these components add them up
to get the current which will be draw from the primary side. And then the equivalent
𝑉
impedance seen by the source can then be calculated easily without because 𝐼1 I will do
1
and say that, these big circuit to the supply is nothing, but this 400 V supply across which
𝑉
you have connected an impedance of 𝐼1 .
1
But what I am asking you to do, in terms of 𝑍2 and 𝑍3 find out that equivalent impedance.
It will be some 𝑎2 𝑍2 this turns ratio square into 𝑍2 , this turns ratio square into 𝑍3 and so
on. You think about it. We will discuss it in the next class.
Thank you.
65
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 08
Modelling of Practical Transfomer - I
Welcome to this lecture number 8 on Electrical Machines I.
(Refer Slide Time: 00:35)
And you remember that in my last lecture; we were considering a problem interesting
problem that is suppose a transformer has got 3 coils wound on it that is like this; this was
the transformer core.
66
(Refer Slide Time: 00:49)
So, one common magnetic circuit what which 3 coils have been wound; then I told volt
per turn is a useful concept; in such cases there are two secondaries; one primary. If you
apply some voltage 𝑉1 number of turns 𝑁1 and it is ideal transformer; no leakage flux no
winding resistance; therefore, 𝑉1 and 𝐸1 are same.
𝑉
Then what you do is you calculate voltage per turn 𝑁1 and you can quickly calculate if that
1
𝑉1
𝑉1
1
1
number is known 𝑁 𝑁3 = 𝑉3; 𝑁 𝑁2 = 𝑉2 . And our problem was we know if there is only
one secondary if I close this what will be the current and how to draw phasor diagram. But
here the problem is when both the switches will be closed, both the secondaries will supply
their respective loads 𝑍2 and 𝑍3 how to calculate the primary current that I told you.
67
(Refer Slide Time: 01:59)
In a simplified way, this previous thing could be drawn; instead of drawing elaborate
diagram involving core; this two lines will indicate core ideal; transformer it is nonetheless
and then I was considering this problem.
So, part 1 and part 2 of the problem were simple, but part 3 was a little bit involved in the
sense that both the switches are closed how to calculate the current. Then what I told that
if 𝑉1 is known then 𝑉2, 𝑉3 all the voltages will be co phasor and they are all shown on the
vertical lines. Then if you know 𝑉2; then you can calculate 𝐼2 , impedance is known it can
be position.
Similarly 𝑉3; 𝐼3 can be known and each of this current 𝐼2 will have its reflected component
𝐼2′ ; 𝐼3 will have its reflected component 𝐼3′ . So, 𝐼1 = 𝐼2′ + 𝐼3′ ; that is what exactly we did.
While calculating the reflected current be careful about the turns ratio for this it is
200
50
with
200
that number you have to divide. Similarly, for this it will be 25 we have to divide here;
like that you have solve the problem.
Now, one thing of course, should not be; the same problem could be done in a much
simpler way by making the; making use of the energy balance equation or power balance
equation on both the sides ok.
68
(Refer Slide Time: 04:01)
You will know that in any AC circuit suppose air is the source ok; that is power. From
power also it is much more easier and you need not go to the; I mean intricacies of
transformer, flux is constant these that like that. What I am telling is suppose this is 𝑉1; the
same problem here is your transformer ideal transformer primary.
And there are two secondaries and there are impedance is connected here; in general I am
telling now and there is another impedance connected here. Suppose this is 𝑍2 , this is 𝑍3
and I know the number of turns 𝑁1 , 𝑁2 and 𝑁3 . And suppose these are the dots of the
terminals; then what you do is this and my problem is to calculate this currents 𝐼2 then 𝐼3
and the current drawn from the supply. This is the problem and this is the transformer code
which is ideal transformer mind you ideal.
Now, in this case I can also do it in this way; see this is the two points; I have injected
power into the circuit. There is no other source which is pumping power into the circuit
and these are the things; two loads; I am consuming power on the secondary side.
Therefore, the power drawn from the supply must balance the sum of the powers given to
𝑍2 and 𝑍3 ; that is the very simple way of looking at it.
And you know that in AC circuit suppose we have a source and you have an impedance
here; 𝑍 a source is there 𝑉 and this is suppose the current the deduction of the current 𝐼
have assumed. Then I will write it like this; power complex power, delivered by the source
is given by
69
𝐶𝑜𝑚𝑝𝑙𝑒𝑥 𝑃𝑜𝑤𝑒𝑟 𝐷𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑆𝑜𝑢𝑟𝑐𝑒 = 𝑉𝐼 ∗ = 𝑆
See in language I have written complex power delivered; delivered is the keyword because
here is a source through the positive terminal of the voltage; current is leaving. So, like a
battery; so it is really delivering power and take voltage phasor and current phasor multiply
with 𝐼 ∗ . And you must be knowing, but still I am very quickly telling; it is not that it is 𝑉𝐼
do not and this is usually denoted by the letter 𝑆. Real part of it will give you real power,
imaginary part of it will give you reactive power.
But only thing is do not just multiply 𝑉𝐼 take; the real part, it is simply because of the fact
that in general voltage phasor could be 𝑉̅ = 𝑉∠𝛼 and current phasor could be 𝐼 ̅ = 𝐼∠𝛽 and
that is the reference phasor; reference phasor. So, this angle is 𝛼 suppose this angle is 𝛽
therefore, real power as you know it should be
𝑃 = 𝑉𝐼 cos(𝛼 − 𝛽)
This angle is (𝛼 − 𝛽). That is the reason you should do like this and reactive power will
be
𝑄 = 𝑉𝐼 sin(𝛼 − 𝛽)
So, if you simply; so 𝑉̅ = 𝑉∠𝛼, this is 𝑉 I mean I am omitting this bar all the time, but this
is the phasors. Therefore, if you simply do 𝑉̅ 𝐼 ̅ in this case; that is if you simply multiply
this two complex number; it will be
𝑉̅ 𝐼 ̅ = 𝑉𝐼∠(𝛼 + 𝛽)
And a real part of it is not real power that is the reason for general case; you do not know
voltage might have also angle, current might have also some angle with respect to some
other reference phasor.
Therefore, more correct thing will be; so real part of this
𝑅𝑒(𝑉̅ 𝐼 )̅ ≠ 𝑉𝐼 cos(𝛼 − 𝛽)
Therefore, you should just remember this; anyway so this is just a side remarks of this one.
Therefore, 𝑉𝐼 ∗ you have to take mind you 𝑉 and 𝐼; although I am not putting bars to they
are complex numbers and this gives you the complex power delivered by the source.
70
Now, after calculating this if both real parts and imaginary parts become positive numbers;
then I will tell real power is delivered by the source; reactive power is delivered by the
source. Because in any circuit you know current direction is your choice and if even if you
have to chosen it wrongly then the it will be reflected in this equation.
It might so happen 𝑉𝐼 ∗ after you calculate; it will give you real part becomes becoming
negative. So, I should not be surprise then I am telling complex power delivered by the
source its some minus real power (-100); means it is really absorbing real power, you know
all these things. So, this is the thing equation; I will be using here to balance the power and
I will demand that.
(Refer Slide Time: 10:57)
See what is the power delivered by the source; it is 𝑉1 𝐼1; I am so sorry; this is 𝐼1 ; 𝑉1 𝐼1∗ is
the power delivered by the source to the system. And this must be equal to see to the load
if you see this is the voltage drop because 𝐼2 direction, I have assumed; I cannot then say
and voltage drop then has to be upper one plus, lower one minus.
You can select the direction of 𝐼2 in whichever way you like, but once you select that
direction of 𝐼2 ; remember the voltage drop polarity across that element is decided then plus
minus I can do. Similarly, this fellow voltage drop across 𝑍3 is 𝐼3 𝑍3 upper one plus lower
one minus. And through the dots current come out that way in the transformer you will
assume, so, that MMF will be compensated by the current drawn by the primary through
the dot that is the whole implication.
71
But what I am telling this must be true; 𝑉1 𝐼1∗ must be equal to this voltage is suppose 𝑉2;
this voltage is 𝑉2 this side upper one plus lower one minus. So, power delivered; power
absorbed by 𝑍2 = 𝑉2 𝐼2∗ plus 𝑉3; this is 𝑉3 plus minus 𝑉3 𝐼3∗ very simple; it has to be from
power balance. Even if you do not go into the complexity of transformer how flux remain
same, but this has to be if it works like that you cannot do. So, in this case I know 𝑉1, 𝑉2
and 𝑉3 because number of turns are known, those voltages are known. What else I know?
I know 𝐼2 , I know 𝐼3 therefore, 𝐼1∗ can be calculated.
𝑉1 𝐼1∗ = 𝑉2 𝐼2∗ + 𝑉3 𝐼3∗
Because my problem is what is the current drawn from the supply it will be equal to
𝐼1∗ =
𝑉2 ∗ 𝑉3 ∗
𝐼 + 𝐼
𝑉1 2 𝑉1 3
But 𝑉2, 𝑉1, 𝑉3, 𝑉1 all are co phasor numbers; so this will give you the turns ratio that is it
is essentially another way of showing that reflected current of this one is
𝐼1∗ =
𝑉2 ∗ 𝑉3 ∗
𝐼2∗
𝐼3∗
𝐼2 + 𝐼3 =
+
𝑉
𝑉
𝑉1
𝑉1
(𝑉1 ) ( 1 )
𝑉
2
3
𝑉
𝑁
𝑉
𝑁
2
2
3
3
𝐼
this is nothing, but 𝑉1 = 𝑁1. Similarly this is 𝑉1 = 𝑁1; 𝑎2 is this one.
So, I will get 𝐼1∗ ; is not, but I want to know 𝐼1 ok; if you have calculated 𝐼1∗ once again take
conjugate of both the sides. So, 𝐼1 then will be equal to 𝐼1∗ ; take once again conjugate and
the conjugate of the respective terms it will simply become
𝐼1 = (𝐼1∗ )∗ =
𝐼2
𝐼3
+
𝑉
𝑉
(𝑉1 ) (𝑉1 )
2
3
Mind you this 𝐼1 you will get magnitude an angle therefore, what is the effective impedance
𝑉
seen by the source it will be simply 𝐼1 and so on; everything is known now. Therefore,
1
although it looks like if there are several coils not even 2; there may be 3, 4 coils on the
secondary side and each one of those coils are supplying impedances; here is a simple way
of calculating the current distributions in the transformer. And also the power factor at
which the transformer will be operating and still we are in ideal transformer.
72
So, ideal transformer otherwise is so simple to handle with that is what I want to point out.
And in one stroke you will get both to real and reactive power supplied by the source and
this one. So, it is a very useful method; if you know the transformers are ideal, one way of
calculating the currents is to apply this power balance. Mind you power balance means
both reactive and real power should be balanced like that.
So, here is a nicer way of handling the situation; you have understood this. Only thing is
if I want to draw the equivalent circuit; refer to the primary side what it will be? This
equation will help me out equivalent circuit; refer to source side; what it will be? Source
side, so 𝑉1 will remain 𝑉1 that is there; then this current
𝐼1 = 𝐼2′ + 𝐼3′
𝐼2′ =
𝐼2
𝑉
(𝑉1 )
2
𝐼3′ =
𝐼3
𝑉
(𝑉1 )
3
Therefore, this impedances must appear in parallel such that one is 𝐼2′ and everything will
be this is the 𝐼3′ . And this impedance I will write 𝑍2′ this first one as this one; I will write it
𝑁
2
as 𝑍2′ = (𝑁1 ) 𝑍2 ; this impedance. And this impedance reflected impedance on the primary
2
𝑁
2
side; I will write it as 𝑍3′ = (𝑁1 ) 𝑍3 .
3
So, you see if there are several secondary coils and each one of them is supplying load;
then the equivalent impedances referred to the primary side; they will be not in series they
will be in parallel. So, this a lot of problems; we will be able to handle when ideal
transformers are connected in different ways supplying different kinds of load.
So, this one is very useful power balance; try to use that power balance means complex
power balance, both real and reactive power delivered by the source. That must be equal
to sum of the real powers delivered to 𝑍2 , 𝑍3 plus some of the reactive powers based on
that this equation is written and you get everything. And please go through this concept
the 𝑉1 𝐼1∗ ; solve these get 𝐼1∗ , but do not forget to take once again the complex conjugate so
that it becomes 𝐼1 and in the system ok.
73
So, this is the thing I wanted to tell about ideal transformer. So, after doing this then we
ask ourselves that what about a practical transformer. So, what I will do now? I will now
bring the realities of a practical transformer into an ideal transformer. I will go on adding
the realities what are present in a practical transformer; start thinking that initially it was
ideal transformer; then you have a practical transformer what are the things there.
For example, there will be rotor resistor I mean resistance of primary secondary bindings;
which I assume 0, there may be leakage flux which I assume 0 and there may be core
losses which may be 0 and there will be a finite magnetic magnetizing current.
(Refer Slide Time: 21:17)
So, a practical transformer will start with. Add realities to an ideal transformer. So, first
reality what I will add is suppose magnetizing current; current is finite; is finite and cannot
be neglected.
Suppose it is like this what I mean by this? This is my practical it is somewhat a practical
transformer still lot of ideal qualities are present, still resistances are neglected core loss is
not there, but only I am telling the core material is having a finite 𝜇𝑟 values. So,
magnetizing current cannot be neglected and I want to model it. So, this is a somewhat
practical transformer having a finite magnetizing current.
Now, let us try to understand what is happening with this switch opened; this we discussed
earlier with this switch is opened secondary impedance is 𝑍2 , what will be the ammeter
74
reading now? It will be magnetizing current; in case of ideal transformer absolutely ideal
transformer with secondary no current; primary current will be 0, but with this switch when
even if it is open, primary will draw a current 𝐼𝑚 finite magnetizing current.
So, it is like this then in the phasor diagram if I draw this is suppose 𝑉1; there is now finite
magnetizing current 𝐼𝑚 . And which will be lagging the supply voltage 𝑉1 by 90° after all
inductance and this is the axis along which flux phasor can be shown. So, this is the open
circuit equivalent circuit and if this is 𝑉2; 𝑉2 = 𝐸2 because still drops are neglected 𝑉1 =
𝐸1 .
So, all voltages will be in phase; so this is 𝑉1 = 𝐸1 same phasor and depending upon the
ratio your 𝑉2 phasor will be here; if it is a step down transformer this is your 𝑉2. What is
the secondary current with S opened nothing 𝐼2 ; what is the reflected current? Nothing, but
what I am telling the moment you close it; there will be a secondary current 𝐼2 , what was
the MMF? Acting in the circuit before S was closed, it was 𝑁1 𝐼𝑚 .
What will be the MMF acting when S is closed? Net MMF once again has to be 𝑁1 𝐼𝑚
because flux remains same; therefore, if you close it dependent upon the power factor
angle 𝐼2 will be here. But primary immediately will draw additional current 𝐼2′ ; I have
assumed low voltage side to be this. So, 𝐼2′ will be of lesser length; high voltage side current
is less 𝐼2′ .
And 𝐼2′ is such that 𝑁1 𝐼2′ which produces flux in the clockwise direction, if it is 𝑁2 𝐼2 which
will produce flux in the counter they will balance of and once again MMF will remain
𝑁1 𝐼𝑚 ,
𝑁1 𝐼2′ = 𝑁2 𝐼2
but the question is what will be the current drawn from the supply? So, current drawn from
the supply will now have two components.
So, one is the magnetizing component 𝐼𝑚 plus this 𝐼2′ . And this I must say this will decide
about 𝐼1 are you getting? So, there was 𝐼2′ plus 𝐼𝑚 you add and you get the primary current;
this is a crucial point open the switch ammeter will read 𝐼𝑚 , close the switch ammeter will
read (𝐼𝑚 + 𝐼2′ ) this plus this.
75
𝐼1 = 𝐼𝑚 + 𝐼2′
Now, the question is this is fine, but how to model this, how to put this to an ideal
transformer I should add something to the ideal transformer so that this model; this
happening whatever is happening in this transformer is correctly modelled in a; what do I
mean by modeling? I means that I will add some parameter some components in an ideal
transformer so that whatever I have seen is happening in a practical transformer.
I will add some components in that ideal transformer which will explain that thing
correctly that is the idea. See it can be very nicely drawn; so this is practical transformer
mind you, this is practical transformer what I will do now?
(Refer Slide Time: 29:15)
I will say consider this is ideal transformer; I will tell, I will put a dotted mark around it.
This is ideal transformer and what I will do across it; I will connect a reactance 𝑗𝑋𝑚 fixed
reactance. Mind you this is not primary coil; this is a reactance added to an ideal
transformer and I am saying it will represent correctly a practical transformer which is
having only finite magnetizing current; still it is not having any resistance, leakage flux,
any core loss.
So, this is a reactance; added by me to an ideal transformer which I have shown in a box
and with same turns ratio 𝑁1 : 𝑁2 like this. So, 𝑗𝑋𝑚 is external to this ideal transformer and
it is no winding I am emphasizing this; this is the primary coil. That is what I am telling
76
that your practical transformer is like this fine and here you observed you; nothing is
connected you will see ammeter you showing magnetizing current connect a load it will
draw a current 𝐼1 = 𝐼𝑚 + 𝐼2′ ; that is what we have seen.
Let us see whether really it can be; it can really faithfully follow the rules of this practical
transformer, it can faithfully represent the happenings whatever is happening in a practical
transformer. So, here is your 𝑉1 fine applied voltage; you see if there is a load connected
like this 𝑍2 with S opened; with S opened this as I told you is a reactance I have model the
transformer thought I will represent.
𝑉
So, what is this current? 𝑗𝑋1 ; that is 𝑉1 is here with S opened I know magnetizing current
𝑚
90° lagging. And this 𝐼𝑚 ; I will choose the value of 𝑋𝑚 means such a way that it is
𝐼𝑚 =
𝑉1
∠90°
𝑋𝑚
So, 𝑋𝑚 is called magnetizing reactance; therefore, how do I find out that? With S opened
𝑉
in a practical transformer with S opened measured 𝐼𝑚 ; 𝐼 1 gives you the value of 𝑋𝑚 and I
𝑚
write it like this with S opened; with S closed listen carefully. Suppose I have closed S;
this current is 𝐼2 fine, but this portion is an ideal transformer with S opened, there was no
current here, there was no current here; this transformer which is ideal has no magnetizing
current. And we know that when S was even opened, current drawn from the supply was
vanishingly small for this transformer 0.
But when I close this; this current has to be 𝐼2′ because it is an ideal transformer we have
I
𝑁
spent so much time on that and this is 𝐼2′ = 𝑎2; where 𝑎 = 𝑁1. I mean it should not disturb
2
you or to anybody that I have applied a voltage; something is connected between these
two; why there is no current when S was opened.
This we have discussed that length previously; this is an ideal transformer whatever little
current I mean ideal means ideal me what is infinitely large; you do not require any
magnetizing current for this portion. And we know ideal transformer how it behaves
connect load there will be reflected current here; 𝐼2′ .
77
And then the current drawn from the supply will be 𝐼𝑚 + 𝐼2′ . And this will be your 𝐼1 you
have to add this and this is exactly correctly representing what I have observed in a
practical transformer with S opened practical. S opened current drawn is 𝐼𝑚 ; is it?
Yes, in this model this is model mind you; this model comprises of an ideal transformer
with a reactance with S opened; current drawn is 𝐼𝑚 because there was no 𝐼2 there is no 𝐼2′ ,
𝐼1 = 𝐼𝑚 . With S closed this current in a practical transformer we have seen it is 𝐼𝑚 + 𝐼2′ .
Yes, it is correctly representing; close the switch from the knowledge all of this portion is
ideal. So, here now we will appear a reflected current 𝐼2′ and 𝐼2′ + 𝐼𝑚 will give you 𝐼1 .
We will continue with this in the next lecture, but please go through this portion very
carefully.
Thank you.
78
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 09
Modelling of Practical Transformer - II
(Refer Slide Time: 00:26)
Welcome to the 9th lecture and you recall we were gradually going towards a practical
transformer. And in my last class, this was the slide, where this is a practical transformer.
Here there is no 𝑋𝑚 connected; you must understand that this is a practical transformer in
the laboratory ok; two secondary, two primary, there was nothing like 𝑋𝑚 connected there.
But it was our thought process, which led us to believe that this whatever is happening to
a practical transformer, what is happening with S opened, current drawn from the supply
is finite, magnetizing current 𝐼𝑚 and when S closed, it must be 𝐼𝑚 + 𝐼2′ , because net MMF
in the circuit has to be 𝑁1 𝐼𝑚 , why? Because flux in the core is decided by supply voltage
and frequency, nobody has any say on that.
But if you are trying to disturb the secondary, I mean if you are trying to pass some current
through the secondary, primary cannot be a mere spectator to this happening. It will react
immediately by drawing an extra current 𝐼2′ , and that 𝐼2′ cannot be of any magnitude; it has
79
𝐼
to be 𝐼2′ = 𝑎2, so that 𝑁1 𝐼2′ = 𝑁2 𝐼2 , so that net MMF once again is 𝑁1 𝐼𝑚 and this is what we
have discussed last class.
So, this is a reactance, this is an ideal transformer and it correctly models it. So, 𝑉1, 𝐼𝑚 I
have drawn, then I am completing the phasor diagram once again. And suppose, this is 𝑉1
to the ideal transformer applied voltage is 𝑉1, so this is 𝐸1 , this is 𝐸2 and this is 𝑉2; 𝑉2 =
𝐸2 .
So, depending upon the turns ratios your, so this 𝑉1 is nothing but 𝐸1 , they still remains
nothing in between, and your this will be your 𝑉2, which is same as 𝐸2 . And this is suppose
the load power factor decide what is the current here 𝐼2 , then current drawn from the
primary will be 𝐼2′ , 𝐼2 we have got, get 𝐼2 .
And then this 𝐼2′ + 𝐼𝑚 if you add, you will get primary current 𝐼1 , understood. This is how
magnetizing current can be taken into account of a it is somewhat a practical transformer,
still not a full practical transformer. I have neglected, so many other things till now, only
magnetizing current, I have been incorporated and for that an external element 𝑗𝑋𝑚 is to
be selected.
If you like you put 𝑗𝑋𝑚1; side 1 ok, because we have seen parameter value changes from
side this side to that side and so on. Therefore, what will be the equivalent circuit looking
from the primary side, it will be 𝑉1, here is a reactance; no windings this one 𝑗𝑋𝑚 and here
will be the impedance 𝑎2 𝑍2 , you know this will be the equivalent circuit refer to the
primary side 𝑗𝑋𝑚1 .
So, whatever impedance, voltage, this that are there and this current I will show as 𝐼2′
reflected current and this current I will show as magnetizing current and this current is
your 𝐼1 . Mind you I am not drawing, but these are dots that is very important with respect
to this. So, this is the equivalent circuit refer to the primary side.
What will be the equivalent circuit refer to the secondary side, refer to source side; refer
to primary or source side, primary side. And refer to load side or secondary side, it will be
secondary voltage remain secondary voltage; this will be 𝑍2 , secondary things I should not
disturbed, they are already there 𝑍2 , this voltage is 𝑉2, this current is 𝐼2 , I will show it.
80
′
And this fellow the transfer of impedance from this to that side will be 𝑋𝑚1
, what is X m1
dashed
′
𝑋𝑚1
=
𝑋𝑚1
𝑎2
𝑉2 =
𝑉1
𝑎
what is this 𝑉2?
So, this is the equivalent circuit refer to the secondary side. So, you solve this you write
′
reflected current. And then you will what current you will get, I will get 𝐼1′ .
𝐼𝑚
So, you either solve this circuit get everything, because if this somewhat practical
transformer, solve this circuit get 𝐼2′ and then predict what will be 𝐼2 . So, people always
refer to work on a equivalent circuit instead of drawing some coupled coils, then
individually calculated. We can do that, but this is a better way of doing things.
(Refer Slide Time: 07:36)
After I have done this, our next reality which is present in a practical transformer is let us
assume, I will assume winding resistance. Till now, I have neglected winding resistances
𝑟 not 0, that is 𝑟1 is present, 𝑟2 is present that is the what is 𝑟1, this winding; one side, side
𝑙
1 resistance and this winding has also got 𝑟2 , because resistance is after all 𝜌 𝑎.
81
So, there is some resistivity of this conducting material, some cross sectional area, so many
turns are there, so many length. Of course, the resistance will be pretty small, it is made of
very good material for example, copper. But none the less there will be some finite
resistances of primary side and secondary side.
And it looks like, this is resistance which is of course, distributed can be considered to be
lumped and I can show it like this. Similarly for this coil, this resistance, I will considered
it to be lumped and represent it like this, this is how I can represent.
Now, the question is how this 𝑟1, 𝑟2 should be shown in this transformer. In this
transformer, there was finite magnetizing current; no resistance, no leakage flux, fine and
this is 𝐼𝑚 magnetizing current taking care of by an 𝑋𝑚 .
So, now once again I will add some parameters to this circuit that is 𝑟1, 𝑟2 in appropriate
place that is very important, so that effect of 𝑟1 and 𝑟2 will be addressed by this model.
Question is where should I put 𝑟1, should I put it here, should I put it there, where that is
an interesting question.
This portion is ideal mind you therefore, if there is winding resistance, what it is going to
do. There will be whenever this coil will carry current your practical transformer, there
will be a voltage drop here in the resistance. Applied voltage minus this drop is going to
create flux, not this full voltage is going to produce your magnetizing current, because a
portion of the voltage will be dropped in 𝑟1, got the point.
Therefore, in this circuit I will spoil this circuit, do not mind; what I will do now if I add
something here, our previous thing will also get disturbed. So, what I am telling is this
what I will do now listen carefully, here I will draw the ideal transformer and I will put a
dotted box around it in order to indicate that and then these are the terminals of the ideal
transformer ok.
And then I am telling to this ideal transformer, this winding is not purely this thing. So, so
you are there will be resistances in series, I will connect it here; which is small resistance
nonetheless 𝑟1, it will come in series.
Now, the question is should I put that magnetizing current branch 𝑋𝑚 , before this or after
this, I must put it after this 𝑗𝑋𝑚 . Similarly here of course, there is no magnetizing branch,
82
𝑟2 is simply comes here. I have shown only one way I am showing not like this resistance;
because these resistances are small, it only indicates that a small resistance in series.
Therefore, you know this is this thing and here is your supply voltage, frequency 𝑓.
So, whatever current it supplies which is decided by it may be the load connected here,
ultimately some current is drawn and when that current flows through the winding, there
will be a voltage drop 𝑟1; and that voltage drop must be subtracted from your supply
voltage and the remaining voltage is responsible for creating flux and giving you the
transformer action, so that is why 𝑟1 should not be shown here, it must be after this.
Therefore, this is 𝑉1, this is ideal transformer, this is 𝑟2 .
Now, the moment 𝑟1 is present I must also distinguish between 𝐸1 and 𝑉1, there will be a
drop here in series. 𝑋𝑚 what do you think; its value will be low or high, its value will be
high you should not choose a magnetic material which requires very large magnetizing
current. See 𝑋𝑚 is what,
𝑋𝑚 ≈
𝑉1
𝐼𝑚
I should not choose a magnetic material which requires very high value of magnetizing
current, then 𝑋𝑚 will be low. Better and better material I use, which is not certainly ideal
its 𝜇𝑟 → ∞, may be 𝜇𝑟 = 5000 quite a large number.
So, 𝐼𝑚 will be small, 𝑉1 is fixed, so 𝑋𝑚 is general high that is why, I have written capital
letter and we so many turns, just to indicate that ok, 𝑟1 is small, small 𝑟2 . Therefore, the
𝑉
magnetizing current which will be flowing here, 𝐼𝑚 is not 𝑋 1 ; 𝑉1 minus this drop divided
𝑚
by 𝑋𝑚 .
And as you know, depending upon the degree of loading the magnitude of the current
drawn from the supply will change. Therefore, drop in this resistance, 𝑉1 minus this drop
is the voltage what is coming here across the ideal transformer. Therefore, the magnitude
of the voltage apply to this ideal transformer is will also we will not remain constant, as
we were thinking in case of ideal transformer apply 𝑉1. I was telling the level of flux 𝜑𝑚𝑎𝑥
is equal to applied voltage in an ideal transformer is equal to 𝜑𝑚𝑎𝑥 =
voltage and frequency is constant, 𝜑𝑚𝑎𝑥 gets decided.
83
𝑉1
√2𝜋𝑓𝑁1
and if applied
But now I come to know ok, the applied voltage to this ideal transformer strictly speaking
will not remain constant because of the presence of 𝑟1, this is constant mind you; 𝑉1 is
constant no doubt, but 𝑉1 minus this drop is what is applied here, what is this drop, this
drop depend on the magnitude of 𝐼1 this your practical transformer is carrying and the
magnitude of 𝐼1 = (𝐼2′ + 𝐼𝑚 ), 𝐼2′ depend on 𝐼2 and depend on 𝑍2 .
So, by as you change 𝑍2 , 𝐼2 is going to change, 𝐼2′ is going to change therefore, 𝐼1 is also
going to change, therefore drop in 𝑟1 is not constant as you change loading. Therefore
applied voltage to this ideal primary winding of this ideal transformer; strictly speaking is
not constant, only consolation is this 𝑟1 is quite small.
Therefore, what people say is this ok, 𝜑𝑚𝑎𝑥 will be approximately constant, because you
have not certainly designed a transformer with high value of 𝑟1 and 𝑟2 , then no one is going
to buy your transformer I mean why, there will be unnecessary power loss in the windings.
Therefore, you must see that very good material is used for example, copper whose
𝑙
resistivity is very low. So, 𝜌 𝑎 the winding resistances are small, so that that way this
assumption that 𝜑𝑚𝑎𝑥 practically remain same from no load to full load is good enough.
Anyway none the less, let us see try to draw the phasor diagram of this one and try to
understand the implication of this. Now, what I will be doing here listen carefully; here I
will start with 𝐸1 and 𝐸2 ok, this voltage and this voltage. Mind you, here is now that 𝑉2
and 𝐸2 will not be same is not, similarly 𝑉1 and 𝐸1 will not be same, because in between
this two sources some 𝐼1 𝑟1 drop here, some 𝐼2 𝑟2 drop here will come, they cannot be same.
But nonetheless, we can do this things. Suppose how to start the phasor diagram drawing,
I will do it like this. Suppose, I draw it will be slightly clumsy, but let it be, but follow my
argument ok; you will draw 𝑉2 first whatever it will be; I draw 𝑉2 arbitrarily, vertically. If
I know 𝑉2, then I can fix up where the 𝐼2 will be is not, because load power factor angle is
known. Suppose, power factor angle of the load is 𝜃2 , 𝑉2 and 𝐼2 , I have drawn with S
closed.
Then I will say, look here your
𝐸2 = 𝑉2 + 𝐼2 𝑟2
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follow the logic, the diagrams may be a bit clumsy, but this is what is going to happen.
Suppose, 𝑉2 is known, 𝑉2 and 𝐼2 you draw; then I have to add to 𝑉2 , this 𝐼2 𝑟2 drop. Then I
will add it what is there and 𝐼2 𝑟2 drop will be very small, because 𝑟2 is small, 𝐼2 𝑟2 follow
the logic that is all.
If you do that, then what I am telling you will get your 𝐸2 . The induced voltage in this
ideal transformer, which I will draw by a red line here, this will be your 𝐸2 . It is an ideal
transformer, this is 𝐸2 , then you can confidently draw your 𝐸2 . If number of turns of this
side is higher, then it length of 𝐸1 will be higher than 𝐸2 , 𝐸1 is drawn. So, this will be your
𝐸1 .
And if this is 𝐸1 , this is 𝐸2 , your magnetizing current will be perpendicular to 𝐸1 , 90°
𝐸
lagging. Here is your finite magnetizing current, 𝐼𝑚 is not? I have got 𝐸1 , so 𝑗𝑋1 = 𝐼𝑚 , I
𝑚
get; so, this is the magnetizing current. And it is an ideal transformer with S opened, 𝐼2
was 0 this current was 0 with S closed, 𝐼2′ will appear here, nowhere else it is an ideal
transformer this portion.
So, 𝐼2 is known and 𝐼2′ will be in same phase with that so I will get 𝐼2′ ; whether its length
will be higher than 𝐼2 or not that depends upon the ratios. So, this will be your. .
Student: (Refer Time: 23:33).
It will be less. So, what I will do is I will make these as 𝐼2 , and this is 𝐼2′ , because I have
shown 𝐸1 𝐸1 is higher. So, reflected current must be lower 𝐼2′ , so this is your 𝐼2′ .
Then I will say that this current 𝐼1 = 𝐼2′ + 𝐼𝑚 . So, 𝐼𝑚 is known, 𝐼2′ is known; so I will add
this two, mind you I have shown 𝐼𝑚 slightly higher length it is not so, 𝐼𝑚 is small, anyway
whatever I have drawn. So, this will be your 𝐼1 and if this is 𝐼1 , then I will say your 𝑉1 =
𝐸1 + 𝐼1 𝑟1. So, 𝐸1 is known to this add 𝐼1 𝑟1 parallel to this and from this to this, then
wherever you will end up that will be your 𝑉1.
I think you have got the idea, see life will be will not be so much complicated as we
proceed, but what I am telling this is exactly what, how to draw phasor diagram. Start with
I could start with 𝑉1, but it is better you, because the primary current drawn decided by
load, it is much more easier. So, I will quickly go through this step, so that you can
understand.
85
Suppose, switch is closed ok, i will clean this and once again redraw. So, I will just I will
(Refer Time: 26:00), but the idea is very important you must keep this in mind. So, I am
redrawing once again very quickly, what I am telling ok, you have close the switch the
circuit is operating, choose 𝑉2, you start with 𝑉2 ok.
(Refer Slide Time: 26:26)
Voltage applied across the load is 𝑉2, then no one can contest me, then the current is
𝐼2 =
𝑉2
𝑍2
𝐼2 suppose lagging power factor load, 𝐼2 will lag here, 𝐼2 will be higher low voltage side;
I assumed I mean, while drawing ok, so 𝐼2 . The moment you know 𝐼2 you can find out 𝐸2 ,
because
𝐸2 = 𝑉2 + 𝐼2 𝑟2
These lengths are small.
And then you get this length to be 𝐸1 , I am simply repeating, because for the first time you
𝑁
are doing this you get 𝐸1 , 𝐸2 sorry. If you get 𝐸2 , this ratio of voltage is 𝑁1 absolutely no
2
𝑉
𝑁
𝑁
2
2
2
doubt; mind you this 𝑉1 may not be 𝑁1 strictly speaking, but this induced voltages are 𝑁1.
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So, 𝐸2 is this, then I know 𝐸2 and 𝐸1 are in phase, so I can get 𝐸1 assuming 𝑁1 is greater
than 𝑁2 , it will be this.
Then once I know 𝐼1 , I know the magnetizing current which will be 90° lagging. And this
current is small, let me draw it now correctly 𝐼𝑚 small current; 𝐼𝑚 and once I know 𝐼𝑚 and
𝐼2 is known. So, 𝐼2′ will be this one here will be your 𝐼2′ . So, current drawn from the source
will be 𝐼1 = 𝐼2′ + 𝐼𝑚 , this is your 𝐼1 ; and if this is 𝐼1 , use this one 𝑉1 = 𝐸1 + 𝐼1 𝑟1, whatever
it is, parallel to this 𝐼1 𝑟1 and you get your 𝑉1.
Anyway I am stopping now, but try to understand. So, this will now truly represent, it is
somewhat practical circuit with the finite magnetizing current and winding resistance. In
the next class, we will bring other realities into an ideal transformer to get a somewhat
better picture of the model.
Thank you.
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Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 10
Modelling of Practical Transfomer - III
Welcome to this 10th lecture on Electrical Machines-I and we were trying to get a realistic
model of a practical transformer. Ideal transformer is real ideal where no winding
resistance, no core loss, no leakage flux and things are very simple and I know how to
calculate secondary current, primary current, draw the equivalent circuit so on. Then what
I started telling you that to an ideal transformer.
If we go on adding some parameters some elements properly then that will be the model
of a practical transformer. We are trying to model a practical transformer. In terms of
model of a practical transformer, I am trying to get by an ideal transformer plus some
circuit elements, we are adding circuit elements.
(Refer Slide Time: 01:19)
And I told you that till now what we have done is, we have added the winding resistance
𝑟1 of the primary and this is my ideal transformer and here is your actual load on the
secondary side 𝑍2 , and here also I showed the secondary resistance 𝑟2 . And then the
88
magnetizing current was also taken into account and it was not a winding, it is 𝑋𝑚 , it is
my thought process which brought me here. This is my applied Voltage.
And these are the current when it is closed it is 𝐼2 , remember this is 𝐼2′ and this is 𝐼𝑚
magnetizing current, which is very small 𝑋𝑚 is high. Mind you I write it here 𝑋𝑚 high, 𝑟1,
𝑟2 are small that is why people do while analyzing suppose this circuit is given practical
transformer ok. Apply the rules of ideal transformer suppose you do not have any
parameter values known 𝑋𝑚 𝑟1 𝑟2 .
You for example, if I say a it is useful to know, so that suppose the same transformers, one
transformer is rated like this 200V 100V 1 kVA 50 Hz transformer is given single phase.
Suppose I say that it is suppose a practical transformer ok. You apply 200 V here you and
I say you connect an impedance say 10 Ω. Practical transformer it is, but what I am telling
suppose you want to quickly calculate. Assume this transformer to be ideal because after
all this 𝑟1 𝑟2 are small 𝑋𝑚 is high.
And then calculate the currents this side 100 V 10 Ω 10 A sorry. This current is 10 A and
say that this current is 5 A and this Voltage is close to 100 V if we have apply 200 V here,
if we have apply 200 V here 200 V.
So, while estimating this is a practical transformer suppose, but for quick calculations you
do this and to this practical transformer you really connected 10 Ω and 200 V what I am
telling is you will not be very far away from this 10 A this 5 A, it will be very close to that.
So, the concept of ideal transformers in many ways help me or very complicated circuit so
many transformer. Each one will have 𝑟1, 𝑟2 , 𝑋𝑚1, 𝑋𝑚2, but I know 𝑋𝑚 is high 𝑟1 𝑟2 are
small neglect them. For getting a quick idea what will be the order of the current, what
will be the order of the Voltage across the load?
It may be in a practical transformer instead of 100 V if you have applied 200 V because of
this drop it may be 96 V. But you care, you get the ideal transformer helps you to calculate
the number so easily and you can always say it is very close to the correct answer, it is not
a main achievement using the concept of ideal transformer wherever possible you do that.
89
So, that is there, but still now let us see whether those parameter values can be taken in to
account. If it can be taken how and to do this I told you this portion is ideal transformer
and go on adding parameters do it like this.
Next thing is I have assumed in the transformer that all the fluxes are leakage, leakage flux
for example, consider the core of the transformer. Suppose we have a transformer this is
the thing and what I have assumed earlier is that here is your some winding, and I assume
that all fluxes were confined to the core. In reality in a practical transformer it is not going
to be like this.
There will be flux lines, most of the flux lines will be confined to the core because it will
be highly permeable material. So, most of the lines of forces will be confined may be 2
percent, 3 percent, 4 percent flux line will complete their path through the air gap.
Therefore when the conductor carry some current, it produces flux a little amount of flux
will be leakage flux. So, this is called the mutual flux, this is called the leakage flux. It is
𝑑𝜑
this mutual flux which is going to give you transformer action is not 𝑑𝑡𝑚 this is the Voltage
induced in the coil no leakage flux, the flux which is common to both the windings. So,
this is leakage flux here 𝜑𝑙1 .
𝑑
Therefore the induced Voltage in the coil of the transformer will be 𝑁1 𝑑𝑡 (𝜑𝑚 + 𝜑𝑙1 ) say
this is the induced Voltage in the coil and this can be separated out
𝑁1
𝑑
𝑑
𝑑
(𝜑𝑚 + 𝜑𝑙1 ) = 𝑁1 (𝜑𝑚 ) + 𝑁1 (𝜑𝑙1 )
𝑑𝑡
𝑑𝑡
𝑑𝑡
Now, this portion this term is not going to transfer energy from the primary to the
𝑑
secondary side. It will simply cause a Voltage drop and it can be written as 𝑁1 𝑑𝑡 (𝜑𝑙1 ) =
𝑑𝑖
𝐿𝑙1 𝑑𝑡𝑙1 . I mean something like that primary current whatever is flowing leakage flux.
A detail analysis I have done it in my machine two course in the first few lectures, electrical
machine two course, in the first 2, 3, 4 lectures in the same NPTEL domain. If you please
go through that it will, I will put a note on that, but it looks like the applied Voltage total
flux created by the winding currents will comprise of two portion, one is the mutual flux
which will be totally on the core, another is the leakage flux. This leakage flux will
90
complete their magnetic path through the primarily through the iron. Their reluctance is
high leakage flux itself will be small.
But nonetheless it will cause an extra Voltage drop. Therefore, it looks like applied Voltage
I have given, there will be a Voltage drop in 𝑟1, but there is a small inductance drop which
will also take place to take care of this Voltage here. In other words what I am telling out
of this total applied Voltage a portion will be drop in the small resistance 𝑟1 and also there
should be a small inductance drop taking place and the remaining Voltage is as if going to
the ideal transformer giving you mutual flux alone.
So, it looks like it should be like this then let me first draw. So, leakage flux the term I
have written leakage flux, you write it leakage flux causes an extra Voltage drop, extra
small Voltage drop in the coil in the practical coil. What does that mean?
(Refer Slide Time: 13:01)
It means that if first I draw the ideal transformer 𝑁1 , 𝑁2 and this portion is suppose ideal
transformer, it is no connection dotted line and then we have seen we have got 𝑟1.
Now, what I am telling in series with 𝑟1 there must be another small inductance 𝑥𝑙1 and
then this is your 𝑗𝑋𝑚 and then this is your ideal transformer this is your 𝐸1 and this is your
𝐸2 . In the secondary side too it will carry current it will produce flux a portion of it may
be leakage flux and therefore, winding resistance as well as leakage reactance 𝑥𝑙2 and then
this two are the terminals and here is the secondary impedance you have connected.
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Therefore this is ideal transformer here all the fluxes are mutual mind you, all the flux here
𝑑𝜑
it is 𝜑𝑚 mutual flux plus 𝜑𝑙1 , 𝑁1 𝑑𝑡𝑙1 is shown as a drop here because this is your supply
Voltage now we know out of this total supply Voltage a portion will be dropped in 𝑟1
absolutely no doubt and there will be another small Voltage drop because of leakage flux
which is not going to create your magnetizing current, no that Voltage is consumed by the
leakage flux.
As in any inductance you know inductance whatever is the flux that is the drop here and it
is not mutually coupling to another coil therefore, no energy transfer because of 𝑥𝑙1 from
primary to secondary side. In the same way as in the primary secondary side to then will
have winding resistance another I have used by the small letters 𝑥 to indicate leakage flux
also will be small in a well designed transformer who wants more Voltage drop takes place
elsewhere and less Voltage is available for transformer action.
It is this Voltage 𝐸1 which is nothing but the applied Voltage minus this drop must appear
here. If 𝑥𝑙1 is high once again it is a very bad transformer out of this total Voltage
transformer winding impedance consumes. So, much and little Voltage, less Voltage is
available for energy transformer energy transfer. Therefore, remember that this is the
situation now.
So, leakage flux effect of leakage flux can be taken care of once again leakage flux I should
not write here no because applied Voltage that will be only consumed in the winding and
it must be in series with the applied lines 𝑗𝑋𝑚 this one. And once we do that I will repeat
this once again in that I do not know whether I have gone it a bit faster without defining
what is leakage inductance and magnetizing inductance. But I request you please go
through the first few lectures of electrical machine 2 course which is already uploaded in
NPTEL website. You will get a better feeling of this one.
Nonetheless so this is more or less the all the parameters we have added to an ideal
transformer to represent a practical transformer. This is the ideal part add these two things
and pretend that this is the model of practical transformer.
Only one thing I have yet to take that is called the core losses that I will discuss, but so far
as this one core loss not yet taken into account; not yet core loss to be taken into account
to be effect of; core loss to be taken into a account, that we will do. But before that the
92
addition of this one mind you I write it emphasis this point this is the small numbers, this
is high number high with respect to whom with respect to these numbers quite high,
similarly these are small .
Now, in this case once again let me do that because suppose s2 we closed, what will be
the thing? There will be a some current 𝐼2 here. The moment through the secondary of the
ideal transformer 𝐼2 flows through the primary of the ideal transformer 𝐼2′ will flow and
this current is 𝐼𝑚 and this current is 𝐼1 that is what is going to happen. How to draw the
phasor diagram? Suppose the power factor are same way.
Suppose, I will sketch once, so let us start. So, suppose you draw first let us assume this is
LV side. So, 𝑉2 is first sketch 𝑉2, I know load power factor angle; I will sketch 𝐼2 . Let me
draw it larger because low Voltage side current is higher. Suppose this is 𝐼2 , all are phasors
I have not put in those bars 𝐼2 ok.
So, 𝑉2 is there, but 𝑉2 and 𝐸2 are not same in between comes a small impedance. So, what
will be 𝐸2 ?
𝐸2 = 𝑉2 + 𝐼2 (𝑟2 + 𝑗𝑥𝑙2 )
The drops here like this is plus minus. So, 𝑉2 to this add this drops get 𝐸2 plus minus, this
dot I have put. So, 𝑉2 plus 𝐼2 𝑟2 to parallel to this and plus 𝑗𝐼2 𝑥𝑙2 and if you add this you
will get 𝐸2 .
If you get 𝐸2 then your 𝐸1 in phase with 𝐸2 , so your 𝐸1 I will draw a different color. So,
your 𝐸1 , I will fix up suppose this is your 𝐸1 . Once I get 𝐸1 , I will say oh my magnetizing
current is this Voltage by this 𝑗𝑋𝑚 . So, I will draw a magnetizing current, which is 90°
lagging 𝐸1 and if this is 𝐼𝑚 your 𝐼1 = 𝐼𝑚 + 𝐼2′ so, add this 𝐼𝑚 .
Student: Sir dashed.
𝐼2′ I am so sorry. So, 𝐼2 is this one so, 𝐼2′ will be less. So, 𝐼2′ correct. So, suppose this is 𝐼2′ .
So, it will be like this, then this was 𝐼2 mind you this was 𝐼2 . So, from 𝐼2 you get 𝐼2′ then
𝐼2′ + 𝐼𝑚 you add 𝐼2′ plus 𝐼𝑚 parallel to this, this 𝐼𝑚 and if you do you will get 𝐼1 ; and if you
get 𝐼1 then
93
𝑉1 = 𝐸1 + 𝐼1 (𝑟1 + 𝑗𝑥𝑙1 )
My target is to show all the variables. So, 𝐸1 I have reached. So, to this 𝐸1 you now add
𝐼1 𝑟1; 𝐸1 add 𝐼1 𝑟1 parallel to this wherever it is 𝐼1 𝑟1 and to this you add 𝐼1 which is 90°;
𝑗𝐼1 𝑥𝑙1 and get your supply Voltage 𝑉1 this will be the and here is the flux along this line.
I think you have got the idea it looks like oh it is quite involved, no not really. We will see
later how things will become much more easier with few approximations here and there
things can be further simplified. So, what I told you that it is almost a practical transformer
accept a core losses yet to be taken into account where I have considered the winding
resistances 𝑟1 𝑟2 , where I have considered the finite magnetizing current and the effects of
this are easily seen.
See I can make one comment. So, far I was telling that the moment you apply a certain
Voltage at the certain frequency the flux level in the core is decided and that is constant,
but when the transformer will be operating when the windings are carrying current applied
Voltage is not creating flux. Flux is created by this Voltage and this Voltage 𝐸1 is 𝑉1 minus
this drop, but the thing is 𝑟1 and 𝑥𝑙1 is small. So, 𝐸1 and 𝑉1, I can say magnitude of 𝑉1 will
be not exactly equal, but there magnitudes will be very closed by because of what because
this elements are dropped.
|𝑉1 | ≈ |𝐸1 |
Similarly, 𝐸2 and 𝑉2 will be also very close by because 𝑟2 and 𝑥𝑙2 are small. Therefore, we
can say roughly that the flux level even in a practical transformer it will change as you
change the value of 𝐼2 by varying 𝑍2 , 𝐼2 will change. So, this drop now with another value
of 𝑍2 it will be different. Similarly here 𝐼2′ will come 𝐼2′ + 𝐼𝑚 and so on.
So, this drops are small for a well designed transformer. Therefore, it is not very incorrect
to say that flux level does change appreciable from no load to full load condition the
change in flux level is pretty small because 𝑟1 and 𝑥𝑙1 small and 𝑟2 and 𝑥𝑙2 what, small.
Anyway in the next class what will do is this, we will tell you about the core loss which is
very important. Core loss we will see that it comprises of two components, one is eddy
current loss, another is hysteresis loss and when a practical transformer is operated, if you
touch the core you will find core is becoming hot ok. Although there is no electrical
94
connection between the conductors and the core. So, that is another interesting topics that
we will take up in my next class.
Thank you.
95
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 11
Core Loss - Eddy Current Loss
(Refer Slide Time: 00:19)
Welcome to lecture 11 on Electrical Machines-I and we have started discussing about
modeling a practical transformer. Earlier we started with ideal transformer which means
that windings has got no resistances, leakage flux is negligibly small, there is no core loss
and then it was so easy only thing when you connect a load on the secondary side there
will be a reflected current and that will decide the primary current. Similarly we found out
what will be the equivalent circuit refer to the source side that is primary side and to the
refer to the load side or secondary side.
Then we started discussing about practical transformer. I first told that let us first assume
that only there exists finite magnetizing current, it is not fully practical transformer, but
only a transformer having a finite magnetizing current, then that can be taken in to account
by connecting a an external considering an external resistance to be connected across the
primary side of the ideal transformer 𝑋𝑚 we called it magnetizing reactance.
Then we considered resistance as well as leakage flux and finally, last time we obtained
the equivalent circuit of a practical transformer as, this is suppose the practical transformer
96
where you apply voltage 𝑉1 and frequency 𝑓, then these are the terminal voltages you get
at the terminal 𝑉2 and 𝑓 same frequency voltage and these are dots and. So, with respect
to this two terminals which are available to me I can model it as an ideal transformer ok.
For example, you will model it by an ideal transformer like this, we will assume this is the
ideal transformer and that portion is ideal transformer.
And then you have connected some resistance and leakage reactance of primary and here
is your magnetizing reactance 𝑗𝑋𝑚 and then here also the secondary winding resistance 𝑟2
and 𝑗𝑥2 and this is 𝑟1 and 𝑗𝑥1 . To simplified matter I am using now a I am replacing 𝑥𝑙1 .
So, which is leakage reactance standing for I will simply write it by 𝑥1 in order to always
writing 𝑙.
So, 𝑥1 and 𝑥2 small letters I have used because there values are small leakage flux is small
and this portion is ideal transformer. And this voltage mind you these are the two terminals
if you call it A B the terminals of the practical transformer C D, then here is your A, here
is your B, here is your C, here is your D and within the transformer it can be then thought
of as if there is an ideal transformer with this external parameters I have connected.
And then we have seen how to this is 𝑉1 and 𝑓1 and this is 𝑉2 and 𝑓2 what is this voltage?
This is your 𝐸1 and this is your 𝐸2 mind you there is now distinction between 𝐸2 and 𝑉2
similarly with 𝐸1 and 𝑉1 these two are dots. And if you connect load here, then this current
is 𝐼2 phasor current this is 𝑍2 and if this ideal transformer we know if it carries current 𝐼2
coming out from the dot, immediately on the primary of the ideal transformer this is the
primary of the ideal 𝐼2′ 2 will come and this is magnetizing current 𝐼𝑚 .
Magnetizing current is produced by 𝐸1 and then to this the this current is the total current
which will be this current. Because I cannot go inside the practical transformer to see all
this things this things are not there in fact, it is my way of modeling so, that everything
matches it predicts correctly. So, this is 𝐼1 this is 𝐼2′ this is 𝐼𝑚 and this is your 𝐼2 and then
we drew the phasor diagram which I am not once again drawing.
Now, only one thing is now left for the which exists in a practical transformer, but we have
not taken note of that that is called the core loss; core loss. And core loss has two
components; one is called eddy current loss; eddy current loss and the other is called
hysteresis loss. We will first see why there will be eddy current loss and why there will be
97
hysteresis loss? We try to understand on what factors this losses depend and then how to
take that loss into account by incorporating a resistance in this equivalent circuit? That is
the goal because after all something becoming the as I told you this core of the transformer
will become hot, whenever there will be a changing magnetic field perpendicular to the
area of cross section exists in the core.
So, this will be the topic of study today. So, first we take the case of eddy current loss that
is what we will do. First we will understand the eddy current loss, then we will try to
impose it on a; mind you in a transformer the this was the case.
(Refer Slide Time: 08:17)
Here is the magnetic material and this material has got a you know a depth in three
dimension I am trying to draw, this is the core material and there will be coils here it goes
like this I am so, sorry. Anyway I will take, this one I will take it.
So, this is the coil and here we are connecting an AC source that is the thing. Secondary
coil I am not drawing to begin with. If this is excited from an AC voltage source 𝑉1 with
frequency 𝑓 then we know there will be current drawn from the supply which will be
flowing like this. Of course, this current will be time varying time varying current lagging
the approximately the voltage by 90° and so on.
But none the less there will be fluxes produced in the core that is 𝜑(𝑡) that too will be time
varying ok. And the cross sectional area of the core is a rectangle that is this areas, this is
98
the area and this area let us assume is constant throughout the core. So, flux will be flowing
perpendicular to the area any area you consider that flux is crossing perpendicular to that
area, this is called core area sectional area of the core this is the thing.
Mind you core is a say ferromagnetic material made of iron and it is solid like this and this
core material is also conducting. Core material can conduct electricity because iron not a
good material like your copper, but it also carries current it can allow current to flow core
material can conduct current ok. Core material means say iron say iron or ferromagnetic
material. So, they are good conductor of electricity.
Now, if you look at this cross sectional area the left hand column, I will draw it separately
in this way. This is the sectional area and here is the this side I am drawing understand this
is the thing. What I told through this flux lines are coming now, many flux lines see these
are 𝐵 or 𝜑. 𝐵 is also function of time because
𝜑(𝑡) = 𝐵(𝑡) × 𝐴
𝐴 is the cross sectional area.
Now, so, there is a time varying flux sometimes flux will go left hand side incidentally,
this vertical part of this core material they are called limb. This is also limb and this
horizontal portions are called yoke. So, in the limbs this is left limb, left limb I have drawn
and this winding I am not drawing it is there. So, I have cut it and trying to figure out what
is happening how flux limbs are flowing perpendicular to this cross sectional area. So, this
area is here I have the sectional area, got the idea. So, this is this.
Now, the moment there is a time varying flux, we know that if you have a closed turn.
Suppose a rectangular turn you imagine and there is a perpendicular to this there is a time
varying flux 𝜑(𝑡). Suppose a single turn you considered single turn of conducting material
single turn and suppose you have a time varying flux perpendicular to its area.
Then there will be induced voltage in this turn, what will be the value of this induced
𝑑𝜑
voltage? Number of turns that is 1 × 𝑑𝑡 and what you what will be the RMS voltage?
𝑅𝑀𝑆 𝑉𝑜𝑙𝑡𝑎𝑔𝑒𝑆𝑖𝑛𝑔𝑙𝑒 𝑇𝑢𝑟𝑛 = √2𝜋𝑓𝜑𝑚𝑎𝑥 × 1
99
See this turn is not this turn which makes the coil what I am telling, I am just considering
a separate problem all together suppose there is a single turn like this and you have a
induced volt flux which is crossing this as coil area in a perpendicular direction, then the
𝑑𝜑
induced voltage is 𝑛 𝑑𝑡 and its RMS value will be just like a transformer and that value
will be √2𝜋𝑓𝜑𝑚𝑎𝑥 × 1.
And if this coil is closed; you first imagine as if this coil I have not they I have connected
a switch suppose then across this coil too suppose it is opened that is let me do it like this
suppose there a imagine there is a. So, there is a flux here. So, between this two terminals
then and RMS voltage exist of this magnitude. Therefore, if the this turn is closed on to
itself that is this is closed, then I will say there was an induced voltage and now this induced
voltage will drive a current in this loop why not?
Because it has become a seat of emf as it happens in any turn in the primary and secondary
coil and it is if it is closed then I will say this RMS voltage whatever is induced here that
divided by the resistance of this turn will given the current. And that current too will be
time varying AC current will be flowing what will be the direction of the current? If
suppose 𝜑(𝑡) is going up and increasing then from Lenz’s law, I know the direction of the
induced current in this loop should be such that it will oppose this increase in flux that is
it will be by it went on like this. So, this was my turn. So, the direction of the current has
to be like this.
Who will drive this current? This induced voltage and the circuit is already closed if it is
a closed loop and this current has to be like this so, that it can oppose the increase in 𝜑(𝑡)
for which the induced voltage is there that was the cause. So, that will be the introduction
of the current it will try to oppose the cause.
And if I want to know what will be the if the current flows here, then there will be power
loss in this loop and that power loss will be 𝑖 2 into resistance of the path if you neglected
inductance of this loop a single turn inductance is neglected. In fact, to be very correct I
should say induced voltage divided by the impedance of the turn that decides the current,
but I am telling neglect the inductance of this coil why? Because single turn inductance is
proportional to 𝑛2 and so on let us neglect.
100
So, that is a information which is true when you are studying transformer or not. The
problem I have stated here a single rectangular turns and let us imagine there is a time
varying flux going up and down crossing this coil, then there will be RMS voltage induced
single turn that is why 𝑛 = 1 and this voltage will drive a current, but direction of the
current is you have to apply Lenz’s law arguing that let us imagine that flux is suppose
increasing in the upward direction, then the current induced in this loop that is this current
must flow in this direction. So, that it will try to oppose the very cause for which it is drew
that is the Lenz’s law.
With this background let us now come here in the transformer core what is happening. In
the transformer core as I told you this 𝑖(𝑡) magnetizing current has produced this 𝜑(𝑡)
here and imagine any section either here or there where not. And every section the same
flux is crossing and it is a solid piece of iron suppose solid iron.
Now, the question is where this fits this thing here it is like this. If I simply draw this way
try to understand first this is the suppose I am looking from the top, this thing I am looking
from the top, this is the section the flux lines are coming out. So, flux lines are suppose I
this is the direction of 𝐵, I have just shown one line 𝐵 is coming out and if it is coming out
and it is time varying doing going up and down.
Now, what you do, you consider a closed path like this. So, this closed path, I can imagine
there is a single turn here through which there is a time varying flux like this therefore, in
this path in this loop which I have considered there will be an induced voltage whose value
will be √2𝜋𝑓𝜑𝑚𝑎𝑥 enclosed by this area, not this full 𝜑𝑚𝑎𝑥 .
If you consider a single loop like this, then you have to take this area and then calculate
what is the 𝜑𝑚𝑎𝑥 × 1 and this switch is already closed you have nothing to do imagine any
closed path through this time varying flux perpendicular to this area and there will be
induced voltage and there will be if it is going up your eddy current loss direction of eddy
current in the single path consider should be like this, because your 𝐵 is coming out and it
will go inside because of this induced. It will try to oppose that cause 𝐵 was increasing
upward direction. So, it will try to oppose the cause your and this current is called 𝑖𝑒𝑑𝑑𝑦
eddy current.
101
Now, the big question is this thing does it happen to this selected so, called closed loop?
No it will happened in all the possible suppose this is the cross sectional area, you can
select a path like this, you can select a path like this are you getting these are paths of eddy
current eddy paths.
You can select a path like this any closed path you imagine there will be induced voltage
in that loop. You can figure out how much it will be and that divided by the resistance of
this path will give me the current in this loop, next loop, next loop I mean therefore, this
currents eddy currents will be I cannot avoid if it is a solid piece of iron and if we are
claiming that it is a time varying field exist here perpendicular to this cross sectional area,
then eddy currents will be flowing.
It is not in the coil induced voltage in the core and there will be eddy currents and because
of that this each of this loop current, 𝑖 2 × 𝑟 of this path 𝑖 2 × 𝑟 of this path. So, if it is a
solid piece of iron you will find it has become hot if you touch this one with your hand
core material if you touch it will become hot.
Therefore, there will be power loss, where the power loss takes place in the transformer in
the winding resistance coil resistances 𝑟1 𝑟2 whenever they will carry current there will be
power loss taking place that is incidentally called copper loss, that will come discuss in
detail.
But what I am telling if you touch the core it will become hot why it will become hot?
There is no electrical connection between the coil and the core which is made of say iron,
solid iron the reason one of the reasons will be because of this, because through the
sectional area a time varying flux exist and you choose any conceivable closed path there
will be induced voltage and eddy currents. It may not be rectangular any arbitrary path
loop now after looking at it if I say the path is like this will there be some eddy current like
this? Yes it will be.
Now, we shall try to understand what should I do to reduce this loss. No mathematics first
physically I mean try to understand why this eddy current loss occurs because so much
sectional area, so many area it looks like it will be very difficult to calculate the eddy
current loss it is not. In fact, it is not so, easy you have to write field equation this that
solve the current eddy currents I am try to estimate the eddy current power loss that we
have to estimate.
102
But I can understand by apply a common sense that this eddy losses can be reduced if I
instead of using a solid block of iron like this what I will do? I will use thin thin plates.
Suppose I say there is a iron plates I am drawing it still thicker this is I will go to next
page.
(Refer Slide Time: 27:47)
What I will do is this, this core material the this was the solid sectional area of the core
this is solid iron. Instead of doing like this what I can do? I will take plates thinner plates
and keep them side by side thin thin plates are you with me? That is the plates are like this
you have you chop this thing with it cutting tools slices like your pages of book thickness
of the book will be achieved.
So, you this is the width of this iron, this was the length whatever it is sectional area, the
same sectional area can be realized by instead of using a solid block of iron thinner plates,
that is you take one plate here, take another plate there and stack them together clamp them
to get I think you have got the idea.
So, the sectional area from the top if you look, it will look like this thin thin plates I will
show you. Suppose this is a thinner plates like this magnetic circuit, there able to see no
thinner plates. You iron plates and you put another like this I have brought only two, you
can put several such plates to have a height here that is the width of this magnetic circuit
this portion.
103
So, many thinner plates you use of small thicknesses and stack them together instead of
using a solid iron. But then not only that also what you do that is very important step. You
simply it is a solid iron you simply cut it in different pieces stag them together no not like
that then what you have to do is you have to give a varnish coating to each plate give a
varnish insulation paint it varnish insulation.
So, that each plate is electrically insulated from the next plate. Although they are side by
side kept like this the one plate, this is another plates you place it, but between these two
plates there will be no current can flow because this plate is insulated from this plate while
you are putting. So, each plate has its own electrical identity, it is not dictated by the others.
(Refer Slide Time: 31:35)
So, the these one this length and this is the width suppose this is the length, this is the
width of this area, this is the width I can realize that one like this that area will be realized
and I kept several thin plates together. And then where is your coil in this previous diagram
if you see while where is the coil if I draw it like this the sectional view, where is the coil?
Coil must be shown here this coil if you look from the top this is this one. So, coil was
here is not coil. I think you have got the idea this coil this primary coil it should be shown
around it creates flux like that.
Anyway; so, this will be the RMS induced voltage where that fellow has gone next page
this was the thing. Therefore, the coil is here still total flux
104
𝜑𝑚𝑎𝑥 =
𝑉1
√2𝜋𝑓𝑁1
is not approximately at least this is 𝜑𝑚𝑎𝑥 if you give. So, 𝜑𝑚𝑎𝑥 is perpendicular to the
sectional area coil is here, which has created this 𝜑𝑚𝑎𝑥 sin 𝜔𝑡 and it was doing like this
here also 𝜑𝑚𝑎𝑥 was in this direction here also 𝜑𝑚𝑎𝑥 in this direction ok.
Now, what is the motivation of using this number of plates? Motivation is if it is a solid
piece of iron the area you can have many number of closed path as I told you and each one
of them is carrying eddy current loss, eddy current loss will be created losses will be more.
So, by common sense then it looks like instead of using a solid block of iron I will now do
it one this page so, that you appreciate the point. Instead of using a solid piece of iron if I
use thinner plates like this, I am showing the gap larger so, that we understand the point.
So, this is a plate and there is 𝜑 perpendicular to this there will be eddy currents now, but
if you make thinner plates the eddy current paths or number of paths available will become
less for each of the plates. Getting each plate is electrically insulated from the other. So, if
you want to concentrate on a single plate you will see in this plate the eddy current loss
will be less because of the thing.
Therefore we, will use cannot the two path actually (Refer Time: 35:24) law this plates to
make a instead of using a solid iron better use thin plates thin insulated plates instead of
solid iron it will lesser than thing very much because the each plate is electrically insulated
from each other and eddy current will a.
In fact, if I say this thickness of each plate is. So, small you will not find any eddy path
existing of course, that is not the case we will discuss it in the next class this is interesting
and try to understand. My our goal will be to find out an approximate expression of this
eddy current loss and what factors it depends.
Thank you.
105
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 12
Factors on Eddy Current Loss Depends
So, we were discussing Eddy current loss, we will continue that topic in lecture 12.
(Refer Slide Time: 00:21)
And so, we understood no point in using solid block of iron because there will be so, many
Eddy paths, Eddy current loss will increase. Because, that Eddy current loss is in any case
is not going to transfer energy from primary to secondary which is our primary job. It will
be locally there will be heat generated, core will be heated up and there will be efficiency
will be poorer.
So, how to reduce that Eddy current loss? From common sense it look like that instead of
using a solid block of iron you better use thinner plates, stack them together and get the
sectional area. Therefore, it is now true that at least I have to use thin plates ok. Therefore,
in this topic what I am going to do is this to find out an approximate expression for Eddy
loss in a thin plate. So, if in one plate I can find out for all other plates it will be same. So,
this is the thing I am trying to get; to do this what we will be doing we have to do a little
bit of mathematics, but that is very simple no field equation nothing.
106
So, consider a thin plate, let me draw it here and this thin plate I will draw slightly larger
although its thickness is quite smaller. Thickness is of the order of 0.25 mm, 0.3 mm, but
I am drawing it in a larger scale. So, that we can understand what is going on and suppose
this is a single plate. And, it has got a length that side also got the point. In this thin plate
what we will be doing? We will make 𝑥 and 𝑦 axis, I will assign this to be my 𝑥-axis and
this is my 𝑦-axis and this is my origin, and suppose the thickness of the plate is 𝜏 is the
thickness.
Now, I have to find out an approximate expression of Eddy current loss; to find that out
what I will do is this, I will consider at a distance 𝑥 from the origin a loop like this got the
point. Let me complete then I will explain, see within that thin plate I have considered at
a distance 𝑥, this distance is 𝑥. And, elementary thickness 𝑑𝑥, why that 𝑑𝑥 is necessary
will be clear.
Our plan is to find out what will be the Eddy current induced voltage in this loop and this
loop exists in all the section. Here also it exist, all along the length this is suppose the
length of that magnetic circuit whatever it is. If you unfold that core material that length
over which that exist that is the somewhat the you got the point length of the magnetic
circuit anyway. So, this is the thing. Now, in this considered loop of thickness 𝑑𝑥, what
will be the induced voltage RMS induced voltage in this loop will be that is:
𝐸𝑒𝑑𝑑𝑦 = √2𝜋𝑓𝜑𝑚𝑎𝑥
𝜑𝑚𝑎𝑥 is what?
𝐸𝑒𝑑𝑑𝑦 = √2𝜋𝑓𝜑𝑚𝑎𝑥 = √2𝜋𝑓𝐵𝑚𝑎𝑥 (2𝑥ℎ) × 1
number of turns that is 1. Therefore, Eddy power loss sorry, Eddy power loss in this loop
will be equal to if you neglect a reactance of this loop only resistance single turn.
𝐸𝑑𝑑𝑦 𝑃𝑜𝑤𝑒𝑟 𝐿𝑜𝑠𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑙𝑜𝑜𝑝 =
𝐸𝑒𝑑𝑑𝑦 2
𝑅𝑒𝑑𝑑𝑦 𝑝𝑎𝑡ℎ
𝐸2
This will be the power loss 𝑟 what else because, this circuit is closed. Therefore, this I
2
will write it now I will just put these values it will be 2𝜋 2 𝑓 2 𝐵𝑚𝑎𝑥
4𝑥 2 ℎ2 . Now, the question
𝜌𝑙
is what will be these resistance? So, resistance of a path I will apply that basic formula 𝐴 .
107
Now the question is what is 𝜌? 𝜌 is the resistivity of this core material, 𝜌 is resistivity so
𝜌. What is 𝑙? 𝑙 will be how much the length of this path current path, it will be (2𝑥 + 2ℎ)
perimeter of this path divided by the cross-sectional area through which this current is
flowing.
Now, what is the here requires some observation that is what is the cross this loop as I told
you exist all along this length here also it exist. So, and current at a given instant of time
will flow this is the cross sectional area because, the loop exist here also for all the this is
not for this single fellow. Loop is positioned at a particular 𝑥 and it is here also and current
Eddy current path will be what? It will be perpendicular to this area got the point.
𝜌𝑙
So, 𝐴𝑟𝑒𝑎 will be then the area of this rectangle. What is the end of this rectangle? The length
of this iron piece and this will be then if I call this length to be 𝐿 which I wrote earlier
suppose, this length is 𝐿 up to whatever distance is both. So, this is 𝐿𝑑𝑥; 𝐿𝑑𝑥 this is the
cross sectional area therefore, this is the thing.
𝐸𝑑𝑑𝑦 𝑃𝑜𝑤𝑒𝑟 𝐿𝑜𝑠𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑙𝑜𝑜𝑝 =
𝐸𝑒𝑑𝑑𝑦 2
𝑅𝑒𝑑𝑑𝑦 𝑝𝑎𝑡ℎ
=
2
2𝜋 2 𝑓 2 𝐵𝑚𝑎𝑥
4𝑥 2 ℎ2
(2𝑥 + 2ℎ)
𝜌
𝐿𝑑𝑥
Now, you see as I told you we would like to find out an approximate formula ok, to have
a feeling on what factors Eddy current depends like that.
Now, since it is a thin plate the value of 𝜏 is much smaller than your this finite dimension
ℎ or 𝑙 whatever it is. The first assumption is this one, this equal to I will write it as it
2
becomes 8𝜋 2 𝑓 2 𝐵𝑚𝑎𝑥
ℎ2 𝑥 2 , this is these thing divided by this one. I will write it as this is
approximately equal to this 2𝑥; see 𝑥 is a subset of 𝜏, 𝜏 is already small compared to ℎ.
So, there will be no harm if you write it as 𝜌2ℎ, this 2𝑥 term can be neglected compared
to 2ℎ divided by 𝐿𝑑𝑥 which I will put up like this exactly.
𝐸𝑒𝑑𝑑𝑦 2
2
2𝜋 2 𝑓 2 𝐵𝑚𝑎𝑥
4𝑥 2 ℎ2
𝐸𝑑𝑑𝑦 𝑃𝑜𝑤𝑒𝑟 𝐿𝑜𝑠𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑙𝑜𝑜𝑝 =
=
(2𝑥 + 2ℎ)
𝑅𝑒𝑑𝑑𝑦 𝑝𝑎𝑡ℎ
𝜌
𝐿𝑑𝑥
≈
2
8𝜋 2 𝑓 2 𝐵𝑚𝑎𝑥
ℎ2 𝑥 2 𝐿𝑑𝑥
𝜌2ℎ
108
4𝜋 2 2 2
𝑑𝑃𝑒𝑑𝑑𝑦 =
𝑓 𝐵𝑚𝑎𝑥 ℎ𝐿𝑥 2 𝑑𝑥
𝜌
(Refer Slide Time: 14:41)
So, we got from the previous slide this is the thing therefore, this is the thing. Now the
question is how to get
𝑃𝑒𝑑𝑑𝑦
? So,
𝑡𝑜𝑡𝑎𝑙
𝑃𝑒𝑑𝑑𝑦
2
= 4𝜋 2 𝑓 2 𝐵𝑚𝑎𝑥
ℎ𝐿 ∫ 𝑥 2 𝑑𝑥
𝑡𝑜𝑡𝑎𝑙
So, I go to previous page; the question is what will be the limit of integration? See at a
𝜏
𝜏
distance 𝑥, I have considered the full loop. So, 0 to 2 will be the limit this is 2, where 𝜏 is
the thickness of the plate so, that will be the limit. So,
𝜏
2
𝑃𝑒𝑑𝑑𝑦 4𝜋 2 2 2
𝑓 𝐵𝑚𝑎𝑥 ℎ𝐿 ∫ 𝑥 2 𝑑𝑥
=
𝜌
𝑡𝑜𝑡𝑎𝑙
0
So, next step is straightforward
𝜏
2
𝑃𝑒𝑑𝑑𝑦 4𝜋 2 2 2
=
𝑓 𝐵𝑚𝑎𝑥 ℎ𝐿 ∫ 𝑥 2 𝑑𝑥
𝜌
𝑡𝑜𝑡𝑎𝑙
0
4𝜋 2 𝑓 2 2
𝜏3 1
=
𝐵𝑚𝑎𝑥 ℎ𝐿 ×
𝜌
8 3
109
=
=
So, the
1 2 2 2
𝜋 𝑓 𝐵𝑚𝑎𝑥 (ℎ𝐿𝜏)𝜏 2
6𝜌
1 2 2
𝜋 𝐵𝑚𝑎𝑥 𝑓 2 𝜏 2 × 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐼𝑟𝑜𝑛
6𝜌
𝑃𝑒𝑑𝑑𝑦
is given by this.
𝑡𝑜𝑡𝑎𝑙
Now, going to the previous page you also notice I have assumed another thing. When I
varied I have considered the loop not like this, but it covers as you vary 𝑥 it covers the
whole area then you are through ok. But, not only that when 𝑥 varies this portion remains
same as if I have calculated many times that area. But anyway, whatever I am estimating
I will be estimating slightly higher it is no harm. So, there is another approximation are
you getting, another loop you consider; you will consider it like this ok. This power loss
here, power loss here and once again this power loss, this power loss.
But, that portion is always small because I am talking about 𝜏 which is far less than ℎ.
Therefore, this is a very handy approximate expression mind you approximate, it will serve
my purpose approximate expression for Eddy loss. And, I know many things Eddy current
loss then depends upon the maximum value of the flux density squared, supply frequency
squared and thickness square.
Remember what is the fun of expressing it in this way; it is because of the fact I have
considered while deriving this with a single plates. There will be multiple plates and I
know oh it can be expressed in this way.
Therefore, if you have used to more number of plates then total volume is to be multiplied
are you getting that is also crucial to note. So, I will say better write
𝑃𝑒𝑑𝑑𝑦
1 2 2
=
𝜋 𝐵𝑚𝑎𝑥 𝑓 2 𝜏 2
𝑉𝑜𝑙𝑢𝑚𝑒 6𝜌
So, many Watt per unit volume and we say that then Eddy current loss will be proportional
to the square of the thickness of the plate. See all these things the argument was if you use
a solid block of iron Eddy current loss is bound to be more. So, many Eddy paths instead
of that you use the Eddy currents, it will be only confining here.
110
Because, the next plate is insulated from the other, if you do not use insulated plates then
the story remain same. The Eddy paths will be more are you getting, that is why the
insulation is necessary therefore, we know that. So, how to reduce then eddy current loss,
because it is a wasteful energy transformer will be on 24 hours and eddy losses will always
take place. If that be the case you try to use such a material whose resistivity is higher, 𝜌
should be made higher at least. This is apparent use small 𝜏 that is one factor, thickness of
the plate should be 𝜏.
(Refer Slide Time: 22:25)
It cannot be too small otherwise those thin plates you cannot further make it smaller and
smaller, it will become then brittle you know. So, the order of the thickness used is maybe
0.25 to 0.5 mm for 50 Hz transformer 𝜏 order of 𝜏. And, transformer size if you see they
may be of the order of distribution transformer, if you look at their heights may be in
meters.
So, the assumption that ℎ and 𝑙 compared to those values 𝜏 is very small is pretty good I
mean nothing like that. But it gives so, do not use just iron, try to use a better material that
is why people use silicon steel; add some small quantity exact percentage I do not
remember. You can see a design book at silicon to ordinary iron steel and make a special
type of a iron which will give you higher resistivity to reduce the loss. And, also try to do
not try to use too high 𝑓𝐵𝑚𝑎𝑥 which will also reduce and remember it is proportional to
frequency square.
111
So, if the transformer rating is 50 Hz based on that you try to make Eddy loss as small as
possible. So, please remember what was a logic why this was necessary to derive? It was
necessary to derive because in the transformer if you go to this previous page as well
previous page.
(Refer Slide Time: 24:41)
See in this transformer therefore, what I am telling if I spoil this page to this I will tell it
like this.
(Refer Slide Time: 24:49)
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It has come there this is the thing. So finally, it will look like this instead of a solid block
of iron, I will use thinner plates. So, this I am showing it by lines another plate, another
plate, another plate, I think you have got the ideas of this kind got the point. So, many
plates like this you stack them together and try to form a core material like this. These are
also called laminated plates, use laminated plates; in short it is called laminations each
plate laminations.
Then you make your winding not over a solid piece of iron, but like this another side it is
2
like this, make your primary and secondary. So, Eddy current loss it depends on 𝐵𝑚𝑎𝑥
, 𝑓2
and thickness of the plate square, thickness should be 0.2, 0.3 mm thickness and each plate
is insulated from the other. And, this will be also reflected here in this toy I think this is
ok. So, and this is the flux, total flux is 𝜑 by this area you a the cross sectional area; if you
look at from the top it will be now like this and your 𝜑 is coming like that.
So, average flux density is 𝐵𝑚𝑎𝑥 divided by because each plate will contribute to the area
that is there that overall area will give you the 𝐵𝑚𝑎𝑥 anyway so far so good. So, this is
called Eddy current loss. Now, how to take this Eddy current loss effect into the equivalent
circuit coming back to the original question. So, it looks like that even with the secondary
of the transformer opened, I mean you can look at it in this interesting way as well. This
is 𝑉1 𝑓 you have switched on the supply, secondary is open nothing like that, but there is
core material.
Now, when you close it there will be 𝜑 produced alternating 𝜑(𝑡) is produced which is
flowing like this and that are Eddy current losses. You can also think oh then I have applied
the voltage and in the core there were turns existing they had induced voltage in the same
way as your secondary coil had induced voltage and they were driving current. You can
also think like that ok, that is what I am trying to tell even with this switch open it looks
like as if there is a secondary coil existing within the iron core there and the circuit is
closed they are carrying secondary current.
But, flux has to remain same therefore, those tiny Eddy currents too will have reflected
current on the secondary side. I mean what I am trying to tell, I mean another way of
looking at the thing this is your voltage ok. And, suppose only Eddy current I am showing
the flux will be a around this point 𝜑 there is induced voltage in the secondary 𝐸2 that is
there.
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But, what I am telling there are some invisible secondary already present in the core. And,
they had induced voltage also in the same direction why not 𝐸𝑒𝑑𝑑𝑦 . And, I assume that
power factor of that Eddy current path to be unity, leakage inductance therefore, with
secondary open there is no 𝐼2 here that is fine.
But, here was a secondary it has the 𝐼𝑒𝑑𝑑𝑦 and why not it will have its reflected component
here in order that your flux in the core remain same. Therefore, apart from this magnetizing
current 𝐼𝑚 we now find there will be and current due to core loss which will be in phase
with the supply as it is expected.
This is how this current will be drawn in, another way of interpreting the result; got the
point. But, anyway what people say is that this Eddy loss after all the power has to come
from this. Therefore, there will be another small component of current drawn from the
source which must represent Eddy current that will be in phase can be physically
interpreted like this.
Now, after I have done this, I will come to this complete phasor diagram slightly later.
Before that I will tell you something about another component of the portion that is called
hysteresis loss ok.
(Refer Slide Time: 31:43)
What is hysteresis loss? Hysteresis loss is another loss which also takes place in the core
takes place in the core like Eddy current loss. Now, what is this loss? This loss is slightly
114
I mean of the track type things for, but we know why hysteresis loss should take place.
See after all if you have a magnetic material like this over which a you wound a coil and
are you getting.
So, suppose let me draw some coil say a piece of iron and here is a coil like this a coil is
wound and this length is more whatever it is. And, suppose you pass current in this way
some AC current connect current some AC current you will pass. If you pass current then
why this core gets magnetized?
It is because of the facts will not go to the quantum level. But we can understand there are
tiny magnets molecular level small magnets are there in any ferromagnetic material like
this, north south, north south, but they are see north south, north south, they are tiny
magnets already existing in a ferromagnetic material.
So, what you are doing the moment you have apply some I will draw now sectional
diagram, if you pass some current like this. So, inside this there are tiny magnets which
are haphazardly oriented and we do not see any magnetism in it. But, if you pass some
current, some DC current you pass to this fellow iron bar, that is how I make permanent
magnets you know. So, you pass some DC current therefore, there will be you have applied
𝐻 along this and each tiny magnets will try to align itself along n s all the magnets.
If you pass a little current they will slightly move and pass more current they will be further
aligned n s, make it more they will school level also we study like that. So, if you pass
sufficiently large current at the end what will happen is s n all tiny magnets will be moved
like this, such that this will become a north pole and this will become a south pole. That is
how we are taught in school level how to make a soft iron piece a permanent magnet, you
pass some electromagnet north-south.
Now, suppose this current is time varying therefore, what you are asking each molecule to
do if this is time varying when the current is positive each tiny magnets was trying to align
along this way. Then you reverse the current, they will now start moving and finally, when
the negative maximum current flows this side will become south. The other way it will
become it is north, it is south, it is north, it is south I am just drawing is not.
So, each one will move and finally, they will align provided you reverse the direction of
the current I am not drawing all these, these are very basic. Therefore, if it is the 50 Hz
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frequency current changing then you are asking each molecule to move this way, then that
way always they will be moving. Therefore, they will encounter friction in the process and
there will be heating effect not because of Eddy current, Eddy current is separate thing.
But, simply because you are reversing the currents this way that way therefore, there will
be always movement of the molecule.
That way you can imagine whether exact thing perhaps detail analysis of at physics level
you have to do. But, we can understand this is what is going to happen; each molecular
level tiny magnets they will be sometimes align this way, sometimes this way. Therefore,
core will be heated up not only because of Eddy current, but because of something else
here. And, this loss which will take place is called hysteresis loss and we will continue
with this in the next lecture.
Thank you.
116
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 13
Hysteresis Loss
Welcome to lecture number 13.
(Refer Slide Time: 00:21)
And we started discussing on Hysteresis Loss, why it takes place some physical reasoning
only I gave. And now to understand why hysteresis loss takes place we have understood
something, then we will try to see on what factors it will depend, as we did in case of eddy
current loss. But before that you I have told about the B-H characteristics of a magnetic
material.
Suppose you have a magnetic circuit like this let it be like this it does not matter and you
have a coil wound over it exciting coil. Here I will pass current 𝑖(𝑡). Now so far I have
𝑁𝑖
drawn the B H characteristics generally it will be like this, this is 𝐻 = 𝑙 ; 𝑙 is the mean
length of the flux path these we know and this is your B.
Now, this is how I draw and if you increase this current 𝑖 starting from 0 initially it was
not magnetized. So, as I go on increasing the current you will find B is increasing almost
linearly in the initial phase. Then after sometime it becomes saturated that is further
117
increase in H here will not cause any practically any increase in B the reason once again
is we can understand the tiny magnets which are north-south etcetera.
As you are increasing H external magnetizing force you are forcing the tiny magnets to be
aligned in a particular direction and after all the magnets are aligned south north; south
north like that whatever way. Then further increase in H what further to align everybody
has become a disciplined soldier and they are in line, that is why these saturation zone
takes place that way you can think.
Now, suppose I am increasing the value of H, but this we will see that what is the effect of
hysteresis? As you go on increasing the current B increases starting from 0 it goes like this
here. Before I tell you about that hysteresis loss or hysteresis loop let us try to understand
one important thing because I have to after all estimate the hysteresis loss. Suppose this
coil this current is increased from 0 value to a particular value here say some value I, this
axis can be treated also as current H or current because 𝑁 and 𝑙 are constant.
Suppose we have increase the value of the current I, as we have increased how much
energy I have supplied to the coil that I want to study. Since this current we have increased
𝑁𝑖
from 0 to capital I or 0 to some H which is to be multiplied by n. So, 𝑙 this 𝐻 value I want
to know how much energy I have supplied?
First of all have I supplied any energy to it the answer is yes. See because of the fact this
is the coil, I am increasing the current here therefore, flux linking with the coil is changing
gradually increasing therefore, it will become a seat of emf, just like transformer plus
minus 𝑒 and it will try to oppose the cause what is the cause, because current was
increasing. So, it will its polarity will be such that it this induced voltage is allowed to act
it will try to reduce the value of the current that is why you we got plus minus.
Now, in any circuit if an element has a voltage across it of this polarity if current is entering
through the positive terminal of that element, then we say it is absorbing energy. As in the
case of a charging a battery to the plus terminal of the battery you have to inject current
then only battery gets charge.
So, energy will be absorbed and voltage across this coil; so, how much energy is absorbed
in the process when 𝑖 increase the current from 0 ampere to some 𝐼 ampere, that is the goal
of the first part ok. What I will do is this as 𝑖 is increasing during this process 𝑖 is the
118
𝑁𝑖
function of time you know; so, induced voltage at any instant suppose the current is 𝑙 and
the 𝐵 value is 𝐵.
So, the and then the induced voltage will be
𝑒=𝑁
𝑑𝜑
𝑑
= 𝑁 (𝐵𝐴)
𝑑𝑡
𝑑𝑡
𝐴 is the cross sectional area, this is the voltage. And the polarity I have taken Lenz’s law
𝑑
into account do not under the confusion that −𝑁 I have written 𝑁 𝑑𝑡 (𝐵𝐴) is positive and
polarity I have very correctly put on this diagram. So, this is the induced voltage.
Therefore, instantaneous power absorbed by the coil this I am just writing 𝑝, instantaneous
power absorbed by the coil will be
𝑝 = 𝑖𝑁𝐴
𝑑𝐵
𝑑𝑡
Why 𝐴 is constant? 𝐴 is the cross sectional area, 𝐵 is changing. So, at any instant this is 𝐵
𝑑𝐵
ok, so 𝑁𝐴 𝑑𝑡 . Then energy supplied, so what I am essentially doing is let me write it like
this at time 𝑡 current is 𝑖, at time (𝑡 + 𝑑𝑡) which means a particular 𝐻, a particular 𝐵 at
time (𝑡 + 𝑑𝑡) suppose some time elapses current has I have increased from 𝑖 to (𝑖 + 𝑑𝑖).
So, that 𝐻 has increased from 𝐻 to (𝐻 + 𝑑𝐻) and 𝐵 has increased from 𝐵 to (𝐵 + 𝑑𝐵)
and all these changes have taken place over a time interval 𝑑𝑡 and that 𝑑𝑡 I can assume to
𝑁𝑖
be very small. So, 𝐵 it has become (𝐵 + 𝑑𝐵), here it was 𝑙 this point is
𝑁(𝑖+𝑑𝑖)
𝑙
and so on.
So, this is 𝑑𝐻 is not this is 𝑑𝐻, therefore, this is how things are going. Therefore, in time
𝑑𝑡 energy supplied I am write in black energy supplied, energy absorbed by the coil in
time 𝑑𝑡 will be simply
𝑝𝑑𝑡 = 𝑖𝑁𝐴
𝑑𝐵
𝑑𝑡
𝑑𝑡
this is the thing.
this 𝑑𝑡 can be struck off and you will be left with 𝑁𝑖𝐴𝑑𝐵 then what you do, you multiply
this with this mean length of the magnetic circuit 𝑙.
119
So, multiply and divide, so you will be getting
𝑝𝑑𝑡 = 𝑖𝑁𝐴
𝑑𝐵
𝑁𝑖
𝑑𝑡 = 𝑁𝑖𝐴𝑑𝐵 =
𝑑𝐵(𝐴𝑙)
𝑑𝑡
𝑙
(𝐴𝑙) becomes the volume of the core material cross sectional area into the average length.
So, this is volume of the core material this will be the case. So, energy supplied during this
same. So,
𝑝𝑑𝑡 = 𝑖𝑁𝐴
𝑑𝐵
𝑁𝑖
𝑑𝑡 = 𝑁𝑖𝐴𝑑𝐵 =
𝑑𝐵(𝐴𝑙) = 𝐻𝑑𝐵 × 𝑉𝑜𝑙𝑢𝑚𝑒
𝑑𝑡
𝑙
but what is 𝐻𝑑𝐵; 𝐻𝑑𝐵 is nothing but this area; this area is this 𝐻𝑑𝐵.
So, as you have increase the current from 𝑖 to (𝑖 + 𝑑𝑖) or 𝐻 from 𝐻 to (𝐻 + 𝑑𝐻) this is
𝑑𝐻 this one, energy absorbed by the coil has become 𝐻𝑑𝐵. Therefore, what is the total
energy supplied then; total energy supplied to whom? Supplied magnetic circuit as you
have increase the current as 𝑖 changes from 0 to some value 𝐼, that is up to this point will
be nothing, but this whole area are you getting this whole area now 𝐻𝑑𝐵; 𝐻𝑑𝐵 this strips
you have to add.
So, these area; so, if I write it O some A this point let us called C, then as total energy
supplied to the magnetic circuit as 𝑖 changes from 0 to 𝐼 will be equal to area OACO this;
this; this; this is the area shaded area that much energy you have supply. This we will be
utilize to find out more about hysteresis loss in the next one.
(Refer Slide Time: 15:07)
120
So, what I will do is this, I will simply draw this coil now, but in my mind it is there it is
wound on a core material things like that and here I will draw the BH curve. So, what we
have seen as I increase the current it will go like this. So, BH curve now the game starts
about hysteresis loss suppose I am increasing the current to this value. So, this axis can be
treated as 𝐻 or 𝑖, suppose you have increase the current to a level of 𝐼𝑚𝑎𝑥 and this axis is
your 𝐵 and suppose it reaches some 𝐵𝑚𝑎𝑥 ok. What is this 𝑖? This 𝑖 is here, this is the coil
having number of turns 𝑁 it will be like this.
Now, for the first time when I am doing let us imagine that this current I will make it AC
Sinusoidal Current. So, current starts rising, current increases with time suppose I say let
me draw that current waveform also in this axis, this is 𝑖 this is 𝐼𝑚𝑎𝑥 and here it will be
−𝐼𝑚𝑎𝑥 . So, suppose current is moving like this and this axis is your 𝜔𝑡, the variation of
current against time sinusoidal suppose. So, current increases from 0 it reaches maximum
at that time it is 𝐵𝑚𝑎𝑥 , after this current decreases with this current 𝑖(𝑡) which is sinusoidal
varying sinusoidal current sinusoidal time varying current suppose.
Now, after these the moment current starts decreasing, the B-H curve does not trace back
the same path through which it went it will be in fact, doing like this now. This is while
current was increasing it will go like this, but while tracing back when you are decreasing
the current it will follow another path above this that is the molecules will be sluggish after
they have reach that you are applying a negative 𝐻, then they will follow that instruction
they will try to follow, but in a sluggish manner and this is called hysteresis it will lag
behind ok.
121
The way they went up while you are decreasing 𝐻 it will be sluggish sort of inertia or
𝜋
whatever, it is you call it will go like this. So, this is your 2 and this is 𝜋 so, between 0 to
𝜋
2
𝜋
it went like this, then from 2 to 𝜋 current has become 0, but you will find it has come
here and even when 𝐻 has become 0 it is having some field retained and it is called you
know residual field. If somebody stops everything here, it comes here only one pulse of
current passes you will find this iron piece will become a sort of permanent magnet
retentivity and it will have a residual field like this.
This positive pulse of the current let us talk about ok, when the current increased from 0
to 𝐼𝑚𝑎𝑥 how much energy I have supplied to the coil, I know the energy supplied to the
coil is the area between this curve and the ordinate ∫ 𝐻𝑑𝐵 × 𝑣𝑜𝑙𝑢𝑚𝑒 is not energy
supplied. So, as the current increase from 0 to 𝐼𝑚𝑎𝑥 energy you have supplied is nothing,
𝜋
but this area; this whole area the pink shaded during these to this part 2 to 𝜋.
𝜋
So, during 0 < 𝜔𝑡 < 2 , 𝑖 was positive and increasing is not. So, during this zone if you
draw the coil this is 𝑖; 𝑖 was increasing and positive the induced voltage must have been
like this plus minus this is induced voltage and current is really positive. So, it was
absorbing power and what is the amount of power that it has supplied if I call it O, if I give
them name this point suppose some name I give A, this point is C, this point is D and O.
So, I will say energy absorbed during this interval
𝐸𝑛𝑒𝑟𝑔𝑦 𝐴𝑏𝑠𝑜𝑟𝑏𝑒𝑑 = 𝐴𝑟𝑒𝑎 𝑂𝐴𝐶𝐷𝑂
𝜋
During this interval what is happening? 2 < 𝜔𝑡 < 𝜋. During this interval 𝑖 is still positive
𝑑𝑖
𝑑𝐵
no doubt, but 𝑖 is decreasing 𝑑𝑡 < 0 therefore, or 𝑑𝑡 < 0 therefore, polarity of the induced
voltage has reversed here. What this coil is doing now? It is then delivering power to whom
𝜋
𝑑𝑖
to the source, earlier it was absorbing power between 0 < 𝜔𝑡 < 2 this region 𝑑𝑡 > 0 it was
absorbing power from the source now it will give back power to the source.
So, power is returned during this interval power or energy, energy is returned to the source;
energy is returned to the source because polarity has decreased. Now, the question is which
area gives me the energy this is now your B-H curve you do not have to take these because
now when you are decreasing current as I told you because of hysteresis the B-H curve
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will be traced by this path. Therefore, area enclosed by this and this 𝑦-axis gives you the
energy enclosed and this area is this area, this blue sketched area if you call this point as P
𝐸𝑛𝑒𝑟𝑔𝑦 𝑅𝑒𝑡𝑢𝑟𝑛𝑒𝑑 = 𝐴𝑟𝑒𝑎 𝑃𝐶𝐷𝑃
𝜋
Therefore during 0 < 𝜔𝑡 < 2 we supplied so much area, but while the material was having
reduced value of 𝐵 tracing back here when you are decreasing the current the amount of
energy returned is less, this must be understood per unit volume. So, let us go to the next
page and let me now draw this once again.
(Refer Slide Time: 25:59)
So, here was my coil mind you it is wound on a magnetic circuit we are not drawing. Now,
what I will do is this, what I have done last time mean do it like this. So, this was the B H
curve which I will draw with this color and while tracing back here it will be like this. So,
this is my positive direction of 𝑖, but suppose this 𝑖 = 𝐼𝑚𝑎𝑥 sin(𝜔𝑡) suppose this current is
this, then I also draw the current, you can always draw current this is my 𝜔𝑡 axis variation
of current I am showing.
This axis as I told you can be treated 𝐻 or i, this is your 𝐼𝑚𝑎𝑥 and this was the current only
𝜋
one cycle I am showing that is good enough 0, 2 , 𝜋. And suppose I have started the game
when there was no magnetic field present in the core. So, 𝐻 here as current was increasing
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𝜋
came up to I am so sorry this is 2 and this is 𝜋 and this is 2𝜋 correct it. And of course, the
cycle repeats only one cycle have.
So, it has gone like this it has increased, then it reached a 𝐵𝑚𝑎𝑥 corresponding to 𝐼𝑚𝑎𝑥 and
suppose I do not stop here I allow the current to flow in the opposite direction then 𝜋 to
𝜋
2𝜋 what is happening here, 𝜋 and this is 3 2 is not. So, if you do that you will see it will
be like this and I think this length should be slightly I mean whatever is 𝐵𝑚𝑎𝑥 has same
modernity if you come it will come here.
So, it will be somewhat like this, it will go like this reach say −𝐼𝑚𝑎𝑥 this is not correct later
on this is 𝐵𝑚𝑎𝑥 I mean it goes, this is suppose −𝐼𝑚𝑎𝑥 this point. And then once again it
will go it will be always lagging it is like this. See after it is tracing the following the
currents I think this is 𝐼𝑚𝑎𝑥 .
So, slight this way I should correct this is suppose 𝐼𝑚𝑎𝑥 , this is the −𝐼𝑚𝑎𝑥 are you getting
not a very good diagram, but it tells you the picture this is 𝐼𝑚𝑎𝑥 this is −𝐼𝑚𝑎𝑥 . Therefore,
when you are decreasing the current; current reaches 0 residual field is there, then you
further need negative current to make it 0𝐵 and then go on increasing the current up to
−𝐼𝑚𝑎𝑥 then it will reach this −𝐵𝑚𝑎𝑥 this level. And after that no looking back so, this part
this initial part is of no consequence now except after the first quarter.
After that as current changes the locus of B-H values will become confined here; here only
it will trace like this go, go, go like that. So, we will continue with this because it will take
some more time, but I will request you to try to go through this lecture hysteresis thing a
little bit carefully particularly a drawing this curve nicely.
You will be perhaps drawing it much better and we will next time show that the hysteresis
loss per unit volume is nothing but the area enclosed by the path that will show that is my
target is this area enclosed and this is called hysteresis loop in area enclosed by B-H loop
is a measure of hysteresis loss per unit volume. We will continue with this next time.
Thank you.
124
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 14
Exact Equivalent Circuit
Welcome to lecture number 14 on Electrical Machines-I and we were discussing about
core losses. In our last class, we were discussing about hysteresis loss. Before that of
course, we discussed about eddy current loss and the factors on which it will depend. I just
a quick review of this previous thing what I did was just have this.
(Refer Slide Time: 00:47)
So, while discussing hysteresis loss first what I did that suppose there is a single coil at
any time 𝑡, current is 𝑖, corresponding 𝐻 is 𝐻 and 𝐵 is 𝐵. And suppose you are increasing
the current from 0 to some fixed value 𝑖, then what is going to happen. So, in time 𝑑𝑡, if
current increases from 𝑖 to (𝑖 + 𝑑𝑖), 𝐻, (𝐻 + 𝑑𝐻) and so on, then we showed that the
energy will be absorbed by the coil when current is increasing or 𝐻 is increasing. And the
area under this curve BH curve, so 𝐻𝑑𝐵 is the energy supplied and then total energy
supplied can be calculated.
125
(Refer Slide Time: 01:45)
In our case, current will be alternating in nature, I am not saying whether it will be
sinusoidal or not, but it is alternating, current will increase will reach a peak, then it will
decrease and so on. And then when you first switch on and suppose there was no residual
magnetism present in the core, then this is the BH curve, this point will go up. And when
it reaches 𝐼𝑚𝑎𝑥 , this axis is incidentally also represents current in some other scale. So, it
reaches 𝐼𝑚𝑎𝑥 and correspondingly 𝐵𝑚𝑎𝑥 .
So, during this process you have pumped energy into the system, the area of which will
𝜋
be, OACDPO. And when you decrease the current that is between 2 to 𝜋 current decreases,
𝑑𝑖
𝑑𝜑
but still remains positive. And but 𝑑𝑡 is negative 𝑑𝑡 . So, induced voltage polarity reverses,
current direction still remains like this because it is positive current.
And then certainly this coil now delivers energy back to the source. And how much energy
is delivered? Once again ∫ 𝐻𝑑𝑏, but then the BH curve is like this, this one and this area
will be this one. Therefore, the difference of this area represents the energy absorbed by
the coil.
126
(Refer Slide Time: 03:36)
Now, next thing what I did is suppose the current is alternating like this through the coil,
then once again if it is alternating, I have not shown this part once again, because after for
the first time it will go like this reaches 𝐼𝑚𝑎𝑥 , 𝐵𝑚𝑎𝑥 . Then when 𝑖 decreases it will the BH
curve will be somewhat different, not the same curve through which it went up, it will
follow this curve current decreases 𝐵.
So, this is the decreasing thing decreases. Then when current is 0, it will have some
3𝜋
residual field then negative current. After 𝜋 to 2 , it will go like this, then it reaches −𝐼𝑚𝑎𝑥
ok. And once again so if your core where was initially un magnetized except for the quarter
of the first cycle, it will do like this. So, that portion need not be taken into account, no
point in taking into account.
And then we were discussing I was telling that, I will show you that the energy absorbed
by the coil when current makes a full cycle of variation from 0 to 𝐼𝑚𝑎𝑥 to −𝐼𝑚𝑎𝑥 to back
to 0. The area enclosed by the BH curve will represent hysteresis loss per unit volume that
is what we will be doing.
127
(Refer Slide Time: 05:50)
So, let me draw this curve once again, so that we understand what is going on. So, let me
start with this B H curve as usual. So, it will be like this. Suppose, this is my 𝐻 or 𝑖 axis 𝑖
or 𝐻 and this is the, this axis is 𝜔𝑡 to show the variation of the current ok. So, let it be like
this. So, this is the variation of the current.
So, you know here I am just like this. So, your B H curve, it will be like this; it will be like
𝑑𝑖
this. So, this is 𝐵𝑚𝑎𝑥 here and this is this level is −𝐵𝑚𝑎𝑥 ok. So, this is when the 𝑑𝑡 is
positive current is increasing, I have to use this; when
𝑑𝑖
𝑑𝑡
is negative I have to use this part
of the BH curve. This axis is 𝐵; this axis is 𝐻 or 𝑖.
𝜋
So, when the for example, when the current is 0 and increasing in this zone say this is 2 , 𝑖
and 𝐻 is increasing. So, 𝑖 = 0. So, this is the point when 𝑖 = 0 and it will start from this
negative residual field a you just forget me this two are equal ok and it increases like this.
So, during this process the coil voltage will be just like this, applied voltage is positive,
𝑑𝜑
negative. And why the induced voltage will be positive, because 𝑑𝑡 is positive and current
𝜋
is also positive. So, energy will be absorbed. This is between0 < 𝜔𝑡 < 2 .
And what will be the energy absorbed by the coil? Energy absorbed by the coil then will
be the area under this curve that is if I hatch it, it will be like this. So, this area will
correspond to the area absorbed by the coil ok and it will come to this point. Of course
128
𝜋
3𝜋
current is changing on its own. So, during 2 to 2 , it will do like this. And this point is how
much?
𝜋
3𝜋
𝜋
This is 2 and this is 𝜋 and this is 2 . So, between 0 to 2 , current was increasing. And then
𝜋
from 2 this, this point will be reached when I have come here is not. I have reached this
𝜋
𝜋
point so 0 < 𝜔𝑡 < 2 , it is like this. Then what happens between 2 and 𝜋 that is this is the
𝑑𝐵
coil and then the induced voltage as you see 𝐵 decreases, true in this path I am there, 𝑑𝑡 is
negative. So, induced voltage will be plus minus.
And current of course, still remains positive means this is the direction of the positive
𝜋
current. So, this delivers energy. So, 2 < 𝜔𝑡 < 𝜋 returns back energy to supply, is it not.
And what is that area? That area will be this area that is why during this point at the end
of 𝜋, it has absorbs so much energy this minus this, this we discussed last time.
Similarly, if between 𝜋 and 2𝜋, if you consider, it will absorb energy that is let me draw
this. So, this is the first positive current cycle, half cycle, then the negative thing will be
3𝜋
then say between 𝜋 < 𝜔𝑡 < 2 . During this portion of this one current is negative means
current is actually entering through this and coming out from this; this is the negative
𝑑𝐵
current. And induced voltage because current is 𝑑𝑡 is this one positive, it still remains this
one then plus minus. So, it once again absorbs the energy absorbs. This is the 𝑒 because
through the plus current is entering.
3𝜋
And finally, between 2 < 𝜔𝑡 < 2𝜋, the situation is current still negative is it not current
𝑑𝑖
still negative. So, it will be like this, but 𝑑𝑡 is opposite. So, it will become like this, so, it
will returns back supply. So, it returns back energy. So, effectively when current makes a
complete full cycle, so that the area corresponding to that will be this.
So, total energy absorbed in one cycle of current variation will be this area plus this area.
Therefore, energy absorbed in Joule energy absorbed in one cycle is equal to area enclosed
by the BH curve. Therefore, this whole area will represent how much energy this magnetic
material has absorbed.
129
Of course, thinner this area energy absorb per cycle will be also small, because the area is
the crucial term enclosed by the BH curve. Mind you this is per unit volume area enclose
energy absorbed in one cycle per unit volume of the magnetic material is the area enclosed
by the BH curve per unit volume. This we have establish that earlier, it is per unit volume,
one should not forget that so per unit volume the energy absorbed will be this one.
(Refer Slide Time: 15:48)
Therefore, to reduce the hysteresis loss, I must see that the area enclosed by the BH curve
should be as small as possible that is for example, if somebody uses a material like this or
instead of that if somebody uses a hysteresis loss for this material will be much lesser than
this that is the idea. Make the choose the magnetic material in such a way that the area
enclosed by its BH curve is as small as possible. This is 𝐻; this is 𝐵; this is 𝐻; this is 𝐵. In
fact, it depends upon the application in case of transformer application, we will demand
that hysteresis this area is as small as possible.
The material used for to reduce to hysteresis loss is CRGO - Cold Roll Grain Oriented,
cold roll commercial name cold rolled grain oriented grain oriented silicon steel, silicon
steel. This some metallurgical process that is you treat the iron laminations in a particular
fashion, so that the grain oriented means the tiny magnets are locally in some groups they
are oriented already by that process when you roll this material, anyway we will not go
into that. But the idea is you try to see that this area enclosed by the BH curve will be small
130
and silicon if you add to this material it will increase its resistivity thereby eddy current
loss will be reduced.
Therefore, in essence this area of course, as you can see it will also depend upon 𝐵𝑚𝑎𝑥 .
Suppose, the 𝐵𝑚𝑎𝑥 is more even if it is thinner area, this area gets increased up ok. Now,
all said and done, we do not know the factors on which this hysteresis loss too will depend,
it will depend upon the frequency factors on which hysteresis loss depends depend is one
is area enclosed by it, area enclosed by it. Then the second thing is it should depend on
𝐵𝑚𝑎𝑥 and also it should depend on frequency at which this reversals of frequency of
supply, because this area gives you hysteresis loss 𝑃ℎ𝑦𝑠 is a per cycle.
So, more the cycles, more the frequencies, hysteresis loss will increase per cycle per unit
volume. So, it will depend on frequency. All this said and done there is a what is that
formula called a some after doing some experiments on several material, some people say
hysteresis loss in facts it was proposed by Steinways a formula for which there is no
derivation ok. What this?
Student: Empirical.
Empirical formula. And it is this some empirical formula, empirical formula, empirical
formula for calculating formula is like this that hysteresis loss is
1.6
𝑃ℎ ∞𝐵𝑚𝑎𝑥
𝑓
Frequency we know it has to be directly proportional to and 𝐵𝑚𝑎𝑥 is the peak value of the
maximum flux density.
Therefore, and recall that
2
𝑃𝑒 ∞𝐵𝑚𝑎𝑥
𝑓2
and thickness square, once the plate is chosen thickness of which may be of fraction of
millimetre say 0.25mm or so, I am showing it just proportional to it is also proportional to
plate thickness square.
But all these things are once you write frequency here, you then write per unit volume.
131
1.6
𝐵𝑚𝑎𝑥
𝑓
𝑃ℎ ∝
𝑉𝑜𝑙𝑢𝑚𝑒
2
𝐵𝑚𝑎𝑥
𝑓2
𝑃𝑒 ∝
𝑉𝑜𝑙𝑢𝑚𝑒
Volume of what? Magnetic material core per unit volume of the core material. So, this is
the at least this results we should remember. And also if you have the BH curve of the
material available to you, then you can much more accurately say about the amount of
hysteresis loss that is going to take place within the core of the transformer.
We will refer to this particular aspects by solving several problems. So, we will give you
in the tutorial. Mind you whatever I have done till now about eddy current and hysteresis
loss, I have written some write up and it will be uploaded. So, read the lecture, listen to
the lecture and also go through those lecture notes where somewhat detail analysis of this
thing whatever I have done is done somewhat neatly, anyway this is the thing.
Now, coming back to the practical transformer, in fact, we have digressed it from this one
coming back to equivalent circuit of a practical transformer circuit, of a practical
transformer.
(Refer Slide Time: 23:26)
Recall that we have got the equivalent circuit of the practical transformer like this. On the
primary side, it will be like this 𝑟1, 𝑥1 leakage reactance small values, then I showed a
132
magnetizing branch here 𝑗𝑋𝑚 . And then here there will be 𝑟2′ = 𝑎2 𝑟2. Then you have 𝑥2′ =
𝑎2 𝑥2 . And whatever load impedance you have connected to the secondary side 𝑎2 𝑧2. What
is 𝑎?
𝑎=
𝑁1
𝑁2
And this is your 𝑉1.
So, it is almost the correct equivalent circuit of a practical transformer, but here in this
equivalent circuit, how to take now what elements now should add to this equivalent
circuits which will represent the core loss; core loss means heating of the core, heating of
the core will take place because of eddy current loss and because of hysteresis loss. This
is the only source of power which pumps the power into the system. What is the system?
Actual thing practical transformer it is like this. You have applied a voltage 𝑉1 and here is
your load connected. In this one, you will never find this resistance reactance is present.
Therefore, here is your 𝐼2 and here is your 𝐼1 that is what you will observe nothing else.
But this whole thing practical transformer is modelled by an ideal transformer and these
things. Now,
𝑃𝐶𝑜𝑟𝑒 𝐿𝑜𝑠𝑠 = 𝑃𝑒𝑑𝑑𝑦 + 𝑃ℎ𝑦𝑠
2
And 𝑃𝑒𝑑𝑑𝑦 = 𝑘𝑒 𝐵𝑚𝑎𝑥
, 𝑓 being constant it will be like this plus 𝑃ℎ𝑦𝑠 = 𝑘ℎ ×
1.6
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑜𝑜𝑝. Instead of writing that it is proportional to I will say 𝑃ℎ𝑦𝑠 = 𝑘ℎ 𝐵𝑚𝑎𝑥
𝑓
it is one and the more exact result loop area, but this is like this.
𝑃𝐶𝑜𝑟𝑒 𝐿𝑜𝑠𝑠 = 𝑃𝑒𝑑𝑑𝑦 + 𝑃ℎ𝑦𝑠
2
1.6
= 𝑘𝑒 𝐵𝑚𝑎𝑥
𝑓 2 + 𝑘ℎ 𝐵𝑚𝑎𝑥
𝑓
𝑓 being constant therefore, the hysteresis and eddy current loss depends upon 𝐵𝑚𝑎𝑥 . And
value of 𝐵𝑚𝑎𝑥 practically remain same, remain same from no load sorry to loaded
condition or full load condition why because you recall this current is 𝐼2′ , this current is 𝐼𝑚
- magnetizing current and this current is 𝐼1 .
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The voltage which will come between these two points that decides the magnetizing
current hence the flux or 𝐵𝑚𝑎𝑥 .
𝐵𝑚𝑎𝑥 =
𝜑𝑚𝑎𝑥
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝐶𝑜𝑟𝑒
Therefore this voltage will change strictly. When no load is connected, suppose there was
no load, so 𝑎2 𝑧2 , 𝑧2 = ∞ here, it will be open circuit, then the current round will be pretty
small. So, this voltage will be this one.
But when the it is loaded, this current will increase no doubt, but this parameters are small.
The drop which will take place here the voltage across this magnetizing branch 𝑋𝑚 will
definitely not be equal to 𝑉1, but it will only marginally change. So, what is I am just trying
to tell is that
𝜑𝑚𝑎𝑥 =
𝐸1
𝑉1
≈
4.44𝑓𝑁1 4.44𝑓𝑁1
So, a little variation of 𝐵𝑚𝑎𝑥 will take place from no load to full load, but otherwise it will
be there. But any way this then the thing is this loss which depends on 𝐵𝑚𝑎𝑥 which
practically remain same should be shown by a resistance connected across 𝑋𝑚 that is what
I am telling this is 𝑅𝑐𝑜𝑟𝑒 which will take care of the core loss.
And it certainly should not be in series, because it depends on 𝐵𝑚𝑎𝑥 . And 𝐵𝑚𝑎𝑥 is decided
by this voltage. This voltage by this impedance is 𝐼𝑚 that decides 𝐵𝑚𝑎𝑥 , therefore core loss
resistance which will represent core loss must be shown connected in parallel with 𝑋𝑚 .
We will continue with this. And this complete this is the complete exact equivalent circuit
of a single phase transformer referred to the primary. We will continue with this.
Thank you.
134
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 15
Approximate Equivalent Circuit
(Refer Slide Time: 00:22)
So, welcome to lecture 15. And we are discussing about the Equivalent Circuit of a
transformer, equivalent circuit. In fact, this is exact equivalent circuit refer to primary side.
And the equivalent circuit we got last time is this one, primary winding resistance 𝑟1,
primary winding leakage reactance 𝑥1 .
Then you have this parallel branch; one is this resistance I am showing by several lines
𝑉2
indicating that it will be high 𝑅 will be this power. And this is I will call it 𝑅𝑐𝑙1 we referred
to primary side. And here is magnetizing branch of reactance 𝑗𝑋𝑚1 and then the reflected
resistance of the secondary side is 𝑟2′ and 𝑥2′ .
And here there will be the load impedance whichever you have connected to the secondary
side of the transformer 𝑎2 𝑧2 , and
𝑎=
135
𝑁1
𝑁2
So, what are the impedances which will be higher, would like to see they must be very
high. This resistance this square, this voltage square by 𝑅𝑐𝑙1 is the power loss in the core,
I would like to reduce them. Similarly, this reactance I would like to make it as high as
possible so that magnetizing current is less. So, this current is called 𝐼𝑚1 , and this current
is the core loss component 𝐼𝑐𝑙1 refer to primary side this current is your 𝐼2′ .
And this current we call it to be 𝐼0 the no load current, because 𝐼0 is now having two
components no load current is
𝐼0 = 𝐼𝑐𝑙1 + 𝐼𝑚1
these two currents will make 𝐼0 . Why, it is called no load current, when the load is not
connected on this secondary side this will be open circuited, 𝑧2 → ∞. Recall that this is
equivalent circuit of what practical transformer is connected like this.
So, let me put this figure also alongside this is what I am examining. This is 𝑧2 ; this is 𝑉1;
this is your core material. And when this switch is opened, it is under no load open means
infinite impedance. So, here 𝑎2 ∞ = ∞. So, this will be open, but transformer then 𝐼1 = 𝐼0
if 𝐼2′ = 0, this is 𝐼2 . In practical transformer, this is what you will observe, you cannot
separately see this branches, it is our logic which brought us here. It must be something
we are modeling this practical transformer into an equivalent circuit ok.
So, what are the things 𝑟1 and 𝑥1 should be small, 𝑟2′ and 𝑥2′ should be small, but 𝑅𝑐𝑙1 𝑋𝑚1
should be high. In fact, from this if you can easily get the equivalent circuit of an ideal
transformer by making 𝑟1 = 0, 𝑥1 =0, because no resistance, no leakage flux, no leakage
reactance. Similarly secondary coil resistance is 0, 𝑟2 = 0, 𝑥2 = 0, which implies 𝑟2′ = 0,
𝑥2′ = 0 and 𝑅𝑐𝑙1 → ∞, 𝑋𝑚1 → ∞.
So, this equivalent circuit then we will simply reduce to this one, nothing is connected,
and here also nothing is connected. These two are shorted, these two are shorted and only
𝑎2 𝑧2 and this is your 𝑉1 is it not. This is the equivalent circuit of an ideal transformer. So,
always keep this in mind ok. These impedances are small and so on ok. If somebody wants
to draw the equivalent circuit it refers to the secondary side of the same thing that is to the
load side, how it will look like?
(Refer Slide Time: 06:56)
136
Equivalent circuit refer to secondary side or load side it will remain same except that this
one will become 𝑟1′ this one will become 𝑥1′ . Here that those two parallel branches it should
be like this, and here it will these two will should not be disturbed, secondary side we are
′
′
going. And this impedance should be 𝑧2 and this of course, should be 𝑅𝑐𝑙1
and 𝑋𝑚1
.
𝑟
𝑥
Now, what is this 𝑟1′ = 𝑎12 opposite thing. What is this 𝑥1′ = 𝑎12 , 𝑟2 and 𝑥2 will not suffer
𝑋
′
any change similarly actual impedance. What is this 𝑋𝑚1
= 𝑎𝑚1
2 . And what is this one
𝑅
𝑁
1
′
𝑅𝑐𝑙1
= 𝑎𝑐𝑙1
2 . What is 𝑎? 𝑎 = 𝑁 . What is this voltage 𝑉2 , what is this current 𝐼2 , what is this
2
current, 𝐼0′ and what is this current 𝐼1′ reflected here. And what is this voltage should I write
𝑉
it should be 𝑉1′ . And what is this 𝑉1′ , it will be actual 𝑉1′ = 𝑎1 .
Now, in this way, I can draw the equivalent circuit either refer to load side or to the source
side and avoid this coupled circuit present in the circuit to be solved, either way you can
do ok. Your practical transformer is this, this is the real thing 𝐼2 this is 𝐼1 , but these things
practical transformer can be viewed as this one. So, either way you can now solve the
circuit even nothing is now neglected, but as you know the parameters value entered into
the ideal transformer concept to represent a practical transformer that is what the thing is.
After I have told you this one, so this is the equivalent circuit referred to the primary side,
the same thing referred to the secondary side with respect to any side you can work. Now,
in general this 𝑟1 and 𝑥1 refer to the equivalent circuit refer to source side, I will be always
doing this refer to the source side ok. Let us stick to one, no point in just complicating
137
thing, remembering so many things this is what I am always going to do. Refer to primary
side I will always draw the equivalent circuit ok.
Now, in this equivalent circuit I know that 𝑟1, 𝑥1 and 𝑟2′ , 𝑥2′ are much smaller compare to
this parallel branch 𝑅𝑐𝑙1 and 𝑋𝑚1. What I am telling is 𝑟1 , 𝑥1 , 𝑟2′ , 𝑥2′ ≪ 𝑅𝑐𝑙1 , 𝑋𝑚1. Therefore,
people started thinking, this is a circuit if I know all the parameters, I can solve it, but
computationally it will be somewhat heavier.
In the sense that these are all ac circuit, phasor you have to draw, these voltage will be
present and so on. Then add these drop to get these voltage, then to these voltage you add
this one all phasorically, and you will get exact solution no doubt about that.
But knowing this fact people also thought that whether some simplifications still can be
make to this equivalent circuit based on this knowledge because this parameters value
series parameter values are smaller compared to 𝑅𝑐𝑙1 and 𝑋𝑚1. For example, what does
that mean is consider suppose a circuit like this, what i am telling suppose you have a
circuit like this.
Suppose, I say this is say 4 Ω, suppose I say this is 4 Ω, this is 4 Ω, and this is a 80 Ω I say
that. So, this impedance is 20 times higher than this. Suppose, I say I connect a say some
64 volt to the circuit, I am interested to know this current ok. You can parallel this, this is
80×4
64
a very simple circuit (80+4 + 4), 64 it divide with that resistance 80×4
(
80+4
+4)
get this current.
But suppose you are a very practical man, practical engineer, you will be happy to know
what will be the current drawn from the battery and I will sacrifice some accuracy. I will
64
say that this current will be you solve it 8 amp. Looking at this circuit you can say oh this
current, why, because this parallel resistance is 20 times higher than this fellows you solve
64
it this 80×4
(
80+4
80×4
80
, this quantity is close to 4 only, (80+4) ≈ 4, 84 ≈ 0.99 something.
+4)
Therefore, and we always know the equivalent resistance of a very high resistance and a
low resistance is of the order of the low resistance itself. Therefore, is it not, just looking
at the circuit I will tell look here this current in this branch will be close to 8 amp that is
all. He calculates correctly, he will maybe he will come to something very close to 8, but
he will save lot of computation. Of course, in a dc circuit resistances are manipulated just
138
like that, but in ac circuit it is not you have to complex algebra, you have to do to find out
the z-equivalent of this branch, this branch and so on.
So, coming back to the equivalent circuit refer to the primary side, if I know that this
impedance is much higher compare to these values and this values. Then I will rather
sacrifice some accuracy not too much, at the same time I will reduce my computational
burden. What I am trying to tell is this, this parallel branch ok, I will take it better draw it
in a separate page.
(Refer Slide Time: 16:31)
So, this you remember it is exact equivalent circuit; suppose, it is the equivalent circuit.
This is 𝑉1 ok, this is the equivalent circuit we have seen. Now, what I am telling that this
equivalent circuit you just select it and push it here. This is the equivalent circuit. Now, in
this equivalent circuit, this branch which is high impedance value compare to this. So, this
branch I will first remove, I will not take considered this branch. So, 𝑟1 𝑥1 and 𝑟2′ 𝑥2′ and
this one I have removed; I know I am making a mistake. This is the exact equivalent circuit,
but what I will do is this I will remove this. And here I will that load impedance is there,
but this branch I will remove and say that this is approximate equivalent circuit. Somebody
says oh you have you are totally neglecting this branch, then I will say I will not neglect
that branch, but I will show it at the beginning. So that I have got some computational
advantage now that branch present here makes it much more complicated than what it was
139
here that is what I am telling. And this I am just not doing just out of nothing because of
the fact that 𝑟1 , 𝑥1 , 𝑟2′ , 𝑥2′ ≪ 𝑅𝑐𝑙1 , 𝑋𝑚1 for a well designed transformer.
I will presume that whoever has design this transformer, they must have kept in mind that
core loss should be made as small as possible, magnetizing current should be made as
much less as possible which indirectly means that 𝑅𝑐𝑙1 is high and 𝑋𝑚1 is high. If that be
the case this being smaller, you forget about this branch and show it on this side, just bring
that branch forward.
So, you are neglecting approximating something, but at the same time you are not totally
ignoring this part to take into account, like the previous problem. In this problem,
somebody says ok, this will be something like this 4Ω, 4Ω you connect load resistance.
And you are skeptic also would like to improve upon your result simply by ignoring this
totally, show it here, 4, 4, 80. And the current supplied from the battery will be not 8amp,
it will also take this into account, but the result will be improved upon, this is also
approximation and totally neglecting this is also an approximation that is fine.
But what I am telling you do not neglect too much, but get a tremendous computational
64
advantage, why, because this current I can easily calculate 8 = 8. And this current
64
whatever it is 80, these two can be added series parallel business goes, and still you will be
very closer to the correct value what I am telling. They will not be exact I agree with that.
So, with that, so this circuit keep in mind what exactly we are doing then you will get
pleasure. And all said and done nowadays powerful calculator is available everything is
there. So, one can leave with the exact equivalent circuit as well, but nonetheless even with
calculators you have to this then parallel these, these, these are to be paralleled complex
numbers, lot of computational effort then that should be added with 𝑟1 𝑥1 get the values.
So, I think I have been able to put in your mind that why that approximation is done, simply
to reduce the computational effort. So, this is called approximate equivalent circuit. So,
the parallel branch in front of the supply 𝑉1, and then this current is your 𝐼2′ . As if we are
pretending 𝑟1 and 𝑥1 only 𝐼2′ is flowing, how does it matter, and this current is no load
current.
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Let us see how much difference it will have. Ok, 𝑉1 minus this drop that voltage divided
by this branch gives you 𝐼0 , now I am telling that this drop you neglect, why you will
neglect that drop 𝑟1 𝑥1 is small. So, these voltage is approximately equal to this. These are
the anyway think about it and try to do. So, it reduces lot of computational burden.
Once you draw the circuit like this, then I then this is the key word approximate equivalent
circuit. We will almost all the cases we will refer to. Now, I will come to the phasor
diagram which I have not drawn for this circuit. For example, I will draw once you, in fact,
can avoid that after learning approximate equivalent circuit, but nonetheless let me at least
draw the equivalent circuit of this exact transformer.
(Refer Slide Time: 24:32)
For example, suppose this is the, suppose these could not be copied this was 𝑟2′ 𝑥2′ , and
here was a line, this is the exact equivalent circuit. Let us see how to draw the phasor
diagram, which will not be doing in fact but let us see suppose this voltage is 𝐸1 . You start
with this voltage to draw the phasor diagram this is 𝐸1 , refer to the primary side. If this is
𝐸1 you know your 𝐼𝑐𝑙1 will be in phase with this. This current resistive, magnetizing current
will be like this 𝐼𝑚 . And the sum of these two currents is called no load current 𝐼0 ; this is
the no load current. I have got 𝐼0 ok.
Then 𝐸1 divided by this whole impedance will give you the current 𝐼2′ is it not. So, 𝐸1
divided by this whole impedance will give you this current. So, suppose that current this
is lagging power factor let us assume, so 𝐼2′ will be suppose somewhere here I will just
141
give you the idea so that so, suppose this is 𝐼2′ . What is this angle, this angle is not the
power factor angle of the load, it is the power factor angle of 𝑟2′ , 𝑥2′ and 𝑎2 𝑧2 , whatever it
is.
So, 𝐼2′ is suppose this one. If this is this, then from this voltage is 𝐸1 . From 𝐸1 if I subtract
𝐸1 − (𝑟2′ + 𝑗𝑥2′ )𝐼2′ = 𝑉2
That is this 𝐸1 minus 𝐼2′ 𝑟2′ I will make it dirty, but you just try to understand how
complicated it becomes, but it can be done. then minus 𝑗𝑥2′ 𝐼2′ , now 𝑗𝑥2′ is perpendicular to
this. So, subtract that whenever you will end up that will be minus 𝑗𝑥2′ 𝐼2′ . And then what I
am telling this will be your 𝑉2, is it not.
Then this angle will be your power factor angle and so on. This is the load power factor
angle whatever it is. This diagram we get. Then
𝐼0 + 𝐼2′ = 𝐼1
I am looking for where is my 𝑉1?
𝑉1 = 𝐸1 + 𝐼1 (𝑟1 + 𝑗𝑥1 )
I have got I know 𝐸1 , therefore, 𝐼1 𝑟1 parallel to 𝐼1 , then ( 𝐼1 𝑟1 + 𝑗𝐼1 𝑥1 ). And this you add
you will get 𝑉1. I mean this is the idea what I am telling if you know this voltage you can
completely draw the phasor diagram.
So, it can be done in several ways, all though we will you will see later this complicated
diagram is not necessary if you have a powerful calculator and if we have decided, we will
use the approximate equivalent circuit, things will be much more easier. And we will see
it in the next class.
Thank you.
142
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 16
Determination of Equivalent Circuit Parameters - No Load Test
Welcome to lecture 16 and we have been discussing about the exact equivalent circuit, its
phasor diagram, then approximate equivalent circuit and I will also draw its phasor
diagram.
(Refer Slide Time: 00:33)
143
(Refer Slide Time: 00:38)
Recall that the approximate equivalent circuit was like this where this parallel branch has
been brought right across the supply. The approximation made is under no load condition
as if there is no current through the winding, but nonetheless core loss will always take
place. So, it was like that now if you see this 𝑟1 𝑥1 and 𝑟2′ 𝑥2′ , they become in series ok. So,
this is one of the greatest advantage of looking at the transformer equivalents circuit in
terms of its approximate representation.
(Refer Slide Time: 01:31)
144
So, what will happen is this equivalent circuit, then can be further simplified. So,
approximate equivalent circuit refer to primary it will then boil down to this one. These 𝑟1
and 𝑟2′ can be represented by a single resistance as
(𝑟1 + 𝑟2′ ) = 𝑟1 + 𝑎2 𝑟2 = 𝑟𝑒1
That is equivalent resistance in terms of primary winding; we see equivalent resistance.
Similarly, the two reactances can be grouped together and written as
(𝑥1 + 𝑥2′ ) = 𝑥1 + 𝑎2 𝑥2 = 𝑥𝑒1
and this is called equivalent reactance. Equivalent reactance referred to primary side 𝑥𝑒1
and that is all. These two are the secondary terminals where an impedance will be
connected if it is connected 𝑎2 𝑧2 .
And this will be 𝑅𝑐𝑙1 and this is 𝑗𝑋𝑚1 and this is your 𝑉1 and mind you this current is 𝐼2′
and this is no load current 𝐼0 and 𝐼0 comprises of two parts; core loss component 𝐼𝑐𝑙1 and
𝐼𝑚1 . Then this voltage if I show it by different colours. So, that you understand, this voltage
is
𝑉2′ = 𝑎𝑉2
This current
𝐼2′ =
𝐼2
𝑎
This is how things will go, then the phasor diagram drawing becomes so simple.
Let us see; suppose we start drawing the phasor diagram with respect to 𝑉2′ . Now as you
can see
𝑉1 = 𝑉2′ + 𝐼2′ (𝑟𝑒1 + 𝑗𝑥𝑒1 )
This will be the phasor diagram.
145
So, we start the phasor diagram drawing in this way. First draw 𝑉2′ and then draw 𝐼2′ .
𝐼
Suppose it is supplying, what is 𝐼2′ ? 𝐼2′ = 𝑎2 and this is then the load power factor angle 𝜃2
whatever impedance you have connected. Then to get 𝑉1 you add to 𝑉2′ ; this drop simply.
So, this will be 𝑉2′ plus 𝐼2′ 𝑟𝑒1 plus 𝑗𝐼2′ 𝑥𝑒1 and you will get 𝑉1. No point in showing 𝐸1 𝐸2
etcetera. So, as I am telling you as you go to the approximate equivalent circuit, you on
the plea that the series impedances are much smaller compared to the parallel impedances
of this core loss resistance and magnetizing impedance. Then it is the phasor diagram that
is all and you this is the angle 𝜃2 and this is your 𝑉2′ .
So, to 𝑉2′ , you add this ok. Now where is your 𝐼1 ? If this is 𝑉1 you see this 𝐼0 will be lagging
this 𝑉1 by whatever that current is. Are you getting? After you see after you get the 𝑉1 to
𝑉1, you have in this phasor in this approximate phasor diagram your 𝐼𝑐𝑙1 will be in phase
with your applied voltage. So, this will be 𝐼𝑐𝑙1 and your magnetizing current you show 90°
lagging in this circuit.
So, this angle will be then 90° as per the approximate equivalent circuit 𝐼𝑚 is not and if
you wish to calculate where your 𝐼1 will be that is what I am doing. So, 𝐼𝑚1 and then add
this to get your 𝐼0 to 𝐼2′ you add your this 𝐼0 to get 𝐼1 an input power factor will be angle
between 𝑉1 and 𝐼1 . I think you have got the idea, but this is the thing. Similarly the same
transformer if you draw the equivalent circuit referred to I will do that because you practice
them.
146
(Refer Slide Time: 08:27)
So, suppose approximate equivalent circuit referred to secondary side or load side;
secondary side. I will draw it very quickly because I know what is what. So, equivalent
first you draw the equivalent circuit. Equivalent circuit referred to secondary side, we have
already drawn.
All parameters now it will look like same thing. Only thing this is 𝑅𝑐𝑙2 . This is 𝑋𝑚2 referred
to secondary side and this
𝑅𝑐𝑙2 =
𝑅𝑐𝑙1
𝑎2
from this side to that side things are to be divided in terms of
𝑋𝑚2 =
𝑋𝑚1
𝑎2
And I will now show it and leave it to you to verify it is indeed true; I will be right here as
𝑟𝑒2 and 𝑥𝑒2 where, this
𝑥𝑒2 =
𝑥1
+ 𝑥2
𝑎2
𝑟𝑒2 =
𝑟1
+ 𝑟2
𝑎2
And this voltage is certainly not 𝑉1 now it is
147
𝑉1′ =
𝑉1
𝑎
and this voltage will be your 𝑉2. And this current is actual current 𝐼2 and this is your 𝐼0′
which is equal to
𝐼0′ = 𝑎𝐼0
This is 𝐼0′ ; I can transform this current. So, it will be 𝐼0′ = 𝑎𝐼0 .
So, get used to this transformation multiplying any impedance transfer from this side to
this side will involve 𝑎 and any current transformation or voltage transformation will
involve 𝑎; whether multiplication division you should be very careful. I gave you some
hints also that is you just simply calculate the turns ratios and you must know that
impedance values in the lv sides are lower; voltage value on the low voltage side is low
definitely and so on current value in the lv side is higher.
So, all these things will you get used to it. So, this is the thing and in this case this current
will be 𝐼1′ and then the phasor diagram for this in the same way. Here this you start with
𝑉2 actual secondary voltage 𝑉2. This is suppose the current delivered by the transformer to
the load which am not drawing. Henceforth I will not draw because 𝑉2, it is supplying
current means some impedance is connected.
This is the actual power factor of the load, then to 𝑉2 you add these drop 𝐼2 𝑟𝑒2 plus this
angle is 90° 𝑗𝐼2 𝑥𝑒2 and then you will get this 𝑉1′ . You will not get 𝑉1 mind you; 𝑉1′ and
𝑉′
then you can fix up where your 𝐼0 will be because 𝑅 1 will give you magnetizing current.
𝑐𝑙2
This will give you magnetizing current
𝑉1′
𝑋𝑚2
and the core loss component of current will be
′ ′
there 𝐼𝑚
𝐼𝑐𝑙 . These two will give you I mean same stuff; I do not want to also mix.
So, 𝐼0 so, to 𝐼0 you add 𝐼2 to get 𝐼1′ and so on; I will not continue. I have done is this I have
represented a practical transformer modelled it in terms of some external parameters
connected either in series or in parallel. For example, 𝑟1 𝑟2 𝑅𝑐𝑙 core loss component of
resistance magnetizing reactance to an ideal transformer and got these things and after
getting the exact equivalent circuit we did another approximation not without any reason.
148
You cannot approximate anything just like that the argument is the winding resistance and
leakage reactants are much smaller compared to the resistance representing core loss and
the reactance representing the magnetizing current and that comes in parallel. So, it can be
approximately pushed in this way.
Now, after doing this, I will now tell you given a practical transformer how do you know
this parameter values? By doing some tests it can be simple test, these parameters can be
evaluated.
(Refer Slide Time: 15:14)
But before that let us you would take some rating of a practical transformer. I took one
like this say 200V/100V, single phase, 50Hz transformer and suppose it is practical
transformer. And what I have missed KVA rating single phase. Suppose the KVA rating
is 1KVA ok; 1KVA 200V/100V, 50Hz single phase transformer; TFO means transformer
short way I am writing; this things given to you.
What does this mean to a practical transformer is this? This is practical transformer, I will
assume this way that if you apply 200V 50Hz source here across this winding with nothing
connected on the secondary side, you will get 100V 50Hz. And this will be turns ratio will
200
be 𝑎 = 100 = 2. This is what will happen this is the turns ratio I can get and this one. I also
told you that HV side rated current which I will call side 1 and this is side 2 HV side and
this is LV side that is equal to
149
1000
𝐼𝐻𝑉
𝐼1
=
=
= 5𝐴
𝑅𝑎𝑡𝑒𝑑 𝑅𝑎𝑡𝑒𝑑
200
Also I know
1000
𝐼𝐿𝑉
𝐼2
=
=
= 10𝐴
𝑅𝑎𝑡𝑒𝑑 𝑅𝑎𝑡𝑒𝑑
100
200
Of course, these I could in one stroke right straight away 5 × 𝑎 = 5 × 100 = 5 × 2 = 10
is the rated current mind you; these things I know.
So, given the transformer rating we know about this numbers or this turns ratio I can make
out calculate rated current that can be allowed to flow through the HV winding and through
the LV winding. KVA remain same approximately. After calculating this, I now make one
very interesting point that no load current of a transformer; no load current that is 𝐼0 you
recall
𝐼0 = 𝐼𝑐𝑙 + 𝐼𝑚
No load current of a transformer is of the order of say 5% of the rated current 2 to 5% of
the rated current. As a practicing engineer, the this idea must be there that is it must be a
well designed transformer whoever has designed this transformer he must have seen that
eddy current, hysteresis losses are reduced is using very good magnetic material.
So, that magnetizing current is also small and it is 𝐼𝑐𝑙 + 𝐼𝑚 . These two together phasor sum
gives you the no load current and that number, I am telling for a well designed transformer
for a well designed transformer. Therefore, without knowing the equivalent circuit this
that I will say that maybe for this transformer, 𝐼01 no load current referred to side one will
be about say
𝐼01 = 0.05 ×
𝐼𝐻𝑉
= 0.05 × 5 = 0.25𝐴
𝑅𝑎𝑡𝑒𝑑
Same transformer; if you are energized from this side no load current LV side; it will be
𝐼02 . If energized from LV side,
𝐼02 = 0.05 ×
𝐼𝐿𝑉
= 0.05 × 10 = 0.5𝐴
𝑅𝑎𝑡𝑒𝑑
150
Anyway exact value I do not know what I am telling given a transformer, you can guess
what will be the order of the no load current.
So, no load current value is much smaller compared to the rated by maybe of the order of
5% of the rated current. See this transformer could be energized from the 100V side as
well that is why LV side, I call it and your no load current can be estimated. That is if I
say suppose you are energizing the transformer from the HV side with secondary no load
means no impedance which is open here no load means s is opened s opened. Then I am
telling suppose I want to measure this current no load current. Then how to choose the
ammeter reading? It guides you from the HV side better take an ammeter which can read
whose range is 0.05Amp do not connect it 0 to 30Amp range ammeter to record a current
of 0.25Amp. So, I can decide upon the range of the ammeter; if I know this information
that is no load of this current is of the order of 5% and this 5% is not unique value that is
given a transformer.
You expect it is it you can expect it will be close to this current, but exact current in any
case I can calculate because the no load current can be about 2 to 5% of the rated current.
You must understand that. After knowing this let us start discussing how to determine the
equivalent circuit parameters.
(Refer Slide Time: 23:43)
Determination of equivalent circuit parameters. So, a transformer practical transformer is
given I want to know what will be the values of 𝑟𝑒2, what will be the value of 𝑥𝑒2 , what
151
will be the value of 𝑅𝑐𝑙2 and what should be the value of 𝑋𝑚2. On any side if you know,
other side can be calculated or in the previous diagram my god.
Student: (Refer Time: 24:51).
That is this is with respect to the secondary side; I drew with respect to the primary side.
Similarly our goal is to calculate 𝑟𝑒1, 𝑥𝑒1 , 𝑅𝑐𝑙1 and 𝑋𝑚1 that is how can I calculate it?
The answer to this is we have to do some testing and simple testing; I will do. So,
determination of equivalent circuit parameters by conducting from two test; open circuit
test, you conduct two test; open circuit test and short circuit test short circuit test. You
conduct these two tests in the laboratory and you can get those to all the parameters ok.
How do I do the test? First will take up this open circuit test, what it is all about? What is
done? You take the transformer and connect an ammeter and connect a Watt meter and
this is the practical transformer primary.
And generally to do the open circuit test from the LV side, this test is carried out. The
reason will tell you later and suppose this is the other side HV side which is open circuit
or sometimes called no load test or no load. No load means no impedance is connected
across the secondary and also you connect a voltmeter across this. And here you apply the
𝑉1 rated voltage rated voltage at rated frequency.
So, what I will do? LV side, I will energize with voltage 𝑉1 at rated frequency 𝑓 and
connect an ammeter and Wattmeter ok. And keep in mind this is an approximate equivalent
circuit I will be referred to what is the approximate equivalent circuit then? Here 𝑉1 at
frequency 𝑓, you have given 𝑟𝑒1 and 𝑥𝑒1 are there, but these are open circuited because
secondary nothing is connected infinite. So, this is that open circuit will be reflected at the
open circuit in the equivalent circuit also.
So, there is 𝐼2′ = 0 nothing is there because no impedance is connected. So, 𝐼2 is not there
𝐼2′ is not there and this is 𝑅𝑐𝑙1 and this is 𝑋𝑚1. So, under this condition this condition, you
see this is 𝐼0 this is also 𝐼0 and this is 𝐼𝑐𝑙1 and this is 𝐼𝑚1 magnetizing current. And what is
the phasor diagram in this approximate equivalent circuit? Phasor diagram is very simple
applied voltage 𝑉1, your magnetizing current will be here 90° lagging and your core loss
component of current will be in phase with the supply voltage and this will be your no load
current; 𝐼0 .
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So, your ammeter is going to read 𝐼0 ok. And this is called no load power factor angle. If
this is 𝜃, this is 𝜃; 𝜃0 will be in an ideal transformer 90°, but it cannot be because it is a
practical transformer having core losses ok. And this is the thing and the Watt meter
suppose read 𝑊0 , then I will say that this Watt meter reading must be equal to the voltage
applied across its pressure coil that is 𝑉1 current flowing through its current coil that is 𝐼0 .
Let us assume wattmeter to be ideal; I mean those losses of Watt meter neglected
𝑊0 = 𝑉1 𝐼0 cos(𝜃0 )
This is what I will get.
And from this, I can get the value of
𝐼0 cos(𝜃0 ) =
𝑊0
𝑉1
and note that
𝐼0 cos(𝜃0 ) =
𝑊0
= 𝐼𝑐𝑙1
𝑉1
from this right angled triangle. So, 𝐼𝑐𝑙1 is known. Then what you do follow? This point
very carefully some students make very I mean big mistake. It is better you try to calculate
𝐼𝑚 also; magnetizing branch ok. Instead of trying to calculate
𝑉1
𝐼0
that will give you the
equivalent impedance things become complicated; it is much more easier as you see.
What do you do now? Since so, from these also I can calculate cos(𝜃0 ).
cos(𝜃0 ) =
𝑊0
𝑉1 𝐼0
So, from this 𝜃0 is known. If 𝜃0 is known, then from this triangle I can say
𝐼𝑚1 = 𝐼0 sin(𝜃0 )
𝐼0 is known ammeter reading. And also
𝐼𝑐𝑙1 = 𝐼0 cos(𝜃0 )
153
Therefore, what we have done I have got this current 𝐼𝑐𝑙1 and 𝐼𝑚1 and my problem is to
calculate 𝑅𝑐𝑙1 ;
𝑅𝑐𝑙1 =
𝑉1
𝐼𝑐𝑙1
𝑋𝑚1 =
𝑉1
𝐼𝑚1
This is known this is known and
So, these two values can be calculated.
𝑉
Instead of trying to calculate the impedance 𝐼1 , do not do it because that impedance will
0
then be the series representation of 𝑅𝑐𝑙 and 𝑋𝑚 equivalent. And then you are gone and do
not say the series part of this equivalent circuit is the core loss resistance. Then it will
define the physical reasoning that is the magnetizing current requires a fixed voltage and
this one.
So, you should be very careful. So, mind you then I have been able to calculate 𝑅𝑐𝑙1 and
𝑋𝑚1. So, doing this open circuit test, I can very quickly calculate by noting down the
ammeter reading and wattmeter reading and these voltmeter reading that is that will read
equal to 𝑉1. I can calculate 𝑅𝑐𝑙1 and 𝑋𝑚1. And then am telling, if it is needed, you can then
after knowing 𝑅𝑐𝑙1 and 𝑋𝑚1, I can always calculate 𝑅𝑐𝑙2 and 𝑋𝑚2.
For these, I do not have to carry out the experiment once again because I know it will be
either divided or multiplied by 𝑎2 . So, correctly if I do that I will get 𝑅𝑐𝑙2 as well as 𝑋𝑚2.
So, in this equivalent circuit, I have got this one only just one point. As I told you, this is
the equivalent circuit approximate equivalent circuit approximate equivalent circuit. But
if you honestly look at the circuit, you will find this no load current is flowing through
winding resistance as well, is not.
Because the exact equivalence circuit is what? 𝑟1 𝑥1 , then your this parallel branch is not
suppose and then of course, these two are open circuited exact equivalence circuit. And
you are doing on a practical transformer whose exact equivalent circuit is this what we
have done. Here we have applied a voltage and this current you are telling 𝐼0 fine, but then
I will say look here this wattmeter reading 𝑊0 ok. We have done some approximation that
154
I have understood, but this Watt meter reading strictly speaking, you will read the core
loss in 𝑅𝑐𝑙 as well as copper loss in 𝑟1.
Some of these two will be read is not, but what happens is this how to justify that why we
neglect this copper loss. It is because of that in open circuit test, winding current is pretty
small; maybe 2 to 5% of the rated current. But applied voltage is rated at rated frequency
level of flux in the core is rated and it is the level of flux that decides the core loss. So,
core loss will be much higher than these very small current causing a power loss in 𝑟1.
So, we neglected the copper loss note that 𝑊0 practically records the core loss only; core
loss. The copper loss which is also called the winding power loss 𝐼02 𝑟1 is much less
compared to core loss, why? Because flux is rated and 𝐼0 is only about 2 to 5% of the rated
current. Anyway please go through this part and try to understand each bit of it because
this is the approximate equivalent circuit ok. If you forget about the exact thing from your
mind erase it, then watt meter this copper loss; I mean core loss only.
But that is not the case, there will be a little bit of copper loss in the winding which we are
neglecting. We are attributing the wattmeter reading to be fully the core loss. We will
continue with this with the SC test to determine these this parameters in the next class.
And remember that generally the reason, I have not told yet the open circuit test is carried
out from LV side; from LV side means there you connect the supply HV side is kept open;
anyway.
Thank you.
155
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 17
Short Circuit Test
(Refer Slide Time: 00:24)
Welcome to lecture number 17 on Electrical Machines-1. And we were discussing about
the determination of equivalent circuit parameters. Equivalent circuit for example, has
been drawn in this diagram referred to primary side. And you know it is approximate
equivalent circuit ok, this is approximate equivalent circuit. And 𝑟𝑒1 = 𝑟1 + 𝑎2 𝑟2, and
𝑥𝑒1 = 𝑥1 + 𝑎2 𝑥2 . Now, the question is this equivalent circuit can be therefore used to
analyze a circuit which involves a transformer? And it is easier. So, therefore, equivalent
circuit is always drawn referred to a particular side.
So, for example here, now the big question is how to determine the parameters of this
equivalent circuit? I will very quickly tell you one of the test open circuit test I discussed
in the last class. From that test this parallel branch 𝑅𝑐𝑙1 and 𝑋𝑚1 can be determined. What
we did there is this. Generally from the LV side, the open circuit test is carried out,
whichever side you energize for testing the parameter values we will get referred to that
side. So, referred to LV side, suppose 𝑉1 is the rated voltage you have to apply rated
voltage, connect an ammeter, wattmeter and the volt meter and this is the practical
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transformer. And the other side is kept open no impedance is connected or load is
connected across the secondary. And what you all you have to do is you have to take the
readings of ammeter, wattmeter and voltmeter ok.
Only thing is that during no load test or open circuit test, if we know the rating of the
transformer, we can have some idea about the order of the no load current as I told you 2
to 5% of the rated current. So, before doing the test, what do you do from the kvA rating
and voltage ratings, calculate the rated current of the LV side and take 5%, 7% of that. To
have some idea that what is the order of the current at the transformer is going to draw
during no load test that will help me to decide about the range of the ammeter, and the
current coil rating of the wattmeter that is important.
For example, the no load current if you calculate it comes out to be 0.8Amp expected to
be of that order, then you connect an ammeter which can record current up to say 1.5Amp
that will be a good choice. Similarly, current coil of the wattmeter should be of the order
of 2Amp range. Of course, 5Amp range ammeter and 5Amp range current coil of the
wattmeter will do, but accuracy will be more, if you use lower ranges of ammeter and
current coil rating of the wattmeter. Pressure coil rating of the wattmeter is of course the
rated voltage of the transformer.
In any case the approximate equivalent circuit is this. So, all the powers the reading of the
𝑊0 will practically record the core loss taking place in the transformer. Because no load
current is very small there will be some copper loss in the primary of the winding 𝐼02 𝑟1 and
that will be neglecting, because flux is rated mind you. Rated voltage, rated frequency flux
will be rated.
And therefore, referring to this equivalent circuit 𝑉1, 𝐼0 you draw 𝜃0 and sometimes it is
tempting to represent it in series combination better do not do it although from circuit point
of view, it is fine. But whatever resistance real part of this equivalent 𝑅𝑐𝑙 and 𝑋𝑚 will not
give in Ω the resistance representing core loss, because core loss the flux is constant
therefore, it is better you represent in parallel and always talk about this resistance. So,
there is a shortcut way, then 𝐼0 is known, 𝜃0 is the power factor angle which I can calculate
from the wattmeter reading, cos 𝜃0 I will calculate. Then from this triangle I can calculate
magnetizing current, this current and this current.
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𝑉
𝑉
𝑚
𝑐𝑙
And once this current is known 𝐼 1 will give you 𝑋𝑚 ok. So, here I have written and 𝐼 1 will
give you 𝑅𝑐𝑙 ok, once I know that. Now, if I ask you the question what is wrong, if I carry
out the test from the HV side that I will decide slightly later. But what I am telling it, I can
after knowing these 𝑅𝑐𝑙1 and 𝑋𝑚1, I can tell you that what will be 𝑅𝑐𝑙2 and 𝑋𝑚2 if you had
carried out the test from the HV side.
You simply divide by 𝑎2 this quantities and we will get 𝑅𝑐𝑙2 and 𝑋𝑚2. Therefore,
remember always that if you have determined some parameters with respect to a particular
side, you need not carry out the test once again from the other side to determine the same
parameters, only thing there will be a multiplying or dividing factor of 𝑎2 depending upon
the situation.
So, 𝑅𝑐𝑙1 and 𝑋𝑚1 , the moment you have calculated what I want to say, I know also 𝑅𝑐𝑙2
and 𝑋𝑚2. I can always calculate by multiplying or dividing with 𝑎2 as the case may be.
Now, today the next test is short circuit test, we will discuss. And in the short circuit test,
if you carry out what happens is this you will get these two parameters 𝑟𝑒1 and 𝑥𝑒1 ok, let
us see what you have to do. Now, during short circuit test what happens is this, although
it suggested always that better you carry out the short circuit test from the high voltage
side ok. So, I will first discuss the test, then I will tell you what do I mean by this.
(Refer Slide Time: 08:06)
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So, this was your transformer ok so, the same transformer that is here it was I should
review it like this LV, HV. So, what I will do during short circuit test, I will connect source
on the high voltage side and the other side I will just put a short circuit. I will short these
two terminals of the LV side. And here what I will do, I will connect once again those
three meters, one is a voltmeter you connect, another is a wattmeter you connect and also
connect an ammeter ok. And here I will connect the supply.
Ok, so you give some supply and note down these meter readings that is the wattmeter
reading which is 𝑊𝑠𝑐 I will write ammeter reading 𝐼𝑠𝑐 I will write and voltmeter reading
𝑉𝑠𝑐 . Now, the question is how much voltage should I apply during open circuit test which
I carried out from the LV side, I applied rated voltage, mind you rated voltage was applied.
Suppose, let us take the same transformer which I am telling that suppose we have a
transformer whose rating is 200V/100V 50Hz and 1 KVA. Now, the moment this is 1
KVA, then as I told you I can calculate the
anyway
1000
𝐼𝐻𝑉
𝐼1
= 200 = 5𝐴 is it not, and
=
𝑅𝑎𝑡𝑒𝑑 𝑅𝑎𝑡𝑒𝑑
1000
𝐼𝐿𝑉
𝐼2
=
= 100 = 10𝐴. So, let us talk in terms of this transformer then
𝑅𝑎𝑡𝑒𝑑 𝑅𝑎𝑡𝑒𝑑
things will be much clearer while doing LV open circuit test, I should apply suppose the
rating of the transformer is this. I should apply 100 volt here LV side I have to energize
100 volt and that is all.
Now, what I was telling
𝐼𝐻𝑉
= 5𝐴 therefore, I expect the no load current to be of the
𝑅𝑎𝑡𝑒𝑑
order of 5% of this that is 5 × 0.05 = 0.25𝐴. So, perhaps in ammeter of range 0 to 0.5Amp
will do. Similarly, the current coil range of this wattmeter I will select as 0 to 1Amp that
will be a good choice and pressure coil will be 100 volt that is what we need.
So, suppose the same transformers; for the same transformer, I am now going to do the
short circuit test. So, this time I, so what is the rating of the transformer rated voltage is
100V/200V ok, 1KVA. This rated current is 10Amp of this side and this side rated current
is 5Amp we have just seen. Now, in this case, how much voltage should I apply here that
is the question. So, from the HV side what is the rated voltage of the HV side 200 volt. So,
should I apply 200 volt 50Hz on the HV side, the answer is a big no, never apply the rated
voltage when the other side of the transformer is short circuited.
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The answer can be easily understood, recall that in a transformer if you anyway I will draw
something here. Suppose, this is the transformer when it will be operated normally this
transformer what do you do, you apply suppose 200 volt here, you get 100 volt and then
you connect the load. I told you in my previous lectures how much load impedance you
will connect is decided by the rated current of the secondary coil. What is the rated current
of the secondary coil, LV side 10Amp.
Therefore, it looks like I should not connect an impedance whose value is less than 10 Ω,
if you connect the transformer will be overloaded. But maybe 10 Ω if you connect then
this current will be 10Amp that is fine, and then this current will be approximately 5Amp
is not that is this scenario. So, less than 10Amp if you connect, the rated current will not
cross its limit.
Now, during short circuit test what you are doing the LV side, if you apply 200 volt, if you
get 100 volt here and if you short circuit it this current is going to be pretty high 100 volt
by almost zero resistance, you are short circuiting and that will definitely be many times
larger than the rated current. And you are there will be problem, I mean transformer may
burned if you have not taken proper precaution that is you are not connected proper fuse
these that. So, but during no load test no such problem apply rated voltage rated current
and the current drawn from the supply will be only 5% of the rated current of the LV side,
but here you cannot short circuit it by applying 200 volt you are putting a short circuit.
Then the question is how much voltage should I apply to the HV side, the answer I hope
you have got. I can only say this much that I should not apply a voltage across the HV side
which will cause the current drawn from the supply to exceed its rated current of 10Amp.
I should not apply a voltage, I will apply such a voltage under this scenario it will be like
this. So, you must have an arrangement here in the form of an auto transformer which you
must have used. So, this is auto transformer, do not forget to connect an auto transformer
ok, where output voltage can be varied starting from zero.
So, what do you do, here your maybe this side here this supply voltage is 200 volt 50Hz
ok. So, set the auto transformer pointer at zero, so that output is zero here, no current flows
even when it is short circuited. Now, gradually increase this pointer, so that you are
applying more and more voltage certainly not 200 volt a fraction of that you go on
increasing it, but keep a strict watch on the ammeter reading. The moment ammeter
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reading is 5Amp, I will stop further increasing this pointer because that is the maximum
voltage you can apply ensuring that the currents in the windings will not exceed the rated
current that is the whole idea. Therefore, a little voltage is required to do this, got the point.
So, the applied voltage here is less 𝑉𝑠𝑐 , mind you applied voltage 𝑉𝑠𝑐 ≪
𝑉𝐻𝑉
whatever
𝑅𝑎𝑡𝑒𝑑
200 volt is the rated of you do not apply 200 volt under short circuit. But a little voltage as
you can see and I have no problem I will connect an auto transformer a variable output
voltage I will get and therefore, I will only go up to that voltage which will make this
current predicted. Of course, one can go 5%, 10% current higher than the rated current that
is not a big issue, but certainly not apply 200 volt ok.
So, during short circuit test I am applying a little voltage compared to the rated voltage.
Now, during short circuit test the level of flux 𝜑𝑚𝑎𝑥 during short circuit test in the core of
the transformer, it will be then
𝜑𝑚𝑎𝑥
𝑉𝑠𝑐
(𝑆𝐶) = √2𝜋𝑓𝑁
𝐻𝑉
≪
𝑉1
√2𝜋𝑓𝑁𝐻𝑉
⏟
𝜑𝑚𝑎𝑥
𝑅𝑎𝑡𝑒𝑑
𝜑𝑚𝑎𝑥
𝜑𝑚𝑎𝑥
(𝑆𝐶) ≪ 𝑅𝑎𝑡𝑒𝑑
This must be understood because you are applying very little voltage, of course, frequency
is high. Which; obviously, then means that
𝐵𝑚𝑎𝑥
𝐵
≪ 𝑚𝑎𝑥
(𝑆𝐶)
𝑅𝑎𝑡𝑒𝑑
because you have to divide it with a constant cross sectional area of the core to get 𝐵𝑚𝑎𝑥 .
It that be the case, so a very little amount of flux will be produced inside the core of the
transformer under short circuit test although current in the windings will be rated values.
So, a interesting situation, now during this test, therefore, I will say the core loss will be
very small compared to the winding copper loss that is current is 𝐼 2 𝑅2 or 𝐼 2 𝑅1 whatever
you call the sum of these two winding copper losses that will be high because rated current
2
you are passing. But the core loss as we know eddy current loss depends upon 𝐵𝑚𝑎𝑥
.
161
2
And by a large factor you have reduced that 𝐵𝑚𝑎𝑥 , frequency same, 𝐵𝑚𝑎𝑥
, 𝑥 2 , 𝜏 is of
course, same thickness of the plate. So, eddy current loss, similarly hysteresis loss depends
upon the area enclosed by the cores by the by the BH curve of the material area enclosed
by the BH curve, but during short circuit test if this is the BH curve during rated I am just
trying to give you a pictorial thing. If this is corresponding to rated condition, then BH
curve may look like this during short circuit test SC and this is rated condition.
So, both hysteresis and eddy current loss will be less. And also from that empirical formula
1.6
we know hysteresis loss depends upon 𝐵𝑚𝑎𝑥
. So, 𝐵𝑚𝑎𝑥 itself is quite small, therefore,
during short circuit test; note, during it is worth writing during SC test core loss will be
much much less than the winding copper loss that is 𝐼 2 𝑟 loss in the winding which is 𝐼12 𝑟1 +
𝐼22 𝑟2 . So, I am not writing that thing got the point. So, this, this is must be understood.
At this I have told you from what is called physical interpretation, why core loss can be
neglected during short circuit test. In other words what I am telling, so during the test when
this 𝐼𝑠𝑐 , I will I will apply such a voltage here 𝑉𝑠𝑐 that
𝐼𝑠𝑐 =
𝐼2
𝑅𝑎𝑡𝑒𝑑
HV side which I am calling as side two compared to my previous thing same transformer
I am working with. And also in this the current whatever will flow is
𝐼1
, because
𝑅𝑎𝑡𝑒𝑑
current ratios are related by that factor 𝑎. Therefore, this is the rated current, but flux is
not rated it is only a small fraction of the rated flux.
Therefore, this wattmeter reading strictly speaking will record both the core loss and the
copper loss in the winding, sum of these two, but because core loss is very much smaller
than the winding copper losses. Therefore, we say that 𝑊𝑠𝑐 is practically equal to the coil
or winding copper losses; coil copper losses. You must know 𝑖 2 𝑟 loss is copper loss, it is
occurring in the copper. Core loss is in the iron, it is occurring that is also some 𝑖 2 𝑟 loss,
we have seen that, but people do not say oh that is called iron loss fine, everything is fine,
because in the iron the eddy current flows.
So copper loss, copper is used to make the primary and secondary so, copper loss is a and
copper loss will be 𝐼12 𝑟1 + 𝐼22 𝑟2 . And you must understand by this time this is same as 𝐼12 𝑟𝑒1
if you express everything in terms of 𝐼1 , and which is same as 𝐼22 𝑟𝑒2 .
162
𝑊𝑠𝑐 ≈ 𝑊𝑖𝑛𝑑𝑖𝑛𝑔 𝐶𝑜𝑝𝑝𝑒𝑟 𝐿𝑜𝑠𝑠 = 𝐼12 𝑟1 + 𝐼22 𝑟2 = 𝐼12 𝑟𝑒1 = 𝐼22 𝑟𝑒2
This portion I think you I told you, you show it you can easily show, so this is the thing;
so, this is the situation. Now, I will go to the from the equivalent circuit then what do I get.
So, this is the thing I will note this things ok.
(Refer Slide Time: 26:51)
Now, the equivalent circuit referred to the high voltage side will be because I have
energized it from the high voltage side. Now, equivalent circuit referred to HV side and I
remember we are discussing short circuit test under that condition. What is the equivalent
circuit referred to the HV side it will be approximate equivalent circuit, it will be you have
applied some voltage on the HV side 𝑉𝑆𝐶 . And then this parallel branch will be there, is it
not, this will be the parallel branch. And then you have if HV side I denote it by 𝑟𝑒2 and
𝑥𝑒2 , this will be I have energized the transformer from the HV side mind you. On LV side,
load is connected.
Now, what load is connected on the LV side, this is HV, this is LV, I have kept it shorted.
So, 𝑧2 = 0 therefore, here also in the equivalent circuit this will be shorted because 𝑧2 =
0 not 𝑧2 , 𝑧1 = 0. So, 𝑧1′ = 0, is it not, it is shorted. And if I want to write this two, I should
write 𝑅𝑐𝑙2 and 𝑋𝑚2 ok. And 𝑅𝑐𝑙1 , 𝑋𝑚1 I have found out.
Now, these impedance now coming to the equivalent circuit and trying to establishing the
same thing. These 𝑟𝑒2 and 𝑥𝑒2 are quite small compared to 𝑅𝑐𝑙2 and 𝑋𝑚2, these are high
163
here. And also I am telling this core loss power can be neglected power in this resistance
is your core loss during short circuit test which is quite small and that is also reflected in
the equivalent circuit. What is the core loss now,
2
𝑉𝑠𝑐
𝑅𝑐𝑙2
, got the point. but this voltage is only
a fraction of the rated voltage that is 200 volt for the transformer we have considered
maybe 5, 10 voltages, you will find in the laboratory.
Therefore, in essence what I am telling is during short circuit test and this impedance quite
small compared to these. So, this branch can be removed approximate. So, what I am
telling is this branch as if it is not there, parallel branch can be shown to be not present.
𝑣
Magnetizing current also will be too small 𝑋 , but 𝑣 I am applying only 𝑉𝑠𝑐 which is much
𝑚2
smaller than the rated voltage. So, both 𝐼𝑚 and 𝐼𝑐𝑙 are quite small compared to this current,
which is now rated current on the HV side for the transformer I have considered I will
apply such a voltage such that 5Amp close now. So, this will be the thing.
Now, therefore, you have this thing you have the short circuit current 𝐼𝑠𝑐 ok. And this is
𝑟𝑒2 and 𝑥𝑒2 . Now, what happens is this I will then say the wattmeter reading whatever you
have got 𝑊𝑠𝑐 must be equal to approximately equal to
2
𝑊𝑠𝑐 ≈ 𝐼𝑆𝐶
𝑟𝑒2
straight. And here you note it is series, applied voltage is known, power drawn by the
circuit is known, 𝑊𝑠𝑐 had a voltage 𝑉𝑠𝑐 𝐼𝑠𝑐 . Therefore, 𝑟𝑒2 can be calculated.
𝑟𝑒2 ≈
𝑊𝑠𝑐
2
𝐼𝑆𝐶
𝑉
And similarly the ratio 𝐼𝑠𝑐, is this impedance, magnitude of this impedance
𝑠𝑐
𝑉𝑠𝑐
2
2
= √𝑟𝑒2
+ 𝑥𝑒2
𝐼𝑠𝑐
it will be this; 𝑟𝑒2 is known. So, from this, this will give you 𝑥𝑒2 can be found out. So,
from the short circuit test, the value of 𝑟𝑒2 and 𝑥𝑒2 can be found out. I will continue with
this in the next class.
Thank you.
164
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 18
Choosing Sides to Carry Out O. C/S. C Test
Welcome to lecture number 18 and we have been discussing about two simple tests,
performing those two tests one can determine the equivalent circuit of a transformer.
(Refer Slide Time: 00:41)
For example, you first do the open circuit test and open circuit test is carried out from the
LV side and you will be if you call the LV side as 1, then you will be getting this 𝑅𝑐𝑙1 and
𝑋𝑚1. Similarly of course, I have not told you why it should be done on from the LV side,
that I will discuss today.
165
(Refer Slide Time: 01:11)
And, then if you carry out the short circuit test better carry it out from the HV side the
reason we will discuss. But, if it is carried out from the HV side then what parameters you
will get is 𝑟𝑒2 and 𝑥𝑒2 and the calculations are pretty simple.
(Refer Slide Time: 01:31)
In this case the wattmeter reading, then approximately it records the copper loss, core loss
can be neglected. During short circuit test some ways I right during SC test core loss it
takes place no doubt is much much smaller then copper loss. Iron loss sometimes core loss
166
is also called iron loss whatever it is much smaller. So, that can be neglected and we say
the wattmeter reads the copper loss only.
And, this saying in language in the equivalent circuit it means you can neglect this branch;
both magnetizing current can be neglected as well as core loss component of current can
be neglected whatever power is drawn. That wattmeter reads real power only, real power
𝑉
2
loss will take place here in 𝑟𝑒2, 𝐼𝑠𝑐
𝑟𝑒2 and 𝑠𝑐 from the 𝑧𝑒2 these two can be found out.
𝐼𝑠𝑐
So, now we have carried out two test, but my final objective is to draw the equivalent
circuit; referred to a particular side. Referred to which side? It is my choice, the user’s
choice he decides with respect to side 1, I want to draw the equivalent circuit. Or,
somebody else would like to draw the equivalent circuit referred to the secondary side, but
nonetheless whichever side you draw your end results will come same. So, you need not
worry about this side ok, you choose this side with respect.
(Refer Slide Time: 03:35)
Now, after doing this test you see that I just draw here LV side, from LV side I carried out
the OC test. And, whatever values you have got in the open circuit test this is the HV side;
HV side was open. You got 𝑅𝑐𝑙1 the parallel branch and 𝑋𝑚1 this two values I got. And,
while doing short circuit test which I told you for reasons not yet told that you suppose
carry out the short circuit test from the HV side and during short circuit test you keep it
167
shorted. So, you are looking at the equivalence circuit referred to this side then you will
be getting 𝑟𝑒2 and 𝑥𝑒2 .
Therefore, to draw now I want to draw the equivalent circuit; to draw the equivalent circuit
referred to and suppose LV side I am calling it side 1, HV side calling it side 2; so, this is
1, this is 2. So, to draw the equivalent circuit refer to, LV side it should be drawn like this
now 𝑉1 rated voltage you will apply and approximate equivalent circuit I am telling this is
𝑅𝑐𝑙1 and this is 𝑋𝑚1 parallel. And these values are already known so, I will put these values
and then I have to that series impedance which I know it must be 𝑟𝑒1 and 𝑥𝑒1 .
But, from short circuit test I have got 𝑟𝑒2 and 𝑥𝑒2 . So, do not put 𝑟𝑒2 and 𝑥𝑒2 here then that
will be a blunder. So now, you have a duty to change these 𝑟𝑒2 to 𝑟𝑒1 and 𝑥𝑒2 to 𝑥𝑒1 and
that can be easily done since 𝑟𝑒2 and 𝑥𝑒2 is known. So, a turns ratio from the rating of the
transformer I know it is
𝑎=
𝑁1
𝑁2
So
𝑟𝑒1 = 𝑎2 𝑟𝑒2
𝑥𝑒1 = 𝑎2 𝑥𝑒2
So, these values should be transformed correctly, then write it here this is the equivalent
circuit. What should I write, should I write 𝑉2 here? No, I should write 𝑉2′ and so on. Then
wherever this circuit is now connected I can replace that transformer by this equivalent
circuit and calculate all the things needed. Similarly, this test although has been done from
LV side and HV side to draw the equivalent circuit. So, once you know this then you know
equivalent circuit refer to HV side will simply be this one.
Structure remains same here with respect to HV side they you should right here 𝑟𝑒2 and
𝑥𝑒2 do not disturb them, because that is with respect to HV side those parameters unknown.
But, the only thing then do not right here 𝑅𝑐𝑙1 , you should write 𝑅𝑐𝑙2 and 𝑋𝑚2. And what
is 𝑅𝑐𝑙2 ?
𝑅𝑐𝑙2 =
168
𝑅𝑐𝑙1
𝑎2
𝑋𝑚2 =
𝑋𝑚1
𝑎2
Therefore, you can do the test either from LV side or HV side then you can of course, draw
the equivalent circuit refer to any side you like. But, you should be careful about these two
points.
Now, I will tell you why people say that generally you carry out the LV test from the short
circuit test from the LV side and open circuit tests from the LV side and short circuit test
from the HV side. The reason is rather more practical than ah saying that it cannot be done
I mean. For example, in this transformer the rating of the transformer I chose to clarify
some of the points.
(Refer Slide Time: 10:07)
Suppose the rating of the transformer is our this transformer 200 volt/100 volt, 50 Hz,
single phase and kVA rating is suppose 1 kVA. And, as I told you the rated current of this
side is 5Amp and this side is 10Amp. And, these numbers you can easily calculate straight
away; once a problem is given be ready with these numbers what are the rated currents of
these two sides. Now, for this transformer of this kind of rating there is really no restriction
because, in the laboratory these voltages are low voltages and available in the lab.
So, so one should not make a issue or insist that for this transformer you carry out the test
from OC test from the LV side and SC test from the HV side. Reason is simple these two
voltages are available in the lab, we can always get these two voltages of 50 Hz using a
169
transformer, auto transformer always you can connect to get these voltages. Similarly,
these current values are also reasonable values 5Amp, 10Amp is no large values at all no
issues. You can have Watt meters of this ranges. Therefore, if you carry out the open circuit
test from the HV side, you can do it what is wrong.
So, for this kind of transformer of this kind of rating no restriction on carrying out a test
from a particular side. Remember let us come to the circuit diagram of open circuit test, it
need some telling open circuit test. Suppose the rating of the transformer is this it is written
there fine. Now, if I carried out the test open circuit test for this transformer from the high
voltage side, I have to apply rated voltage 200 volts 50 Hz is available, I will apply that.
Rated current of the HV side is 5Amp only is not, HV side is 5Amp.
So, is no load current will be small ok, but meters are available to record that, similarly
wattmeter with rating 200 volt and that current rating estimated current is 2 to 5%. I can
easily carry out the test either from this side or then that side; no problem at all in the
laboratory because all the ranges of the meters are available and so on and no safety issues
also is not. Now, similarly during short circuit test, if you had carried out the short circuit
test energizing from the LV side, keeping the HV side shorted also for this transformer no
issue. Because, rated current is only 5Amp and 10Amp; LV side you can do you will.
So, all the voltages and currents to be measured, power to be measured or all those meters
are available and there is nothing wrong in that. But, you imagine suppose I say you that I
have to test a transformer, I asked you to find out the rating of a transformer whose rating
is 100 kVA. I am just taking some 100 kVA, then say 1000 volt/100 volt, 50 Hz
transformer single phase. This is the transformer whose parameters I want to find out.
Suppose, somebody says that ok I will carry out the open circuit test, he suppose thinks
that open circuit test from the HV side, OC from HV side suppose somebody decides.
Can I not do it? Open circuit test demands that whichever side you energize you have to
apply rated voltage because, you have to create rated flux. So, all the core loss should take
place therefore, suppose somebody this is the HV side and LV side he keeps open circuit.
So, he has to apply 1000 volt here and he has to connect an ammeter and a wattmeter and
a voltmeter. First thing is to carry out the open circuit test from the HV side in the
laboratory you then require a 1000 volt, 1000 volt is not easily available.
170
Although perhaps 200 volt is available, connect another transformer get 1000 volt that is
also not a big issue. But, the point is whoever will be doing this experiment his safety will
be in question or the heat a table of this equipments we have kept, you will switch on 1000
volt supply, 1000 volt supply is dangerous in the laboratory. You cannot work unless it is
a high voltage laboratories specially designed therefore, immediately you see to apply
rated voltage I come and say you do not do this a that will be a big safety issue cannot
apply. So, open circuit test carrying out from HV side that is why people in general say
carry out the open circuit test from the LV side understood.
So, 1000 volt you cannot apply. Similarly, also you see then he will require a pressure coil
whose rating is 1000 volt current coil rating of course, will not be a big problem. Because
what is the rated current? As I told you given a rating of the transformer of this kind, what
is rated current of the side
𝐼𝐻𝑉
is how much?
𝑅𝑎𝑡𝑒𝑑
3
100 × 10
𝐼𝐻𝑉
=
= 100𝐴
𝑅𝑎𝑡𝑒𝑑
1000
What will be
𝐼𝐿𝑉
? LV side rated current will be higher. What is the turns ratio? 10 so,10
𝑅𝑎𝑡𝑒𝑑
times higher
𝐼𝐿𝑉
= 10 × 100 = 1000𝐴
𝑅𝑎𝑡𝑒𝑑
Now therefore, you see I am doing the test; trying to doing the tests from the HV side then
I see
𝐼𝐻𝑉
is this one. What will be the order of no load current? About say 5%. So, 5%
𝑅𝑎𝑡𝑒𝑑
of 100Amp is how much? 5Amp therefore, the current drawn will be less, that ammeter I
can have. Similarly, current coil of the wattmeter will not be an issue from the HV side,
but the voltage is high which is dangerous for the personnel will be carrying out the test.
So, this should not be carried out from the HV side.
But, from the LV side if you do what during open circuit test, how much voltage you need
to apply? 100 volt only, very safe available voltage in the lab. What will be the pressure
coil of the wattmeter? 100 volt. What will be the order of the no load current? 1000 into
say 5% of this one, how much it is? 50 is not for this particular transformer; so, 5% of this
is 50Amp ok, 50Amp large current voltage is not high. So, I can use CT PT or ammeters
171
of measuring a 50Amp is available do not worry, even if it is not available I can use CT to
scale down the current, pressure coil no issue.
Therefore, it is this thing you do LV side so, for large transformer with ratio of turns ratio
are very large then really it becomes an issue high voltage. So, carry out so, OC test from
LV side always LV side. Now what about SC test? Short circuit test people say carry on
carry out from the HV side. Why? Because, during short circuit test you know you will be
suppose somebody says, no I will carry out the short circuit test from the LV side. If you
want to carry out the short circuit test from the LV side, what are the things you require?
What are the meters you require LV side suppose? You require ammeter, you require
wattmeter and a voltmeter. Suppose from the LV side, how much voltage you will be
needing? What is the LV side? 100 volt very little voltage is needed that is not a big issue
with an auto transformer not 100 volt you apply and HV side is shorted, is not HV side is
shorted.
But, what is the rated current? Rated current is 1000Amp. So, you have to measure that
current 1000Amp, range ammeter may not be so easily available or of course, you can use
CT PT and voltage rating is small. Similarly, the current coil of the wattmeter should be
of the rating of the same in the laboratory such a wattmeter may not be available.
Therefore, all these point to the fact, but if this was HV side and this was LV side you see
high voltage side what is the rating 1000 volt. But, do not require 1000 volt to carry out
SC test only little voltage maybe 10% of that which will circulate the rated current.
So, 100 volt 50 volt etcetera and the current will be also small rated current is only
100Amp. So, these are the issues why people say carry out for transformer with high kVA
rating large voltage ratings, if voltages ratios are one side is kV another side is volt. For
example distribution transformer 6.6 kV of course, those are three phase transformer, but
6.6 kV and 440 volt. This is the kind of transformation ratios, in that case to carry out the
open circuit test you carry it out from the LV side. And, similarly to carry out the short
circuit test carry it out from the HV side.
And, in generally this is the statement of course, since it is expected you will carry out the
test from two different sides. Therefore, at the end after getting the parameters parallel
branch parameters referred to the LV side and equivalents series impedance parameters
which you will get from the short circuit test carried out from the HV side.
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Then you stop you think a bit before drawing the equivalent circuit referred to a particular
side, you have to do a little bit of transformation either that parameter or this parameter
depending upon from which side you want to draw the equivalent circuit of a transformer.
So, there are a lot of examples, consult any machine book I have not yet told any machine
books names yet. I will tell you will somewhat later, I will insist that first go through the
lectures and whatever study materials I give you go through that then I will give you names
of some books. There are so, many books good books are there, open the chapter
equivalence circuit see the data of open circuit test and short circuit tests, on your own try
to find out the equivalent circuit parameters. Only one comments about these two testing,
see two comments.
(Refer Slide Time: 26:31)
Two or one anyway let us see whether two comments can be made, what I am telling note
that during open circuit test no matter from which side you do theoretically transformer
will have core loss only practically, core loss only is not. Because, copper loss is neglected
only little current is drawn and during short circuit test transformer will have only copper
loss only. I have put this symbol to indicate that little bit of core loss copper loss will be
there, here also core loss will be there. Why?
Because, during open circuit test rated flux, rated flux means rated flux density; rated flux
density means core loss will be at its rated value. And, we will have only copper loss during
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short circuit test because flux is much much smaller than rated value. Why? Because, you
require very little voltage to carry out short circuit test and it is copper loss only.
Therefore, if you take a transformer and while carrying out this two test we find this is the
situation. But, certainly you have purchased a transformer for example, the same old good
old transformer I am always referring for easy calculation 1 kVA etcetera, rated current is
5Amp, rated current is 10Amp, 50 Hz etcetera is there.
Now, you have purchased a transformer you do the open circuit short circuit test, but you
will be putting this to use. So, that transformer is fully loaded, it should operate under full
load condition. For example, you would like to see transformer this transformer in
operation, what will be the situation? In operation situation will be you apply 200 volt
here, approximately you will get 100 volt there and you will connect an impedance here.
So, that you have purchased 1 kVA transformer, you would always like to see the
transformer operates under full load condition. So, you must apply rated voltage, get
almost rated voltage on the secondary side.
Voltage across this terminal maybe a little less that is we will see that, but what I am telling
for normal operation I would like to see this side carries 10Amp current. This side carries
current 5Amp and you will be happy because based on that only you have purchased a
transformer. You have not purchased a transformer to carry out open circuit and short
circuit test and that is there to find out the parameters. But, you have purchased these
transformer you know you have to supply a load at 100 volt 10Amp current. So, this
transformer should be put to use under this condition, applied voltages are on this side,
that side, rated voltage and windings are also rated current.
Then only you are really utilizing the transformer properly, these the transformer in
operation you would like to have this situation prevailing. You have spent so, much money
to install this transformer to supply the load. Certainly not you have connected a
transformer and supplying a load which is 2Amp secondary, what is the plan then why you
purchased a transformer of secondary current rating 10Amp. So, this will be the situation.
We would like to see this thing happens you will be happy, oh transformer is now fully
utilized.
And, it is called rated condition windings are carrying rated current and also. So, in this
case transformer in operation under full load condition, under full load condition I will be
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very happy. So, the flux is rated as well as copper loss is also rated because, windings are
carrying rated current is not. Both these losses unlike open circuit and short circuit test
during open circuit test only core loss rated value. During short circuit test only copper
loss at rated value, but when the transformer will be in operation both these losses will be
at its rated values you must understand this.
So, both the losses are at the rated values. So, if I ask you a simple question that open
circuit tests there is power loss temperature will rise. There will be some temperature rise
in open circuit test, there will be some temperature rise during short circuit test you
measured them ok. This is the maybe 30 degrees, the temperature some number I am just
telling. Here it is suppose 35 degree copper loss, but when the transformer will be operating
at rated voltages and also windings are carrying rated current, that is what I would like to
happen.
I would like to see this thing happens, then only I will be satisfying myself I purchased a
transformer of this rating whose windings can carry rated current 10Amp 5Amp, voltages
can have 200 100 volt under this condition; only I would like to see the transformer is
operating. But, when the transformer will be operating under this condition both these
losses will be present and temperature arise will be much higher than this; under these two
conditions. Anyway this you would just note, we will refer to this information later in the
next class.
Thank you.
175
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Khargpur
Lecture – 19
Efficiency of Transformer – Losses
(Refer Slide Time: 00:23)
Welcome to lecture number 19 and in our last lecture as you know I was telling you that
when a transformer will be in operation both copper loss and core loss will be present. You
cannot neglect during operation one loss with respect to another loss. Because, rated flux
is there and hopefully you are efficiently using the transformer, so that rated current are
carried by the coils. Our target will be to see such a condition prevail then only it will the
transformer is in full operation, rated currents it is carrying, rated flux it has produced this
is the way.
176
(Refer Slide Time: 01:13)
Now, today I will tell you, so equivalent circuit parameters I have found out. I will tell you
now our next important topics will be one is called a efficiency (𝜂) and another is called
regulation (ℜ) this two are very important. It is usually denoted by this curly letter, these
are the topics we will be discussing. Efficiency of a transformer what do I mean by it and
another thing is called regulation, and I will be talking this in terms of the equivalent circuit
of a transformer.
But listen to me very carefully first I will take up efficiency. So, a transformer is as I told
you it will be operating with load connected not under open circuit or short circuit and here
you will apply rated voltage frequency during it is operation core is there, core loss
practical transformer ok.
So, it is expected when the coils now to before I tell you about the efficiency.
𝜂=
𝐾𝑊 𝑂𝑢𝑡𝑝𝑢𝑡
𝐾𝑊 𝐼𝑛𝑝𝑖𝑡
In whichever unit you do you always express in the same unit. This is the efficiency of a
transformer.
𝜂=
𝐾𝑊 𝑂𝑢𝑡𝑝𝑢𝑡 𝑃𝑂𝑢𝑡𝑝𝑢𝑡
=
𝐾𝑊 𝐼𝑛𝑝𝑖𝑡
𝑃𝐼𝑛𝑝𝑢𝑡
177
So, what is 𝑃𝑂𝑢𝑡𝑝𝑢𝑡 , 𝑃𝑂𝑢𝑡𝑝𝑢𝑡 is here; what is 𝑃𝑂𝑢𝑡𝑝𝑢𝑡 , 𝑃𝑂𝑢𝑡𝑝𝑢𝑡 is here from the supply this
is the overall efficiency of the transformer. This efficiency of course, if it is ideal
transformer it will be 100% because there is neither core loss winding resistances are
neglected, copper loss is also absent. So, whatever power will be drawn from the supply
will be dumped to the load, so this is your load, efficiency would have been 100%. But in
real practical transformer we have seen that when the transformer will be operating it will
have both core loss and copper loss will be present.
So this can be written as
𝜂=
𝑃𝑂𝑢𝑡𝑝𝑢𝑡
𝐾𝑊 𝑂𝑢𝑡𝑝𝑢𝑡 𝑃𝑂𝑢𝑡𝑝𝑢𝑡
=
=
𝐾𝑊 𝐼𝑛𝑝𝑖𝑡
𝑃𝐼𝑛𝑝𝑢𝑡
𝑃𝑂𝑢𝑡𝑝𝑢𝑡 + 𝐿𝑜𝑠𝑠𝑒𝑠
𝜂=
𝑃𝑂𝑢𝑡𝑝𝑢𝑡
𝑃𝑂𝑢𝑡𝑝𝑢𝑡 + 𝐶𝑜𝑟𝑒 𝐿𝑜𝑠𝑠 + 𝐶𝑜𝑝𝑝𝑒𝑟 𝐿𝑜𝑠𝑠
This I will break it up into two parts, core loss plus copper loss. This will be the expression
for efficiency of a transformer.
Let us first so, this losses are there therefore, efficiency will be less than 100%. Of course,
as I told you in a very well design transformer efficiency may be as high as 99% so because
it is a static machine, no friction loss windage loss this things will be absent only core loss
and copper loss. Windings will become hot, core will become hot, in normal operation as
I told you at rated condition these things are there.
Now, first so let us see the losses first what happens to core loss?
2
1.6
𝐶𝑜𝑟𝑒 𝐿𝑜𝑠𝑠 = 𝑃𝑒𝑑𝑑𝑦 + 𝑃ℎ𝑦𝑠 ∝ 𝐵𝑚𝑎𝑥
+ 𝐵𝑚𝑎𝑥
or in terms of area that B max it will be proportional to……….
So, core loss will be proportional to this things of course, the constant of proportionality
will be different
2
1.6
2
1.6
𝐶𝑜𝑟𝑒 𝐿𝑜𝑠𝑠 = 𝑃𝑒𝑑𝑑𝑦 + 𝑃ℎ𝑦𝑠 ∝ 𝐵𝑚𝑎𝑥
+ 𝐵𝑚𝑎𝑥
= 𝑘𝑒 𝐵𝑚𝑎𝑥
+ 𝑘ℎ 𝐵𝑚𝑎𝑥
178
So level of 𝐵𝑚𝑎𝑥 is a crucial factor. Now, we know that 𝐵𝑚𝑎𝑥 remains practically constant
this are also I pointed out earlier remains practically constant from no load to full load
condition.
Is not core loss remains constant because the drop in that 𝑟1 and 𝑥1 is little which gives
you that the value of induced voltage 𝐸1 in the ideal transformer that may not be equal to
𝑉1. Because recall that exact equivalent circuit of the transformer what is the exact
equivalent circuit of the transformer? Let me go to next page.
(Refer Slide Time: 08:30)
Exact equivalent circuit is 𝑟1 and 𝑥1 then your this parallel branch. Suppose refer to side
one I am drawing this is 𝑎2 𝑟2 , this is 𝑎2 𝑥2 is not, this the equivalent circuit and there is the
your reflected impedance 𝑧2′ . And this is 𝑅𝑐𝑙1 and 𝑋𝑚1. Magnetizing current decides 𝐵𝑚𝑎𝑥
and this voltage will decide what is the value of 𝐼𝑚 in the exact equivalent circuit here is
your 𝑉1.
Now, what do I mean by no load first, no load means secondary nothing is connected
therefore, current drawn will be the no load current. And when you connect load this
current will now becomes earlier the current no load current was very little 2 to 5% this
drop could be practically neglected. That is why the approximate equivalent circuit we
drew earlier, but anyway what I am telling so, this voltage decides magnetizing current
hence flux.
179
Now, what is the difference of this voltage and this voltage this is 𝐸1 it is this drop, but 𝑟1
and 𝑥1 are small even when rated current flows when load you have connected. There will
be little drop here that will definitely decrease your 𝜑. If you go by this approximate, what
is 𝜑𝑚𝑎𝑥 ?
𝜑𝑚𝑎𝑥 =
𝐸1
4.44𝑓𝑁1
But this 𝐸1 value will deviate from 𝑉1 a little, no load condition it will be almost same as
𝑉1 and during loaded condition a little drop here.
So, change in this 𝜑𝑚𝑎𝑥 from no load to full load what is no load? 𝑧2′ is infinite it is open
nothing connected, what is full load? You have connected such an impedance that rated
current flows. And when rated current flows there will be a drop here we agree to that
drop. But 𝑟1 and 𝑥1 is small therefore, there will be a voltage drop taking place in 𝑟1 and
𝑥1 which will change 𝐸1 a little, but 𝜑𝑚𝑎𝑥 will decrease maybe by 2% or 1% whatever it
is.
So, the assumption that 𝐵𝑚𝑎𝑥 remains practically constant from no load to full load
condition, full load condition you must have understood now when the coils will carry
rated current and applied voltage is rated at rated frequency. So, flux is almost rated from
no load to full load.
If that be the case then I will conclude that in the expression of the efficiency which is
𝜂=
𝑃𝑂𝑢𝑡𝑝𝑢𝑡
𝑃𝑂𝑢𝑡𝑝𝑢𝑡 + 𝑃𝐶𝑜𝑟𝑒 𝐿𝑜𝑠𝑠 + 𝑃𝐶𝑜𝑝𝑝𝑒𝑟 𝐿𝑜𝑠𝑠
I find that this 𝑃𝐶𝑜𝑟𝑒 𝐿𝑜𝑠𝑠 will then remain constant; practically remains constant from no
load to full load, since 𝐵𝑚𝑎𝑥 ≈ 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 got the point. Therefore, here is a loss which
will be always taking place within the transformer. If it is primary is energized no matter
whether you have connected a load on the secondary side or not, 𝑃𝐶𝑜𝑟𝑒 will be always
there. So, that is why it is the loss has got two components.
So, one is a fixed loss. So, P𝑃𝐶𝑜𝑟𝑒 𝐿𝑜𝑠𝑠 ≈ 𝐹𝑖𝑥𝑒𝑑 we will not break our heads a little change
in 𝑃𝐶𝑜𝑟𝑒 𝐿𝑜𝑠𝑠 from no load to full load no point in doing that because little change will take
180
place. So, this loss is called the fixed or constant loss. Now, the other loss, what is this
loss; 𝑃𝐶𝑜𝑝𝑝𝑒𝑟 𝐿𝑜𝑠𝑠 , is it fixed? The answer is no, why no? Answer is very simple.
As you change the load the current 𝐼2 and 𝐼2′ too will change, what do you mean by loading
I will draw the actual diagram 𝑧2 you are varying to vary the load on the transformer. Here
it is 𝑉1 and 𝑓. At open circuit 𝑧2 → ∞ switch is open. Then I will load it by decreasing this
𝑧2 and your current will go on increasing and ratio of the currents are approximately the
turns ratio.
Therefore, level of current will go on changing as you change the load on the secondary
of the transformer. Approximate equivalent circuit is this, this current is 𝐼2′ here is the
parallel branch. Then what is the total copper loss in terms of approximate equivalent
circuit?
𝑃𝐶𝑜𝑝𝑝𝑒𝑟 𝐿𝑜𝑠𝑠 = (𝐼2′ )2 𝑟𝑒1 = 𝐼22 𝑟𝑒2
I mean several ways you can write this. But the point to be noted here the magnitude of
the current that you are drawing out of the transformer from the secondary side is
proportional to the square of the current. So, as you change the load 𝐼2′ changes therefore,
copper loss in the winding will be a function of current supplied.
So, copper loss cannot be a fixed loss, it depends upon what current you are extracting out
of the secondary of the transformer it depends on that is that clear. After I have learned
this, see there are many books to calculate the efficiency of the transformer in several
ways. I will give you a certain way a particular way of calculating efficiency which will
make your life easier and also it will bring out the concept of degree of loading.
181
(Refer Slide Time: 18:36)
For example during short circuit test we have noted that Watt meter reading, what it was
reading? It was reading the copper loss of the transformer that is winding copper loss of
the transformer at full load current. Because during short circuit test windings were
carrying rated current. Recall, during short circuit test, one increases the input voltage from
HV side such that only apply that much voltage which will cause rated current to flow in
both the windings and whatever is the wattmeter reading I say that is the copper loss. So,
that copper loss was calculated at rated full load current.
And 𝑃𝑐𝑜𝑟𝑒 is equal to the wattmeter reading during no load test is not because I have applied
rated voltage rated frequency. So, flux remains same and wattmeter reading was reading
only the core loss alone neglect the copper loss at that time. So, this is the thing and which
is constant as you change 𝑧2 𝑃𝑐𝑜𝑟𝑒 will practically remain constant.
𝑊𝑎𝑡𝑡𝑚𝑒𝑡𝑒𝑟 𝑅𝑒𝑎𝑑𝑖𝑛𝑔 𝑑𝑢𝑟𝑖𝑛𝑔 𝑆. 𝐶. 𝑇𝑒𝑠𝑡 =
𝑃𝐶𝑢
(𝐹𝑢𝑙𝑙 𝐿𝑜𝑎𝑑)
𝑊𝑎𝑡𝑡𝑚𝑒𝑡𝑒𝑟 𝑅𝑒𝑎𝑑𝑖𝑛𝑔 𝑑𝑢𝑟𝑖𝑛𝑔 𝑁𝑜 𝐿𝑜𝑎𝑑 𝑇𝑒𝑠𝑡 = 𝑃𝐶𝑜𝑟𝑒
Now, the big question is this is the
𝑃𝐶𝑢
then what will be the copper loss when
(𝐹𝑢𝑙𝑙 𝐿𝑜𝑎𝑑)
the windings will be carrying half the rated current. Because the level of the current the
windings will carry depends upon what impedance we have connected on the secondary
side. So, I can adjust the impedance connected such that it carries half the rated current, I
182
will connect across the secondary of the transformer such an impedance that it carries half
the rated current or 75% of the rated current what not got the point.
Now, I am asking you a straight question that if the windings are carrying rated current,
then what will be the total copper loss? That copper loss I am telling
𝑃𝐶𝑢
when
(𝐹𝑢𝑙𝑙 𝐿𝑜𝑎𝑑)
the coils carry rated current. Now, my question is if the transformer is loaded such that
coils carry half the rated current. Then what will be the copper loss now; in terms of this
full load copper loss how much it will be? See after all copper loss is some 𝐼 2 𝑟𝑒 . It is 𝐼12 𝑟𝑒1
or same as 𝐼22 𝑟𝑒2, I will not separately calculate 𝐼12 𝑟1 + 𝐼22 𝑟2 , what is the point after I have
done this.
So, I will try to exploit this simple equivalent circuit no point in separating 𝑟1 and 𝑟2 .
Because after all from open circuit and short circuit test you cannot separate 𝑟1 and 𝑟2 is
not. It comes as a package.
𝑟𝑒1 = 𝑟1 + 𝑎2 𝑟2
𝑟𝑒2 =
𝑟1
+ 𝑟2
𝑎2
See beauty is these that from the equivalent circuit whether it is either 1 or 2 take the
corresponding values these is the copper loss.
So, at full load the copper loss is
𝑃𝐶𝑢
2
= 𝐼𝐹𝐿
𝑟𝑒1
(𝐹𝑢𝑙𝑙 𝐿𝑜𝑎𝑑)
This I get from short circuit test. Now, I am saying that winding current is now made half.
So, what will be the copper loss now?
𝐼𝐹𝐿 2
𝐶𝑜𝑝𝑝𝑒𝑟 𝐿𝑜𝑠𝑠 𝑎𝑡 𝐻𝑎𝑙𝑓 𝑜𝑓 𝑅𝑎𝑡𝑒𝑑 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 = ( ) 𝑟𝑒1
2
If you make the currents half rated current if you have connected such an impedance on
the secondary side such that the coils are carrying half the rated current 𝑟𝑒1 remains same.
183
𝐼𝐹𝐿 2
1 2
𝑟𝑒1 )
𝐶𝑜𝑝𝑝𝑒𝑟 𝐿𝑜𝑠𝑠 𝑎𝑡 𝐻𝑎𝑙𝑓 𝑜𝑓 𝑅𝑎𝑡𝑒𝑑 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 = ( ) 𝑟𝑒1 = ( ) (𝐼𝐹𝐿
2
4
1
𝑃𝐶𝑢
=( )
4 (𝐹𝑢𝑙𝑙 𝐿𝑜𝑎𝑑)
Getting the point this must be understood
𝑃𝐶𝑢
, if you know this is 100 Watt I am
(𝐹𝑢𝑙𝑙 𝐿𝑜𝑎𝑑)
telling you if you are winding currents are half, then I will say the copper loss will become
100
4
= 25𝑊𝑎𝑡𝑡.
3
9
If it is carrying 4 of the rated current, then copper loss would have been 16
𝑃𝐶𝑢
.
(𝐹𝑢𝑙𝑙 𝐿𝑜𝑎𝑑)
𝐼𝐹𝐿 2
1 2
𝐶𝑜𝑝𝑝𝑒𝑟 𝐿𝑜𝑠𝑠 𝑎𝑡 𝐻𝑎𝑙𝑓 𝑜𝑓 𝑅𝑎𝑡𝑒𝑑 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 = ( ) 𝑟𝑒1 = ( ) (𝐼𝐹𝐿
𝑟𝑒1 )
2
4
1
1
𝑃𝐶𝑢
=( )
= ( ) (𝑊𝑎𝑡𝑡𝑚𝑒𝑡𝑒𝑟 𝑅𝑒𝑎𝑑𝑖𝑛𝑔 𝑑𝑢𝑟𝑖𝑛𝑔 𝑆. 𝐶. 𝑇𝑒𝑠𝑡)
4 (𝐹𝑢𝑙𝑙 𝐿𝑜𝑎𝑑)
4
It is not compulsory to write it like this, but this concept is important you define. Then
copper loss directly depends upon the degree of loading. Now, the question is what is
degree of loading? Degree of loading is I will use a number 𝑥 this I must define degree of
loading with respect to that rated current carrying capacity of the coil.
(Refer Slide Time: 28:05)
I will use a number
184
0≤𝑥≤1
𝑥 = 0 → 𝑁𝑜 𝐿𝑜𝑎𝑑
𝑥 = 1 → 𝐹𝑢𝑙𝑙 𝐿𝑜𝑎𝑑
𝑥 can have any values; 𝑥 = 0.5 means it is 50% loaded the transformer is, although I told
you transformer should be operated at full load condition that is why you have purchased
the transformer. So, that it always operates at full load, but sometimes load varies on the
secondary side then it is a nice way of telling the term degree of loading, got the point.
So, we would like to calculate try to estimate efficiency at 𝑥 degree of loading. So, I will
call it 𝜂𝑥 , where 𝜂𝑥=1 corresponds to full load efficiency. Now, this one will be the output,
suppose
𝑆 = 𝑅𝑎𝑡𝑒𝑑 𝐾𝑉𝐴 𝑜𝑓 𝑡ℎ𝑒 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟
Then when it is degree of loading is 𝑥,
𝑃𝑂𝑢𝑡𝑝𝑢𝑡 = 𝑥𝑆 cos 𝜃
𝑥 = 1 means it is supplying rated kVA, cos 𝜃 is the power factor of the load, because in
the definition of efficiency, kilowatt I have to calculate supply to the load.
Suppose the transformer is supplying 𝑥𝑆 kVA to the load having a power factor of cos 𝜃
on the secondary side. So, this is the output then
𝜂𝑥 =
𝑥𝑆 cos 𝜃
𝑥𝑆 cos 𝜃 + 𝑃𝐶𝑜𝑟𝑒 + 𝑃𝐶𝑜𝑝𝑝𝑒𝑟
𝑃𝐶𝑜𝑟𝑒 remains constant it does not depend upon the degree of loading, no matter whether
this is 50% loaded,100% loaded 𝑃𝐶𝑜𝑟𝑒 remain same.
Now, the question is I know
𝑃𝐶𝑢
; at full load means, copper loss at 𝑥 = 1 when
(𝐹𝑢𝑙𝑙 𝐿𝑜𝑎𝑑)
transformer is carrying rated currents. So, what will be the copper loss now at 𝑥 degree of
185
1
loading we have just discussed, if it is 50% loaded then (4)
and
𝑃𝐶𝑢
. So, 𝑥 is a fraction
(𝐹𝑢𝑙𝑙 𝐿𝑜𝑎𝑑)
𝑃𝐶𝑢
must be multiplied then by 𝑥 2
(𝐹𝑢𝑙𝑙 𝐿𝑜𝑎𝑑)
𝑥𝑆 cos 𝜃
𝜂𝑥 =
𝑥𝑆 cos 𝜃 + 𝑃𝐶𝑜𝑟𝑒 + 𝑥 2
𝑃𝐶𝑢
(𝐹𝑢𝑙𝑙 𝐿𝑜𝑎𝑑)
This is the expression for efficiency of a transformer.
So, please try to assimilate this expression very clearly 𝑥 is the degree of loading, cos 𝜃 is
the power factor of the load and output of the transformer at 𝑥 degree of loading should be
written 𝑥𝑆 cos 𝜃, where 𝑆 = 𝑅𝑎𝑡𝑒𝑑 𝐾𝑉𝐴 𝑜𝑓 𝑡ℎ𝑒 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 So, output kilowatt is
how much rated kVA it is delivering at a power factor of cos 𝜃.
So, efficiency is output divided by output plus losses. Losses got two parts; one is the fixed
loss 𝑃𝐶𝑜𝑟𝑒 it does not depend on 𝑥.
𝑃𝐶𝑢
I have defined at rated current then if it is
(𝐹𝑢𝑙𝑙 𝐿𝑜𝑎𝑑)
loaded 𝑥 degree of loading. Then copper loss at that degree of loading must be multiplied
1 3
by 𝑥 2 where, 𝑥 may be 2, 4 40% = 0.4 and so on. So, this is the most important formula,
so far as efficiency is concerned and we will discuss further about this expression in my
next lecture.
Thank you.
186
Electrical Machines - 1
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 20
Efficiency (Contd.)
Welcome to lecture number 20 on Electrical Machines I. And as you know we were
discussing about how to estimate the Efficiency of a transformer.
(Refer Slide Time: 00:40)
And you recall that I assumed the 𝑆 to be the rated KVA of the transformer; and then I
defined a factor degree of loading, I tried to represent by a factor x, which whose value
normally will be 0 ≤ 𝑥 ≤ 1, 0 corresponds to no load and 1 corresponds to rated condition.
What is rated condition? Rated condition is rated voltage applied and also windings are
carrying rated current ok. Then I explained that there are two main losses in a transformer;
one is the core loss core loss and this I denoted it by 𝑃𝐶𝑜𝑟𝑒 and this is independent of 𝑥;
degree of loading that is it is constant, no matter whether no load or full load.
And this 𝑃𝐶𝑜𝑟𝑒 this can be obtained; this is also is equal to the wattmeter reading of no load
test, so this value is known. And then I defined the full load copper loss; as I used this
notation I hope 𝑃𝐶𝑢𝑓𝑙 . Full load copper loss is a constant theme; at full load current when
the coils are carrying full load current, what is the total copper loss? It is 𝑃𝐶𝑢𝑓𝑙 .
187
So, this is constant; then I told you copper loss depends on 𝑥 the degree of loading. And
therefore, copper loss at 𝑥 degree of loading; in terms of this 𝑃𝐶𝑢𝑓𝑙 will be simply 𝑥 2 𝑃𝐶𝑢𝑓𝑙 .
So, this must be kept in mind while attempting to calculate the efficiency of a transformer.
(Refer Slide Time: 04:53)
.
Then efficiency at 𝑥 degree of loading, will be I denote the efficiency by 𝜂𝑥 indicates that
𝑂𝑢𝑡𝑝𝑢𝑡 𝐾𝑊
at 𝑥 degree of loading. 𝜂𝑥 = 𝐼𝑛𝑝𝑢𝑡 𝐾𝑊 ; what is the output kilowatt? At 𝑥 degree of loading,
the 𝑂𝑢𝑡𝑝𝑢𝑡 𝐾𝑊 = 𝑥𝑆 cos 𝜃; what is this cos 𝜃? Power factor of the load, that is; this is
the actual practical transformer, this is your load its power factor angle 𝜃. And this is the
𝑂𝑢𝑡𝑝𝑢𝑡 𝐾𝑊 and 𝐼𝑛𝑝𝑢𝑡 𝐾𝑊 = 𝑂𝑢𝑡𝑝𝑢𝑡 𝐾𝑊 + 𝐿𝑜𝑠𝑠𝑒𝑠.
What are the two losses? One is the fix loss 𝑃𝐶𝑜𝑟𝑒 , plus 𝑥 2 𝑃𝐶𝑢𝑓𝑙 . This is the expression for
the efficiency, very important expression.
𝜂𝑥 =
𝑥𝑆 cos 𝜃
𝑥𝑆 cos 𝜃 + 𝑃𝐶𝑜𝑟𝑒 + 𝑥 2 𝑃𝐶𝑢𝑓𝑙
Therefore, I can calculate the efficiency of a transformer at any degree of loading and at a
given power factor of the load because this 𝑃𝐶𝑜𝑟𝑒 is known to me from the open circuit
test.
And 𝑃𝐶𝑢𝑓𝑙 is also known to me from the short circuit test, because this 𝑃𝐶𝑢𝑓𝑙 ; here I write
is the wattmeter reading; reading of short circuit test got the point? Therefore, this is the
188
thing; therefore, I will be able to suppose I want to sketch the efficiency versus degree of
loading, suppose I want to sketch. Here, I will write degree of loading, loading that is 𝑥
and here I will plot the efficiency (𝜂𝑥 ); how efficiency changes, as you change 𝑥.
What is the domain of 𝑥? It can be under no load condition, as I told you 𝑥 = 0 corresponds
to no load, 𝑥 = 1 corresponds to rated condition that is coils are carrying rated current.
Therefore, it looks like the domain of 𝑥 will be from 0 to 1 is not, the value of 𝑥 ok. If you
exceed the value of 𝑥 greater than 1; that means, the transformer is overloaded.
Transformer of course, we will try to see it is not continuously done under overloaded
condition; maybe for a brief period you can overload it by 10% or 20%, then the rated
current the value of 𝑥 then go up.
But any way this point corresponds to full load and this point corresponds to no load ok.
Suppose 𝑥 = 0; let us see how this characteristics is expected to look like, if 𝑥 = 0 that is
no load. So, efficiency; obviously, will be 0 because there is no output power; although
there will be input power. So, 0 by some finite number 𝑃𝐶𝑜𝑟𝑒 will give you the efficiency
under no load condition and that is this point, so efficiency is this one; at 𝑥 = 0.
𝑥 = 0 → 𝜂𝑥 = 0
Now let us ask our self ok, suppose you make 𝑥 very large; I want to have some idea how
this efficiency curve changes? Suppose 𝑥 is made very large at least theoretically. I will
not subject the transformer to this high degree of loading, but just let us mathematically
see; what this expression tells me. If 𝑥 tends to very large; ok, on pen and paper let me call
it going to infinity how does it matter? Then what will happen to this efficiency term? See
there is a 𝑥 2 term below; so it will prevail upon all the other 𝑥 terms. Therefore, efficiency
will tend to once again 0, because these terms with respect to 𝑥 2 can be neglected and it
𝑥𝑆 cos 𝜃
will be some 𝜂𝑥 = 𝑥 2 𝑃
𝐶𝑢𝑓𝑙
.
So, 𝑥 tends to infinity means efficiency once again zero.
𝑥 → 0, 𝜂𝑥 → 0
So, at a very large value of 𝑥 once again I know efficiency will be 0. But in between,
whatever is the value of 𝑥; this will give you some finite numbers therefore, from 0; if I
increase the value of 𝑥, efficiency is expected to have some finite numbers; it will grow
189
up, but once again it has to come to 0. So, it is expected the; this curve will be something
like this, here is not? It has to be.
Therefore, efficiency will rise then finally, it has come to once again 0, and it is a very
reasonable expression of 𝑥; function of 𝑥 no complications. Therefore, it will be sort of a
smooth curve and I would then expect the efficiency to rise initially and later it will start
decreasing becomes 0.
In other words, what I am telling I would expect that efficiency will have some 𝜂𝑚𝑎𝑥 , at
some degree of loading got the point, so this is the thing. Now then it is natural to ask that
at what degree of loading; efficiency will be maximum is not? That is the question; so this
value of 𝑥 is how much? Such that efficiency will be maximum; I want to know that. So,
to find out at what value of 𝑥, 𝜂𝑥 will be maximum that is what I want to find out, clear?
Mind you, when I am varying 𝑥, I will keep cos 𝜃 constant, load power factor I will not
change; I will keep it a fixed value of cos 𝜃 and only vary 𝑥. So, this here in this expression
I will vary 𝑥; but other things remaining constant 𝑆 is of course, rated k VA, so cos 𝜃 is
kept constant. So, I will vary only 𝑥; to find out that one is pretty simple because I will
just maximize this function ok.
(Refer Slide Time: 14:57)
So, I will start telling that,
190
𝜂𝑥 =
𝑥𝑆 cos 𝜃
𝑥𝑆 cos 𝜃 + 𝑃𝐶𝑜𝑟𝑒 + 𝑥 2 𝑃𝐶𝑢𝑓𝑙
And only under the condition, you are varying 𝑥; you want to know at what value of 𝑥 this
efficiency is maximum. What you have to simply do is, you have to calculate this you have
to differentiate this expression with respect to 𝑥 and set it to 0 and get the value of 𝑥.
𝑑𝜂𝑥
=0
𝑑𝑥
But that I will do; I will do in a slightly different way, so as to reduce my mathematical
effort to get this value. What I will do is I will divide both numerator and denominator by
𝑥.
𝜂𝑥 =
𝑆 cos 𝜃
𝑃
+ 𝑥𝑃𝐶𝑢𝑓𝑙
𝑆 cos 𝜃 + 𝐶𝑜𝑟𝑒
𝑥
Now by dividing numerator and denominator by a factor 𝑥, numerator has become
constant. Then I will say 𝜂𝑥 will be maximum, when denominator is minimum. When this
fellow is minimum which therefore, tells me that to have this denominator to be minimum;
I should differentiate this like this and set it to 0.
𝑑
𝑃𝐶𝑜𝑟𝑒
(𝑆 cos 𝜃 +
+ 𝑥𝑃𝐶𝑢𝑓𝑙 ) = 0
𝑑𝑥
𝑥
So, if you differentiate this will then become
−
𝑃𝐶𝑜𝑟𝑒
+ 𝑃𝐶𝑢𝑓𝑙 = 0
𝑥2
𝑥 2 𝑃𝐶𝑢𝑓𝑙 = 𝑃𝐶𝑜𝑟𝑒
This is one very important result; whether it is minimum or maximum? Ok, you calculate
these double differentiate; you can show this is the point we are looking for. Only thing
you could also differentiate this expression, then in the numerator there is 𝑥 denominator
𝑥, I want to maximize this means you minimize this one; it is a just one line then derivation
that is why I have adopted this you divide by 𝑥.
191
Because, we are looking for some finite values of 𝑥 for which efficiency will be maximum.
And this is the condition; this must be set in now what is this 𝑥 2 𝑃𝐶𝑢𝑓𝑙 ? This is a variable
loss, so in other words I will say in language that 𝜂𝑥 will be maximum, will be maximum
when variable loss; in this case the copper loss is equal to the fixed loss, in this case the
core loss got the point; this is the thing.
(Refer Slide Time: 20:12)
Or I can say that
𝑃𝐶𝑜𝑟𝑒
𝑥=√
𝑃𝐶𝑢𝑓𝑙
this is the thing very important formula. Therefore, efficiency will be maximum at that
value of 𝑥. Whatever number you get it will give you the degree of loading at which the
efficiency of the transformer will be maximum.
192
(Refer Slide Time: 21:10)
And that is if I go to the next page; as I was telling the efficiency versus degree of loading
curve; it is 0 here, it is here degree of loading is 1, then efficiency curve will be somewhat
like this; it then decreases ok. And this is your 𝜂𝑚𝑎𝑥 and I will write the value of 𝑥 here as
𝑃
𝑥 = √𝑃𝐶𝑜𝑟𝑒 it will give you a number, it is dimensionless 𝑥. Suppose you get this value to
𝐶𝑢𝑓𝑙
be equal to 0.7, transformer rated KVA is 10; then I will say that if you operate the
transformer at 7 KVA; efficiency will be maximum; that is the implication of this number.
And I would like to see this number is between 0 to 1, certainly I should not have a number,
where efficiency will occur at say 1.5; suppose are you getting me? Suppose I say that
suppose the 𝑆 is equal to 10 KVA. And suppose you calculate the degree of loading at
maximum efficiency; how do you calculate?
𝑃
You calculate from this 𝑥 = √𝑃𝐶𝑜𝑟𝑒 do I know 𝑃𝐶𝑜𝑟𝑒 ? Yes, from the open circuit test. Do
𝐶𝑢𝑓𝑙
I know 𝑃𝐶𝑢𝑓𝑙 ? Yes, from the short circuit test watt meter reading. Take the ratios, these are
only two numbers with same dimension take under root and it will return you a number. If
suppose say this becomes equal to 0.7 then I will say if the transformer is operated at 0.7 ×
10 = 7𝐾𝑉𝐴, then maximum efficiency will be achieved, so this number this is the
maximum efficiency.
193
So, this axis is efficiency and this is 𝜂𝑚𝑎𝑥 ; this is general 𝜂𝑥 . Of course, I should not forget
to attach another conditions, that I have imposed upon the efficiency expression, what is
that? This curve is true for a fixed power factor angle; cos 𝜃 I assumed constant is not? I
assumed that is why while differentiating cos 𝜃 was kept constant; recall that this cos 𝜃
was assumed to be constant in this whole derivation.
Therefore, I should attach to this curve; a tag saying that ok this curve we have got and
this is for example, say at cos 𝜃 = 0.8; leading or lagging does not matter, no point in
qualifying it further that is cos 𝜃 leading or lagging because output power is 𝑥𝑆 cos 𝜃. So,
I must attach another parameter to this curve that this curve has been achieved; when you
kept cos 𝜃 = 0.8. What that does that mean? It means that this is your transformer, actual
transformer when you will do this exercise of loading it. What I am telling is, how keeping
cos 𝜃 constant I can vary the current degree of loading ok, any impedance you know such
that you can vary the current; it is |𝑍|, I will vary keeping 𝜃 constant that is what it means;
you vary |𝑍| that can be done keeping 𝜃 constant.
And then you will get different current as the value of |𝑍| will be different, but 𝜃 remaining
constant. Therefore, this is the curve and I should not forget at what power factor I am
plotting this curve ok. After I get this curve, then I will ask myself; what as an user of this
transformer, I would like to have at what value of 𝑥 efficiency should be maximum is not?
Ok, given a transformer; you do this note down the Watt meter reading from the open
circuit short circuit full load reading, very simple these two test; take this ratio and you
will come to know about this number. But I am asking as an user of the transformer, at
what value of 𝑥 we would like to have maximum efficiency to occur? As an user of the
transformer; at what value of 𝑥 efficiency will be maximum?
Do I have a choice for that, to say I have purchased a transformer let its efficiency occur
should I say at 𝑥 = 0.5 or 𝑥 = 0.8. So, these are the things which we will discuss very
interesting point when you purchase a transformer, you know for what use you will be
using the transformer.
And whoever is the manufacturer of the transformer, I will ask while purchasing the
transformer; look here you design a transformer, whose maximum efficiency should occur
at so much value of 𝑥; certainly it will not be greater than 𝑥 = 1.2 or 𝑥 = 1.5 for
194
continuous running. And this is the point I will discuss in the next class. I hope you have
understood this.
Thank you.
195
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 21
Condition for Maximum Efficiency when Load Power Factor Constant
Welcome to 21st lecture on Electrical Machines - I.
(Refer Slide Time: 00:27)
And we are discussing about how to estimate the efficiency of the transformer and if you
see the previous lecture, this is the end result I have got that efficiency versus degree of
loading. And we found out that efficiency will be occurring at maximum value at certain
value of 𝑥; 0 ≤ 𝑥 ≤ 1.
Then finally, I was saying that, if you sketch an efficiency versus degree of loading curve
do not forget to attach at what constant power factor we have plotted this curve. Because
in that derivation I kept cos 𝜃 constant maybe it would be cos 𝜃 = 0.8 if this is constant,
𝑥
vary is the magnitude of 𝑧. Is it not and that is vary 𝑟 and 𝑥 of the load such that 𝑟 ratio
remains constant that will ensure the power factor is constant, but anyway this is the curve
we have got for a given power factor constant. Now I am asking you that while purchasing
a transformer should I also tell the manufacturer that design my transformer in such a way
that the value of 𝑥 at which maximum efficiency occurs this is so much number that is this
was the last thing.
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What is my choice? User’s choice should be what? It depends I told you that how this
transformer is utilized; now the things will be clearer from this.
(Refer Slide Time: 02:55)
.
Suppose I say I have a transformer say let me take that simple example, I will write it as
10KVA. Let me take a single phase transformer whose rating is 10 KVA and say
200V/100V just so that we play with the numbers to easily understand what I am meaning.
So, single phase transformer 10 KVA, 50 Hz. I told you the moment you know the rated
10000
current always do that. So, 200 will give you the rated current of the high voltage side 50
Amp it is and of course, the rated current of the LV side will be twice this amount 100
Amp. Suppose this is the rating of the transformer.
And I have purchased a transformer and I know this transformer I will use at full load
condition for all the time 24 hours a day. Suppose this transformer I will use it,
continuously at full load condition this is LV side this is HV side and this is 100 Amp and
this is 50 Amp, this is the 200V side, this is the 100V. Always sketch something of this
sort to understand what is happening. Suppose I say I have purchased a transformer in
industry I will use it and it will be continuously operating at rated condition. Suppose this
transformer will be operated always at full load condition and that is what is expected, you
have purchased a 10 kVA transformer whose current ratings are 50 Amp, 100 Amp.
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So, you would like to operate the transformer at full load, not that this transformer you half
load it and continuously done, then you should have purchased a transformer having lesser
KVA rating, but since you have purchased it looks like you will be using it always at full
load condition.
So, suppose this transformer is to be operated always at full load condition, that is always
it will be handling 10 KVA then the question is at what value of 𝑥 I should demand that
the transformer should have maximum efficiency? That is the question naturally I should
ask and the answer to this is very simple. If you know which transformer will be operated
continuously at full load condition, then say at 𝑥 = 1 𝜂𝑚𝑎𝑥 should occur. Is not it?
Therefore, a transformer which will be always operating under full load condition say in
industry 24 hours it will be put to use, then while ordering this transformer say to the
manufacturer that look you design the transformer in such a way that 𝑃𝐶𝑜𝑟𝑒 and 𝑃𝐶𝑢𝑓𝑙 they
are equal so that 𝑥 will then be equal to 1, is not it?
For example, you remember one of the uses of the transformer is in power system. See
you have generating I will draw a single line diagram to further emphasize this point. A
generator is there in a generating station then what you do that generated voltage is of the
order of say 10 kV or so 50 Hz, here the voltage level is 10 kV.
And suppose its rating is hundreds of megawatt, you want to transmit. Generally
generating stations are located at a far off place from the place where this power will be
actually utilized. The distance between where this power will be utilized and the generating
station may be hundreds of kilometers or more. Therefore, this 100 megawatt of power
should be transmitted over a long distance.
And suppose the generated voltage is the is of the order of 10 kV , but to transmit this large
bulk amount of power over hundreds of kilometers at voltage level of 10 kV is not
economical at all, why? Because whatever currents see approximately this power divided
by voltage gives you the current ok, power factor will be there in ac circuit make it assumed
to be 0.8.
So, this power divided by this voltage gives you some idea of the current that will be
flowing through the transmission line and transmission line over 100 kilometers of length
will have some resistance. And just to transmit the power from generating station to the
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load center, here the load center where this power will be used there will be considerable
amount of power loss in the transmission line.
And you cannot afford so it will become a very inefficient system of transmitting power
bulk amount of power of this order. So, what is done? Here at the generating station itself
there is a transformer first connected generator then this voltage is stepped up. I am
drawing a single line diagram and this voltage is stepped up to a very high voltage may be
200 kV or even in our country we are having 400 kV level voltage.
So, there here is the transformer who does this. In transformer kVA remains same. So,
current in this side will now become less and then you transmit it. But of course, at the
recipient this 200 kV or 400 kV cannot be utilized directly you have to have another
transformer or multiple transformer to step down the voltage, step down. at appropriate
level of voltage so that that power will be utilized.
Now, therefore, this transformer if you look at these are called station transformer power
station transformer. This transformer perhaps will be operating 24 hours a day almost at
full load condition. Therefore, I will demand that this transformer then should have
maximum efficiency occurring at 𝑥 = 1 reasonable demand; because any device you
would always like to operate it at maximum efficiency condition so that losses are
minimized and so on.
Therefore, in such cases 𝑥 = 1 will be a good choice, but also see in a some industry a
transformer is used to step down the voltage to a; step down or step up a voltage to a
particular level. And here in shift duty things will work and a particular transformer will
be always operating at full load condition whatever the load on the secondary of the
transformer is, ok.
Therefore in such cases I will say 𝑥 = 1, but if that be the case and the argument I put
forward earlier that you are purchasing a transformer and you must see that it is always
operating at full load condition. But what happens is this, this is sometimes not in your
hand. I will give you an example.
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(Refer Slide Time: 13:31)
A distribution transformer for example, a distribution transformer is that transformer from
the secondary of which all the households are supplied power; in your house, in my house
ok.
So, what happens is this, the voltage may be at 6.6 kV of a distribution transformer and
this side voltage is 440 volt line to line and this is also line to line voltage. We will discuss
about the connection this at later, but let us try to understand. Now the secondary of these
between R and neutral a group of houses are supplied power from Y phase and neutral
another group of houses or residential buildings are supplied, offices are supplied, B and
neutral another, so that is the another job that is to see that loads are almost balanced.
So, here this thing is 3 terminals come, 4 terminals come out from this side actually. It is
a single line diagram just to emphasize this point; single line diagram. Here R-Y-B and
supply neutral is there. So, loads will be connected between R and n so that the supply
voltage in our houses are 220 volt level, is not? Residence, all the fans, lights they are
designed for 220 volt.
So, it will be there and those will become the load of the distribution transformer on the
secondary side. Now if you see the nature of variation of load on the secondary of such a
transformer that will change over a day I mean it will be a very large changes taking place
all the time. For example, you can easily see the load on the secondary of such a
transformer will be almost 0 during wee hours of the night that is say midnight and behind.
200
Because no lights loads will be there, all people will be taking rest. Maybe some AC, this
that now a days will run, but compared to daytime loading or evening loading where people
come back home, all electrical loads are switched on, this that. So, there will be a huge
fluctuation of load from no load to some peak loads, then some medium degree of loading
that is the value of 𝑥 will go on changing with some time intervals.
But the point is you do not know whether in the midnight the engineers sitting in the
substation will not take decision like this that in the mid time load is less. Therefore, let
me put this primary side off so that no losses take place. No, you cannot do that. You have
committed that you will supply power to every residential complexes at a fixed voltage
and fixed frequency and at all times during emergency situation you have to switch on the
loads in your house.
Therefore primary is to be kept energized all the time whether secondary of this
transformer is loaded or not it is a compulsion, on the part of the company which is selling
electricity to usual consumers, he cannot say that he has; he is duty-bound to keep its
primary of the transformer energized all the time, although he knows that sometimes load
will be very less sometimes load will be very high.
Therefore in such a scenario, that is what do I mean by load changing; 𝑥 is changing and
it is not in my hand, it is in the hands of the consumers. It also depends on the season ok.
During winter that variation of load on the secondary of the transformer will be level of
load will be much less, but nonetheless you will be prepared with primary energized
thinking that some consumer may switch on load at any point of time of the day.
So, this is the thing. Now the question is in such a situation where 𝑥 is in not in my hand
while purchasing the transformer at what value of 𝑥 should I demand that efficiency will
be maximum? Certainly 𝑥 = 1 at full load condition you should not demand that because
you are fully aware that the load on the secondary of the transformer goes on changing;
sometimes very light load condition almost no load, sometimes full load ok, full load will
be there evening peak and so on.
But some other time it is medium degree of loading maybe 𝑥 = 0.3, or 𝑥 = 0.4 transformer
will be operating. Therefore, perhaps if we asking for efficiency to occur maximum at 𝑥 =
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1 that is at full load condition will not be a very judicious demand at the time of ordering
the transformer, may be 𝑥 = 0.6 will be a good thing.
Anyway for a distribution transformer I told this is a typical situation but for a dedicated
transformer used in industry 8 hours shift duty and all the trying the transformer is on and
under full load condition yes, you can very concretely say that I will buy a transformer
with 𝑥 = 1 that is its maximum efficiency must occur at 𝑥 = 1 that is at rated condition.
But here is another extreme example where a distribution transformer secondary supplies
different consumers and the degree of loading on the transformer goes on changing and
goes on changing widely you do not know, but your primary is to be energized. Now to
qualify a therefore, another efficiency is calculated for such transformer where load
fluctuates over 24 hours with primary energized continuously and that energy that
efficiency is called therefore, a distribution transformer is to be judged from not the
ordinary efficiency that I have told you and there is another efficiency called energy
efficiency, 𝜂𝑒𝑛𝑒𝑟𝑔𝑦 .
That is better judge a distribution transformer by its energy and it is defined as
𝜂𝑒𝑛𝑒𝑟𝑔𝑦 =
𝐸𝑛𝑒𝑟𝑔𝑦 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 𝑡𝑜 𝑙𝑜𝑎𝑑 𝑖𝑛 𝑎 𝑑𝑎𝑦 (24 ℎ𝑜𝑢𝑟𝑠)
𝐼𝑛𝑝𝑢𝑡 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑣𝑒𝑟 24 ℎ𝑜𝑢𝑟𝑠
Therefore, a distribution transformer perhaps efficiency is to be calculated in a different
way. We will discuss about this, this is sometimes also called energy efficiency or this is
called all day efficiency, clear?
We will discuss about this, what this how to calculate energy delivered to the load in 24
hours and input energy over same 24 hours of length. And from this we will try to say
whether this distribution transformer is good or not, that is the idea. But before that one
point about that normal efficiency if you look at this curve and this I will leave as an
exercise to you ok, I will draw like this.
Before I discuss this topic, before this is calculated. Let us go back to power efficiency
expression, once more to highlight another interesting point, ok.
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(Refer Slide Time: 26:03)
So, let us see what is this. We have seen that efficiency curve will be somewhat like this,
it will have a peak and 𝜂𝑥 as you know
𝜂𝑥 =
𝑥𝑆 cos 𝜃
𝑥𝑆 cos 𝜃 + 𝑃𝐶𝑜𝑟𝑒 + 𝑥 2 𝑃𝐶𝑢𝑓𝑙
And I told you you do not forget to attach at what power factor angle this curve has been
drawn may be 0.8 and this is degree of loading 𝑥 and this is efficiency 𝜂.
Now the question I will ask you ok, suppose this is the 𝜂 versus 𝑥 curve at point power
factor 0.8 gladly it does not matter. At a given power factor 0.8 this is the curve and this
is the value of 𝑥, what is that?
𝑃𝐶𝑜𝑟𝑒
𝑥=√
𝑃𝐶𝑢𝑓𝑙
𝑊𝑎𝑡𝑡𝑚𝑒𝑡𝑒𝑟 𝑅𝑒𝑎𝑑𝑖𝑛𝑔 𝑑𝑢𝑟𝑖𝑛𝑔 𝑆. 𝐶. 𝑇𝑒𝑠𝑡 = 𝑃𝐶𝑢𝑓𝑙
𝑊𝑎𝑡𝑡𝑚𝑒𝑡𝑒𝑟 𝑅𝑒𝑎𝑑𝑖𝑛𝑔 𝑑𝑢𝑟𝑖𝑛𝑔 𝑁𝑜 𝐿𝑜𝑎𝑑 𝑇𝑒𝑠𝑡 = 𝑃𝐶𝑜𝑟𝑒
This degree of loading 𝜂𝑚𝑎𝑥 occurs, is it not; for this power factor. Now the question is
suppose I vary the power factor, suppose I do the same exercise what will be the plot? The
plot of 𝜂 versus 𝑥 if cos 𝜃 I change it to say 0.6.
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I can do that once again repeat the exercise, keep cos 𝜃 constant, where this curve will be
will it be above it, will it be below it and things like that; are you getting? Now what I am
telling for a particular power factor this is the nature of the curve. First thing is this one,
where it will be suppose I am not yet very clear about that but about one thing I am clear.
If you repeat the same plot with cos 𝜃 = 0.6, I am sure about one thing; the load at which
the maximum efficiency will occur when cos 𝜃 = 0.6 will once again be this point only.
It is at this point maximum efficiency will occur for all the power factors no matter what
the power factor is. If it is 0.8, it is here. Now I am asking you what will be its value if the
power factor is 0.6; you think about this and we will continue with this.
Thank you.
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Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 22
Family of Efficiency Curve at Various Power Factor and Energy Efficiency
Welcome to lecture 22nd and we were discussing about efficiency of transformers.
(Refer Slide Time: 00:32)
And there are efficiency which is called power efficiency as you know at a degree of
loading 𝑥 is I am writing many a times, so that it really you always remember this. So,
𝜂𝑥 =
𝑥𝑆 cos 𝜃
𝑥𝑆 cos 𝜃 + 𝑥 2 𝑃𝐶𝑢𝑓𝑙 + 𝑃𝐶𝑜𝑟𝑒
And what I saw with this expression of efficiency is this thing that you at a particular
power factor if you sketch it, it will be like this and this is efficiency 𝜂.
And this is the load at which maximum efficiency will occur
𝑃𝐶𝑜𝑟𝑒
𝑥=√
𝑃𝐶𝑢𝑓𝑙
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And I told you also this is important to note that this curve has been obtained at what fixed
power factor, so tag that 0.8 power factor
Now, the question is power factor of the load may also change. Suppose I repeated the
same exercise I want to get the 𝜂 versus 𝑥 curve for say 0.6 power factor or unity power
factor or 0.2 power factor. All these 𝜂 versus 𝑥 curve will change will it is position will
remain same I am not sure right now. But I am sure about one thing whatever be it is
position the value of 𝑥 at which 𝜂𝑚𝑎𝑥 will occur it only depends on 𝑃𝐶𝑜𝑟𝑒 and 𝑃𝐶𝑢𝑓𝑙 it will
be here only.
Now, let us see how to then find out what happens, if cos 𝜃 also varies. A how to find out
this that is what this problem can be framed at this way I will frame the problem. This is
your efficiency curve at a given power factor say 𝑃𝐶𝑢𝑓𝑙 , what I am going to do is now I
will fix the load on the transformer that is I will see that kVA I want to see the effect of
variation of power factor angle of the load on efficiency that I want to do.
So, to do that what I will do this is my 𝑥 no doubt I will keep this 𝑥 fixed and vary 𝜃. So,
at a given power factor angle suppose at a given 𝑥 this I will now this vertical line I will
keep fix, line fixed means what 𝑥𝑆 this I have fixed I am not now going to vary 𝜃 in this
expression.
Got the point that is what I mean efficiency is now
𝜂𝑥 =
𝑥𝑆 cos 𝜃
𝑥𝑆 cos 𝜃 + 𝑥 2 𝑃𝐶𝑢𝑓𝑙 + 𝑃𝐶𝑜𝑟𝑒
This is the efficiency expression what I am telling now to understand how this curve how
efficiency will vary if you vary 𝜃 what I have done is I will keep this constant at a given
kVA I want to see I will vary power factor at this kVA.
What is this kVA? A fixed value of 𝑥𝑆, 𝑥 is fixed now. And then vary 𝜃 that is what I will
do power factor I will vary and I want to see at a given fixed power factor at a given kVA
then I will be able to tell at what power factor efficiency will be maximum.
𝑑𝜂
Now, this thing that is what I will do next is this expression I will do 𝑑𝜃. In this expression
in this modified problem statement What I am telling 𝑃𝐶𝑜𝑟𝑒 is fixed 𝑥 2 𝑃𝐶𝑢𝑓𝑙 fixed.
206
Because I am not going to change 𝑥 I will be on this line only fixed, I am not going to vary
𝑥. At a given 𝑥𝑆 kVA the transformer is handling tell me at what power factor efficiency
will be maximum. So, to do this 𝑥𝑆 I will keep fixed that is this line 𝑥 is fixed, and vary 𝜃
other things are fixed 𝑥 being fixed.
Now, So, it will be not very tough if you set
𝑑𝜂
=0
𝑑𝜃
and see that prove that. Show, that not proved, 𝜂 will be maximum, when cos 𝜃 = 1 this
is the thing. You go on vary for a given kVA this is corresponding to say 0.8 power factor.
What this expression will tell you this efficiency will be maximum when cos 𝜃 = 1 that is
at this point it must be above this curve.
Therefore, I can now draw, so this you please try to prove on your own you have to simply
differentiate it and, so that at a given kVA that is what I am telling this is the transformer.
You always pass a given fixed current by varying the load, but in this case I will vary only
cos 𝜃 to change the magnitude of the load. And then I am telling if it is unity power factor
then efficiency at this fix load as you vary 𝜃 power factor efficiency will be maximum
here.
(Refer Slide Time: 09:52)
207
In other words next page I will draw this is quite interesting to note that now I can complete
this 𝜂 versus 𝑥 curves by drawing a family of curves. Suppose this is at cos 𝜃 = 0.8, what
I am telling the curve will be above it and we draw a vertical line here.
This will be corresponding to cos 𝜃 = 1, it will be above cos 𝜃 = 0.8. If cos 𝜃 = 0.6, it
will be this you can draw a family of curve. Now, showing really that is I am now
restricting that cos 𝜃 = 1 now the full picture is clear to me.
𝑃
But nonetheless the that is this thing remains this one 𝑥 = √𝑃𝐶𝑜𝑟𝑒. And this maximum
𝐶𝑢𝑓𝑙
efficiencies are corresponding to different power factors 𝜂𝑚𝑎𝑥 at 0.6 𝜂𝑚𝑎𝑥 at a 0.8 it is this
much. And this is 𝜂𝑚𝑎𝑥 at one unity power factor, so this axis is 𝜂 efficiency.
Now, we come to know oh the transformer will have maximum efficiency possible
efficiency this much at 0.8 power factor if that 0.8 is kept fixed that will give you some
maximum. But you can still operate the transformer at higher efficiency than the load these
provided a load is unity power factor got the point. This is the highest maximum efficiency
possible 𝜂𝑚𝑎𝑥 highest this is an interesting point to note that is at unity power factor.
But nonetheless whatever be the power factor the load at which this maximas will occur
that is fixed. So, this curves a family of curves now bring out the total picture what is going
on in a transformer when you are changing load both power factor and impedance.
So, highest possible efficiency is when you say the load on the secondary of the
transformer is resistive unity power factor. Then only you will get the highest efficiency
for a given power factor if the load power factor is 0.8. It will still give you maximum
efficiency here, but that will be lesser than the maximum efficiency which is still possible
if the impedance is purely resistive that is what I want to tell.
But here is a catch, the catch is the load on the secondary of a transformer will be several
kinds of motors in industry this will supply some motors. Motor means RL there will be
other RL type of load, RL type of load are most common load that is not in my hand you
know.
You must be knowing that the load power factor people try to maintain it at say 0.8 is a
good number ok. You connect RL load this that load, but try to see the power factor is
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close to unity, but at least 0.8. And if you load power factor is a below that then what
supply authority says that you will be penalized.
Because you will be drawing you are utilising lesser power out of the total kVA you are
for drawing from the lines. Therefore, it is the condition imposed by the supply authority
that you are the user of electricity try to see that load power factor is not below this maybe
0.8 say. And if you are found to draw power for loads which is below 0.8 say 0.6 you will
be charged extra, these are the things which come into to play now
But as I told you unity power factor is not the usual case. So, that is why even if your load
is having low power factor that is why people connect capacitor at the beginning capacitor
at the beginning of your supply. So, that power factor is improved you know those
techniques from your circuit analysis and this that. Therefore, although resistive load will
be a desirable load to be connected across the secondary of a transformer to have always
the highest efficiency.
For any kVA you mind you if you are operating the transformer at this kVA fixed kVA,
then you also you see efficiency will be maximum among these maximas among these
available efficiency at different power factor that will be maximum that is a very good
thing. So, it is an important point to be noted nothing is better than unity power factor load.
At whatever kVA it is discharging you will always get highest this thing highest efficiency
with unity power factor load. And the highest most efficiency will be when the kVA is this
much at this kVA you are operating and the unity power factor load it is. You must think
about this you think yourself what I am telling.
So, that things will these are quite interesting I mean there is some logic in that nothing
out of nothing something is being told to you. So, so this curves then I have drawn a family
of curves with different power factors. Then I told you now I will keep the kVA fixed and
would like to see these are the efficiency which of these will become maximum.
And you have been asked to find it out that at cos 𝜃 = 1 you have to differentiate this
expression by noting that now you will not change kVA only power factor you will change
differentiate it set it to 0 and come to the conclusion that if cos 𝜃 = 1 that efficiency will
be maximum it is very interesting to note, so that is all.
209
(Refer Slide Time: 18:51)
Now, we will come to the as I told you if in a transformer. Suppose here is a transformer
a distribution transformer say, what is going on in a distribution transformer here are loads,
how the loads are connected I am a consumer I will connect my load. All loads are
connected in parallel you know, here is another consumer he will connect load like this.
There is another consumer you connect load like this dot and, so many consumers and
perhaps when all the consumers, because my habit of using electricity will not be matching
yours.
So, I will sometime switch on my load or not at any point of time that may change many
complicated situation. So, far as the loading pattern on the secondary of the transformer is
concerned we really do not know. But what happens is this with large number of
consumers it is found that there is a general type of variation of load on the secondary of
the transformer, when you close these two consumers then current supplied is this plus
this,.
So, load increases as consumers are connecting their loads you understand that. Now, the
question is a in distribution transformer there will be a variation of load pattern over 24
hours. There will be some evening peak, there is some morning peak offices are open
electricity’s are used. In the evening once again as I told you there is a peak load demand,
because most of the users will switch on their lights and fans and what not.
210
So, so, there will be a variation of load. And if there is a variation of load then certainly I
am sure about this point that no point in asking that for this transformer I will demand
power efficiency maximum power efficiency for this transformer should occur at 𝑥 = 1,
because at full load condition this is will be operating at some point of time over certain
time period.
But certainly not for 24 hours that must be understood before I calculate that is why this
type of transformer should be judged their performance should be judged not from the
point of view of power efficiency. But from the point of view of what is called energy
efficiency or called all day efficiency.
And as I told you I wrote last time that eta all day is equal to,
𝜂𝑎𝑙𝑙 𝑑𝑎𝑦 =
𝐸𝑛𝑒𝑟𝑔𝑦 (𝐾𝑤ℎ) 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑡𝑜 𝑙𝑜𝑎𝑑 𝑖𝑛 𝑎 𝑑𝑎𝑦 (𝑜𝑣𝑒𝑟 24 ℎ𝑜𝑢𝑟𝑠)
𝐼𝑛𝑝𝑢𝑡 𝑒𝑛𝑒𝑟𝑔𝑦 𝑡𝑜 𝑡𝑓𝑜 𝑖𝑛 𝑎 𝑑𝑎𝑦 (𝑜𝑣𝑒𝑟 24 ℎ𝑜𝑢𝑟𝑠)
You must be knowing energy is in kilowatt hour, very simple calculation nothing very
complicated ok. Now, so what it is what are the information’s needed to find out the energy
efficiency.
(Refer Slide Time: 23:55)
So, suppose I say that this is the primary mind you a distribution transformer this a primary
is energized for 24 hours 𝑉1 whatever it is the voltage at frequency 𝑓 in distribution
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transformer. Primary must be kept energized for 24 hours continuously for 24 hours it must
be.
Because you do not know when your consumer wants power that is not known to me I
must be ready at their door steps with the power available. Whether they will use it or not
that is their botheration I do not know. Now, suppose I say I have to do some statistical
observations how the load pattern changes etc, but we will not go to that level of statistical
distribution.
But I will say this much suppose for this distribution transformer this information is
known, what is that information? That is for the first I will make a column time intervals.
Very simple calculation mind you, see that is first 𝑇1 hours.
Suppose day starts after midnight or say 6 AM say, from 6 AM we start counting. See first
𝑇1 hours. Next column degree of loading. I say it is 𝑥1 , because it is not full load always.
So, first 𝑇1 hours interval 𝑥1 is the degree of loading and 𝑆 is the kVA rating. Similarly I
will write another column my power factor of load.
Suppose it is given as cos 𝜃1 . In the same way for the next 𝑇2 hours let the degree of
loading is 𝑥2 and power factor is cos 𝜃2 . Then 𝑇3 , 𝑥3 is the degree of loading and cos 𝜃3
is the power factor.
In this way I will say that the last 𝑇𝑛 hours 𝑥𝑛 is the degree of loading and cos 𝜃𝑛 is the
power factor angle. This
𝑇1 + 𝑇2 + 𝑇3 + ⋯ 𝑇𝑛 = 24ℎ𝑜𝑢𝑟𝑠
over a complete day that I must see. First 6 hours, next 12 hours, next 6 hours such that it
is like this, this will be the thing.
So, what I have to calculate and 𝑆 is the rated kVA, so energy output over 24 hours I have
to calculate. Next, what I do is the copper loss, in the next column, in the first 𝑇1 hours
degree of loading is 𝑥1 . So, what is the copper loss? It is 𝑥12 𝑃𝐶𝑢𝑓𝑙 , what is the copper loss
during 𝑇2 ? 𝑥22 𝑃𝐶𝑢𝑓𝑙 , next is 𝑥32 𝑃𝐶𝑢𝑓𝑙 . In this way this is 𝑥𝑛2 𝑃𝐶𝑢𝑓𝑙 , where 𝑃𝐶𝑢𝑓𝑙 is constant.
At full load condition what is the copper loss, which I get from short circuit test data. So,
this is the copper loss at various time intervals taking place, what is core loss? Core loss
212
as you know is independent of the degree of loading. So, it will remain 𝑃𝐶𝑜𝑟𝑒 all the time
no matter whether 𝑥 = 0, or 𝑥 = 1 or 𝑥 = 0.8 no.
So, it is always 𝑃𝐶𝑜𝑟𝑒 fixed, got the point. These informations you need.
Time Intervals Degree
of Power
Factor Copper Loss
Core Loss
(Hours)
Loading
of Load
𝑇1
𝑥1
cos 𝜃1
𝑥12 𝑃𝐶𝑢𝑓𝑙
𝑃𝐶𝑜𝑟𝑒
𝑇2
𝑥2
cos 𝜃2
𝑥22 𝑃𝐶𝑢𝑓𝑙
𝑃𝐶𝑜𝑟𝑒
𝑇3
𝑥3
cos 𝜃3
𝑥32 𝑃𝐶𝑢𝑓𝑙
𝑃𝐶𝑜𝑟𝑒
⋯
⋯
⋯
⋯
⋯
⋯
⋯
⋯
⋯
⋯
⋯
⋯
⋯
⋯
⋯
𝑇𝑛
𝑥𝑛
cos 𝜃𝑛
𝑥𝑛2 𝑃𝐶𝑢𝑓𝑙
𝑃𝐶𝑜𝑟𝑒
Then you can straight away write down I will write it here 𝜂𝑎𝑙𝑙 𝑑𝑎𝑦 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 as equal to
what I told I must write energy output over 24 hours. What is the output energy in the first
𝑇1 hours, what is the kilowatt output? It will be 𝑥1 𝑆 cos 𝜃1 𝑇1 . In the first 𝑇1 hours, so much
kilowatt hour. next 𝑇2 hours, what is the energy output 𝑥2 𝑆 cos 𝜃2 𝑇2 .
In this way you go on adding and finally, you add to this 𝑥𝑛 𝑆 cos 𝜃𝑛 𝑇𝑛 , so over 24 hours
this is the energy output what is the unit kilowatt hour.
𝑥1 𝑆 cos 𝜃1 𝑇1 + 𝑥2 𝑆 cos 𝜃2 𝑇2 + 𝑥3 𝑆 cos 𝜃3 𝑇3+⋯ 𝑥𝑛 𝑆 cos 𝜃𝑛 𝑇𝑛
This is the output energy.
213
So, this bracketed term I am not rewriting this numerator this thing this is the energy output
that is this thing here plus the losses energy losses. Now, the question is what will be the
energy losses? See of which you see the core loss remains constant all the time it does not
depend on degree of loading.
So, it will be 24 × 𝑃𝐶𝑜𝑟𝑒 . How this 24 comes? In fact, 𝑇1 + 𝑇2 + 𝑇3 + ⋯ 𝑇𝑛 = 24ℎ𝑜𝑢𝑟𝑠
So, core loss will always take place in the distribution transformer plus the copper losses.
Copper losses energy dissipated will be how much? I will write it in different colour it will
be
𝑥12 𝑃𝐶𝑢𝑓𝑙 𝑇1 + 𝑥22 𝑃𝐶𝑢𝑓𝑙 𝑇2 + 𝑥32 𝑃𝐶𝑢𝑓𝑙 𝑇3 + ⋯ 𝑥𝑛2 𝑃𝐶𝑢𝑓𝑙 𝑇𝑛
So, this divided by this will give you the energy efficiency or all day efficiency of the
transformer.
𝜂 𝑎𝑙𝑙 𝑑𝑎𝑦
𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦
=
(𝑥1 𝑆 cos 𝜃1 𝑇1 + 𝑥2 𝑆 cos 𝜃2 𝑇2 + 𝑥3 𝑆 cos 𝜃3 𝑇3+⋯ 𝑥𝑛 𝑆 cos 𝜃𝑛 𝑇𝑛 )
( ) + 24 × 𝑃𝐶𝑜𝑟𝑒 + (𝑥12 𝑃𝐶𝑢𝑓𝑙 𝑇1 + 𝑥22 𝑃𝐶𝑢𝑓𝑙 𝑇2 + 𝑥32 𝑃𝐶𝑢𝑓𝑙 𝑇3 + ⋯ 𝑥𝑛2 𝑃𝐶𝑢𝑓𝑙 𝑇𝑛 )
And it is based on this number that is while purchasing a distribution transformer you
should rather tell the manufacturer this is my degree of full loading pattern on the
secondary of the transformer will be and I would like to have energy efficiency of this
much. Rather than, so your transformer has a specific rating here 𝑆 kVA is the rated kVA
fixed voltage ratings are there.
But you will say I do not like to have the maximum efficiency to occur at rated condition,
knowing fully well that the load on the secondary of the transformer is not in my hand it
is going to change over time. And after doing some statistical observations I have seen on
an average first 𝑇1 hours so much demand of power degree of loading is 𝑥1 then degree of
loading 𝑥2 then. And power factors of the expected load on the secondary of the
transformer at that time will be cos 𝜃1 for the first 𝑇1 hours and so on.
And as degree of loading changes copper loss in terms of rated 𝑃𝐶𝑢𝑓𝑙 will go on changing
and I know how to calculate that I will calculate, core loss remains same. Only thing
𝑂𝑢𝑡𝑝𝑢𝑡 𝐾𝑊
𝑂𝑢𝑡𝑝𝑢𝑡 𝐾𝑤ℎ
instead of 𝐼𝑛𝑝𝑢𝑡 𝐾𝑊 it is 𝐼𝑛𝑝𝑢𝑡 𝐾𝑤ℎ . What is the 𝐼𝑛𝑝𝑢𝑡 𝐾𝑤ℎ? This bracketed term once
214
again I have not repeat it, it is this one numerator plus core loss remains same. So, watt in
to hour 24 hour plus this, mind you 𝑇1 + 𝑇2 + 𝑇3 + ⋯ 𝑇𝑛 = 24ℎ𝑜𝑢𝑟𝑠.
Thank you we will continue with this next time.
215
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 23
Load Description and Energy Efficiency
Welcome to lecture number 23. And as you know, we were discussing about how to
estimate the efficiency of transformers, power transformers and we found that the normal
efficiency, we usually use power efficiency. And we found that efficiency at any degree
of loading is like this;
𝜂𝑥 =
𝑥𝑆 cos 𝜃
𝑥𝑆 cos 𝜃 + 𝑥 2 𝑃𝐶𝑢𝑓𝑙 + 𝑃𝐶𝑜𝑟𝑒
(Refer Slide Time: 00:44)
𝑂𝑢𝑡𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 (𝐾𝑊)
This is called power efficiency; 𝐼𝑛𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 (𝐾𝑊) and we found that efficiency curve will
be like this, this is some efficiency curve and I must attach to each curve the power factor,
cos 𝜃 is equal to say 0.8 and I showed last time that for a given kVA this axis is 𝑥 degree
of loading and this is your efficiency and this is the value of 𝑥 at which maximum
efficiency occurs
𝑃𝐶𝑜𝑟𝑒
𝑥=√
𝑃𝐶𝑢𝑓𝑙
216
And if you repeat the same thing for other power factors, these curves will be some this
may be power factor cos 𝜃 = 1 and lower power factor it maybe cos 𝜃 = 0.6 no point in
attaching, leading or lagging. Efficiency will remain same and the maximum efficiency
points of course, will lie always at this, this we have discussed.
And therefore, highest efficiency of a transformer will be occurring, when the secondary
impedance or load connected is at unity power factor and the value of 𝑥 is this number. At
𝑥𝑆 kVA maximum efficiency will occur. Then we told you that the importance of energy
efficiency, particularly with respect to distribution transformer, where the load on the
secondary of the transformer will change as day progresses. For some interval of time it
will be operating at degree of loading of 𝑥1 , power factor cos 𝜃1 for a duration of time, 𝑇1
and we found the and that efficiency is called all day efficiency or energy efficiency.
And we found that this is, will be equal to the x i S i in a short form now, I am writing,
because cos theta i, where i is equal to say 1 to N and the output so, this is into T i. So, you
understand this and this one will be same thing here that is x i S i cos theta i, output energy
this is energy. T i is the interval of time summed over i and plus the copper loss, which
depends on the x should be equal to x i square P cu full load into T i. This is a energy loss,
because of copper loss and plus 24 into P core that is all, because core loss remains
constant.
∑𝑛𝑖=1 𝑥𝑖 𝑆𝑖 cos 𝜃𝑖 𝑇𝑖
𝜂 𝑎𝑙𝑙 𝑑𝑎𝑦 = 𝑛
∑𝑖=1 𝑥𝑖 𝑆𝑖 cos 𝜃𝑖 𝑇𝑖 + ∑𝑛𝑖=1 𝑥𝑖2 𝑃𝐶𝑢𝑓𝑙 𝑇𝑖 + 24𝑃𝐶𝑜𝑟𝑒
𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦
So, this is summed over some 𝑛 interval such that
𝑇1 + 𝑇2 + 𝑇3 + ⋯ 𝑇𝑛 = 24ℎ𝑜𝑢𝑟𝑠
Anyway this complicated way, it can be I mean in a compact form it can be written, we
have discussed all these things. Therefore, to judge whether, the transformer is good or not
particularly, a distribution transformer, it is better you calculate all day or energy efficiency
and see that that is quite high, that is how it is decided.
Before we proceed to discuss about regulation, only one thing I will tell see the a loading
of a transformer I told you how loading is specified. I told you, if the rated kVA is 𝑆 kVA
217
then 𝑥 is the degree of loading degree of loading and this value is 0 ≤ 𝑥 ≤ 1, 1 corresponds
to full load and so on.
0 corresponds to no load therefore, transformer is delivering a KVA of I have assumed
here delivering KVA is equal to 𝑥𝑆, if it is at power factor cos 𝜃 then I say output kilowatt,
that was the essence of calculating that thing output in kilowatt. If 𝑆 in kVA output kilowatt
is 𝑥𝑆 cos 𝜃, cos 𝜃 is the load power factor, this we have discussed.
Now, sometimes load that is in other words what I am telling, the load on the secondary
of a transformer is not, although I am drawing like this practical transformer, I was telling
you to make you understand ok; there is some effective impedance connected this 𝑍2
maybe, because of some parallel loads. Anyway, in terms of 𝑍2 I told, but this load
connected across the secondary therefore, is not really explicitly mentioned in terms of a
complex impedance ok. If it is effectively, it will be some complex impedance.
In other words one way of a specifying this load indirectly is that you specify what is the
kVA of the load and power factor as a here I have done or it could be one way of specifying
load in terms of kVA and power factor, another way to specify load in terms of kilowatt
and power factor, are you getting; that is load is 100 kilowatt 0.8 power factor lagging.
You have fully describe the load and voltage is this voltage out of these and voltage is of
course rated voltage like that of that side where you are connecting, rated voltage of this
side where load is connected.
Therefore, say 𝑉2 therefore, kVA power factor and 𝑉2, from this I will be able to calculate
what is 𝑍2 effective, but generally kVA power factor and the voltage at which the kVA is
delivered will be specified. Similarly, it will be kilowatt power factor and 𝑉2. So, if it is
KVA, power factor and 𝑉2 for example, then this value of 𝑥 why I am telling this, how to
calculate the value of a degree of loading.
If you know the rated KV of the transformer and kVA it is handling at a particular interval
of time, take the ratio get the value of 𝑥. However, there is a student sometimes make
mistake, if the load is specified in terms of kilowatt, power factor and 𝑉2 to calculate 𝑥
mind you must bring it to KVA that change we have to do. If kilowatt is given; so, 𝐾𝑉𝐴 =
218
𝐾𝑊
𝑃𝑜𝑤𝑒𝑟 𝐹𝑎𝑐𝑡𝑜𝑟
, you calculate then the corresponding 𝑥 at which the transformer is operating
(
𝐾𝑊
)
𝐹𝑎𝑐𝑡𝑜𝑟
is this 𝑥 = 𝑃𝑜𝑤𝑒𝑟
this kilowatt by power factor divided by rated kVA.
𝑅𝑎𝑡𝑒𝑑 𝐾𝑉𝐴
This is the value of, this you remember and there are very nice problems given in a book
which is so popular you know Parker Smith’s book problems on electrical engineering
very famous book and very nice problems are there.
The beauty of this book is in each problem, if you solve open circuit, short circuit test,
efficiency calculations, you will learn something new. So, please solve as many problems
as you can from that book apart from the fact that we will provide you some tutorial
problems, whenever it will be required, but this is the thing. So, you must remember this
load is not specified in terms of explicit value of 𝑍2 .
How it will be specified? I will say a load consumed so much kVA at this much power
factor at this much voltage. Therefore, from this I will be able to calculate 𝑍2 , but I will
not do that, because to calculate efficiency, I do not require 𝑍2 in terms of kVA and power
factor and degree of loading this 𝑥 is to be known. However, from kVA whatever kVA
here, it is operating what is the value of 𝑥, if the load is specified? It will be simply this
𝐾𝑉𝐴
𝑥 = 𝑅𝑎𝑡𝑒𝑑 𝐾𝑉𝐴, that will be your 𝑥 here.
For this load, if load is specified in terms of kilowatt, power factor and 𝑉2 then 𝐾𝑉𝐴 =
𝐾𝑊
𝐾𝑉𝐴
and then 𝑥 corresponding to this load will be 𝑥 = 𝑅𝑎𝑡𝑒𝑑 𝐾𝑉𝐴, but I will insist
𝑃𝑜𝑤𝑒𝑟 𝐹𝑎𝑐𝑡𝑜𝑟
that better do not memorize this formula, never do it. Each time if the load is totally
described over the 24 hours, for 𝑇1 hours, it is 𝑥1 degree of loading, 𝑇2 hours 𝑥2 degree of
loading.
So, my target will be to calculate 𝑥1 , 𝑥2 , 𝑥3 depending upon the load, it is handling at
different intervals of time and I hope you have understood how to calculate that degree of
loading and use it here, this is energy efficiency kilowatt hour on the top divided by
kilowatt hour on the down. Core loss remains constant and finally, once again repeating
that what is this 𝑃𝐶𝑢𝑓𝑙 , when the windings will carry rated current what is the power loss
that you will get it from your short circuit test.
219
What is 𝑃𝐶𝑜𝑟𝑒 ? 𝑃𝐶𝑜𝑟𝑒 is the reading of the wattmeter under no load test. So, with that
comments we conclude this part of this lecture that is the efficiency; how to calculate,
estimate, this that. Now, what I will do is, I will try to calculate another important thing
that is called regulation of a transformer.
(Refer Slide Time: 14:39)
So, next topic is regulation. We know that the transformer has got internal impedance.
These are practical transformer that is the winding has resistance, there will be leakage
flux, 𝑟1, 𝑥1 magnetizing impedance all things will be there.
Now, what do I mean by regulation? Suppose, you apply the rated voltage at rated
frequency, always apply rated frequency and note down this voltage with no load
connected with 𝑆 open suppose, I what I am telling is what is regulation of a practical
transformer and suppose, I am, I want to find it out in the laboratory what does that really
mean that is what I am telling.
What you do? You apply rated voltage at rated frequency in practice ok, in lab. We have
energized with rated voltage and frequency with S open, measure this voltmeter reading,
you connect a voltmeter here. With S open this voltmeter reading whatever you will get I
call it 𝑉20.
Note down this reading, is equal to reading of voltmeter. Note down this reading
magnitude of what is this voltage voltmeter reading across the secondary here, is the
220
voltmeter connected. Then connect a particular load; for a particular load that is 𝑍2 or in
terms of kVA, kilowatt power factor, etc in whichever way it is specified connect that load
close S and measure once again the terminal voltage.
Reading of voltmeter, close S, 𝑉2 is the reading of voltmeter with S closed, what do you
think will there be a difference in the readings of the voltmeter because I have not done
any mathematics to get some expression for regulation nothing I have done, I have
connected a load in the laboratory, applied a fixed voltage, rated voltage at rated frequency.
What I have done is with S open I have noted this voltmeter reading with S closed same,
voltmeter reading I have noted with load connected and will there be a difference; it is
expected to be, because of the fact it is a practical transformer. Therefore, there will be
voltage drop in the series impedances in the model of the equivalent circuit of the
transformer.
That is in 𝑟𝑒1 and 𝑥𝑒1 there will be a voltage drop and hence, it is expected there will be a
change in the difference in the magnitude of this voltage. So, what you do you calculate
this change and divided by the open circuit voltage, with respect to that how much voltage,
the change in voltage how much it is and you get regulation, this is in per unit. So, multiply
it with 100 get regulation percentage regulation, this is the thing.
%ℛ =
|𝑉20 | − |𝑉2 |
× 100
|𝑉20 |
So, in a practical setup, you can easily calculate regulation easily we will see, but this is
what you have to do imagine at least you can do this; apply rated voltage rated frequency.
Generally, regulation is to be calculated with the rated current it is handling rated kVA and
load should be very clearly specified. So, I will say this is the value of the regulation at, I
must add to it at the when the load is 𝑍2 I mean let me write like that. When load is 𝑍2 at
a given load, because if you change 𝑍2 itself it might give you some other value the change
in dropping voltage may give you other values, we will see that.
So, for a given load you can calculate regulation of the transformer. Regulation is best
understood what exactly it means is suppose, you have a battery what I am telling you
have a battery, you have its internal resistance 1Ω and suppose a source of dc voltage.
221
Suppose, this voltage is 10 volt and its internal resistance is 1Ω suppose internal and these
are the terminals of the battery.
And suppose, the rated current, the battery can supply is 5 ampere equal to rated current
of the battery, but so, this is the rated current of the battery I know. So, I have connected
the switch I will connect here a load. What should be the value of this resistance, because
the battery can supply and you connect a voltmeter here. With S open the reading of the
voltmeter will be 10 volt, it is open circuit.
So, regulation is this 10 volt open circuit voltage minus when you close the switch. Now,
how much resistance can I connect so that current will be rated current, at rated current,
because battery can supply rated current I will be always trying to draw rated current that
is what I told. So, so 10 by another 1Ω suppose, load resistance is 1Ω it is a bad way.
Anyway, whatever number I have taken let us stick to that.
So, what will be this resistance 1Ω. So, when the battery will supply rated current with S
closed the current in the circuit will be how much; 5 ampere, is not? 5 ampere will be the
current and what will be your output voltage; 10 minus this drop that is another 5 volt here.
Therefore, with S closed this voltage will be 5 and divided by 10 and I will say regulation
is 50 %, are you getting?
%ℛ =
10 − 5
× 100 = 50%
10
This is into 100 and I will say oh, this is a very bad source. You thought you will supply
your load with rated voltage, is not; but at the moment you connect load the voltage drops
from 10 volt to 5 volt, it is not a good source. A good source will be perhaps oh sorry, you
take another battery.
Suppose, there is another battery which has got internal resistance say a 0.1Ω are you
getting and here, I will connect load, same battery can supply 5 ampere current, internal
resistance is small then what will be the voltmeter reading here. With S open same 10 volt,
with S closed and here how much resistance should I connect so that it will be 1.9Ω.
If you connect, you close it. So, once again you are supplying 5 ampere to the load but the
voltage drop here it will be now 0.5 volt. Therefore, output voltage available will be 10.
So, with load connected this voltmeter reading will be, 9.5 volt. So,
222
%ℛ =
10 − 9.5
× 100 = 5%
10
Therefore, you have connected you have purchased a source to supply load and whatever
this your source is capable of delivering current that you will draw and you would like to
see across the load almost this open circuit voltage comes.
So, this source is better than this source, are you getting? Therefore, regulation value unlike
efficiency it should be as small as possible. For example, if it is an ideal battery nothing is
better than that so far regulation is concerned. If it is an ideal battery no internal impedance
then you can whether you connect load or not this voltage will be 10V always.
So, regulation is 0. We will address to this problem whether it is better to have a source
having zero internal impedance or not, that is the another thing to be discussed but it looks
like that from no load to full load if at all a change takes place in the magnitude of the
voltage across the load that should be small. So, that it can maintain voltage there.
So, in this case of transformer it is nothing but the load sees what load will see; load will
see a series impedance and there is a source of voltage here. Therefore, there will be
voltage drop in equivalent resistance and leakage reactance of the transformer and
therefore, the magnitude of the voltage available across the load from open circuit to that
loaded condition is bound to change and that change we would expect will be less. Now,
we will go to find out an approximate expression of regulation. So, we will continue with
this.
Thank you.
223
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 24
Regulation: Its Expression
(Refer Slide Time: 00:20)
Welcome to lecture 24 and we were discussing about Regulation and goal of this lecture
will be to obtain an approximate expression for regulation of a transformer.
I will do this before that it is now timed as slightly I will review the equivalent circuit, how
we are drawing this way that way with respect to primary side, secondary side. So, it is
like this; reviewing quickly, I mean no, quickly the equivalent circuit and approximate
equivalent circuit. See this was my transformer, here were my coils primary, secondary;
this is the practical transformer, these are the terminals available to me.
Then and here will be some impedance connected across the secondary 𝑍2 . Now this can
be taught in various ways one way is here is your 𝑉1 some deeper look into this equivalent
circuit, so that it will enhance your understanding I hope. So, what I told there will be 𝑟1,
𝑥1 I can show it like this and then here there will be end 𝐸1 .
Because this ideal transformer part is this one and parallel branch is here, that I am not
drawing as you can easily see that parallel branch will not come in the expression of the
regulation. In other words what I am telling why it is approximate we will neglect the no
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load current. Because no load current which comprises of core loss component of current
and magnetizing current is only about 5% of the rated current and drop because of that in
𝑟1 𝑥1 or 𝑟2 𝑥2 they are very small.
So, this is the thing and this is 𝐸1 a source of EMF this becomes a source of EMF as there
is a time varying flux polarity I know this is my 𝑉1. And secondary side it is like this, there
is it becomes a source of EMF this is plus minus that is the dot so fine. And here was your
𝑟2 𝑥2 and here is your load. This is the ideal transformer primary this is the ideal
transformer secondary, no load and magnetizing I have neglected so, this is like this.
Now interesting thing is this one because this portion is ideal; you can view it, I will better
tell you one of them only you adopt. What I am trying to tell, this is same as this thing oh
sorry this is load, that is if you wish you show no impedance on the secondary side this
you punch here. I mean push it here on the primary side this plus minus and this circuit
and this circuit will be identical no problem if you solve it this is equivalent to this.
Or this thing can be also thought of although space is less, but I will draw it here this way
you can also think it; you show 𝑉1 here. It means that you are assuming secondary coils
have no leakage flux and resistance all resistance and leakage fluxes are there on the
primary side many ways of looking at things. This one could also be drawn like this, you
can show all the impedance on the other side that is here it is 𝑉1 here only you show 𝐸1
you pretend this is your 𝐸1 this part your ideal. And here is another source 𝐸2 plus minus
and here you do not neglect 𝑟1, 𝑥1 but push it here.
𝑟
𝑥
It will be 𝑎12 , 𝑎12 and then 𝑟2 , 𝑥2 do not disturb and then 𝑍2 this is also correct and finally,
this thing if you have drawn referred to this side, I think you have got this idea this is one
block also this 𝑍2 can be this circuit this is the circuit we will look at. This circuit can be
shown to be let me use different color and I will show it 𝑉1. And here 𝑟𝑒1 = 𝑟1 + 𝑟2′ , then
sorry 𝑥𝑒1 = 𝑥1 + 𝑥2′ and here I will forget about this ideal part and connect 𝑎2 𝑍2 .
In fact, I could draw it straight away because we have come up to this only thing
magnetizing branch, I am as we will see it can be because so this is the equivalent circuit
finally, I will be using one of them you use always use this one. I will I am just requesting
you it does not mean that you cannot have other options either this or that. Similarly this
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one means what? And you will show here it is 𝑉2′ . Here similarly this part will be shown
𝑟
𝑥
as this can be here I have shown it can be shown as 𝑟𝑒2 = 𝑟2 + 𝑎12, and 𝑥𝑒2 = 𝑥2 + 𝑎12 .
And then you say that ok, this is the thing and here you say 𝑍2 but here you write
𝐸1 𝑁1
=
𝐸2 𝑁2
that is strictly. So, it is 𝑉1′ , so this is the equivalent circuit referred to load side, secondary
side and it is the equivalent circuit refer to primary side ok. So, either of this or you can
leave with these, but this is somewhat complicated because complicated in the sense
computationally it is difficult. You first know, this is your 𝑉2 this is whatever current it is
delivering things like that anyway you know how to handle it.
(Refer Slide Time: 11:14)
So, now, coming to the expression for regulation. So, I will refer to equivalent circuit refer
to primary side is this one. Primary side this will be simply this 𝑟𝑒2 and 𝑥𝑒2 . And here is
your load reflected load, 𝑎2 𝑍2 and here is your 𝑉1 and the voltage here I must write 𝑉2′ is
not and we have defined the regulation as magnitude of secondary voltage that is where I
wrote
%ℛ =
|𝑉20 | − |𝑉2 |
× 100
|𝑉20 |
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That was the regulation and all quantities here were with respect to the secondary side
what was this? Here was the transformer here I am measuring all the voltages.
Oh sorry that is it will be 𝑟𝑒1 and 𝑥𝑒1 , thank you. Now come to this, this is the load side,
now what I do; I multiply both the sides numerator and denominator by turns ratio
𝑎=
ℛ=
𝑁1
𝑁2
|𝑉20 | − |𝑉2 |
|𝑉20 |
=
𝑎|𝑉20 | − 𝑎|𝑉2 |
𝑎|𝑉20|
=
′ |
|𝑉20
− |𝑉2′ |
′ |
|𝑉20
This will be the thing with load disconnected; with load disconnected this is open. So, what
will be this voltage whatever will come here this voltage with nothing connected is 𝑉20;
that voltage will be simply this voltage reflected.
Student: (Refer Time: 15:12).
This will be the thing got the point, where 𝑉2′ is the reflected terminal voltage when it
supply some load current 𝐼2′ . Some people will also draw the approximate equivalent
circuit means this it is there.
But this drop as you can see it only depends on 𝐼2′ . So, 𝐼0 I can be neglected in this
approximate equivalent circuit. So, this is the thing you have to calculate. Now how can I
calculate? I will first draw the phasor diagram of this circuit. So, the phasor diagram is
very simple, I will start from here, the terminal voltage across the load is 𝑉2′ . I will draw,
then I will draw the current suppose it is supplying some current 𝐼2′ , load power factor
angle is 𝜃, load power factor angle does not change if you multiply 𝑍2 with 𝑎2 ; 𝑎 is a
scalar number.
′
′
So, it is like this; then you can easily see 𝑉20
is what? 𝑉20
; that means, with this switch
′
open in this equivalent circuit also when this is open 𝑉20
= 𝑉1 this voltage will come here
as there will be no current. I think this point must be understood. So, to this 𝑉2′ if I add
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these two drops, which I am drawing on a larger scale so that we understand what we are
doing into 𝐼2′ (𝑟𝑒1 + 𝑗𝑥𝑒1 ) and you must understand it is not drawn to scale if this is 𝑉2′ how
this can be, but there will be some geometrical thing I have to show that is why I am
drawing in a larger scale.
′
So, and this will be how much? This is 𝑉1 and this 𝑉1 = 𝑉20
you must understand this
point. So, this angle is 𝜃 therefore, this regulation means
|𝑉1 | − |𝑉2′ |
ℛ=
|𝑉1 |
Therefore I will say suppose I name these various points like this is suppose O; let this
point be A this point will be B very simple calculation C; it is like this ok. Then |𝑉1 | =
𝑂𝐶, this length |𝑉2′ | = 𝑂𝐴. Now I am writing
ℛ=
𝑂𝐶 − 𝑂𝐴
𝑂𝐶
this will be regulation; clear?
So, it will be like this ok, in this page let me let it be a bit dirty but you will be continuing
without any disruption that these I have to find out. Now the argument you listen. The
argument is this drops this and this both these drops are much smaller compared to these
lengths; you must understand. Because it is a well designed transformer 𝑟𝑒1 and 𝑥𝑒1 are
smaller much smaller compared to whom? Compared to these voltages, compared to 𝑉1 or
𝑉2′ ; do not forget to attach these.
These are quite small although I have drawn very large here so that some geometrical
concept I will apply. That is, in other words what I am telling it will be something like this
if I draw it to this scale maybe it will look like if this is your 𝑉2′ , it will be like this; are you
getting? This is where I am making this. So, these lengths are really small compared to
these lengths. So, you have got the idea it will be like this.
Now rest of the thing is pretty simple therefore, if these two lengths are small this angle,
suppose I call this angle to be 𝛿 compared to 𝑉1 or 𝑉2 then 𝛿 will be also small. That is, as
I was telling it is 𝑉2′ I wipe that out but it is like this. If this be the case then this 𝛿 will be
pretty small because we have added only small small thing here and there. So, 𝛿 is small.
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So, keeping this in mind is it possible then to get the difference between OC and OA; OC
and OA this difference. So, I will now go to next page and start with this phasor diagram
rather neatly it requires like this.
(Refer Slide Time: 22:50)
So, what you do? This is your 𝑉2′ , this is your suppose it is supplying a lagging power
factor load 𝐼2′ , this is theta load power factor and the here are once again I will draw larger,
so that, so this is suppose 𝐼2′ 𝑟𝑒1 and then 90 degree here and this is suppose 𝐼2′ 𝑥𝑒1; and this
length is your 𝑉1 and this angle is 𝛿 and I named it as if I remember correctly OC, OA, B.
Now knowing fully well, that 𝛿 will be small for a well designed transformer. Then and
this one I have got
ℛ=
𝑂𝐶 − 𝑂𝐴
𝑂𝐶
Now by using geometry I will try to find out this length minus this length.
Now, since 𝛿 is small, if you drop a perpendicular on this extended line here, drop a
perpendicular. And also drop a perpendicular from this side to this side ok. Suppose this I
call point M since 𝛿 is small, then I can say that 𝑂𝐶 ≈ 𝑂𝑀; is not? This length will be
approximately equal to this length because 𝛿 is small. You know, if the angle is small this
arc is this one if it is a radius of a these two lengths are almost equal 𝑟𝑑𝜃 = 𝑑𝑠.
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Therefore, I can say that
(𝑂𝐶 − 𝑂𝐴) ≈ (𝑂𝑀 − 𝑂𝐴) = 𝐴𝑀
So, numerator is this length, so this one is approximately equal to
ℛ=
𝐴𝑀
𝑂𝐶
Now the question is what will be the value of M? Pretty simple, you know this angle is 𝜃,
this angle is 𝜃 because 𝐼2′ 𝑟𝑒1; angle between two straight lines is same as angle between
their perpendiculars. This line is perpendicular to this and this line is perpendicular to this,
so this too will be 𝜃.
Therefore, if you drop a perpendicular from point B to N then, now AM can be written as,
𝐴𝑀 = 𝐴𝑁 + 𝑁𝑀
so in the small right angle triangles. So, it will be simply 𝐼2′ 𝑟𝑒1 cos 𝜃 plus NM, NM will be
same as this length B say Q; 𝑁𝑀 = 𝐵𝑄 it is a rectangle. Therefore, this will be BQ and
BQ from this right angle triangle is this one 𝐼2′ 𝑥𝑒1 sin 𝜃.
So, this will be
ℛ≈
𝐼2′ 𝑟𝑒1 cos 𝜃 + 𝐼2′ 𝑥𝑒1 sin 𝜃
𝑉1
Now I have converted it to once again voltage thing, so 𝑉1 I can now invoke upon. So,
regulation is now approximately equal to why I am telling approximate? Because I have
used this approximation 𝛿 is small and so on. So, regulation is approximately equal to this.
And this is the expression we are looking for now what is the use of this expression?
Because regulation is so simple otherwise when even without doing mathematics as I was
telling just energize it with rated voltage frequency in the lab and then connect the load,
measured with this switch open what is the voltmeter reading close this switch with this
load present, take the difference of these two voltmeter readings divided by the open circuit
voltage will give you regulation.
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Then why I have found out an expression; this is because of the fact the rating of the
transformer maybe very large maybe 100 KVA transformer whose current rating is maybe
100 Amp. So, 100 KVA load you will be requiring in the lab to find out the regulation at
full load that is a 0.8 power factor that is not available for example, a distribution
transformer which is 200 KVA and it caters power to various consumers I mean totaling
200 KVA can you imagine that load in your lab, no.
Therefore, you do simple test open; circuit short circuit test, get these values and on pen
and paper you calculate this. You ask that oh what load you have connected; it is delivering
rated current at what power factor, 0.8 power factor lagging. Then on pen and paper
calculate regulation into 100 if you want to get percentage regulation.
%ℛ ≈
𝐼2′ 𝑟𝑒1 cos 𝜃 + 𝐼2′ 𝑥𝑒1 sin 𝜃
× 100
𝑉1
So, you are predicting a regulation without actually loading the transformer that must be
understood ok; because open circuit short circuit test are very simple test; I will carry them
out, find out these values then tell me any load current the transformer is delivering of
course, up to rated current and at what power factor you give me these two information
and I will be able to give you the value of regulation how much voltage will drop; we will
continue with this.
Thank you.
231
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 25
Regulation: Its Expression (Contd.)
(Refer Slide Time: 00:17)
Welcome to lecture 25 on Electrical Machines - I, we are discussing regulation and in our
last class we found out an approximate, but very useful expression for regulation.
(Refer Slide Time: 00:38)
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And it was like this that finally, this is the thing that regulation will be
𝐼2′ 𝑟𝑒1 cos 𝜃 + 𝐼2′ 𝑥𝑒1 sin 𝜃
%ℛ ≈
× 100
|𝑉1 |
Mind you, this is an algebraic equation ok. And if you neglect 𝐼0 you can say this is
%ℛ ≈
𝐼1 𝑟𝑒1 cos 𝜃 + 𝐼1 𝑥𝑒1 sin 𝜃
× 100
|𝑉1 |
Very neat formula approximately equal to and I told you that determination of regulation
by actual loading the transformer is almost an impossibility because of the fact that if the
rating of the transformer is very large kVA, large voltage ratings.
To get that value of load in the laboratory is out of question. Therefore, these expressions
comes very handy and it is often used very widely used formula. So, this is the
approximate, what are each term?
𝐼2′ ≈ 𝐼1
because you can neglect no load current, 𝐼0 . Recall that
𝐼1 = 𝐼0 + 𝐼2′
So, this if you neglect this is 𝐼0 , 𝑟𝑒1 is equivalent resistance of the coils referred to primary
side. So, this expression I have found out with referred to primary side, 𝑥𝑒1 is equal to
equivalent leakage reactants of primary side
𝑟𝑒1 = 𝑟1 + 𝑎2 𝑟2
𝑥𝑒1 = 𝑥1 + 𝑎2 𝑥2
So, this is the expression and what is cos 𝜃? cos 𝜃 is equal to power factor of the load.
Mind you, I have found out this expression for a lagging power factor load. Similarly, I
could find out the expression of the power factor for a leading power factor load.
In that case, what it would; what would happen? That is I will draw 𝑉2′ and load power
factor is suppose; leading it is like this 𝐼2′ then 𝑉2′ dashed plus 𝐼2′ 𝑟𝑒2 will be like this which
is small, I am drawing in a larger scale plus this is 𝑗𝐼2′ 𝑥𝑒2 and this would have been your
233
𝑉1, is not; and what I am telling this angle is small like this it will come, it will be very
small that is 𝑉2′ then this, then this.
This is 𝑉1 this is the actual thing it will be, but in a larger scale so that I understand every
component and 𝛿 is small same argument and once again you can approximate it, you can
derive this expression and what is 𝜃? Theta is this angle load power factor angle of 𝑉2′ and
if you do the same exercise regulation expression will then be equal to
ℛ≈
𝐼2′ 𝑟𝑒1 cos 𝜃 − 𝐼2′ 𝑥𝑒1 sin 𝜃
|𝑉1 |
This I leave it as an exercise.
Mind you, these expressions are algebraic expressions; note this is algebraic expression no
phasor, algebraic equation. Similar is this one algebraic equation here no phasor, 𝐼2′ is only
number it is the magnitude, 𝑟𝑒1 is number; no 𝑗 etcetera. Therefore, it will be like this.
So, for leading power factor, this is the expression, this for lagging power factor and this
is for leading power factor and combining these two you can say that regulation is it would
same thing I am writing
𝐼2′ 𝑟𝑒1 cos 𝜃 ± 𝐼2′ 𝑥𝑒1 sin 𝜃
ℛ≈
|𝑉1 |
This is worth remembering and easy to remember. Plus sign for lagging power factor load
and minus sign for leading power factor load.
Ok, what this regulation actually means? 𝑉1 is constant primary voltage it means change
in secondary terminal voltage, expressed as a percentage of rated voltage. Numerator gives
you change in magnitude of voltage divided by denominator which gives you rated
voltage, that is the regulation is and all these expression after I get this, I will be able to
transform it with respect to the secondary side as well if I please. For example, I will say
that I mean these I am just pointing out it will only take some time, but you must be
understanding this one.
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(Refer Slide Time: 08:04)
ℛ≈
𝐼2′ 𝑟𝑒1 cos 𝜃 ± 𝐼2′ 𝑥𝑒1 sin 𝜃
|𝑉1 |
This is equal to refer to primary side. Everything could be written in terms of secondary
side ok, how it can be written? Because of the fact,
𝐼2′ =
𝐼2
𝑎
𝑟𝑒1 = 𝑎2 𝑟𝑒2
𝑥𝑒1 = 𝑎2 𝑥𝑒2
Power factor does not change.
𝐼2
ℛ≈
𝑎
𝐼
𝑎2 𝑟𝑒2 cos 𝜃 ± 𝑎2 𝑎2 𝑥𝑒2 sin 𝜃
|𝑉1 |
Now these a goes one 𝑎.
So, I will bring it down. So, I can write it as
ℛ≈
𝐼2 𝑟𝑒2 cos 𝜃 ± 𝐼2 𝑥𝑒2 sin 𝜃
|𝑉 |
( 𝑎1 )
235
|𝑉 |
And what is 𝑎1 , is the a no load voltage |𝑉20|. So, this is same as
ℛ≈
𝐼2 𝑟𝑒2 cos 𝜃 ± 𝐼2 𝑥𝑒2 sin 𝜃
|𝑉20 |
same expression but the only thing you must be careful this is |𝑉20 |.
So, either of them you use, but as you know you stick to one, you always transfer it to the
primary side and get these values and these are algebraic equation. So, the numerator
actually is the difference of the magnitude of no load and full load voltage. Before that, I
will just try to sketch some curve, what will be the order of value? This into 100 will give
you percentage regulation ok; this is per unit regulation.
%ℛ ≈
𝐼2 𝑟𝑒2 cos 𝜃 ± 𝐼2 𝑥𝑒2 sin 𝜃
× 100
|𝑉20 |
How this curve will look like, if I sketch it?
(Refer Slide Time: 11:56)
So, you sketch this curve in this way at least one curve I will sketch. So, this is the value
of regulation, I will sketch percentage regulation. The regulation curve and this is suppose
you sketch power factor and at rated current. Rated current means if secondary current is
rated primary too will have rated current; you do not have to bother. So, load current is
rated means primary side current is also rated.
236
So, the a typical regulation curve will look like this one, this is I will just sketch it.
Something like this and this point is suppose unity power factor cos 𝜃 against 𝜃 I am
sketching. This side is lagging power factor and this side is leading power factor and this
is regulation curve.
I have plotted for example at rated current I must specify that at rated current and as I was
telling you at rated current, power factor suppose 0.8 lagging, regulation will be small or
high? Like efficiency is 99% or 98% that is common but regulation should be low it may
be about say 5%, got the point? At unity power factor also there is regulation, it is positive,
what is regulation? Regulation is no load voltage minus the voltage with load connected
divided by the no load voltage.
(Refer Slide Time: 15:03)
Therefore, it is positive means the numerator is positive. No matter either you use this
expression or that one and the same thing you will get same result. Therefore, as you can
see if you want to calculate regulation at unity power factor; that means, 𝜃 = 0 then it will
be simply
ℛ≈
𝐼2′ 𝑟𝑒1 cos 𝜃
|𝑉1 |
Put the corresponding power factor, how I plotted that. Fixed up 𝜃 at rated current I know
the rated current, knowing the rating of the transformer 𝑟𝑒1 and 𝑥𝑒1 I know from OC and
237
SC test, 𝑉1 is rated voltage I know. So, I can calculate and this is how I will get this curve.
Only one thing you note that, regulation this side is positive and this is negative regulation.
So, it looks like for capacitive loading; for some capacitive loading that is leading power
factor, regulation maybe negative indicating that with respect to the voltage when no load
is connected when you connect a leading power factor load terminal voltage will rise.
Unlike a battery, battery with an internal resistance 𝑟 you connect a load always the voltage
will fall, positive regulation means voltage falls.
So, for leading power factor load, if you draw the phasor diagram correctly there will be
some power factors for which the length of this phasor, will be more than this, got the
point? Therefore, regulation may be negative sometimes and voltage may rise and if you
see this expression and this curve, there may be some power factor you will be able to say,
you will be able to predict if the transformer is.
So, this curve must be attached with this tag that at what current you are doing at full load
current and you are varying power factor. So, for rated current there exist a point, when
the regulation may become 0 indicating that at no load condition record this voltage, this
is your voltmeter. No load condition, record this voltage. Load it at rated current and at
this power factor which is leading, you will find there is no change in voltmeter reading it
means that.
So, regulation may be come 0 and I can predict at what power factor regulation will
become negative because of the fact regulation is with respect to the primary side. If it
becomes 0, this is finite it cannot be 0, finite and rated value applied voltage.
So, this will become 0 if
𝐼2′ 𝑟𝑒1 cos 𝜃 + 𝐼2′ 𝑥𝑒1 sin 𝜃 = 0
you can easily see
tan 𝜃 =
sin 𝜃
𝑟𝑒1
=−
cos 𝜃
𝑥𝑒1
So, 𝜃 will be negative indicating it is leading power factor ok. So, it is this point. Therefore,
is regulation in case of a DC source with the internal resistance regulation will be always
238
positive voltage will always drop and drop, but here in case of a transformer which is
giving you AC supply and you can choose a load power factor of
𝜃 = tan−1 (−
𝑟𝑒1
)
𝑥𝑒1
𝑟
So, load power factor if you choose tan−1 (− 𝑥𝑒1 ) you can say regulation may become 0.
𝑒1
Anyway it is only a statement that is what it will happen, but unfortunately I cannot operate
a transformer under this condition because load is always lagging type.
Anyway, this is a point what should be what noting that is all and regulation maybe
negative sometimes. So, one should not get surprised I have connected a load and voltage
has risen, maybe you have then connected a capacitive load whose leading power factor is
greater than 1 not for all capacitive load it will be negative or 0 because if the capacitive
load is of this power factor, regulation will become positive.
So, one can get a number of curves at different current levels. For example, at lower current
level the curve will shift below which I am not drawing. It will become regulation will
become less. So, it will be above.
It will, that is what I am telling this is this one. Suppose on the same curve I want to show
regulation at reduced current then curve will be like this.
Amplitude will get reduced, you have understood; you can scale this thing. So, it will
change in its amplitude, but it will follow the same curve. Anyway this you can easily
verify therefore, the regulation of a transformer we have understood.
The difficulty of regulation concept, it is otherwise very simple, it can be directly loaded,
but directly loading a big transformer in the lab is out of question. Therefore, these simple
expressions will come very handy because it is an algebraic expression, easy to remember.
Tell me; what is the current the transformer is delivering at what power factor I will be
able to tell you what will be the change in voltage.
239
(Refer Slide Time: 23:40)
Some more words about this regulation; for example, regulation is
ℛ≈
𝐼2′ 𝑟𝑒1 cos 𝜃 + 𝐼2′ 𝑥𝑒1 sin 𝜃
|𝑉1 |
This is regulation. Now, in this expression if you neglect no load current 𝐼2′ ≈ 𝐼1.
So, you can calculate regulation given a particular current and particular power factor of
the load. Now this one therefore, one must say along with this number whatever you get
say ℛ = 2.5% I get, I must write at rated current and at 0.8 power factor lagging. I must
specify these two things along with this number, you cannot simply say regulation is ℛ =
5% at what load current and at what load power factor angle.
So, this is at 0.8 power factor lag; you must specify this then only it becomes complete. I,
this expression can be written as only one point I will tell,
𝐼2′ 𝑟𝑒1
𝐼2′ 𝑥𝑒1
ℛ≈
cos 𝜃 +
sin 𝜃
|𝑉1 |
|𝑉1 |
what will be the regulation of an ideal transformer? 𝑟𝑒1 = 0 and 𝑥𝑒1 = 0 it will be 0. For
ideal transformer regulation is 0 because there is no internal drops which will be taking
place. So, it can be written like this.
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Generally, as I told you regulation is calculated at rated current and at 0.8 power factor ok.
So, I know this number is very important, calculate regulation at rated current and 0.8
power factor lag. Now, look at this term; this equation once again I am repeating, it is
algebraic equation; no phasor involved.
Suppose, let us take that transformer 1 KVA for easy calculation or say 10 KVA,
10𝑘
200V/100V 50Hz transformer. What is the rated current of the HV side? 200 = 50𝐴, what
10𝑘
is the rated current of the LV side? 100 = 100𝐴. What this number 𝐼2′ 𝑟𝑒1 will mean? It will
mean, this 𝐼2′ 𝑟𝑒1 = 50 × 𝑟𝑒1 it will mean that. Suppose this is my primary side 1 and this
is side 2.
So
𝐼2′ 𝑟𝑒1 50𝑟𝑒1
=
|𝑉1 |
200
That is the equivalent circuit if you look at, here is your 𝑉1 = 200𝑉 and this is your 𝑟𝑒1
and 𝑥𝑒1 and rated current is 50A, is not? Load is connected here whatever it is.
Therefore, what this 𝐼2′ 𝑟𝑒1 mean? It means this voltage drop at that rated current 𝐼2′ 𝑟𝑒1 =
50 × 𝑟𝑒1. So, with respect to the primary side or HV side in this case, this number tells you
voltage drop in the internal resistance of the transformer in volts. This is the; if I write in
language, I will write it is the equivalent resistive voltage drop in the transformer.
Similarly, 𝐼2′ 𝑥𝑒1 is the equivalent leakage reactance voltage drop in transformer, refer to
your primary side if you want to write that also you better write, refer to primary side and
what is the denominator? Denominator is rated voltage of the primary side.
So, this number this ratio is actually how much of this total rated voltage is consumed by
the resistance of the transformer; if it is comes out to be say this number after calculating
if it comes out to be 5%, I will say oh, in the resistance 5% of the this 200V out of these
at rated current some voltage will be dropped across here, some voltage will be dropped
across here.
That magnitude of that voltage divided by the rated voltage is nothing but how much of
this total voltage rated voltage is consumed by the resistance, how much of this total rated
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voltage is consumed by this reactance. And this one in books, they will write this is called
the per unit resistance of the transformer and they will write it like 𝜖𝑟 , what is 𝜖𝑟 ? 𝜖𝑟 is
this one per unit resistance.
So, what is per unit resistance of a transformer? You simply calculate transformer is
operating how much of the rated voltage, is dropped in the resistance this number you
would like to have more or less definitely less. Out of this total rated voltage lesser and
lesser get consumed by the transformer is better. So, that regulation will improve.
Similarly, this number this is the 𝜖𝑥 , this is called 𝜖𝑥 is the per unit reactance of a
transformer.
𝜖𝑟 =
𝐼2′ 𝑟𝑒1
𝑉1
𝐼2′ 𝑥𝑒1
𝜖𝑥 =
𝑉1
ℛ ≈ 𝜖𝑟 cos 𝜃 + 𝜖𝑥 sin 𝜃
(Refer Slide Time: 33:10)
Suppose I say for last statement suppose I say that 10 KVA transformer and 200V/100V
50Hz etc and rated current I know, it is 50A this side rated current is 100A.
242
If I say instead of giving you the value of 𝑟𝑒1, I say 𝑟𝑒1 = 2% I say that, what that does
that mean? It means per unit resistance percentage. So, 𝑟𝑒1 |𝑝𝑢 = 0.02, is not? And this I
am telling what it is? It is nothing but that is 𝑟𝑒1 value is not explicitly given in ohm
suppose, it is given in terms of per unit values then 𝑟𝑒1 |𝑝𝑢 = 0.02. Means what is the
absolute values of 𝑟𝑒1 I want to calculate then I will say oh,
50𝑟𝑒1
= 0.02
200
and from which 𝑟𝑒1 can be calculated, are you getting? Similarly, with 𝑥𝑒1 .
So, in some problems instead of giving you absolute values of 𝑟𝑒1 and 𝑥𝑒1 per unit values
are given, but physically try to understand per unit values means that at rated current what
will be the voltage drop in 𝑟𝑒1 or 𝑟𝑒2. Similarly, this is
100𝑟𝑒2
= 0.02
100
That will be also 0.02 because of that ratio business.
Therefore, given knowing the rated currents and the equivalent per unit values. So, per
unit values will never be specified in terms of primary or secondary because from
whichever side you calculate that value will remain same. Please think about this and read
it, solve problems. We will continue with this next time.
Thank you.
243
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 26
Auto Transformer - Introduction
(Refer Slide Time: 00:24)
Welcome to lecture number 26 on Electrical Machines - I and you recall that we were
discussing about regulation of a transformer.
(Refer Slide Time: 00:27)
244
And we found that the regulation is given by this expression ultimately we derived. It could
of course, be expressed in terms of either primary and secondary. So, with respect to
primary side regulation turns out to be
𝐼2′ 𝑟𝑒1
𝐼2′ 𝑥𝑒1
ℛ≈
cos 𝜃 +
sin 𝜃
|𝑉1 |
|𝑉1 |
These are magnitudes this is an algebraic equation and cos 𝜃 is the power factor of the
𝐼′ 𝑟
2 𝑒1
load, then I told that this quantity |𝑉
is called the per unit resistance.
|
1
In fact, it means that how much of the rated voltage is dropped across the equivalent
resistance of the transformer at rated current. Similarly,
𝐼2′ 𝑥𝑒1
is the per unit leakage
|𝑉1 |
reactance. It denotes how much of the rated voltage is dropped in the leakage reactance of
the transformer and so, we start from here and then I told the meaning of this, suppose I
say that this is the rating of a transformer 𝑟𝑒1 = 2%, what does it mean? 𝑟𝑒1 |𝑝𝑢 = 0.02, it
simply means that 200 × 0.02 will be dropped in the resistance when rated current will be
flowing that is 50 × 𝑟𝑒1 = 200 × 0.02, that way you interpret.
Extending these you can also say per unit impedance of the transformer can be written as
𝑃𝑒𝑟 𝑢𝑛𝑖𝑡 𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 =
𝐼2′ 𝑍𝑒1
|𝑉1 |
𝐼2′ ≈ 𝐼1 when no load current is neglected. All are numbers only where 𝑍𝑒1 is the actual
impedance in ohm and
𝑍𝑒1 = √𝑟𝑒1 2 + 𝑥𝑒1 2
And per unit values, per unit resistance and impedance or reactance, they are same no
matter whether you are calculating the per unit values from the high voltage side or from
the low voltage side, this will come out to be same ok. Another interpretation of per unit
resistance I will just point out, it is I told you it is
𝐼2′ 𝑟𝑒1
𝜖𝑟 =
|𝑉1 |
And these 𝐼2′ ≈ 𝐼1 .
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So,
𝜖𝑟 =
𝐼1 𝑟𝑒1
|𝑉1 |
What you do is you multiply the numerator and denominator by 𝐼1 to get
𝜖𝑟 =
𝐼12 𝑟𝑒1
𝐶𝑢 𝐿𝑜𝑠𝑠
=
𝑉1 𝐼1
𝑅𝑎𝑡𝑒𝑑 𝐾𝑉𝐴
So, per unit resistance can be told in language that amount of voltage drop in the equivalent
resistance expressed as a per unit value of the rated voltage and this current is rated current.
And this is equal to therefore, copper loss as a percentage of kVA rating of the transformer
ok. So, there are various ways of telling so anyway. So, if you know the per unit values
regulation expression then comes out to be
ℛ ≈ 𝜖𝑟 cos 𝜃 ± 𝜖𝑥 sin 𝜃
Minus sign to be used for leading power factor of the load. So, this is the thing.
(Refer Slide Time: 05:33)
Now, continuing our discussion on regulation what do you like to have as a good source
of power from the secondary side? Secondary side will be supplying load it is load I will
connect on the secondary side. Therefore, I would like to see that as the value of load
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impedance changes from no load to full load condition, the voltage applied across the load
that should remain constant, but unfortunately that is not going to be with a practical
transformer because 𝑟𝑒 is present 𝑥𝑒 is present.
Therefore, there will be some change in magnitude of the voltage and all, frequency of
course, remain same. So, so regulation should be small maybe 2% to 5% of the rated
voltage that is one thing. Now, therefore regulation should be small no doubt.
So, that constancy of voltage across the load is not disturbed that much. Now, how this
should be made small? Suppose, you have a transformer like this and as I told you the
windings are wound on the limbs this horizontal portion is yoke. So, suppose because the
leakage reactance is to be reduced, to reduce the regulation of a transformer 𝑟𝑒1 and 𝑥𝑒1
you cannot play much with them because number of turns is fixed based on the voltage
ratings and you will be using copper.
So, resistivity is there 𝑟𝑒 will be there, but 𝑥𝑒 is the you know is because of the fact that as
I told you this I will draw roughly, this is suppose the one winding primary I drew while
we started discussing then I told there will be mutual flux which links both the primary
and secondary, this is the mutual flux 𝜑𝑚𝑢𝑡𝑢𝑎𝑙 and there will be leakage flux which will
be linking only primary or only secondary.
So, this leakage fluxes will not contribute to the energy transfer from primary to secondary
except, that they will cause some extra voltage drop as voltage drop takes place in the
resistance. So, it was represented by leakage reactance. Now, therefore, to reduce the value
of regulation, it depends on 𝑟𝑒1 and 𝑥𝑒1 . So, 𝑟𝑒1 copper you use that is ok, but can I reduce
the value of 𝑥𝑒1 ; leakage flux.
So, this winding is the say LV these winding is HV whatever it is. You know the two coils,
we will just discuss in terms of, suppose there are two coils if you take the coils far away
leakage flux will increase mutual flux will decrease from common sense we can say that.
Therefore, if LV side coil is wound on limb1 and HV side is totally wound on limb2, there
is separation and leakage flux is there, but this can be reduced to a great extent if this
windings are distributed both the windings are distributed in both the limbs, what do I
mean by this is whatever is the total number of turns of the LV winding you arrive at, you
take half of the turns here only LV I am drawing this is suppose LV winding.
247
And this also I called LV; half of LV turns on this limb and half of LV turns on this. So,
when you pass current through this and these two coils you connect them in series, got the
𝑁
𝑁
point? So, 𝐿𝑉 number of turns, you divide like this 2𝐿𝑉 and 2𝐿𝑉 and distribute them on both
the limbs instead of trying to put all the turns 𝑁𝐿𝑉 on the single limb.
Then I told you also that HV winding are wound over, the first LV winding then HV
winding, then what you do; I am using different color to indicate the HV winding then you
also wound the HV winding like this the separate coil, are you getting? So, you make the
LV winding half of LV winding here, then half of LV winding there over the HV winding.
So, this red one is
𝑁𝐻𝑉
, this red one is
2
𝑁𝐻𝑉
2
and these two halves, you connect them in series
properly. So, that fluxes are produced in the same direction when they carry current ok,
then you say ultimately after connecting series for example, LV winding you will get two
terminals here, these are the two terminals of the LV winding. Similarly, HV winding also
can be connected in series ultimately two terminals that is, you distribute the low voltage
winding on both the limbs.
Similarly, HV winding on both the limbs and these two halves of each of LV and HV
windings will be connected appropriately in proper way in series to give you LV terminals
and HV terminals. By doing so, what you will be reducing; that is these two coils will be
now brought in close proximity. So, leakage flux will be reduced because they are close
by.
So, if you see the section of this limb for example, what I mean this is the cross sectional
area. First, you see the LV winding; it will be wound like this. What is this diagram I am
drawing? Your sectional diagram looking from the top, this is LV. So, this is LV winding
the turns which I am showing by shaded line and this is HV winding like that.
So, this is HV and this is LV this inner one is LV, outer one is HV. Therefore, the leakage
flux will be reduced to a great extent. So, this is how at the design level people try to adjust
the leakage reactance. Anyway, this is just no mathematical derivation just from common
sense we are trying to tell how leakage reactance can be reduced.
Now, the next question is, is it that I should go for trying to have leakage impedance, 0;
that is use better and better material, make 𝑟𝑒 close to zero. Similarly, 𝑥𝑒 tight coupling
248
between the two coils, leakage flux is also reduced; so that the leakage impedance is
brought to very low value, is it like that? No, if you bring it down to low value as I told
you that is our goal then the constancy of voltage across the load will be maintained,
irrespective of degree of loading of the transformer.
(Refer Slide Time: 16:37)
But there is one problem, the problem is from this fact because after all these power
transformers are integral part of a power system you know. I drew it there is generator,
there is transformer here, then transmission line and high voltages and so on.
So, for example, look at this transformer there are three lines three phase lines RYB, if a
direct short circuit occurs if a fault occurs in the form of direct short circuit across the
secondary of a transformer, what will be the order of the current? Now, mind you the
primary is excited with full voltage ok. If primary of a transformer is excited with full
voltage and secondary shorted, how much will be the current? Current will be very large.
In the short circuit test of course, this voltage we apply is very little that is why we apply
little voltage so that rated current flows.
But you think of a situation, a transformer is ready to supply load rated voltage frequency
we have applied secondary voltage is also rated, secondary suddenly short circuited, then
what happens is there will be large short circuit current, who will limit that short circuiting
current? That is in terms of equivalent circuit 𝑟𝑒1 and 𝑥𝑒1 this is shorted and here it is now
rated voltage of that particular side, 1.
249
So, current will be limited by the equivalent leakage impedance of the transformer and if
you make it very low, this current will be very high of course, protections are taken in the
circuit, but suppose somehow that fails for some reason or other then both the transformer
and the your source maybe at danger because it will supply almost infinitely large current
and if that current sustains for long time and your relay circuit breaker does not operate
there will be problem.
So, it is better this leakage impedance try to see it is small, but at this same time not too
small, ideally 0, you should never attempt to do like that. Although from regulation point
of view this is fine, but you should be prepared you do not know there may be a direct
short circuit at the secondary of the transformer and then huge current will flow and that
is if some leakage impedance is there the short circuit current will be at least limited by
𝐼𝑠𝑐 =
𝑉𝑅𝑎𝑡𝑒𝑑
𝑍𝑒
is not; by this, instead of 0; have some finite small values small it is, but anyway it will try
to limit the current.
So, these are some of the points I wanted to tell you about the regulation of a transformer.
Now so solve lot of problems as I told you from open circuit test and short circuit test then
efficiency of a transformer and how to calculate regulation, but last word about regulation
I must tell you that, if you look at the regulation formula it was nothing but ℛ ≈
𝐼2′ 𝑟𝑒1 cos 𝜃+𝐼2′ 𝑥𝑒1 sin 𝜃
, is not? That is the thing and if you remember the numerator was like
|𝑉1 |
this |𝑉1 | − |𝑉2′ | this was the thing change in magnitude this was the thing. Therefore, you
see that this
|𝑉1 | − |𝑉2′ | = 𝐼2′ 𝑟𝑒1 cos 𝜃 + 𝐼2′ 𝑥𝑒1 sin 𝜃
What this thing means? This expression is often used to calculate one important thing apart
from regulation. For example, what I am trying to tell what is this thing equivalent circuit
𝑟𝑒1 and 𝑥𝑒1 this is 𝑉1 and this is 𝑉2′ = 𝑎𝑉2 . actual voltage across the load this side load is
connected some 𝑍2′ .
Therefore, from this formula suppose they I pose the problem in this way. Suppose, you
have a transformer say 5 KVA, 200V/100V, 50 Hertz transformer single phase, this
250
transformer and if it is 5 KVA, its rated current of this side 𝐼1 will be how much?;It will
5000
be 𝐼1 = 200 , is not?
5000
So, 𝐼1 = 25𝐴𝑚𝑝 this rated current is and LV side current, rated current will be 𝐼2 = 100 =
50𝐴𝑚𝑝. Now, suppose I say that I have applied rated voltage I have applied 200V on the
primary side, please be with me what I am telling listen carefully. Suppose, I have applied
rated voltage 200V then I ask myself, if you deliver rated current and what is the actual
scenario?
Actual scenario is this is the equivalent circuit, here 200V you have applied, here you have
connected the load 𝑍2 , is not? And then this is equivalent to this, from this I have got that.
Now, what I am telling; suppose the transformer is delivering here 50 ampere current rated
current at some known power factor angle cos 𝜃 you are delivering.
If I simply ask you, what will be this terminal voltage 𝑉2 with load connected? Without
load this voltage would have been equal to 100V with this S open 200V you have applied.
Voltage across the terminals of the secondary of the transformer would have been 100V,
but you have now connected load, how much do you expect? I will expect this voltage will
drop it will be less than 100 volt because of these drop here; it is 200 volt. So, there will
be drop here.
Now, if rated current is flowing 25Amp there, 50Amp is flowing there. So, I will use this
formula and tell that
|𝑉2′ | = |𝑉1 | − (𝐼2′ 𝑟𝑒1 cos 𝜃 + 𝐼2′ 𝑥𝑒1 sin 𝜃)
Bring it to this side you will get |𝑉2′ |.
So, you will put all the values, that is 𝑉1 = 200𝑉 blah blah blah. I mean 𝐼1 = 25𝐴𝑚𝑝,
cos 𝜃 if it is known say 0.8. I will put that and get by what amount this difference is how
much? I will be able to calculate maybe it will come out to be 5V, then you will be able to
calculate 𝑉2′ and then I want to know what is 𝑉2. So, I can easily calculate 𝑉2 how much it
will be? From 𝑎𝑉2 , I will divide and get it. Therefore, this formula is very useful in getting
the secondary voltage of the transformer rather quickly, if you know the degree of loading
on the secondary by how much the voltage will fall. So, try to use it efficiently.
251
On the other hand, I can also say suppose I would like to get 100V across the load of the
transformer secondary of the transformer then how much voltage is to be applied, it must
be greater than slightly more than 200V. That also can be estimated if I insist, no across
the load when the load at rated current I must get 100V here, then certainly this should not
be 200V.
If you give 200V you know the fact, it will be less than 100V maybe 95V, but if I insist
that no, I will when the load is connected I want to get 100V then how much should be
applied voltage? Then once again I will use the same formula which is algebraic in nature,
no phasor equation.
So, I will then calculate
|𝑉1 | = |𝑉2′ | + (𝐼2′ 𝑟𝑒1 cos 𝜃 + 𝐼2′ 𝑥𝑒1 sin 𝜃)
So, currents, 𝑟𝑒1, power factor is known 𝑉2′ = 100 I want to get there; that means; here it
is 200V so
|𝑉1 | = |𝑉2′ | + (𝐼2′ 𝑟𝑒1 cos 𝜃 + 𝐼2′ 𝑥𝑒1 sin 𝜃)
= 200 + ( )
So, you have to apply to the primary then a voltage higher than 200V slightly higher I
mean, one should not think oh, higher voltage but its rated voltage is 200V, but as I told
you 𝑟𝑒1 and 𝑥𝑒1 are in general small.
So, regulation maybe of the order of 5%. So, its slightly more voltage applied you will be
able to maintain 100V across the secondary of the transformer, you can have several nice
problems on it. So, regulation formula is useful.
So, also useful is this formula that is the |𝑉1 | − |𝑉2′ |, the magnitude of 𝑉1 and magnitude
of 𝑉2′ can be easily computed for a given load and power factor of the transformer. Now, I
will just mention because as a continuation of this that; this I will leave for later discussion,
but our next topic will be what I have thought is Auto transformer.
252
(Refer Slide Time: 29:36)
Our next topics will be auto transformer. Only the idea I will tell, we have so far seen that
a two winding transformer essentially means that you have a core and there are two
separate coils at least two separate coils and these coils will be distributed, but that is not
generally shown here to reduce the leakage flux, but the this is understood, but this is the
side1 and side2; two coils are there and you can then change the level of voltage from 𝑉1
to 𝑉2, 𝑁1 turns 𝑁2 turns. Mutual flux is common to both of them.
Then we know that if you have applied the rated voltage, it is better you use voltage per
turn concept.
𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑝𝑒𝑟 𝑇𝑢𝑟𝑛 =
𝑉1
𝑁1
and that remains same for both primary and secondary. Therefore,
𝑉1
𝑉1
𝑁2
𝑉2 = ( ) 𝑁2 = = 𝑉1 ( )
𝑁1
𝑎
𝑁1
𝑎=
𝑁1
𝑁2
So, you can then either step up or step down a voltage depending upon this ratio 𝑉1 may
be stepped up or stepped down. So, this is one way of doing. Now, this is two winding
transformer this one is two winding transformer, two separate coil. We will see that the
253
same transformation of voltage AC voltage from 𝑉1 to 𝑉2 level can also be done by another
𝑑𝜑
kind of transformer, work on same principle 𝑑𝑡 that is all and that transformer is called
auto transformer and we will continue with this in the next lecture.
254
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 27
Auto Transformer Versus Two winding Transformer
(Refer Slide Time: 00:19)
Welcome, to lecture number 27 our earlier classes were very important we derived the
expressions for efficiency and regulations. And we then told at the end that our next topic
will be Auto Transformer. Now auto transformer also does the same thing, it changes
voltage from one level to another keeping the frequency same. So, this was two winding
transformer of course, the basic working principle remain same as we will see now.
What is done is in auto transformer instead of using two separate coils a single coil is used
the idea is like this. For example, you say that, your the core is like this; and you use single
a number of turns ok. I am showing it to be wound on one limb, but it may be distributed
over the whole core length.
So, suppose 𝑁 number of turns are wound ok, these are the two terminals; A and say C
where I will apply first get the idea how it works? I will apply some rated voltage at rated
frequency. The moment you apply a known voltage at a known frequency what gets fixed,
the flux in the core gets fixed. And first I will tell about ideal auto transformer, ideal auto
transformer that is all flux are confined to the core little magnetizing current is required all
255
these things. In another words I will neglect magnetizing current, no load current, eddy
current hysteresis loss in the core.
Therefore, as the current flows little magnetizing current alternating flux (𝜑) will be
produced and it will move like this. And let me say that, number of turns of this 𝑁𝐴𝐶 = 𝑁1
total number of turns. So, you know the applied voltage and induced voltage between these
two points will be
𝑉1 = 𝐸𝐴𝐶 = 𝐸1 = √2𝜋𝑓𝜑𝑚𝑎𝑥 𝑁1
we know this is what happens and that is it.
Now, this flux mind you will link all the all the turns. Suppose I take a voltmeter then, let
us assume there is ohmic contact I can make so I connect a voltmeter here. Now this
voltmeter if I connect between these two points will it read? Yes it will read why not?
What will be the magnitude of the voltage here if you applied 𝑉1 here? You will then tell
how many turns are there between these two points? If the number of turns are 4, then you
will be able to calculate because voltage per turn remain same.
So, if there are 𝑁2 turns here between these two tappings, the voltage induced will be
𝑉1
𝑁
𝑁1 2
𝑉
Voltage per turn is 𝑁1 and that into 𝑁2 will give you the voltage here .
1
Suppose the number of turns is 𝑁1 = 100 and you have taken tappings at 50% midpoint
tappings then there will be 50 turns there. So, if you have applied 200 volt you will get
100 volt is not? Step down will take place. And therefore, between this tappings, I will
connect the load after changing this available voltage to a voltage level which is required
for the load.
So, I will not use to separate coils, single coil and have suitable tappings from the
secondary and it looks like I can get then any level of voltage. For example, so this
diagram, hence forth I will draw it simply like this; this whole thing core I will not draw I
will simply draw a coil, here you have applied voltage and here you have taken tappings
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suppose these two points. So, these two is your input terminal and these two is your output
terminal. And let us draw a nice picture here, so that we know what is happening.
(Refer Slide Time: 07:05)
So, autotransformer it is like this, this is the coil I will draw this way. And here is your
tappings taken, I will say this terminal as A this terminal is C indicating common between
primary and secondary. And I will say that
𝑁𝐴𝐶 = 𝑁1
and here I will apply voltage of frequency 𝑓.
And this point let me call this point to be B and
𝑁𝐵𝐶 = 𝑁2
Then, voltage available here I will call it 𝑉2.
𝑉2 =
𝑉1
𝑁 = (𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑝𝑒𝑟 𝑇𝑢𝑟𝑛) × 𝑁2
𝑁1 2
and that is it. So, so by changing 𝑁2 , I will be getting output voltage. So, in this
arrangement if B is very low at this point if 𝑁2 is very low you get high voltage to low
voltage very low voltage transformation is possible.
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So, in a two winding transformer this is precisely we do we want to change the level of
voltage can we step down step up the voltage? Here it is step stepping down the voltage
yes you can step up the voltage. For example, this is step down autotransformer step up
autotransformer will look like this what you do? You do you give input voltage here, 𝑉1
this you call your 𝑁1 ok.
And here from you will take the output, voltage per turn will remain same this is 𝑁2 number
of turns. So, available voltage here on the secondary side will be
𝑉2 =
𝑉1
𝑁 = (𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑝𝑒𝑟 𝑇𝑢𝑟𝑛) × 𝑁2
𝑁1 2
but since 𝑁2 > 𝑁1 you will get higher voltage on the secondary side. That is this is the if I
draw it here like this, this is suppose the transformer core it is like this. In this case what I
do single winding no two separate coils; you give supply voltage to this number of turns,
flux will be linking all the turns mind you.
So, this is your A in this case this is your C this is your B and this is your A C and B. So,
𝑁𝐵𝐶 = 𝑁2 turns is more than 𝑁𝐴𝐶 = 𝑁1 and you can also step up the voltage. Therefore, it
looks like one can use a single coil like this instead of two winding transformer, where you
are using two coils therefore, it is natural then that is I now have two option, suppose
somebody says me that I have 200V supply and my load requires 100V.
Then, I will have now two options what are the two options? I will use a two winding
transformer 200V/100V this will be the thing. Or I can also use an auto transformer, where
the number of tappings here will be high half of this full number of turns here so a 200
here, 100 there. So, either of them you can then use to supply your load at 100V when
200V supply is there.
Now the moments two options come in we must examine which one to choose should I
choose a two winding transformer or should I choose a an auto transformer? Here realize
the question. Same thing can be done apparently by both the options ok. So, we will see
examine that thing right now.
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(Refer Slide Time: 13:08)
Therefore comparing I will use different color, comparing a 2 winding transformer, and
an auto transformer. Now the moment you say I will compare two transformers, their
ratings must be same then only you can compare two things at same level.
So, what is the ratings? I demand that your 2 winding transformer should do like this 𝑉1 it
must transform it to a voltage 𝑉2. And you will neglect the no load current that is pretty
small ideal conditions to understand what is the implication of this which one to choose.
So, I want to transform an available voltage 𝑉1 to a level 𝑉2 and I must see that the current
delivered to the load is 𝐼2 that is k VA rating I must also specify, so the transformer will
deliver KVA of 𝑉2 𝐼2 .
So,
𝑉1 𝐼1 = 𝑉2 𝐼2
that is what I want to do got the point. And this is your 2 winding transformer. Now the
same thing I want to do with an auto transformer. So, I will draw the auto transformer like
this, suppose it is a step down case. So, 𝑉1 you apply and you take a tapping here and this
terminal is A this terminal is C and this terminal is B. So, 𝑉1 I will apply I should get here
𝑉2 and your load should get same current 𝐼2 got the point; flux in the core should be same.
So, how the flux in the core should can be same? If this is 𝑁1 turns this is 𝑁2 turns then I
will demand that 𝑁𝐴𝐶 = 𝑁1 hm. And 𝑁𝐵𝐶 = 𝑁2 is it not? This is what I will do. Therefore,
259
when this current is delivered here also the current should be 𝐼1 ; output volt ampere must
match with the input volt ampere that is what is this so this 𝑉1.
Now, the number of turns present here in this portion it is
𝑁𝐴𝐵 = (𝑁1 − 𝑁2 )
These are the number of turns present here between A and B. Mind you, the dots are here
this is dot this is dot.
Now, if you look at this connection very critically you will immediately see that this two
coils this coil and this coil that is coil existing between terminals A and B. And coil B C
there is no common turns present they are separate coils like this. For example, as a whole
in this particular configuration, which one to call primary? You have applied between A
and C, a voltage 𝑉1 perhaps this is primary where the whole number of turns is 𝑁1 .
Where is secondary, between B and C how many turns are there? 𝑁2 , but this 𝑁2 turns also
a member of the primary coil number of turns 𝑁𝐴𝐶 is not this is common to both primary
and secondary this causes problem I mean initially oh what to do then. So, first thing is I
say that coil AB and coil B C are two separate coils totally separate are two separate coils.
And this portion this must be understood that is very important point to be noted. And they
are linking same flux therefore, MMF and voltage ratios will follow the rules of a two
winding transformer what else, flux is common same flux is not? Previous diagram is the
core flux is common to both the parts. So, same flux and I have identified coil A B and
coil B C ok, if you connect some load on the secondary primary you apply voltage this
coil is supposed to carry some current coil B C and coil AB is also supposed to carry some
current.
But I am sure about one thing these two coils will carry current and they have to follow
the rule of a two winding transformer what else. In other words what I am telling if this is
𝐼1 , suppose current is 𝐼1 current delivered to the load here load is connected delivered is 𝐼2
then apply KCL at this point, this current I will write 𝐼2 − 𝐼1 is not?
(𝐼2 − 𝐼1 ) + 𝐼1 = 𝐼2
KCL I have applied this is the situation.
260
In a two winding transformer remember this two are dots here, if current delivered is 𝐼2
and this is having 𝑁2 turns and if your primary is 𝑁1 . Then what I told you that if it is an
ideal transformer
𝐼1 = 𝐼2′ =
𝐼2
𝑎
MMF must be balanced whenever you show current coming out from the dot, then only
transformer primary will invite current from the source through the dot terminal such that
𝑁1 𝐼1 = 𝑁2 𝐼2
Ideal transformer no load current you neglect that is only 2% to 5% forget about that and
this must prevail. In this transformer also the moment you have applied a certain voltage
𝑉1 at certain frequency 𝑓 across the turns 𝑁𝐴𝐶 then the flux level gets fixed. In the core
KVL is to be satisfied here the same arguments therefore, for any reason if this two parts
of the windings carry current their MMFs must cancel out, so that the net MMF remains
because of that 𝑁1 multiplied by the small magnetizing current, which is negligibly small.
Therefore the moment I do this I will say that look here this is (𝐼2 − 𝐼1 ) coming out from
the dot this is the direction very important. So, I will write that MMF balance from MMF
balance I will write
(𝑁1 − 𝑁2 )𝐼1 = 𝑁2 (𝐼2 − 𝐼1 )
This must prevail no way, that is the moment you are trying to draw a current 𝐼2 in the load
some current 𝐼1 will follow will flow in the primary coil through A and if this is 𝐼2 and this
is 𝐼1 this current is I can correctly show as (𝐼2 − 𝐼1 ) phasors these are from bottom to top.
And these two MMFs must cancel them out so that this will remain there.
Now, in this case
(𝑁1 − 𝑁2 )𝐼1 = 𝑁2 (𝐼2 − 𝐼1 )
𝑁1 𝐼1 = 𝑁2 𝐼2
We have seen same ratings. See, I will compare a 2 winding transformer with an auto
transformer with same ratings once again what do I mean by same ratings? I will change
261
the voltage level from 𝑉1 to 𝑉2 total number of turns 𝑁1 here secondary turns 𝑁2 here. And
I will delivered if current 𝐼2 to the load here also I will deliver a current to the load 𝐼2 . But
incidentally this current deliver to be load is not the current flowing through this portion
of the coil that is different issue. Overall, as a black box I must be able to deliver 𝐼2 current
as I am doing it here.
So, 𝑁1 and 𝑁2 for both this transformers the number are same because flux level is same
and 𝑁1 𝐼1 = 𝑁2 𝐼2 . But in the case of an auto transformer I get this relation or I will say that
(𝑁1 − 𝑁2 )𝐼1 = 𝑁2 (𝐼2 − 𝐼1 ). So, this term 𝑁2 𝐼1 will cancel out once again giving you this
relationship same as this fellow. But the reason that this will be true I must get it by
balancing the MMFs of the two separate portions of this coil that I have identified as AB
and BC. Then only I can invoke all the things I did for 2 winding I can do it here also
thinking that these two coils are separate coils common flux induced voltage there induced
voltage there and so on.
So, this is the relationship which will hold good ok. What is the KVA? KVA rating is 𝑉2 𝐼2
or same as 𝑉1 𝐼1 same KVA we must compare it that is there. Now after getting this, the
interesting thing is that if you look at this portion of the winding BC whatever current will
flow is not equal to 𝐼2 current secondary current it is not it is in fact, the difference of these
two currents. Unlike this thing in the secondary of these two winding transformer whatever
is the load current 𝐼2 that flows through the winding all along primary of course, 𝐼1 current.
But here it is not like that current in the load is 𝐼2 , but current in the portion BC is the
difference of (𝐼2 − 𝐼1 ). Therefore, it looks like the sectional area of this portion of the coil
copper section required will be less compared to the copper sectional area here because
more current 𝐼2 mind you 𝐼1 and 𝐼2 all will be in phase. Therefore, their difference will be
real difference when you take the even the phasor difference because 𝐼1 and 𝐼2 will be in
phase.
Therefore the difference of current only flows, suppose 𝐼2 is the rated current 𝐼1 will be
rated current no doubt this will be high whatever it is. But in this portion magnitude of the
current will be quite small. And therefore, the sectional area of copper whatever copper
you are using for 𝑁2 number of turns it will be much thinner compared to the sectional
area of this portion. That is what I am telling is, if I draw it like this suppose I use a bigger
this one.
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Suppose this is the copper wire I am using for this portion AB I am sorry thicker wire it is
like this. And for this portion, that is this is your A and this point is B. Now for this portion
BC you use a thinner wire getting thin wire this is crucial to understand, this should we
thick unlike a 2 winding transformer 𝐼2 whatever load current you are delivering that we
will decide what it is if it is well beside this sectional area will be more that is all there.
But here in this portion the sectional area of the upper portion will be same as the sectional
area of this 2 winding transformer here.
But for this portion I can make it thinner, so it looks like this point is C thin wire this will
be thick wire. Therefore, it looks like I will be transmitting a voltage 𝑉1 to a voltage 𝑉2
and it will deliver a current 𝐼2 no doubt it will deliver same volt ampere it is handling. But
it looks like you will be able to save some copper volume I will continue with this in the
next lecture.
Thank you.
263
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 28
Auto Transformer Versus Two Winding Transformer ( Contd.)
(Refer Slide Time: 00:23)
Welcome to lecture number 28. Ad and we were discussing about Auto Transformer ok.
Remember in our last class, so with the help of an autotransformer it looks like I can
change the voltage from one level to some desired level of voltage as we have done with
the help of it two winding transformer and I could transmit a certain amount of kv here.
Therefore, I now have two options. Suppose, I want to change the voltage level from 𝑉1 to
𝑉2 and it will deliver a KVA of 𝑉2 𝐼2 or the transformer will handle a KVA of 𝑉1 𝐼1 = 𝑉2 𝐼2 ,
they are same. Same thing can be done with an auto transformer.
The moment you have two options, therefore which one to choose if I want to change the
level of voltage from one value to another, whether should I go for a for whether should I
select it two winding transformer to do this or an auto transformer to do this? So, after
looking at these things, what I have assumed, they are same kva rating 𝑁1 turns are same,
so that flux level in the core will be same, so they are that way similar.
264
And then we thought, but here you note the current supplied to the load will be 𝐼2 , because
what happens in an auto transformer there is a common portion of the winding and this is
dot, this is also dot here. Therefore, this current should be 𝐼2 . So, we must bring them at
this same platform before comparing, so same KVA rating, same voltage ratios and so on.
But looking carefully here we note that if this current is 𝐼2 and this current is 𝐼1 that is what
it should do like this, then current in the common portion BC is the difference of this two
current. And the currents as you know are in phase ideal transformer I am considering, no
load current let us neglect to understand the major thing in an auto transformer, how it
distinguishes itself from a two winding transformer is that ok, 𝐼2 current then 𝐼1 current
must be drawn in and therefore this is (𝐼2 − 𝐼1 ).
Then I know that here once again I have applied voltage 𝑉1 at some frequency, therefore
flux level in the core will remain same. And then we argued that this portion of the winding
AB and BC this two coils are two separate coils, no common part in between them.
Therefore mmf produced by this coil and this coil must balance, through the dot if (𝐼2 − 𝐼1 )
is coming out and through the dot if 𝐼1 is inputed in the upper portion of the coil, this two
mmf must balance ok. And that thing from this one also it is interesting to note we get that
𝑁1 𝐼1 = 𝑁2 𝐼2 that is there.
Then I argued without doing any mathematics perhaps this current will be much less
difference of these two current (𝐼2 − 𝐼1 ) going up and therefore, a thinner coil can be used
compared to its two winding counterpart. And therefore, you can save some copper that is
what I told. Now, today, we will try to find out some estimate that how much copper will
be saved, copper segments, volume of the copper ok. So, we start with this.
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(Refer Slide Time: 05:09)
So, I draw the transformer once again. So, this is the auto transformer which step downs
the voltage 𝑉1 AC voltage. This point I call A, this point is C and here is the tapping B,
and here is the load connected get to secondary load. And this was the current distribution
𝐼2 , this was 𝐼1 and then this current was (𝐼2 − 𝐼1 ). This number of turns 𝑁𝐴𝐶 = 𝑁1 this total
number of turns 𝑁𝐵𝐶 = 𝑁2 turns.
And from these two, we get 𝑁𝐴𝐵 = (𝑁1 − 𝑁2 ). And I told you that these are the dots here.
Therefore, if (𝐼2 − 𝐼1 ) leaves there, 𝐼1 comes in, so mmf balance. This is what we did last
time mmf balance. Gives me (𝑁1 − 𝑁2 )𝐼1 = 𝑁2 (𝐼2 − 𝐼1 ), these are all phasors you know.
So, 𝑁1 𝐼1 = 𝑁2 𝐼2 . This is the thing.
At a volume of copper required for a coil, volume of copper required for a coil, required
for a coil will be simply the length of wire, length of wire into its cross sectional area, wire
cross sectional area, is it not, very pretty simple, area into length gives you volume. Now,
this length of wire; length of wire is proportional to the number of turns of the coil; number
of turns of the coil, is it not? This is the core each turn. So, for a given core size, this thing
number of turns would be proportional to length because each turn has got a fixed
perimeter, therefore, number of turns of the coil it will be equal to 1.
266
(Refer Slide Time: 08:56)
So, now what I will be doing is this, if this is the case then let me go to the next slide. I
will draw the two-winding transformer of same rating 𝑉1, here it is 𝐼1 , here it is 𝑉2, here
the load current will be 𝐼2 , this is the thing. And here it is your auto transformer, where
this is the load same load 𝑍2 , same load 𝑍2 here. So, that same current flows 𝐼2 . And here
this point is A, C, B, let me repeat I mean 𝑁𝐴𝐶 = 𝑁1 , 𝑁𝐵𝐶 = 𝑁2 and 𝑁𝐴𝐵 = (𝑁1 − 𝑁2 ) that is
there. And this is 𝐼2 , this is 𝐼1 and this is (𝐼2 − 𝐼1 ).
Now, what I will be doing is this, I will find try to find out these ratios. Here it is 𝑁1 here
𝑉
𝑉
1
1
it is 𝑁2 , because voltage per turn is 𝑁1 and output voltage will be 𝑁1 𝑁2 . Here voltage per
𝑉
𝑉
1
1
turn is once again 𝑁1 , and this turn 𝑁𝐵𝐶 = 𝑁2 . So, 𝑉2 = 𝑁1 𝑁2 , got the point. Now, what I
will do, so this two transformers are of same rating. If this two are put in a black box only
two terminals are available, two terminals are available, you will not be able to distinguish
whether a two-winding transformer is doing the job or an auto transformer is doing the
job, so far as the output terminals are concerned ok.
Now, what I will do is this, I will find out the ratio of these two quantities copper required
in auto transformer. I am writing it short divided by total copper required in two winding
transformer, this ratio I will try to find out ok. In the previous slide, I have told you length
of the wire is proportional to the number of turns of the coil. What about cross sectional
area?
267
Cross sectional area, let me write here, cross sectional area of the wire cross section which
is forming the coil cross sectional area of the wire will be proportional to the magnitude
of the current because magnitude of the current based on that only thicker or thinner wires
are chosen, is it not.
So, cross sectional area of the wire is proportional to the current it is carrying based on
that I have decided upon the cross sectional area. So, cross sectional area is proportional
to the current. And we are talking about rated current. At rated current, based on that the
cross sectional areas are chosen.
(Refer Slide Time: 12:53)
So, it looks like copper required in autotransformer rather
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐶𝑜𝑝𝑝𝑒𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑖𝑛 𝐴𝑢𝑡𝑜𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐶𝑜𝑝𝑝𝑒𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑖𝑛 2 − 𝑊𝑖𝑛𝑑𝑖𝑛𝑔 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟
I am trying to finding. Now, what is the number of turns, length of these wire will be
proportional to 𝑁𝐵𝐶 , is it not and 𝑁𝐵𝐶 = 𝑁2 into the cross sectional area proportional to the
current.
What current this coil is carrying, magnitude of the current (𝐼2 − 𝐼1 ), it is this portion plus
this portion plus what is the length of the copper required in portion 𝑁𝐴𝐵 , number of turns
existing between A and B terminals that is 𝑁𝐴𝐵 = (𝑁1 − 𝑁2 ). And cross sectional area of
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this portion should be decided by the amount of current it is carrying that is the 𝐼1 , all these
currents are rated values corresponding to that. So, this will be this thing.
Now, because these two transformers are doing the same job come here. Now, here I will
write volume of copper required for this portion of the winding will be 𝑁2 𝐼2 , volume of
copper required for this portion is 𝑁1 𝐼1 , that is all.
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐶𝑜𝑝𝑝𝑒𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑖𝑛 𝐴𝑢𝑡𝑜𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐶𝑜𝑝𝑝𝑒𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑖𝑛 2 − 𝑊𝑖𝑛𝑑𝑖𝑛𝑔 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟
=
(𝑁1 − 𝑁2 )𝐼1 + 𝑁2 (𝐼2 − 𝐼1 )
𝑁2 𝐼2 + 𝑁1 𝐼1
Now, we have already established that 𝑁2 𝐼2 = 𝑁1 𝐼1. Is 𝑁1 and 𝑁2 different in the
autotransformers? No, same, same, same thing, 𝑁1 , whatever is 𝑁1 here. 𝐼1 and 𝐼2 is
different? No, 𝐼2 is the current delivered to the load; 𝐼1 is the current flowing in the primary
same thing 𝐼1 and 𝐼2 . So, those things are there.
But we have shown in case of a two winding transfer these already known. In case of auto
transformer also we have shown that it so happen that 𝑁2 𝐼2 = 𝑁1 𝐼1. Although I should not
say that 𝑁1 𝐼1 is the mmf like that which portion I do not know, but I found out very
meticulously this relationship is not from this (𝑁1 − 𝑁2 )𝐼1 = 𝑁2 (𝐼2 − 𝐼1 ) from that also
something was obtained.
Therefore, it looks like 𝑁2 𝐼2 = 𝑁1 𝐼1 is also true ok. Then and that constant of
proportionality will be same that will cancel out, this proportional mind you and I am
taking the ratios. Then what you do is this
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐶𝑜𝑝𝑝𝑒𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑖𝑛 𝐴𝑢𝑡𝑜𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐶𝑜𝑝𝑝𝑒𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑖𝑛 2 − 𝑊𝑖𝑛𝑑𝑖𝑛𝑔 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟
=
(𝑁1 − 𝑁2 )𝐼1 + 𝑁2 (𝐼2 − 𝐼1 ) 𝑁2 𝐼2 − 𝑁2 𝐼1 + 𝑁1 𝐼1 − 𝑁2 𝐼1
=
𝑁2 𝐼2 + 𝑁1 𝐼1
𝑁1 𝐼1 + 𝑁2 𝐼2
Just open the brackets. But 𝑁1 𝐼1 = 𝑁2 𝐼2 from both this. So, these two can be combined and
simply written as
269
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐶𝑜𝑝𝑝𝑒𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑖𝑛 𝐴𝑢𝑡𝑜𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐶𝑜𝑝𝑝𝑒𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑖𝑛 2 − 𝑊𝑖𝑛𝑑𝑖𝑛𝑔 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟
=
(𝑁1 − 𝑁2 )𝐼1 + 𝑁2 (𝐼2 − 𝐼1 ) 𝑁2 𝐼2 − 𝑁2 𝐼1 + 𝑁1 𝐼1 − 𝑁2 𝐼1
=
𝑁2 𝐼2 + 𝑁1 𝐼1
𝑁1 𝐼1 + 𝑁2 𝐼2
=
2𝑁1 𝐼1 − 2𝑁2 𝐼1 𝑁1 − 𝑁2
𝑁2
1
=
= 1−
=1−
2𝑁1 𝐼1
𝑁1
𝑁1
𝑎
𝑎=
𝑁1
𝑁2
So, copper required in auto transformer will be this times the copper required in it two
winding transformer. If this fraction is less than 1. Then there will be savings of copper.
And mind you I am considering the step down case.
1
So, value of 𝑎; value of a for step down case 𝑁1 > 𝑁2 , which means that 𝑎 > 1 or 𝑎 < 1
1
is not this will be a number which is less than 1 fraction. So, suppose I say that 𝑎 is that
𝑎
𝑁
𝑁
2
2
is 𝑁1 suppose, suppose I say that 𝑁1 = 2 = 𝑎, then copper required in autotransformer will
be equal to half of copper required in two winding transformer that is 0.5 of copper
required in two winding transformer.
So, if you require say 10 kg of copper to make it two-winding transformer, you will be
requiring only 5 kg of copper to make a auto transformer is it not. You save copper it looks
like savings of copper can be easily calculated by taking the difference of these two I am
not going to do that it can be easily done. But I will just see how much copper is required
in terms of how much copper is required in a two-winding transformer, I find for step
down case if turns ratio is 2, it is
𝑁1
𝑁2
= 2 you are stepping down the voltage. Here also you
are stepping down the voltage.
Then I find oh my god you will get advantage here because if you require 10 kg copper,
there you will be requiring half of that copper required 5 kg copper. So, weight of the
transformer will be less initial investment will be less and so on. And this auto transformer
will do the same job as this two winding transformer is doing. And it looks like then that
1
no matter what is the ratio if it is greater than 1, 𝑎 is always to be a number less than 1.
And you will be always saving copper, getting?
270
4
For example, I want to suppose I say that 𝑎 = 3, then what will be the savings of copper
1−
1
1
3
= 1−
= 1 − = 1 − 0.75 = 0.25
4
𝑎
(4⁄3)
Only one-fourth of the copper will be required. See if you want to change the voltage
ratios, in this ratio
𝑁1
𝑁2
4
= 3, then I find savings of copper is much more, because copper
1
required will be only 4, it looks like that is it depends on 𝑎, this is the crucial equation.
But in any case no matter what is the value of 𝑎, if you are stepping it down, it looks like
it will be saving copper. Savings of copper will be more if 𝑎 → 1. If you decrease the value
4
of 𝑎, savings nonetheless will take place, but at not at high rate as it was doing for 𝑎 = 3.
So, we understand now that in an autotransformer savings of copper will always take place.
Whether it is more or less that depends on the ratio. If 𝑎 → 1, savings of copper will be
more and more; lesser and lesser copper will be required.
So, if that be the case, then we understand then wherever you require a transformation of
voltage, two candidates are there you can choose a two-winding transformer or an auto
transformer, it then looks like always go for auto transformer, because less copper will be
required. Whether that less is really less or not that is secondary issue, but mathematically
we note that savings will always take less, no matter what is the value of 𝑎. What is the
value of 𝑎? 𝑎 > 1 here, is it not?
Therefore, it looks like savings of copper will be there in an auto transformer of same
rating compared to a two winding, same KVA it will transmit, same voltage ratios.
Therefore, this exercise then prompts me look whenever you want to use a transformer,
out of these two always go for auto transformer because savings are there that is what we
get. But the story is not completed, if that would have been the case, then two winding
transformer would not have existed is it not, no one would have used two winding
transformer, but that is not the case. There are other issues we have to consider.
271
(Refer Slide Time: 27:15)
For example, I will tell you that suppose you have an two-winding transformer, 𝑉1, 𝑉2 you
are getting. And a similar rated auto transformer where 𝑉1 and 𝑉2 you are getting that is
fine. A two winding transformer, suppose let us consider a situation, it is supplying load
ok, it is supplying load, connected rated current it is supplying that is what I would like to
operate the transformer because I have purchased a transformer of given KVA let it
handled given KVA why operate it under no load.
Of course, there are situations where things are not under your control like a distribution
transformer, load varies we have talked about that. But here let us see suppose they are
delivering rated current primary current is also drawn rated, here also doing the same thing,
here it is 𝑉1 like that. And here the supplying load everything is fine, same job is being
done two winding and auto transformer.
Let us imagine that an open circuit fault has taken place open circuit fault in the primary.
Primary or secondary, say on the secondary, open circuit fault in secondary coil. What has
happened is this, you do not know a fault has occurred which creates an open circuit here,
excessive current was flowing because of some reason and winding has become opened
and which you cannot see from outside. Primary remains energized. Will I get any voltage
between these two points? No, because the moment it becomes open circuit current will
vanish current cannot flow. But there will be voltage induced in this portion and in this
portion, but voltage between these two points will be zero as there is no current flowing.
272
Therefore, what will happen if a open circuit takes place here, your load will suffer, it will
not get any voltage now, no current flows through the load that is everything on the
secondary side will not get power. There this story ends. Let us come here suppose it is
doing like this, but what happens is this a open circuit fault takes place in the common
portion of the winding that is BC portion. Suppose, open circuit fault in the common
portion of the winding that is BC, a open circuit has taken place.
What happens to this current this time we find that current can flow 𝑉1 is there through
this current will flow which was not possible here, and not only that this voltage is no
𝑁
longer it is 𝑁2. This voltage minus this drop. And if this point is closer to this point this
1
voltage will become of same level as that of this one is it not? Therefore, across the load
perhaps the voltage will go up which is not in your control via this connection.
And there will be problem on the load if you apply high voltage. Just because of what,
because some open circuit fault has taken place which is not in your hand, it may happen
in a two-winding transformer as well as in an auto transformer. Then what happens by this
ohmic contact in this path, there may be higher voltage coming of the order of 𝑉1 may
creep across may come across 𝑍2 . And if it is a critical load your load goes which requires
a more or less constant voltage of level 𝑉2, suddenly we will see oh 𝑉2 has gone up. Trying
to go up to 𝑉1 via this path and things will be, and this problem is called isolation.
So, an auto transformer does not provide isolation here if such a thing happens it is isolated
from the primary. So, isolation to this is most biggest point two-winding transformer
provides isolation. Here no isolation. In case a open circuit fault takes place in the common
portion of the primary and secondary, therefore autotransformer gives you copper saving
no doubt, but at the same time it will never be able to give you a proper isolation across
the source side between the source side and the load side, load may be critical you cannot
just apply high voltages.
We will continue with this next class. And this is very important and interesting I mean
how to choose auto transform, we will continue it is not yet concluded.
Thank you so much.
273
Electrical Machines-I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 29
Numerical Problems on Ideal Auto Transformer
(Refer Slide Time: 00:24)
Welcome to lecture number 29 on Electrical Machines I and we have been discussing Auto
Transformer. In fact, we have been comparing auto transformer versus two winding
transformer, two winding transformer.
274
(Refer Slide Time: 00:50)
Recall in the last class I told, because if you want to change a transfer a certain amount of
KVA from primary source site to the load side, at two fixed voltage levels 𝑉1 and 𝑉2.
(Refer Slide Time: 01:09)
You can do that by using either a two winding transformer or also by an auto transformer.
Now therefore, now we will find that there are two options I can do the same thing change
the level of voltage from 𝑉1 to 𝑉2 and transmit the same KVA.
275
Now, which one to choose? Two options are there it looks like either of them can be done,
but I told you that if you use an auto transformer then copper required will be less
compared to the copper required in a two winding transformer always less. So, savings of
copper will be can be easily found out subtracting these two. But, the point is that this ratio
1
1
𝑁
is (1 − 𝑎) copper required in an auto transformer is (1 − 𝑎) where, 𝑎 = 𝑁1 . And, we also
2
showed with some numerical examples, that as the turns ratio is close to 1 savings of
copper will be more and more, but nonetheless no matter what is this ratio either large or
small there will be always savings of copper.
Then we argued that if that be the case then it should be all the autotransformer all along
the winding transformer should not be used, what is the fun? Because you will be always
saving copper to transmit same KVA from this source to the load side at two fix different
voltages, same job can be done by an auto transformer. But, there were further issues then
I told you that look a two winding transformer is a better option, if you require isolation,
some critical load is present, if there is an open circuit on the secondary of the two winding
transformer ok. No problem at most the load will not get any supply there ends the matter,
but that is not to be the case if an open circuit fault occurs in the 𝑁2 turns that is common
portion of this windings of an auto transformer.
Even if it becomes open, primary energized this high voltage may appear across these of
course, these voltage minus whatever is the drop here that will appear across the load.
Therefore, on the load side there may see a sudden rise of voltage if such a fault that is
open circuit in the secondary that is in the BC part of the winding takes place and auto
transformer does not provide isolation. Therefore, where isolation is a must no, two
winding transformer should be preferred I mean, autotransformer should not be used the
loads may be critical. That is why two winding transformer is also there, because it
provides that isolation that is what I told in my last class.
Now, the question is then when to use auto transformer? Auto transformers can be used
see the I write this thing
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐶𝑜𝑝𝑝𝑒𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑖𝑛 𝐴𝑢𝑡𝑜𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟
1
= 1−
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐶𝑜𝑝𝑝𝑒𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑖𝑛 2 − 𝑊𝑖𝑛𝑑𝑖𝑛𝑔 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟
𝑎
And, we have observed that savings will be more, savings of copper will be more if 𝑎 ≈ 1.
276
In fact, if 𝑎 = 1 and this mathematical expression gives in you autotransformer does not
require any copper, but I will use an autotransformer. It simply means, that no transformer
is necessary then supply voltage the same voltage in one, but there may be situations when
for example, you want to step down the voltage from 1000V to say 900V, here this ratio
is close to 1, is not 1000 to 900.
In this type of transformer we find savings of copper will be more that is what we
4
established here in our last class, Turn’s ratio close to 3 copper required is only 25%,
Turn’s ratio is 2 and savings of copper is only 50%. So, closer you go to 𝑎 = 1 savings of
copper will be more and more. If that be the case, then I will say better go for an auto
transformer that is when this voltage 𝑉1 and this voltage 𝑉2 their orders are nearly same
transformation ratio for example, 1000V to 900V then savings of copper will be quite
large.
Because for this portion you require thinner, where (𝐼2 − 𝐼1 ) and for this you recall this is
𝐼1 this is 𝐼2 and this is (𝐼2 − 𝐼1 ) savings of copper will be much more. But, then the problem
of isolation still remains, the problem of isolation you cannot avoid that is there. But, look
if this is of the order of 1000V and your load requires 900V and now, an open circuit takes
place here in the common fault of the winding then the voltage which will be coming
across the load will certainly not be 900V; it will shoot up by a this path, supply will come
here by a this path. But that voltage rise will be not too high, because it is already these
two are of the same order we must understand that.
So, not only the savings of copper will be more, but also isolation is not a big issue in the
sense, that when this voltage and this voltage are at comparable level same level then
voltage will shoot up from 900V to at best maximum to 1000V only, what percentage of
increase it is 10% is it. Therefore, but imagine I am using an autotransformer 1000V to
say 100V, I want to get 100V here and I have decided I will use an auto transformer. Now,
first of all in this case this ratio being too high savings of copper will be there compared
to a two winding transformer, but savings of copper compared to this configuration will
be much less.
1
Because, 𝑎 = 10 so, 1 − 10 = 0.9 only. So, copper required in auto transformer will be
90% of what? Of the copper required for making it two winding transformer, but
nonetheless is having little saving is there. But, here the issue is if there is a open circuit
277
here then you see the load side gets connected to this high voltage side. It will not shoot
up to 1000V, but it may shoot up to 600V or 700V. Why not 1000V? That will be a drop
here in this case.
But nonetheless so far as the load on this side is concerned it will fill that rising voltage or
the personnel whoever is working here on this side they might get shock, if there is a open
circuit takes place here and no isolation is provided by auto transformer. I think you have
understood this point therefore, for the safety of the personnel who expects always here
100V he will suddenly see it has gone up to 800V, this is dangerous and also it may be
dangerous to the load, load may not cannot accept that large voltage.
But in this case 900V, load working whoever is working here, 900V it will go up to he
must have taken enough precaution to work at this voltage length, but voltage you will
only shoot up a little maybe by 10% or 5% like that if there is open circuit occurs. So,
isolation problem when 𝑎 → 1 isolation is not that urgent; is not that urgent this is one and
second thing is savings of copper is substantial; savings of copper substantial. Whereas, in
this case in this ratio I will say isolation is urgent, isolation must be provided, under what
condition? Open circuit in the common and secondly, savings of copper is also not very
high.
Savings of copper is little, savings of copper with respect to whom? With respect to two
winding transformer therefore, for such a large transformation ratio of voltages you better
go for a two winding transformer, where isolation will be there I think you have understood
this point ok. So, this is the thing. Now, before I proceed further that is see to understand
what is going on in an auto transformer. So, far I have taken an ideal case that is
transformers are having no leakage impedance this that and so on, ok.
278
(Refer Slide Time: 15:14)
I will just point out, one thing that that suppose yeah in the form of a problem so, that you
understand what I am trying to tell. Suppose you have an auto transformer ok, here is and
it is stepping down, it could also step up. See I have done all the derivation assuming it to
be stepping down the voltages. Now, the same sort of equation I can write and come to the
same conclusion even if it is step up, when the ratio will be I better do not do it things like
that ok, that I leave it as an exercise.
Suppose, I say that derive whatever things I have done if I apply a voltage here and
secondary voltage is 𝑉2 whether steel copper savings; why not copper savings will be there
anyway primary and secondary can be interchanged that way you can always anyway that
is the thing ok. So, you know it is there. So, this is the thing now suppose, let me solve one
problem ideal autotransformer, how you should be handling this problem ideal auto
transformer.
An example, numerical problems; so, winding impedances are neglected core losses are
not there, we will discuss this little later. Suppose I say it is 400V I will apply here, a
problem is given it is 400V and suppose it is 200V I say. I tell that the transformation ratio
is such that 400V to 200V and I say that you have connected an impedance here of value
say 𝑍2 = 10∠30°Ω. I ask you to find out calculate currents in all the branches of the
circuit, a very straight forward problem but, you should do it like this.
279
One way of doing this because this is 200, this is 400 given otherwise number of turns
could have been given suppose, I say that this is C 𝑁𝐴𝐶 = 100. Then immediately I will
400
calculate volt per turn which will be equal to 100 = 4𝑉. And, suppose I say that 𝑁𝐵𝐶 = 50,
then I will say this voltage 𝑉𝐵𝐶 = 4 × 50 = 200𝑉 is not, that is how I got the voltages.
Now, I have been asked to calculate currents in all the parts of the auto transformer and
mind you this is dot, this is dot here. First thing what I can see, that I can calculate this
current 𝐼2 . So, 𝐼2 will be all voltages will be in phase, because same flux links all the
voltages and there is no distinction between 𝑉2 and 𝐸2 , no distinction between 𝑉1 and 𝐸1
they are one and the same and all the voltages are in time phase. Why? Because common
flux all these part of these coils same flux is crossing, only thing number of turns are
different. Therefore, 𝐼2 will be taking this voltage in reference 𝐼2 I will calculate simply
like this.
𝐼2 =
200∠0°
= 20∠ − 30°𝐴
10∠30°
Now I will argue you as I did in case of a two winding ideal transformer that the this is 𝐼2
and this is so, 𝑉2 first I write 𝑉2 = 200∠0°V, 𝐼2 is this. Then output power output complex
power is equal to 𝑉2 𝐼2∗ that is what we will be doing. So, this will be
𝑉2 𝐼2∗ = (200∠0°) × (20∠ + 30°) = 4000∠30°
That is the output complex power. And this must come in from this input, because there is
only one source of power where I am feeding supply 400V AC, rms value is 400V.
Therefore, if I assume this current to be 𝐼1 which is still now unknown I will say and mind
you 𝑉2 is this therefore, 𝑉1 = 400∠0°, but I do not know 𝐼1 . So, what I will do is I will say
that
𝑉1 𝐼1∗ = 𝑉2 𝐼2∗
Whatever you dump on the load must be drawn in from this source side this is source, it
must be like that. Therefore,
𝐼1∗ =
𝑉2 𝐼2∗ 4000∠30°
=
= 10∠30°
𝑉1
400∠0°
280
But mind you this is 𝐼1∗ therefore,
𝐼1 = 10∠ − 30°
Therefore, balancing the power on the load side and that is the power drawn from the
source, you can avoid several physical things what is going on that is I know 𝐼1 now. The
moment I know 𝐼1 then I can calculate this current, this current better show in this way,
this current then will be (𝐼2 − 𝐼1 ) that is how we are specifying. 𝐼2 is known 𝐼1 is known.
Therefore, current in this portion in this direction it must can be calculated.
So,
(𝐼2 − 𝐼1 ) = (20∠ − 30°) − (10∠ − 30°) = 10∠ − 30°
So, the current distribution of various parts of this circuit can be easily calculated, provided
of course, the transformer is ideal, winding resistances not there, leakage reactants are not
there, such that 𝑉1 = 𝐸1, 𝑉2 = 𝐸2 and so on. Another example I will ask you to do is this
one.
(Refer Slide Time: 24:59)
Suppose you have same problem sort of thing, suppose here I have applied 400V 50Hz
and once again, I say that 𝑁𝐴𝐶 = 100, then immediately I will calculate volt per turn is
400
100
= 4. Now, what I am doing is I am taking a tapping from point B, such that 𝑁𝐵𝐶 = 75.
Therefore, I will immediately calculate 𝑉𝐵𝐶 voltage available between these two. So,
281
𝑉𝐵𝐶 = 4 × 75 = 300𝑉
I will be getting and suppose I have two loads. So, one requires 300V supply another
requires 200V supply.
So, from the same sort of auto transformer suppose, you have another tapping say at point
D such that 𝑁𝐷𝐶 = 50. Then therefore,
𝑉𝐷𝐶 = 4 × 50 = 200𝑉
And, mind you all these voltages are in phase.
𝑉𝐵𝐶 = 4 × 75 = 300∠0°𝑉
𝑉𝐷𝐶 = 4 × 50 = 200∠0°𝑉
Your supply voltage 𝑉𝐴𝐶 is the supply, that is also
𝑉𝐴𝐶 = 400∠0°𝑉
Now, I say that I will connect here load. So, here you will get 200V and here you will get
300V. And suppose, 200V I terminate it on a load whose impedance is say 10∠30°Ω like
the previous one.
And between these two points I supply another load where you are getting this and
suppose, I connect here an impedance of 30∠ − 30° like this that is from the same auto
transformer by two tappings at B and D, C, being the common terminals I will supply two
loads which requires two different voltage levels. Then the question is this: Show the
current distribution in various parts of the coils and this way. One way of doing this is
going by physical argument trying to balance the MMF, but there is a shortcut method,
what is that? This is the only source of input power and power are consumed only here
and there transformer being ideal so, these two must match. So, this is your 𝑉1 supply
voltage.
Therefore I will first say that let this be 𝐼2 , let this be 𝐼3 and let this be 𝐼1 . So, I will say
that
𝑉1 𝐼1∗ = 𝑉2 𝐼2∗ + 𝑉3 𝐼3∗
282
𝑉1 𝐼1∗ total complex power absorbed by through the terminal AC must be equal to suppose,
this I say as 𝑉2 must be equal to 𝑉2 𝐼2∗ plus suppose, I call it as 𝑉3 , 𝑉3 𝐼3∗ this is the thing.
And, 𝐼2 and 𝐼3 are known.
𝐼2 =
300∠0°
= 10∠30°
30∠ − 30°
𝐼3 =
200∠0°
= 20∠ − 30°
10∠30°
Of course,
𝐼2∗ = 10∠ − 30°
𝐼3∗ = 20∠30°
Everything is known including 𝑉1. Then you calculate 𝐼1∗ which I am not going to calculate
put the values, all are phasors
𝐼1∗ =
𝑉2 𝐼2∗ + 𝑉3 𝐼3∗
𝑉1
Of course, after getting this
𝐼1 = (𝐼1∗ )∗
Do not forget to take this complex along with it.
Anyway this is how no matter how many tappings you have taken, what different kinds of
load you have connected very easily it can be done. And, after I get this 𝐼1 , 𝐼2 and 𝐼3 all
the things, what I will do? I will go on applying this KCL here at B and also do not forget
about these this way or that way it does not matter. So, far you are correctly writing the
KCL, you can show this current to be (𝐼1 − 𝐼2 ), 𝐼1 coming 𝐼2 going (𝐼1 − 𝐼2 ); here,
(𝐼1 − 𝐼2 ) coming then another 𝐼3 is going (𝐼1 − 𝐼2 − 𝐼3 ).
Interesting thing you will observe the MMF of this portion, whatever turns may be MMF
these are the isolated coils, this portion, this portion, this portion. So, some of this MMF
will vanish to 0, it will be automatically taken care of. Anyway, we will continue with this
next class.
283
Thank you.
284
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 30
Two Winding Transformer Connected as Auto Transformer
Welcome to 30th lecture. And we are still in Auto Transformers. We have compared an
auto transformer with two winding transformer, then I solved some numerical problems in
last class that auto transformer, several voltage levels also you can get and you can find
out the currents in various parts of the auto transformer that I did.
(Refer Slide Time: 00:42)
Today, first I will tell you that a two winding transformer I am having two winding
transformer.
285
(Refer Slide Time: 01:00)
In short form w d g I am writing, two winding transformer as an auto transformer.
Remember any transformer a two winding transformer, it is like this. And I was drawing
the primary winding this way, they are separate coils, is it not, this is the thing. And we
know from this windings sense, these are the polarity. And suppose for easy calculation, I
will assume that this transformer is rating of the transformer, voltage rating of the
transformer is say 600V/200V suppose. And KVA is say 12 KVA, suppose this is a single
phase two winding transformer, understood, this is the thing.
What I told you the moment this data is with you, you should immediately calculate what
12000
is the rated current of this side which will be 600 = 20𝐴. And this side low voltage side,
the current will be more. How much?
12000
That is I calculate it, you must be, 200 = 60𝐴, is it not. This I should be ready the moment
KVA ratings and voltage ratings are known and it is a two winding transformer ok.
Now, after having this suppose somebody comes and this is what does this mean, if you
apply 600 V, you will get here 200 V roughly. Now, somebody comes and tell that ok,
what has happened in the lab 600V supply is available voltage, but you have a peculiar
load which requires 800 V, load requirement is 800 V, in the lab 600 V supply is there
which can be of course, transformed to 200 V to supply 200 V load. But suppose I say that
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no, it has so happened that you require 800 V, is it possible to use this two winding
transformer whose input voltage will be 600 V, but output voltage will be 200 V?
The answer to this question, yes, it can be used provided you suitably connect this to coils.
For example, in a simplified diagram, I will draw it like this, instead of drawing this two
separate coils, this is the 600 V side. I will just draw them vertically, so that this I can draw
it anywhere I like so long I know these two are these two coils electrically nothing wrong
in that. And suppose this two are dots. So, these two coils I have drawn here. This is the
high voltage coil and it can be supplied with 600 V 50 Hz, I will supply a all are RMS
values here 600 V. But the moment you apply 600 V, you are sure you are going to get
200 V here, why that true rule will change, because same flux links.
Therefore, you are having this two voltages with this polarity is also known. If it is plus at
any instant, these also plus, this is minus, this is minus. Therefore, it looks like if I connect
this two in series, this two points I join, then what will be the voltage available here, 800
V, why not, you will get 800 V here. Therefore, this two winding transformer, no doubt,
it can be used to change a 600 V to 200 V or vice versa. If 200 V is there, you can get 600
V connect loads.
But it I now after learning not yet I have connected it with auto transformer, but I am
simply telling that if you apply 600 V 200 V here this coils only I have drawn in a
convenient way, so that I understand it much better. So, if you apply 600 V here, there
must be induced voltage here connect it here. So, you will get 800 V, this two winding
transformer can be also used to change a voltage from 600 V to 800 V, why not, provided
you connect like this. And the moment you connect like this, you can immediately tell,
then it is like an auto transformer, where you are stepping up the voltage, this way you
have connected that is the thing I want to tell. It can be connected in this way 800 V ok.
Now, one interesting thing I must tell you here that rated currents of the windings are
known LV side rated current is 60, HV side is 20 ok. Now, the question is how much
current I will allow to flow, what sort of impedance should I connect, what will be the
rated I mean the maximum current I can draw from this combination, here what should be
the current I must allow to flow through the load, this is load, how much? The rated current
of 200 V is 60 Amp.
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So, pass 60 Amp, I can pass no problem, these winding. Because the from the dot if 60
Amp comes out I know through the dot here 20 Amp must flow, this I know from my two
winding transformer. Whenever in this is dot, this is dot, this fellow carries 60 Amp rated
current, this will draw 20 Amp ideal transformer consider first, it will draw 20 Amp. Why
it should draw 20 Amp, so that it can balance this mmf, so that the flux is created only by
that magnetizing current which is very little, because KVL is to be satisfied here, flux will
remain same, no matter what you are connecting across this two terminals that is the good
old explanation.
Therefore, simply because you have connected in this fashion those rules are still not
existing is not correct, it is there. Through the dot if you want to draw some current,
through this dot, do not show it here, 20 Amp through this high voltage winding there must
be 20 Amp in this direction.
I will come, just let me see whether any mistake I have done, but let me carry on. So, this
is 60, this is 20. Then what is the input current, input current must be 80 Amp. So, 80 Amp
goes there, 20 Amp comes, 60 comes here. So, KCL is satisfied here, there and so on,
everything is fine ok. And I will say I will be very happy, flux will be at the rated value as
in a two winding transformer when it is used, coils are carrying rated currents that is also
very good.
But one interesting thing has happened that is if you now look with a an open mind, you
will see this way if you connect, how much KVA the transformer is handling, how much
KVA I write it how much KVA the transformer is handling that is the question? How much
KVA it is supplying to the load or how much KVA it is drawing from the source, this is
source? How much KVA it is delivering?
800 × 60 = 48𝐾𝑉𝐴
And how much KVA it is drawing from supply, KVA drawn from supply is
600 × 80 = 48𝐾𝑉𝐴
Ideal two winding transformer.
As I told you that better start with ideal transformer things will be closed to this numbers
nothing. So, and it is fine, it is delivering 48 KVA as I told you in the previous example,
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nothing doing it must also absorb 48 KVA from the supply. But here lies the some fallacy
is there, when you use these two winding transformer, as a two winding transformer that
is you will connect load here, this current 60 Amp. And at that time this current HV side
current will be 20 Amp. What is the KVA handled? If it is used at the rated two winding
transformer,
600 × 20 = 200 × 60 = 12𝐾𝑉𝐴
That is what it is rated two winding transformer.
But the moment you connect it in this fashion, we find that this way of connecting the
transformer can handle 48 KVA without overloading the transformer that is the most
important part. It is now handling 48 KVA. So, KVA handled now is 48 KVA and I must
write along with it without overloading the transformer, it is not overloaded rated current
of this 60 Amp fine, rated current of this 20 Amp it is fine, rated voltage of this winding
600 V, rated voltage of this winding is 200 V.
So, everything is in rated condition so far as windings or coils are concerned. And still it
is handling as two winding transformer it could handle only 12 KVA, but to much of
surprise it looks like 4 times more it is handling, is not what is the thing which makes it to
deliver 48 KVA. The answer to this is that in an auto transformer, power is delivered from
the primary to the secondary side is by two means.
(Refer Slide Time: 17:04)
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Why, for example, in an auto transformer which is stepped down we have considered I
have not told anything about that till now this an auto transformer. Give some input power
get some output power that is fine. And in this case, it is a step up, and I told you nothing
is sacred about 𝑉1 and 𝑉2 that is fine in whichever way you go.
So, this input power is transmitted to this secondary by two mains; one is power is
transferred to the secondary side. Both for step down and step up case, it does not matter
here as you can see power is transferred to this secondary side by two means to the
secondary side from primary side, from primary side by two means. What are the two
means? One by transformer action and two by direct electrical connections that is how we
explain this by direct electrical connection, connection.
What is this direct electrical connection? In this case through this, whether there was some
induced voltage or not still power would have delivered here also there is a direct
connection. But as a two winding transformer that is that electrical conduction transfer
from primary to secondary is not there, because there is no connection between these two,
but here this is how this thing.
Now, the question is how much, therefore, it looks like out of this 48 KVA, a portion is
transferred by because of magnetic induction by transformer action. And the remaining
portion then must have been delivered because of direct conduction of power from primary
to secondary side. Now, the question is how to calculate the power delivered by
transformer action?
Listen here, come here in this slide. This two windings this two are separate windings this
fellow is carrying 60A and then this fellow is drawing 20A from the dot in the winding.
Therefore, by these two coils can be treated as isolated coil and the Volt Amp handled by
this because of transformer action, I write by different color that is by transformer action
KVA transmitted should be equal to
20 × 600 = 200 × 60 = 12𝐾𝑉𝐴
In fact, by transformer action, it cannot do anything better 12 KVA only and then I will
say 48 KVA is total transferred. So, I will say KVA transferred by electrical conduction
must be
290
48𝐾𝑉𝐴 − 12𝐾𝑉𝐴 = 36𝐾𝑉𝐴
So, this is quite interesting to note, that is a two winding transformer will carry rated
currents will have rated voltage across it and still it can transfer a much more KVA
provided you connect it in this fashion ok. And this is always done like this. For example,
I will just the same problem I will do, so that you understand where we are going.
(Refer Slide Time: 22:43)
Suppose, this two coils the same transformer can be connected in different ways, suppose
what was the rating, one was 600 V, another is 200 V ok. So, suppose I say that this is the
600 V, this is the 200 V, this two windings have connected in series. I can say that this can
be also used. And suppose your supply voltage is 800 V that I can apply, because it is then
𝑁
the voltage will be divided induced voltage in this ratio 𝑁1, whatever it is a 800 V. And I
2
can take tapping here and supply a 200 V to the load is that clear? So, the same two winding
transformer this is 600 V, this is 200 V, this two are dots. I have connected them in series
with regard to dot polarities and supplied it with 800 V.
So, if it is 800 V, then here the voltage will be 600 V, even when load has not been
connected, because voltage for turn remains same and 200 V here. And I can supply a 200
V here. Now, here once again I pose the same problem it is a step down connection. So,
the question is how much current should I now allow to I mean what should be my load
impedance, so that the transformer winding currents will be rated. Are you getting? You
291
cannot just connect any impedance. So, without satisfying yourself that ok, if I connect
these value of impedance then winding currents will be rated.
It is true when the HB side will carry rated current, you do not have to worry LB side will
automatically carry rated current, so that the mmf will be balanced. Now, the big question
is how to handle this situation this I will handle in this way now. What is the rated current
of the HB side, it was?
Student: 20.
20 Amp this side rated current is?
Student: 60.
60 Amp low voltage side higher current, high voltage side low current. Therefore, I can
say that look here under no circumstances, the maximum possible load maximum current
you should draw here that will yield 60 Amp here, because that is the rated current that is
the rated current of this winding, I am so sorry, this is 20 Amp, is not it? So, maximum
KVA of the load should be
800 × 20 = 16𝐾𝑉𝐴
And this 16 KVA will come here I know the voltage. So,
16000
= 80𝐴𝑚𝑝
200
So, I will draw 80 Amp, I am because this is not the winding current. And you see yourself
this is 20Amp, this is 80Amp apply KCL here, is it not? What will be this current 60Amp?
Then only this will be 20 + 60 = 80. Here also 80A, 60A goes up 20A comes here. So,
once again it is KVA handled by the transformer this time is not 48 KVA, it is 16 KVA
which is greater than 12 KVA. How much KVA is conducted by conduction? KVA
conducted by magnetically or by transformer action will be
20 × 600 = 200 × 60 = 12𝐾𝑉𝐴
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And I know what is the maximum KVA it can handle that KVA is 16 KVA, this number.
So, where from this extra KVA is coming, I will conclude that KVA transferred by through
electrical conduction is this
16𝐾𝑉𝐴 − 12𝐾𝑉𝐴 = 4𝐾𝑉𝐴
Therefore, I will request you, you try several combinations of connecting this two coils in
series, apply voltage, stepping up, stepping out down, it may be it may so happen that you
your dot is here, this dot is there. So, I leave it that has an exercise and find it out. Except
one thing I will tell, this could be this numbers I calculated like this, also it could be done
like this very quickly I will tell this is the thing. Here was the load.
Now, first thing I told you that this two are dots given is it not? And I told you that this is
the HV side. And I will not allow this 20 Amp first I decide 20 Amp. But then I will argue
if this is 20 Amp instead of calculating in input KVA, I could also say like this. If this is
20 Amp coming in through the dot of this transformer at that 20 Amp must 60 Amp flow,
then only this two mmfs will be balanced and then I could arrive at this 60𝐴 + 20𝐴 = 80𝐴
alternatively also in this way. So, get used to it please 20 Amp, very simple calculation
otherwise.
Thank you, we will continue with this.
293
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 31
Practical Auto Transformer
(Refer Slide Time: 00:22)
Welcome to 31st lecture on Electrical Machines I. In our last class as you know, we were
discussing about how a two winding transformer can be also connected as an auto
transformer and for some two connections, interesting connections. We how to calculate
the distribution of current in several windings? See, a two winding transformer when you
connect in auto transformer form it really this primary and secondary winding is really
unaware of the fact that you have connected in this manner or like a two winding
transformer; it only knows this much that if it has to deliver 60 amp. The moment it
delivers through the dots here 20 amp must come in; that is what duties.
Voltages are rated voltages; based on that everything all the calculations go, And as and
we have also seen another interesting feature that a two winding transformer whose rating
is suppose for the numerical example we have considered 12 KVA. It can be; apparently
it looks something very surprising that the KVA handled by this connection is much more
than the KVA rating of the two winding transformer and we also told that this is possible
in this mode of connection because of the fact that power will be transferred from source
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to the load side either by two actions. One is just by transformer action another is by
electrical conduction.
It is very easy to calculate how much is transferred by transformer action that I will
calculate. And from the total KVA I will subtract that to figure out how much is conducted
by conduction that way you just think. And you can do; and of course, these are the only
two connections I have discussed; you try on your own how differently you can connect
this two coils in, apply voltage appropriate voltage mind you to the windings to get how
much KVA? How much current? The interesting thing is whichever winding is left out
alone sort of thing you first fix up that current I should not allow more than. And the
moment you decide that current the other part of the winding which is which has its own
identities separate no common portion of this and this.
Therefore this current can be fixed up and then, the total current. Anyway this you please
try solve many problems. Now, in today’s lecture I; my plan is to tell you something what
is going to happen if it is a practical auto transformer.
(Refer Slide Time: 03:57)
I will discuss that, but before that one point I must tell you; see, auto transformer is the
thing. What will be its useful applications auto transformer? It is useful applications will
be you must have heard that starting of three phase induction motor; is not? You use an
auto transformer start it at a tapings of 𝑥 then the current drawn from this supply is reduced.
295
Torque is also of course, reduced by a factor of 𝑥 2 current drawn from the supply is
reduced by 𝑥 and so on.
Another interesting application of auto transformer will be to connect two power systems.
You must understand this point that we have in our country or in all other countries also
there are common power grid ok; you have several level of voltages ok. Suppose, the the
one line 400 KV system ok; see, eastern region grid; suppose, there is a grid eastern region
400 KV. Now, the western region grid also should be supposed to be 400 KV ok, another
grid with different colors I will draw; try to understand. This is supposed to be also 400
KV.
And you know this power lines will be ultimately connected in parallel that is this is
suppose A phase B phase C phase, A phase B phase C phase, coming from western region
grid; this is coming from eastern region grid and loads are connected in both the system,
but I want to make them common member; that is power may go when, there is deficiency
in power generation in western grid. Perhaps, this grid will supply power or there and vice
versa.
So, grids, common grids if you want to make all over the country; all the grids are to be
connected in parallel. Suppose, this level of voltage should be also 400 KVA then, only
this two can be; suppose this is A phase; this is B phase; this is C phase and these are
corresponding phases here ABC. This is supposed eastern region, eastern region grid; this
is western region. Who have kept this lines alive? Several generators. They are connected
in parallel you get the voltage. Here also several generators connected in parallel as
generated this grid. Now suppose, you want to make it more flexible that I will connect
them also in parallel and if due to some reason Western Region some generators fail. So,
the eastern region can help to pump power into this side or vice versa that is what I am
telling.
Now, what happens is this? This is 400KV when you connect this to in; if you want to
short this you cannot just do just like that because of the fact the level of the voltage must
be same. It may so happen; it is absolutely this side voltage is suppose not 400 KV because
of some practical reason or otherwise, it is suppose 390 KV; are you getting? Because
loads here also it is not 400 KV; may be 402 KV or say 395 KV because you will be
loading. These systems are loaded; this bus voltage may not exactly match. Suppose it is
296
395 KV, but I want to connect them in parallel. So that flexibility increases this side can
feed power; this side can feed power and what not.
So, in such situations these two systems can be connected. If you connect an auto
transformer in between, three phase auto transformer I will tell what this auto transformer.
Three phase through an auto transformer you will be able to parallel; two winding
transformer can be also used to parallel them, but I will not do that why? Because I see
that the ratio of the voltage is so small; 395 KV to 390 KV I have to connect. This side is
395 KV; this side 390 KV, nothing better than auto transformer you should think about
because turns ratio or the voltage ratio is close to 1.
You will be saving more or very large amount of copper; size of the auto transformer will
be less. So, this is one very practical use of an auto transformer, three phase auto
transformer. Now, what is a three phase auto transformer? I will just this I will take up
when I will take three phase transformer; right now you just try to understand. Here is a
case, where the voltage ratios will be close to 1 and I have to connect these two.
So that I can parallel them not directly because if I connect them directly without
transformer 5 KV difference will cause large current to flow almost like short circuit. So,
this is one case or maybe you can connect an auto transformer to start a three phase
induction motor. Of course, I must add one remark. Now to start an induction motor using
a bulky auto transformer that too I mean is not a good solution; there are very good
solutions because of the use of inverters power electronics there, but earlier it was very
much used.
So, auto transformer is a natural choice when, the voltage of the two sides you require
voltage levels are of this same order ok you really save money copper that is there. Now,
another thing you have used in the laboratory which is called Variac sorry Variac. In the
laboratory you must have used to carry out experiments and I also was telling that is it is
drawn like this and this stepping point you can move you can adjust to anywhere you like
and you can supply your load at various level of voltage.
Suppose, there is 220 V you want to do some experiments to find out the characteristics of
the load starting from 0 V up to 220 V. So, there is a pointer there which you can vary by
rotating a shaft through a carbon brass this is connected. You touch different segments of
the turns and gradually you can increase voltage to the load.
297
But it is called Variac. It will be same the current distribution etcetera we have to find out
from the whatever analysis we have done like that you have to do no doubt about it, but
here while doing this, while making this variac do you think the cross sectional area of this
portion is less than the cross sectional area of this portion no because you do not know
where he will be operating; so, this must be understood.
Variac is strictly not an auto transformer in this sense we design the auto transformer get
some economic advantage. Well, here nothing like that you should be prepared. So, the
gauge of the wire here throughout is fixed; you must understand this distinction in a variac
laboratory equipment that is why you will find variac is too heavy mean there is no savings
this side. Its KVA rating is known; maximum current it can draw and so on. This point I
must bring out to you. So that you distinguish between an auto transformer, dedicated auto
transformer like this where it will be used; it will have a fixed tapping; is not?
This is thinner wire. I drew with different colors, fixed tapping over. Nothing like that this
can be moved this way that way. If you do it between taking lot of risk you do not know
what is the current rating, mixed up current rating of the turns? I hope you have got the
point. So, this must be kept in mind.
(Refer Slide Time: 15:23)
.
Now, I will discuss what to do if the auto transformer; how to analyze a circuit involving
an auto transformer, but which is not ideal, non ideal auto transformer, a practical auto
transformer; what should I do? Practical auto transformer not variac; a practical two
298
winding transformer we; you have to take then 𝑟1, 𝑥1 ; 𝑟2 , 𝑥2 and then, magnetizing
reactance 𝑋𝑚1 and 𝑅𝑐𝑙1 core loss component of resistance; all these things we have to bring
here.
So, similarly in the auto transformer since there is a core; there is time varying field
therefore, there will be core losses and not only that there will be this is the auto
transformer thing where a fixed tapping is taken. So, this portion of the winding will have
𝑟1, 𝑥1 ; this portion of the winding will have 𝑟2 , 𝑥2 and so on. Similarly, the magnetizing
branch and no load current; first what will do is this I will assume that windings will have
some resistance and leakage reactance because leakage flux will be there and this portion
and this portion of the windings have their own electrical identity no mixed up between
them.
So, I will say the resistance of this portion of the winding is 𝑟2 and leakage reactance is 𝑥2
and this portion of the winding is 𝑟1 and 𝑥1 ; I can very well define that ok. So long when
there was this things were and magnetizing and no load current first neglect. I will tell you
how to take no load current, neglect no load current; that means, 𝐼0 which implies that I
am telling that 𝑋𝑚 → ∞ and 𝑅𝑐𝑙 → ∞ it means that we shall see. Now, and I will first draw
as usual; this is my transformer and I use this nomenclature AC and this is BC. And I will
say that
𝑁𝐴𝐶 = 𝑁1
𝑁𝐵𝐶 = 𝑁2
This two are my output where, I will connect load; is not? This is the thing.
Now, I am saying that this portion AB has a resistance 𝑟1, 𝑥1 and this portion BC is having
a resistance 𝑟2 and leakage reactance 𝑥2 and here I am applying a voltage 𝑉1. Earlier, when
this things were not present 𝑉1 is equal to the induced voltage here; 𝑉2 was equal to induced
voltage here. Therefore, we can easily see that if I want to show this resistance and leakage
reactance I must draw it like this. A point I will draw it slightly higher place. So, I will
draw it in this way. A; I start from a point; 𝑉1 is the supply voltage A. Then, there will be
just like two winding transformer 𝑟1, 𝑥1 ; it will only cause extra voltage drop and then, we
will come your 𝑉1 and this two were dots, I am not sure whether it will be 𝐸1 . I will write
what it will begin.
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Now, I reach up to point B. So, it will be; this is the point B and then, I will draw 𝑟2 ; I will
draw I mean 𝑗𝑥2 I am not writing you understand it is 𝑗𝑥2 , 𝑟2 𝑥2 and then, there will be
another induced emf in this portion. This induced emf, I can write it is 𝐸2 no problem why?
Because this is the voltage you have applied; flux is same. So, voltage induced 𝑁𝐵𝐶 = 𝑁2 .
So,
𝐸2 = 4.44𝑓𝜑𝑚𝑎𝑥 𝑁2
𝐸1 = 4.44𝑓𝜑𝑚𝑎𝑥 𝑁1
So, this is the induced voltage correctly I have drawn and voltage per turn is fixed; how
𝑉
much it is? It is approximately 𝑁1 , total number of turns
1
Number of turns here is (𝑁1 − 𝑁2 ). So, what will be the induced voltage here? It will be
(𝐸1 − 𝐸2 ). Are you with me? Because induced voltage in 𝑁1 turns; if I call it as 𝐸1 , induced
voltage in (𝑁1 − 𝑁2 ) turns will be (𝐸1 − 𝐸2 ). So, I let me write 𝐸1 = 4.44𝑓𝜑𝑚𝑎𝑥 𝑁1 ;
between these two terms 𝐸2 = 4.44𝑓𝜑𝑚𝑎𝑥 𝑁2 and if I say what is the induced voltage
between A and B; here, what will be the induced voltage? It will be equal to
𝐸𝐴𝐵 = (𝐸1 − 𝐸2 ) = 4.44𝑓𝜑𝑚𝑎𝑥 (𝑁1 − 𝑁2 )
So, this is the thing and here will be one load. So, immediately we see that there will be;
it is not as simple as that of a two winding transformer where, the equivalent circuit was
𝑟1, 𝑥1 and so on 𝐸1 , 𝐸2 . Here this, this is what it should look like and there will be a
distinction between 𝑉1 and see, if you start from this to this in so for as induced voltage is
concerned it is fine; 𝐸2 and 𝐸2 will cancel and 𝐸1 induced voltage, but there is a mix up
now.
I will tell that this current is 𝐼2 first thing is; first thing first 𝐼2 and suppose, I say that this
current is 𝐼1 . I do not know what are the relationships that I will find out, but I am sure
about one point that is if this is 𝐼2 ; if this is 𝐼1 no matter whether 𝑟1 𝑥1 or 𝑟2 𝑥2 are present;
the current here has to be (𝐼2 − 𝐼1 ) and this current has to be 𝐼1 it does not debar me or
does not confuse me that 𝑟1 𝑥1 is there; can I write these? Yes; KCL I have simply applied.
But only argument is interesting argument; see, the portion AB and portion BC they are
two individual coils as if and these voltage I will call 𝑉2 terminal voltage which is not same
300
as 𝐸2 because 𝑟2 𝑥2 is there, but 𝑟2 𝑥2 is not carrying same current as the load it is different.
So, this current is (𝐼2 − 𝐼1 ); is not? So, therefore, the moment load is connected some 𝐼2 is
flowing it will draw some current 𝐼2 and this time 𝑟1 𝑥1 and 𝑟2 𝑥2 is present I cannot easily
say how much current it will draw by balancing output power and input power because
there will be power loss in this element; in this element as well; is not? It is slightly
complicated, but one thing is clear, the mmf balance that rule must prevail. If this portion
carries winding, this portion cannot remain silent.
Therefore, I will say and this two are dots. So,
(𝑁1 − 𝑁2 )𝐼1 = 𝑁2 (𝐼2 − 𝐼1 )
𝑁1 𝐼1 = 𝑁2 𝐼2
See, the point is I should not start from this I must establish it. So, happens that like a two
winding transformer I mean 𝑁1 𝐼1 = 𝑁2 𝐼2 . I have derived that I will define
𝑁1
=𝑎
𝑁2
Then, I will say that
𝐼1 =
𝐼2
𝑎
This much I am I can confidently say. This I am sure off; got the point; this is the thing.
Now, this voltage is 𝑉2 and this voltage is 𝑉1, applied voltage; terminal voltage, induced
voltage they are not now same there will be a little drop here, but this tells me that the
voltage here across so called across AB portion of the winding, this must be (𝑉1 − 𝑉2 )
with this side plus this side minus.
Now, look at this; today, I will only tell this much look at this portion this coil and this
coil. These two coils are separate; is not magnetically coupled? Therefore, I can say that I
mean what I am trying to tell in the next page I will draw.
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(Refer Slide Time: 29:30)
What I am telling is AB is a coil and I will write BC is another coil; this two coils and I
find that A B has been supplied with a voltage of (𝑉1 − 𝑉2 ) terminal voltage; is not? And
this fellow across 𝑉2 I am not showing the load yet because the problem here is this coil
current and load current as different, but can say this much. See, this winding is delivering
a current of (𝐼2 − 𝐼1 ); this much I am certain.
What things is to be connected to achieve this? Let us not bother right now, but this is what
I can think of ok. So, I will draw once again here. So that we can easily refer to A, B and
C. What I am telling this current is (𝐼2 − 𝐼1 ) and this current is 𝐼1 and this voltage is 𝑉1
and this terminal voltage is 𝑉2; these are all fine 𝑉2; this current is 𝐼2 that is there, but if
you look at this two separate coils it is telling me as if I can draw this diagram and say that
it is a at least these two portions are a two winding transformer across the primary of which
I have applied (𝑉1 − 𝑉2 ) and secondary is loaded such that it delivers a current of (𝐼2 − 𝐼1 )
this much I can say.
And then, I will say; so for as this transformer is concerned what is the number of turns of
this 𝑁𝐴𝐵 = (𝑁1 − 𝑁2 ) is not very interesting, (𝑁1 − 𝑁2 ) is the number of turns and what
is the number of turns of this 𝑁2 ? I will say that this is a transformer whose turns ratio is
(𝑁1 −𝑁2 )
𝑁2
.
I will say whose trans ratio is this
(𝑁1 −𝑁2 )
𝑁2
and this one is equal to then,
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(𝑁1 − 𝑁2 )
=𝑎−1
𝑁2
Because I have defined
𝑎=
𝑁1
𝑁2
Now, please be with me, what I have done; these two are separate windings. This fellow
is carrying a current (𝐼2 − 𝐼1 ); this portion is carrying a current 𝐼1 . So, it is just like a two
winding transformer having applied voltage to this coil
𝑉𝐴 𝐵 = (𝑉1 − 𝑉2 )
𝑉𝐵𝐶 = 𝑉2
So, this is the thing and if that be the case then, I will say that this is a transformer of ratio
(𝑎 − 1): 1. Then I will go another step ahead and today, I will stop it here. Then draw the
equivalent circuit; equivalent circuit of this thing circuit refer to source; refer to I will not
say source, to (𝑉1 − 𝑉2 ) side because this winding will see it across it has been applied a
voltage of (𝑉1 − 𝑉2 ) and what will be the equivalent circuit? (𝑉1 − 𝑉2 ) is the applied
voltage; is not? Then there is 𝑟1; there is 𝑥1 will be there, it is winding; this winding 𝑟1 𝑥1 .
This winding B C, I have assumed that to be 𝑟2 𝑥2 ; so, this will be 𝑟1 𝑥1 and this will be
your what? 𝑟2′ and 𝑥2′ .
Now, the big question is what is this 𝑟2′ in terms of 𝑟2 ? Turns ratio is this. So, this must be
𝑟2′ = (𝑎 − 1)2 𝑟2
not 𝑎2 𝑟2 . Similarly, this
𝑥2′ = (𝑎 − 1)2 𝑥2
What voltage should I write? Here, this 𝑉2 . So, I must write it here
𝑉2 ′ = (𝑎 − 1)𝑉2
Actual voltage 𝑉2 into a times this. See, what I have drawn; I have this two windings I
have drawn separately, they are doing this.
303
So that I can invoke whatever I have learned from two winding transformer those simple,
but important principles I will straightaway apply and say this is the thing. We will
continue with this in the next class; it will take some time, but up to this point be absolutely
clear ok; try to understand it is very logical and interesting. In autotransformer the
equivalent circuit drawing is slightly; I will not say the word tricky, but rather I will say
you have to apply your own minds correctly. So that correct equivalent circuit you can
draw refer to a particular side.
Thank you.
304
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 32
Equivalent Circuit of Auto Transformer
(Refer Slide Time: 00:19)
So, welcome to next lecture on Electrical Machines I. And we have been discussing about
the Equivalent Circuit of a practical Auto Transformer. And this equivalent circuit when
compared to a two windings transformer is somewhat difficult to obtain and that is why
we were discussing it how to get it you recall from my earlier lecture.
(Refer Slide Time: 00:51)
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This was the actual auto transformer which is practical and I assume that this these are the
usual current distribution I have assumed. Here is 𝐼1 here is 𝐼2 . And this part AB which is
having (𝑁1 − 𝑁2 ) turns. And part BC these are two different coils they have nothing in
common between them and this part suppose has got a resistance 𝑟1 and leakage reactance
𝑥1 . And similarly this part BC has got a resistance 𝑟2 and 𝑥2 and we know the current
distribution and 𝑧2 is the load.
Now, without of course, having an equivalent circuit drawn, this circuit as such is also
electrically connected is not by this point. So, it looks like a network problem that way it
can be solved that is if you know 𝑉2 if you know 𝐼2 if you know these parameters, then 𝑉
plus this current into 𝑟2 , 𝑥2 these that two loops are there two meshes these circuit can be
handled. But only transformer business which will come in here to note that the induced
voltage (𝐸1 − 𝐸2 ) here 𝐸2 here this ratio of these voltage is
(𝐸1 − 𝐸2 ) (𝑁1 − 𝑁2 )
=
𝐸2
𝑁2
That is the thing, but it can be handled that way.
But we want to find out equivalent circuit and equivalent circuit of this transformer
referred to say source side that is this side.
(Refer Slide Time: 02:51)
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So, to do this next what we did is that portion AB and BC we noted in my last class is that
these two windings are separate; this winding has a turns (𝑁1 − 𝑁2 ) this winding has a
turn 𝑁2 only. Therefore, and this part is having I am sorry this is having 𝑟1, 𝑥1 and this is
𝑟2 , 𝑥2 ; only thing is the as if the voltage applied to this winding is (𝑉1 − 𝑉2 ) and this side
it is 𝑉2. And the current this winding is delivering is (𝐼2 − 𝐼1 ) and not this 𝐼2 mind this we
understand. Therefore, between these two windings I can apply all the rules that I have
applied for a two winding transformers having trans ratio (𝑎 − 1): 1, what is 𝑎?
𝑎=
𝑁1
𝑁2
That we know.
Therefore, it looks like the equivalent circuit referred to (𝑉1 − 𝑉2 ) side of these two coils
will be like this, (𝑉1 − 𝑉2 ), 𝑟1, 𝑥1 then these parameters will be referred to their turns ratio
is this. So, 𝑟2′ = (𝑎 − 1)2 𝑟2 , 𝑥2′ = (𝑎 − 1)2 𝑥2 and here you will get this actual voltage 𝑎𝑉2
means in this case (𝑎 − 1)𝑉2. And mind you this current here is not 𝐼2 how much will be
𝐼
the current? In case of two winding transformer you know 𝑎2 it comes, so it will be
(𝐼2 − 𝐼1 )
(𝑎 − 1)
This will be the current. So, what we have got here that I will that will be the starting point
here.
307
(Refer Slide Time: 05:37)
So, I will redraw it on a fresh page that is here is your 𝑟1, 𝑥1 was there, then the reflected
value; 𝑟2′ , 𝑥2′ is here.
𝑟2′ = (𝑎 − 1)2 𝑟2
𝑥2′ = (𝑎 − 1)2 𝑥2
This is the thing.
Magnetizing branch till now we have neglected that can be easily incorporated later. And
this voltage was (𝑉1 − 𝑉2 ) and this voltage is how much (𝑎 − 1)𝑉2 this diagram only I
have redrawn there afresh so this is the thing.
Now, in this equivalent circuit this plus minus usual plus minus usual and this current is
(𝐼2 −𝐼1 )
(𝑎−1)
. But we have seen that
𝐼1 =
𝐼2
𝑎
That we have already established is not from this MMF balance recall this we will always
use. See
(𝑁1 − 𝑁2 )𝐼1 = 𝑁2 (𝐼2 − 𝐼1 )
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𝑁1 𝐼1 = 𝑁2 𝐼2
∴ 𝐼1 =
𝑎=
𝐼2
𝑎
𝑁1
𝑁2
So, knowing these what is our goal? Goal is to replace this equivalent circuit similar to
that of the equivalent circuit of a two widening transformer, that is my input voltage 𝑉1
should be present here alone load side there should be 𝑉2 alone present, current some 𝐼2′
sort of thing should be present things like that we are trying to do.
Now, what is the importance of doing this? If you do this way then this equation by
applying KVL rule I can write down
(𝑉1 − 𝑉2 ) = (𝑎 − 1)𝑉2 +
(𝐼2 − 𝐼1 )
(𝑟 + 𝑗𝑥𝑒1 )
(𝑎 − 1) 𝑒1
𝑟𝑒1 = 𝑟1 + (𝑎 − 1)2 𝑟2
𝑥𝑒1 = 𝑥1 + (𝑎 − 1)2 𝑥2
So, what is 𝐼2 ?
𝐼2 = 𝑎𝐼1
𝑉1 = 𝑎𝑉2 +
(𝑎 − 1)𝐼1
(𝑟 + 𝑗𝑥𝑒1 )
(𝑎 − 1) 𝑒1
𝑉1 = 𝑎𝑉2 + 𝐼1 (𝑟𝑒1 + 𝑗𝑥𝑒1 )
And now I am happy ok, I have got an equation which is really we are looking forward
that is after getting this equation, I will say that this equation prompts me to draw the
equivalent circuit referred to primary from this equation. As if a voltage is there 𝑉1 this
side and here is the impedance (𝑟𝑒1 + 𝑗𝑥𝑒1 ) where 𝑟𝑒1 = 𝑟1 + (𝑎 − 1)2 𝑟2 , 𝑥𝑒1 = 𝑥1 +
(𝑎 − 1)2 𝑥2 and then here is I am having 𝑎𝑉2 .
So, this current is 𝐼1 is not this equation suggests that, that this is 𝐼1 therefore, 𝑎𝑉2 plus the
drops gives you 𝑉1 in the same way as we draw the equivalent circuit referred to primary
309
of a two winding transformer. But there is an important difference what is that important
difference ok, terminal voltage gets multiplied by 𝑎𝑉2 fine.
Only basic differences the 𝑟2 and 𝑥2 are to be multiplied by not 𝑎2 , but (𝑎 − 1)2 got the
point, so this is the thing. Then that one has to remember at least with refer to primary if
you draw then the equivalent circuits will be like this. And then after that everything is
fine that you can draw the phasor diagram start with 𝑎𝑉2 here 𝑎 is a scalar number this is
also direction of 𝑉2 then draw 𝐼1 and then draw this drops at and you get 𝑉1.
So, this is how the equivalents circuit of an auto transformer can be drawn or obtained;
and to do that the method there are several other treatment people write lot of equations
try to prove it. But what I am telling this can be very quickly drawn provided you apply
the rules of two winding transformer for the windings AB and BC because they are
separate winding and so, as if I say that across AB, 𝑉1 and 𝑉2 is applied this one, but only
thing is its turns ratio is this therefore, it is to be multiplied these voltage will come like
that.
And then, defining this 𝑟𝑒1 = 𝑟1 + (𝑎 − 1)2 𝑟2 , 𝑥𝑒1 = 𝑥1 + (𝑎 − 1)2 𝑥2 and so on. You will
come here then, from this you write down the KVL equation of this and that and soon I
find this is the equation this is the reflected current and so on. So, with respect to primary
this is the equivalent circuit. In the same way that is from this equation only if you wish
that I am not doing mind you. I will be able to draw, the equivalent circuit referred to the
load side also got the point I will be getting the equivalent circuit referred to the load side
as well.
Starting from this basic equation here, I will be able to do that clear, but that I leave it to
you to ponder over, but better do not waste too much time on that is also true because one
𝐼
side you have got that is fine only one point in this diagram. This 𝐼1 = 𝑎2 .
𝑎𝑉
What do you think the impedance connected here it will be this voltage 𝐼22 this point is
( )
𝑎
worth mentioning impedance here, load impedance I will at least from this circuit it tells
𝑎𝑉
me it is 𝐼22 , which is equal to
( )
𝑎
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𝑍=
𝑎𝑉2
𝑉2
= 𝑎2 ( ) = 𝑎2 𝑍2
𝐼
𝐼2
( 𝑎2 )
Mind you, this current these two isolated current is (𝐼2 − 𝐼1 ) and 𝐼1 they must balance the
MMF the rules of two winding transformer follows.
But (𝐼2 − 𝐼1 ) is not your load current that is why that complication comes in, that is if I
say across the output terminals of the auto transformer that is here if I draw it to highlight
that; if it is 𝑍2 , this is 𝐼2 and this is (𝐼2 − 𝐼1 ) and this is 𝐼1 is it? Therefore, the load
impedance will be multiplied by 𝑎2 because of the fact 𝐼2 is not part of this current got the
point therefore, this point should be remembered.
So, auto transformer equivalent circuit if you want to draw is slightly difficult means if
you are conceptually ok, you can easily develop, but one should be careful. The winding
leakage impedance will be multiplied by (𝑎 − 1)2 , but the load impedance should be
𝑁
multiplied by 𝑎2 where 𝑎 = 𝑁1 is not. So, the that is all I mean equivalent circuit we have
2
completed.
But only thing is I will just now tell you now suppose, I say that in this equivalent circuit
you have not talked about the no load current. In any case I will consider the approximate
equivalent circuit therefore, the magnetising current and core loss component of resistance
that is the loss component of current; can be shown to be connected right across the supply
terminals as two components; one is 𝑅𝑐𝑙1 , another is 𝑋𝑚1. For example, here I will connect
𝑅𝑐𝑙1 and 𝑋𝑚1 got the point.
So, if I draw it here that I can always do. So, equivalent circuit of an auto transformer, in
fact, a lecture 32 I have written here we were at on the verge of completion last lecture
itself, but anyway. So, this will be then 𝑉1, (𝑟𝑒1 + 𝑗𝑥𝑒1 ) and this is 𝑎𝑉2. And the load
impedance whatever was connected will be multiplied by 𝑎2 𝑍2 and this 𝑟𝑒1 = 𝑟1 +
(𝑎 − 1)2 𝑟2 , 𝑥𝑒1 = 𝑥1 + (𝑎 − 1)2 𝑥2 do not forget that is it.
And if you consider this impedance, refer to this side show it here supply side as 𝑋𝑚1 and
𝑅𝑐𝑙1 like the two winding transformer and this is 𝐼1 is the current here clear. So, 𝐼1 then I
should tell this one perhaps I will tell some reflected current of these primary side that is,
𝐼
𝐼
2
𝐼2′ and once again 𝐼2′ is not (𝑎−1)
it is 𝐼2′ = 𝑎2 .
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I think we have spent enough time go through it and try to solve some problems on this
auto transformer equivalent circuit. We will also give you some problems in the tutorial or
in the hand lecture notes some solve problem, I will try my best to include that, but it is
very interesting to study the autotransformer; because autotransformers are all both the
two winding transformer and autotransformers are used.
Although you can change the level of voltage from one level to another, for any ratios
theoretically it is possible, but there are some advantages of auto autotransformer while
the voltage ratios are not deferring too much. Then you choose auto transformer, then also
isolation problem we can easily see is not that urgent at the same time economically it will
be much attractive. And two winding transformer on the other hand gives you the isolation
where it is needed.
After completing this, our next topic will be a 3 phase transformers. But before 3 phase
transformer one small thing I would like to tell you a tidbits of see two winding transformer
for example, about one test I will just tell. We have talked about open circuit and short
circuit test on a two winding transformer mind you, in auto transformer I am not going to
those tests, but if this equivalent circuit is with you can easily find out these parameters by
doing similar test this.
(Refer Slide Time: 26:30)
But what I am telling is, for 2 winding transformer you recall for 2 winding transformer,
we have done open circuit test and short circuit test, to find out parameters equivalent
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circuit parameters to find out that we have done. And I told you during open circuit test
you recall you apply rated voltage 𝑉1 rated at rated frequency rated frequency secondary
is open.
So, for this test open circuit test when you connect like that flux will be rated flux will be
rated and there will be core loss only core loss only; because the copper loss is negligibly
small secondary winding is not carrying any current and primary current is only two to
five percent. So, only core loss takes place during open circuit test.
On the other hand during short circuit test, you keep the secondary shorted, here you apply
not rated voltage such that rated current flows current is rated here 𝐼1 rated; since applied
voltage is pretty small and flux in the core is directly proportional to the applied voltage.
So, flux will be very less, but current will be rated. So, in this test what happens is this
practically core lose is can be assumed to be zero, core loss is absent, as if is absent
compared to copper loss all the losses will be copper loss to copper loss ok. And when the
transformer will be in use this is during test these I told, but I am once again perhaps
repeating, but it is better to repeat keep these points in mind.
When the transformer will be in operation it will be 𝑉1 rated here also and the windings
will carry rated current both the windings at full load condition, is not here load is
connected rated voltage. And secondary is not silent it is supplying power therefore, during
operation this is during operation at rated condition operation at rated condition both the
losses are present, both core and copper losses are rated are rated values are rated are
present you cannot neglect one from the other that is fine.
Therefore, during open circuit test and short circuit test when you are finding out the
parameter values, the temperature rise during this test or that test will never be close to the
temperature rise of a transformer which will be put to use at rated condition. Because here
both the losses are present temperature rise will be much higher than either of this test and
that test. We will continue discussing on a very simple test, I will just mention because
that test is so nice and this will also enhance our understanding of transformers. So, I will
discuss about sampler test briefly in the next class and then starts topics on 3 phase
transformer.
Thank you.
313
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 33
Polarity Test and Sumpner Test
Welcome to 33rd lecture on Electrical Machines I.
(Refer Slide Time: 00:32)
And you know we will be discussing first another very popular test which is called
Sumpner’s test which is also called sometimes back to back test ok. And as I was telling
if you carry out the open circuit test only core loss will be there and if you carry out short
circuit test only copper loss will be there. Both the losses will be there when the
transformer will be put two operation at rated condition. At that time, so, neither during
the open circuit test nor during the short circuit test, the temperature rise of the machine
will be the actual temperature rise when the transformer will be in operation.
So, what is the solution? Solution is in the laboratory then try to load the transformer in
the laboratory connect the load and apply rated voltage at rated frequency and connect a
load such that rated current also flows and so on. Then temperature rise can be measured,
but unfortunately for a large transformers for example, a transformers of ratings of 200
KVA 500 KVA distribution transformer or more I mean ratings, then such a load will not
be available in the lab.
314
And not only that during open circuit test from the source you are drawing only the core
loss power which is responsible for core loss only. During short circuit test burden on the
supply you are putting is only of what amount of power? Very little power which will be
corresponding to short circuit test. But if you put to actual loading of a large transformer
in the lab, that huge amount of power is to be drawn from the supply this must be
understood.
So, during testing always try to draw less power from the supply at the same time see that
whatever is your objective of the test that is performed. What is the difference? If the
temperature rise is different, because in open circuit test on the very low temperature rise
and so on.
See because of temperature rise, temperature difference the resistance of the windings or
the resistance which are representing losses their values will change. Suppose during short
circuit test you calculate 𝑟𝑒1, equivalent resistance refer to primary, but when the
transformer is put to use that should be corrected because of temperature rise there is
temperature rise coefficient
𝑟𝑡 = 𝑟0 (1 + 𝛼𝑡)
Anyway, so to have a better idea of the temperature rise of a transformer one test which is
called back to back test or Sumpner test is carried out it was suggested by most probably
Sumpner. And it is very interesting test, not only the test is interesting, but at the same
time since you are studying transformer for the first time it will also enhance your
understanding whatever we have discussed earlier that is why I have chosen this topic to
be included here.
In this test what is done do not use actual loading no never uses actual loading, but at the
same time creates such a situation that transformer will feel that it is fully loaded that is
rated voltage applied, rated flux will be there and also windings will carry rated current
ok. But without drawing very large power from the source only it will draw the powers
corresponding to copper loss and corresponding to core loss that is the beauty of the test,
sometimes it is called phantom loading type thing ok.
Now let us see what this test means, only condition for this test to be successful is that you
have to take two similar transformers not with a single transformer you can do that two
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transformer because the transformers are manufactured in large numbers. So, after
manufacturing you test them take two at a time and test them although you can do
individual open circuit short circuit test that is there.
But to carry out the test you require two identical rated, very similar rated transformer. For
example, I will take two transformer whose I will use of course, small rating say 10 KVA
just to give you a feel of what is going on. 2 numbers of 10 KVA 200V/100V 50 Hertz
single phase transformers suppose you take.
And then first thing what you have to do is that you have to carry out the polarity test, I
will also take opportunity to explain to you how to ascertain polarity test. Because that is
essential to carry out the test, unless you do not know the polarity it will be not possible to
carry out the testing it cannot be left to chances. Now, what do I mean by polarity we told
you very simple that suppose you have a transformer whose rating is 200V/100V ok.
Now, what you do here is you apply, because the dot convention is not known. Dot
convention that is if I put a dot here what to put dot there I know what it means
theoretically. And also I can ascertain that without doing any test provided you know the
sense of the winding, but in general for a practical transformers only two terminals will
come out from the primary two terminals comes out from the secondary. And windings
really you cannot see is not it will be inside a sort of black box. I mean steel box everything
will be there two terminals only come out here two terminals will come out there that is
the problem.
But as I told you dots must be known, in fact, for three phase transformer connection that
is also essential to know. In case of open circuit test, short circuit test you do not bother it
is not necessary as you understand now, for this particular test why it is necessary. Now,
as I told you in this case if what you do you apply rated voltage see how to find out; find
out dots.
So, what you do is this? You apply say to this winding 200V you apply and there is no dot
markings here. Now, but you are free to choose one dot terminal of any of these windings;
for example, you say that this is this side you take it chalk and mark this is my dot terminal
of this winding.
316
But that is what you cannot do now here, if this is positive at any one time which terminal
is positive on the other side that is the goal ok. Now, this diagram what you do is this then
you short these two terminals, you make connected take a piece of wire connect these two
terminals and take a voltmeter connect it here, that is what you do.
Now the moment you apply 200V here no matter whether you have connected this or not
by transformer principle 100V will appear here. And this 100V I am not sure, whether this
is dot or this is dot. The thing is suppose let us assume this is dot suppose, suppose this is
also dot. These two coils are connected in series as you can see and I am measuring the
voltage, what will be the volt meter reading? 100V because there in this is plus this is plus
100V. What could be the other possible reading? It could be like this dot may be here also
I am not sure. If that is the case then what will be the volt meter reading 300V.
So, it is this volt meter reading which will help me to put dot correctly either here or there
depending upon the looking at the volt meter reading. If this reads the difference of this
two voltage, then I will put dot there. If it volt meter reading gives sum of these two voltage
I will put dot here very simple logic nothing like that. Only one thing is this about this test
the comment is suppose the ratio of the transformer is very large. For example, you have
a transformer whose rating is say 1000V/100V suppose you have a transformer, in this
case then what will happen is this.
So, if you apply 1000V here, you will certainly get 100V no doubt and you are free to
choose one dot terminals of one of the coils. So, I will suppose choose here and what I told
I will connect these two points here and I will connect a voltmeter there. In this case the
volt meter readings will be either 1100V or 900V one of them will come, but you will see
with a practical voltmeter these two voltages are close by. So, a slight misinterpretation or
somewhat wrong reading of the voltmeter you really cannot ascertain that is the problem.
If the ratios of the voltages are high then difference and addition become of the same order
therefore, it will be very difficult to distinguish these two readings confidently. So, in such
cases when the voltage ratios are high, it looks like this method is not reliable. So, what
people do is this in this case for very large voltage ratios another simple test can be done
and that is called the DC kick test.
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(Refer Slide Time: 14:48)
Very simple test what is done, you do not require also large supply rated voltage those
things may not be required what you do? You connect a switch here much simpler than
this you switch here and take a DC source batteries say. And here you connect a zero
centered moving coil voltmeter mc voltmeter that is all and the positive of the battery
wherever you are connecting put the dot there, one you can you are free to choose. Now,
and it is also a voltmeter connected between these two points, now you close this switch.
The moment you close the switch it is not AC supply, but there will be a sudden change
in flux.
Therefore, we expect there will be instantaneously there will be an induced voltage and if
this is dot happens to be dot and see moving coil meters has got plus minus marking is not
moving coil meters has got plus minus marking if current enters it gives positive. So, zero
centered meter do you taken, zero centered meter. So, you close this switch there will be
a deflection in the moving coil voltmeter a short duration steady state will soon be reached.
If it gives positive deflection you are sure this is also dot, if it gives negative if this is dot
this is dot.
So, anyway this is called DC kick test, you select this voltmeter reading such that steady
state current does not exceeds the rated current of this meter these things, but very
interesting test. So, anyway this test by doing this simple test we can find out the in polarity
of a transformer ok, so you do this. Now, coming back to Sumpner test, so two numbers
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of similarly rated transformer you take and then connect like this that is this is one
transformer.
(Refer Slide Time: 18:12)
And it is secondary is here and suppose I know the dots, even if you do not know then you
have to do those two tests to find out either of this two tests conveniently, this is
transformer one ok. This is one transformer practical transformer and you take the other
transformer which I have drawn here it is core and here also suppose these two are dots.
Now the connection, suppose these are the HV side in this case 200V and 100V both are
available in the lab we will not bother. So, what you do, you connect these two transformer
HV in this case it could be LV also no problem and connect them in parallel. And then
here in this side you connect an ammeter and then wattmeter very interesting test and apply
rated voltage.
That is in this case for this transformer I will apply 200V 50 Hz is not that is what I will
do. Suppose secondary I have not done anything they are open, what do you think this
ammeter will read and this wattmeter will read? If you apply this these two transformers
are manufactured by the same manufacturer of same ratings they are expected to have
same parameters same core loss these that etc. it is a very good assumption.
Therefore, the current the no load current will be drawn, this transformer will draw it is no
load current only no load current secondary are open this transformer also will draw no
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load current. Therefore, I would expect this ammeter will read 2 times the no load current
that is whatever ammeter reads, divide ir by 2 and tell that the no load current of each of
the transformer is half of this that is what.
And what this wattmeter is going to read, wattmeter with the secondary open for both the
transformers wattmeter will only record the core losses, sum of the core losses of this two
transformer, because rated flux is there in both the transformer. Therefore, it will also read
2 times the core loss is not, this wattmeter will read and ammeter will read 2 times the no
load current and this wattmeter will read 2 times the core loss.
Now, as I told you this that is all it is almost like an open circuit test is not secondaries are
open. But now comes the interesting part what is done now and what will be will there be
induced voltage here? Yes it’s value is 100V just I am putting this number, so that things
are very clear what I am talking about this way. Suppose I join by a piece of wire these
two terminals and these two I bring it here, these two terminals. If I connect a voltmeter
across this what will be the reading of the voltmeter you think a bit and then tell, because
polarities are this is plus this is plus, so they are opposing.
So, it will be zero voltage. So, between these two terminals although this is a connected
by piece of wire etcetera, but voltage existing between these two point is zero, so here
voltage is zero. Now, what you do? You take an auto transformer here I mean I should say
variac in the laboratory I should draw some space is necessary. So, you connect it like this
and here once again you connect an wattmeter and an ammeter and make an arrangement,
so that you can apply small voltage here across this.
If the pointer was at this point no voltage you have applied and net emf acting in this circuit
is zero nothing happens. Now, what you do with this primary energized mind you, you
apply little voltage increase slowly ok. Then what will happen this is, what is the net emf
acting in the circuit now this 100V and this 100V will cancel out there as if not there. So,
applied voltage divided by, so apply a little voltage current will start flowing is not current
will flow. If current flows suppose for some low applied voltage suppose current.
So, increase this voltage such that rated current flows what is the rated current of this side?
It is staying as I told you given the transformer rating immediately calculated rated currents
that is always important.
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10000
= 100𝐴
100
That is the rated current of the LV side, perhaps the test would have been better if you
have energized 100V side current anyway; 100Amp for understanding we are doing.
So, I will apply such a voltage such that 100Amp flows here, but the moment 100Amp
flows suppose this way 100Amp is flowing this way. This is a secondary, primary of a
transformer having a magnetic core and this is also primary, secondary having its own
separate magnetic core. But flux in the core cannot be changed because KVL is to be
satisfied here that good old reasoning argument.
Therefore, through the dot 100Amp coming out for this transformer means that through
this dot 50Amp must flow through the dot current commutes through the dot 50Amp must
flow. Now, similarly for this transformer it is just happening in opposite way, as I told you
if these two are dots you can consider these two are dot as well and remove this, see these
are the things you must understand. So, through the dots here 100Amp is coming out;
therefore, in this case through this dot a 100Amp must flow as so, called is not.
So 50Amp must flow is not. Therefore, I force some current to flow in this secondary coil
in this case LV side. I think it would have been better for this transformer even to consider
100V to be the rated supply because ammeter rating and wattmeter, but anyway for just
understanding purposes.
So, this is 100Amp and then reflected current here 50Amp 100Amp, reflected current this
way. Apply KCL here what will happen? 50Amp comes in this direction is it not 50Amp
comes in and 50Amp comes in here. So, this 50Amp business never comes here, this
current still remains 2𝐼0 whatever is the no load current got the point that is the most
important thing. Now, this transformer windings are carrying rated currents therefore,
rated copper loss will take place and rated flux is also present.
So, rated core loss will also take place and also winding carrying rated current, so rated
copper loss, so both the losses are present here. This watt meter draws supply from this
supply which is equal to 2 times the core loss of individual transformers, this watt meter
will record what? Only the copper loss 2 times the full load copper loss is not, because this
current equivalent resistance this set. From this source second source here which can
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supply large current this ammeter reading of course, will be not two times these are in
series to. So, rated LV side current this ammeter then should be based on this one 100Amp.
So, this is equal to rated LV current, in this case LV side current. So, apply such a small
why small voltage will generate large current it is because of the fact that net emf acting
in this circuit was zero, 100V 100V which was caused by this application of 200V here
they always cancel out. And therefore, you can pump current now into this network, so it
will read full load copper loss. Therefore, transformer will feel I am being used under full
load condition is not, because my windings are carrying rated current and my core is also
carrying rated flux both the losses are at their rated values.
So, temperature rise and if you do this test for a long time say 1 hour. So that the thing
attains a steady temperature I mean telling just like that put some thermometer there you I
mean just telling that I can measure the temperature rise at some convenient places to say
temperature rise here. What I am telling? That temperature rise must be higher compared
to either for open circuit test whatever will be the temperature rise or short circuit test.
Therefore, by estimating these copper losses, I will now will be able to estimate the
equivalent resistance
𝑖⏟2
𝑟𝑒𝑞 from this I will be able to calculate new 𝑟𝑒𝑞 . And
𝐴𝑚𝑚𝑒𝑡𝑒𝑟 𝑅𝑒𝑎𝑑𝑖𝑛𝑔
from short circuit test whatever 𝑟𝑒𝑞 I calculated it will be higher than that and from this
variation of the resistance I should be able to predict what is the temperature rise.
So, anyway this is a very nice test for which you do not require 10KVA load, it will only
draw core loss and copper loss. So, and it why it is called back to back test is also
understood you connect the secondaries in back to back manner. But only condition is you
must connect them in after noting down these dots correctly for dots do the previous test.
You should not connect the secondaries just like that, this is dot this is dot connect them
then 200V will appear is not then your basic idea of carrying out the test will be lost. Any
way I stop here today, but this circuit how it works please think physically what is
happening then it will further definitely it will enhance your understanding of transformer
always remember a transformer two winding transformer.
If secondary carries rated current, primary will also carry rated current the direction of the
current is decided by the dots as if through the dots rated current is coming out. Through
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the secondary through the dots current will come out and these concepts I have used not
only for two winding transformer, but also while analyzing the auto transformer ok. So,
with this I stop here I could not start three phase transformer next time we will start.
Thank you.
323
Electrical Machines – I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 34
3 Phase Transformer Using 3 Single Phase Transformers
Welcome to lecture 34 on Electrical Machines I and we will start today 3 Phase
Transformers.
(Refer Slide Time: 00:32)
And just in my last lecture I told you about one very important test called Sumpner’s test
or back to back test and this was the picture. So, here the beauty is the transformer under
the condition of the test will have rated flux as well as rated current. And if you wish you
can find out the equivalent circuit parameters as well, because you know the no load
current of each transformers you know the wattmeter reading or core loss of each
transformer from which the parallel branch 𝑅𝑐𝑙 and 𝑋𝑚 can be calculated. And from this
side also you know what is the voltage needed to supply the rated current, half of that will
be actually appearing if you measure the voltage here if you connect also a voltmeter.
I mean suppose you connect also a voltmeter. This is the rated voltage similarly here you
connect a load and voltmeter. So, once again you have generated the short circuit test data
test as we carried out from the HV side in this particular example another from the LV
side and equivalent circuit parameters once again can be estimated. And these parameter
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values will be corresponding to at rated temperature for example, while carrying out the
test you energize the transformer for a long time maybe hours, so that the steady state
temperature is reached, but it is an interesting test.
(Refer Slide Time: 02:25)
Now, today we will start 3 phase transformers ok; 3 phase transformers. They 3 phase
transformers can be of two type so, one is use 3 numbers of identical transformers. And in
this case when you use 3 numbers of identical transformers to be used in a 3 phase voltage
transformation, it is called a bank of 3 single phase transformers. Or a 3 phase transformer
can be manufactured as a single unit ok.
So, we will actually first examine this one that is 3 numbers of identical transformers I will
take. And the problem is because our system power system is of 3 phase and in 3 phase I
have 3 phase voltages and those voltages will be assumed to be balanced. And if you
require to change that voltage to some high or low 3 phase voltage, then we will be using
3 phase transformers that is the idea, that is suppose you have 3 phase system like this
which is suppose 6.6 KV on one side.
Then connect a 3 phase transformer and at the output you will get suppose 440 volt 3 phase
this is 3 phase 6.6 KV this is 440 volt 3 phase of course, 3 phase 50 Hz in our country and
this will be also 50 Hz because frequency remains same. And all these voltages here
whatever I have written is line to line voltage.
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A typical distribution transformer will have and of course, it will have a KVA rating
associated with it. So, 440 volt 50 Hz 3 phase and this 440 volt once again is line to line
voltage. So, the supply here is ABC and here the supply I may call on the secondary side
abc. So, this box will contain a 3 phase transformer, which may use 3 number of identical
single phase transformer. I must write it like this: 3 numbers of single phase transformer.
Or a 3 phase transformer as a single unit when 3 single phase transformers you use it is
called the bank of 3 phase transformers.
(Refer Slide Time: 07:19)
So, let us start with this. Now suppose let us assume that, I have 3 numbers of single phase
transformer each of rating say 5 KVA 200V/100V 50 Hz single phase transformer; this is
the rating of each of the 3 transformers I have used ok.
Therefore, each transformer will have its primary and corresponding secondary. For
example, so there are 3 three such transformers you take 3 numbers, 3 such transformers.
Now the moment there are 3 transformers for identifying the primary and secondary coils
so far in single phase there was one primary one secondary.
So, terminal markings was not that important although I could do it is only HV side LV
side that is all, but in case of 3 transformers there will be 6 number of coils, each
transformer will have two coils one primary one secondary, one high voltage one low
voltage. Therefore, it is essential to mark the terminals of the transformer appropriately.
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So, what I will do is this suppose this is the transformer one, its primary and this time for
convenience I am sketching the secondary of one transformers like this.
And I will mark the terminals as 𝐴1 and 𝐴2 generally capital letters are used for the high
voltage side and small letters I will tell it is small 𝑎1 small 𝑎2 lowercase letter and this is
capital letters so this is one transformer. And also I will assume that this terminal capital
𝐴1 and small 𝑎1 are dot and that I know how to ascertain by doing some test.
So, polarity test I will do put the markings so this is one transformer and this transformer
I will call transformer A. In the same way I will have the second transformer which I am
drawing below, 𝐵1 and 𝐵2 you understand now and its secondary small 𝑏1 and small 𝑏2 .
And similarly these two fellows are dots, so 1 1 are dots and the third transformer in the
same way it will be 𝐶1 capital 𝐶1 capital 𝐶2 and small 𝑐1 small 𝑐2 . Therefore, I have very
clearly identified the terminals it looks like I named the transformers in A B C, and I will
assume it will be supplied from supply A B C that is why A phase B phase C phase like
that, this polarity marking is essential I am telling you.
Now suppose what I will be doing, so the rating of this winding is 200V, this winding also
200V rated voltage. And similarly rated voltage of this secondary coils are also 100V,
these I know and also I know the rated current of this side as you know
𝐼⏟
𝐻𝑉 =
𝑅𝑎𝑡𝑒𝑑
𝐼⏟
𝐿𝑉 =
𝑅𝑎𝑡𝑒𝑑
5000
= 25𝐴𝑚𝑝
200
5000
= 50𝐴𝑚𝑝
100
So, rated current of these transformers are also known rated voltages are known and I have
the mark the terminals. Now what I will do is this suppose I want to connect the primary
coils in star ok, individually I know there each one these; these; these are individual
transformers. But as I told you my supply is A B and C, 𝐴𝑠 , 𝐵𝑠 , 𝐶𝑠 and I will assume the
supply phase sequence to be supply terminals, which is a 3 phase system I will marked as
𝐴𝑠 , 𝐵𝑠 , 𝐶𝑠 and phase sequence is phase sequence is 𝐴𝑠 -𝐵𝑠 -𝐶𝑠 you understand this.
So, this is about the supply so this is supply. Now these 3 terminals 𝐴2 𝐵2 𝐶2 suppose I
short them, then it becomes a star connected primary star connection means 3 coils are
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there 3 terminals of each of them you short ok, so this is how you short it. And I will give
supply like this 𝐴𝑠 I will connect to 𝐴1 , 𝐵𝑠 I will connect to 𝐵1 and 𝐶𝑠 I will connect to 𝐶1 .
Now, since it is star connected, the supply voltage whatever it is line to line that divided
by √3 will appear here ok; since the rated voltage of the transformer is 200V therefore, I
can apply a line to line voltage of 200√3 safely is not line-to-line voltage 200√3V.
Because after all these transformers are identical transformers they will have equal
parameters, so √3 times less voltage will come across the primary windings of each of
these 3 transformers. If it is two winding because transformer rating was 200V/100V
therefore, I can apply a voltage of 200√3V.
So, that rated voltage will then be applied across these coils primary, secondary coil I have
not done anything they are separate now nothing has been done. Therefore, this is the thing
now let us see what is the phasor diagram of this applied voltage and let us assume this is
the thing now let us see. Suppose the phasor diagram of this side is mind you I have shorted
𝐴2 𝐵2 𝐶2 .
So, this is the point 𝐴2 𝐵2 𝐶2 they are shorted they will be at same potential let this be
called 𝐴1 , which has been connected to 𝐴𝑠 this is same as 𝐴𝑠 supply 𝐴𝑠 ; this point is 𝐵1
which is connected to supply 𝐵𝑠 and this point is 𝐶1 which is connected to supply 𝐶𝑠 ; 𝑠
stands for supply therefore, this point is the neutral point I can say.
So, 𝐴1 𝐴2 is the voltage phasor here, the moment and each length here it is 200V that is
why this length is 200√3 line to line. Now the moment you apply like that this transformer
see the rules of the single phase transformers only we will be using to understand what is
going to happen if you connect in a either star or delta we will see those things.
But I am sure about one thing, that the voltage phasor if it is 200V here it must be 100V
and induced voltages are in phase. Therefore, the secondary voltages if I draw with a
different color it will be half of this length although I am just drawing 𝑎1 𝑎2 and it will be
parallel to capital 𝐴1 𝐴2 because induced voltages or applied voltages are all in phase.
So, for this single phase transformer, this is what is going to happen you will get a voltage
like this. Similarly for the second transformer and mind you a 3 phase balance system
means they are 120° apart is not it lengths are same supply voltages these are 120° apart.
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Therefore, 𝐵1 𝐵2 the second transformer what is the applied voltage which phasor is the
applied voltage 𝐵1 𝐵2 what will be the induced voltage in small 𝑏1 𝑏2 a line parallel to this
and since 𝑏1 is dot so it will be just parallel to this like this.
Although since I have not connected any of this they are allowed to leave alone therefore,
they there will be a voltage parallel to 𝐵1 𝐵2. Similarly there will be another voltage
available parallel to this C phase voltage which will be 𝑐1 𝑐2 then this will be the same
secondary coils I have not collected if you connect volt meters or see in the oscilloscope
you will find 3 voltages existing across the secondary terminals which are 120° apart.
Because 𝑏1 𝑏2 is parallel to this and 𝑐1 𝑐2 parallel to this angle between these two is 120°
here also the angle between these two is 120° and so on. Now what I will be doing is these
three terminals, I will short you imagine this three are shorted the moment you short this
3 secondary also gets connected in star.
So, this is star connection and then the secondary also becomes star connected and the
moment 𝑎2 𝑏2 𝑐2 are shorted this 3 phasor cannot live in isolation because 𝑎2 𝑏2 𝑐2 I have
forced to these three terminals to be at equal potential their potentials cannot be different
if it is connected by a piece of wire like this. Therefore, these 3 phasors, this, this and this
cannot now remain in isolation, but what will happen is this, it will be like this that 𝑏1 , 𝑎2
is connected to 𝑏2 and 𝑐1, 𝑐2 are you getting 𝑎2 𝑏2 𝑐2 I have forced them to be at
equipotential these 3 phasors were in isolation they cannot be. So, it will be like this.
So, what I have done I have applied 200√3 line to line voltage here, which will ensure
that across each of the primaries of these three individual transformers 200V appear of
course, these 200V are fine, but they will be displaced by time because of the supply is
balanced. And but I know if this is 200V it cannot be it must be 100V because of the rating
of individual transformer being known, 100V is known.
And then I draw the primary phasor diagram, then I have told you 𝑎2 𝑏2 𝑐2 a shorted so do
not leave this 3 phasors in isolation as for your connection you put them in. But one thing
is clear 𝐴1 𝐴2 will be parallel to small 𝑎1 𝑎2, 𝐵1 𝐵2 parallel to this one and this length is
200V this is 100V and so on that will be maintained single phase only. See, although we
have connected it in a particular fashion rules of single phase transformers will only I will
apply to carry on further.
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So, this will be the phasor diagram and then I will say this is a 3-phase transformer which
is connected in star-star that I will write like this. And where are the output terminals
output terminals are here, if you bring it in this way these will be the output terminals and
I will write here perhaps small 𝑎 output 3 phase terminal 𝑎 this one is 𝑏 and this was in to
𝑐 this is output terminals going to load going to 3 phase loads.
So, it will be like this therefore, I have applied 200√3 line to line voltage here what will
be line to line voltage across 𝑎1 𝑏1? It is this length line to line voltage once again will be
balanced 3 phase. And the line to line voltage on the secondary side so if I write it like this
𝑉1𝐿𝐿 = 200√3
𝑉2𝑙𝑙 = 100√3
Small letters for the low voltage side.
So, I have been able to change the level of a balanced 3 phase voltage, which is 200√3 V
to a level 200√3 V that is how I have been able to make is that clear. Now the second
thing is we will discuss I mean I will be going a bit slowly do not worry about that. So,
this is the thing so there I will now draw a bit faster to say about loading of this transformer.
(Refer Slide Time: 26:29)
For example what I have done this is 𝐴1 𝐴2 this is corresponding small 𝑎1 𝑎2 these are dots
similarly, this is 𝐵1 𝐵2 the previous diagram only I am drawing 𝑏1 𝑏2 these are dots and this
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is the third phase 𝐶1 𝐶2 and this is small 𝑐1 𝑐2 is it not. And what connection I have made?
I have made this a star connection and here I have connected 200√3 volt three-phase
supply 𝐴𝑠 -𝐵𝑠 -𝐶𝑠 and here the line to line voltage applied is 200√3 why 200√3 because
rating of each of the transformer was 200V each one is single phase 200V/100V 5 KVA
single phase transformer.
And rated current of the HV side was 25Amp this was 25Amp rated current and this rated
LV side will be higher 50Amp two times and what I have done is I have shorted this and
I have taken output here. These are my output terminal 𝑎, 𝑏, 𝑐 who had 100√3 you will
get line to line that is what I have told I am just repeating this repeating this pins.
Now, the question is this supply I will connect a balanced 3 phase load ok, here I will
connect a 3 phase balanced load if you connect a balanced 3 phase load which may be star
or delta connected I do not mind. But the only restriction I will put what is the maximum
current I can what is the maximum line current on the secondary side will be the LV side
rated current is 50Amp and it so happens that this current the line current because it is star
connected happens to be same as the winding current LV side winding current.
Therefore maximum current I will allow to flow here is 50Amp here also 50Amp here also
50Amp ok. And be rest assured the moment through the dot 50Amp goes in for this
transformer, this fellow will draw 25Amp. Similarly this fellow will also drop 25Amp this
is also 25Amp will these currents be in phase no not at all they will be in time phase
difference depending upon the power factor of the load etcetera, but magnitude of the
currents will be same. Each one will start drawing.
So, I have taken 3 numbers of single phase transformer whose rating is this. Now what is
the total k VA total k VA will become then will become as a 3 phase transformer; as a 3
phase transformer is any side you can calculate for example, if you calculate from this side
you must be knowing that in a 3 phase system if the line to line voltage is 𝑉𝐿𝐿 .
If the line current is 𝐼𝐿 for balanced thing the total KVA is
𝐾𝑉𝐴
⏟
𝐵𝑎𝑙𝑎𝑛𝑐𝑒𝑑 3−𝑃ℎ𝑎𝑠𝑒 𝑆𝑦𝑠𝑡𝑒𝑚
331
= √3𝑉𝐿𝐿 𝐼𝐿
irrespective it takes care of both star and delta that is why people love to use this formula
so often, forget about start delta tell me what is the line to line voltage? only thing load is
balanced so that is a condition line to line voltage is 𝑉𝐿𝐿 line current is 𝐼𝐿 total k VA will
be √3𝑉𝐿𝐿 𝐼𝐿 .
So, if you calculate from the secondary side the total k VA then will become
𝐾𝑉𝐴
⏟
= √3 × 100√3 × 50 = 15𝐾𝑉𝐴
𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑓𝑟𝑜𝑚 𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑆𝑖𝑑𝑒
which of course, will be same if you calculate from this side which tells you
𝐾𝑉𝐴
⏟
= √3 × 200√3 × 25 = 15𝐾𝑉𝐴
𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑓𝑟𝑜𝑚 𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝑆𝑖𝑑𝑒
Nothing surprising each transformer was rated at 5 KVA.
So, you have connected it as a 3 phase transformer with star; star connection, then total
KVA will become 15 KVA anyway we will continue with this in the next class.
Thank you.
332
Electrical Machines – I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 35
Varians Connections 3 Phase Transformer – I
Welcome to lecture 35 and we have been discussing about 3 Phase Transformers and 3
single phase transformers I have taken earlier, each of rating this much.
(Refer Slide Time: 00:31)
And I showed you a simple star star connection ok. The terminals markings are essential,
along with polarity marking. So, I have assumed because three transformers are there, I
will name them A transformer B transformer C transformer. Capital 𝐴1 , 𝐴2 small 𝑎1 , 𝑎2
are the are respectively high voltage and LV side terminals and I have presumed that these
ones are the dots for all the phases so that is there, so they here also it was taught anyway.
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(Refer Slide Time: 01:23)
So, and then I told you how to make a successful star connection and the rating of each
transformers were 5 KVA and maximum KVA this 3 phase transformer, which has been
formed by using three single phase transformer each of rating this will become 15 KVA
and it is pretty fine; I mean each one is 5 KVA you can handle 15 KVA that is good. Now
today I will go slightly we will try to tell you this one very important thing why polarity is
so much important. For example when you take three, because we will start delta
connection so, it is better to appreciate this point what do I mean by star connection?
(Refer Slide Time: 02:27)
.
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If suppose three resistances are there, three resistances are there I ask you to connect them
in star means any one terminals from each of these elements you short them. Either you
short it is a star connection or you could short it like this also, this and this will be also star
connection no difference.
So, star connection means two terminal devices three are there. So, one from each you just
short them without asking any polarity or this that is what I am telling similarly what is
delta connection? So, this is also star this is also star suppose their resistance or inductance
or capacitance any way you connect it is fine. Now suppose I ask you to connect this thing
in delta three resistances, what is this connection, connection this connection in language
we tell it that connect them in series first, series this series this series and close this path it
becomes delta connection.
And from the junctions you take the output. Although, we draw it like this that is fine here
also we are doing this second resistance third resistance, you connect and from the
junctions you take the output and it gets delta connected is this different from this? No.
So, this is the way we will be drawing delta connected here coils will be in series and
closed and from the junctions you take the outputs that is it.
If that be the case then suppose I say that, somebody a has a three single phase transformers
like this 𝐴1 , 𝐴2 ; its secondary small 𝑎1 , 𝑎2 . And then and then later three are there, so 𝐵1,
𝐵2 small 𝑏1 , 𝑏2 and this is small 𝑐1 𝑐2 identical transformer 𝐶1 𝐶2 .
Suppose he does not care a careless person for example, he argues that star connections, I
know one from each I will short and other three free terminals will become the terminals
of this star connected devices. So, he does like this. For example, he does not care about
polarity, he tells that this is also my star connection why not? And he tells that you connect
this although I know for discussion purposes, but I know that capital 𝐴1 small 𝑎1 are dots
capital 𝐵1 small 𝑏1 dot, capital 𝐶1 small 𝑐1 are dot.
This is the actual thing which is prevailing, but he does not care what he does he tells that,
this is the thing connect this star in the primary going by this general definition of star; and
decides that he will give supply to supply he has got three terminals 𝐴𝑠 𝐵𝑠 𝐶𝑠 of phase
sequence ABC. He gives supply to here 𝐴𝑠 , then this 𝐵𝑠 supply terminal what he does he
gives it to here are you getting me? 𝐵𝑠 and 𝐶𝑠 supply he gives it to there. And he tells that
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I have connected the primary in star and I have energized; secondary he has not done
anything.
Now, in this case if you look carefully the supply is balanced 3 phase ok, supply is a
balanced 3 phase system and you have connected like this. So, supply 𝐴1 is connected to
𝐴𝑠 𝐵2 terminal has connected to 𝐵𝑠 . So, with respect to neutral it will be just not correct in
terms of phase sequence; because what we know that 𝐴𝑠 𝐵𝑠 𝐶𝑠 is a balanced 3 phase system
for example, 𝐴𝑠 this is this is suppose supply neutral this is your 𝐵𝑠 , this is 𝐶𝑠 .
Now, you have energized the primary of this coil with a voltage this voltage and this 𝐴𝑠 is
𝐴1 is not? And your this 𝑁𝑠 is equivalent to your 𝐴2 ,
1
√3
times. 𝐵𝑠 I have connected to 𝐵2
is not I have connected to 𝐵2 therefore, this 𝐵𝑠 terminal is 𝐵2 and 𝐵1 is connected here.
So, 𝐵𝑠 is 𝐵2, I have connected. And here is your 𝐵1, similarly 𝐶𝑠 I have connected to 𝐶1
and this is 𝐶2 that is fine. This is what I have done.
Now, on the secondary coils there will be induced voltages, following the rules of single
phase transformer capital 𝐴1 𝐴2 , you will get a voltage here because these polarities are
correct polarities are like this. So, you will have a voltage here 𝑎1 𝑎2. For the second
transformer, you will have 𝐵𝑠 is connected to 𝐵2, you will have a voltage phasor parallel
to this; 𝑏2 𝑏1 that is what you will get.
And here 𝐶𝑠 is 𝑐1 𝑐2 you will get ok. Now the point I want to make it without doing polarity
test, you should never try to connect the 3 phase transformer. Either in star or delta and the
difficulty what will come that I am telling. I am telling somebody connects like this
although the polarities actual polarities are like this, but he does not care star connection
ok, I connect supply then what I am telling. Look at the individual applied voltage across
the primaries.
For the A transformer it is 𝐴1 𝐴2 . For the B transformer it is 𝐵2 𝐵1, 𝐵2 is connected to 𝐵𝑠
which is 120° apart. Because, your supply is balanced 3 phase supply with phase sequence
𝐴𝑠 𝐵𝑠 𝐶𝑠 ; for 𝐶1 of course, 𝐶𝑠 is connected to 𝐶1 . Therefore, induced voltage in the
secondary’s of each of the coils, will be parallel to the primary voltages, which for A phase
𝐴1 , 𝑎1 ; 𝐴2 , 𝑎2 parallel to this, for C phase 𝐶1 𝑐1; 𝐶2 , 𝑐2 ; for B phase 𝐵2 𝐵1.
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So, it must be this way getting the point. Now suppose he says that secondary I have to
connect in parallel, he has quite arbitrarily connecting the things. For example, let us
assume he is a crazy man he is telling that star connection, he decides that he will connect
like this. Secondary these voltages are available in this way. And he tells that I will connect
the secondary also in star in this way.
If he does, then these phasors are no longer staying in isolation some of the conditions you
have put. What is this condition? 𝑎2 , 𝑏2 , 𝑐2 you have forced them to be at same potential.
Therefore, this three phasors I will write here so this is the thing.
So, here from this I now say that ok, this is how you have connected then these three
phasors will have their existence like that you cannot do anything. This is the this follows
from single phase transformer fundamentals. But what we have done is, you have made
𝑎2 , 𝑏2 , 𝑐2 to be at forced them to be at same potential; so 𝑎1 𝑎2 and this 𝑏2 is there, but
𝑏1 𝑏2 will be parallel to this fellow that will not you cannot do anything.
So, move it parallel bring it here, here is the most important point 𝑏2 𝑏1. 𝑐1 𝑐2; 𝑐2 has been
joined with 𝑎2 , 𝑏2 . Therefore, it will be like this 𝑐1 𝑐2 that is what you will get. Length of
each of them will be 100V provided you have applied here 200√3. Length of this at 200V,
so 100V; 100V; 100V and then he tells that from this you take the output 3 phase output.
Is he going to get a balanced 3 phase voltage? And that is, no.
You get some three terminals, the voltage of each phase looks like same, but they are not
120° apart. So, a balanced 3 phase voltage you will not you are not getting. Therefore, a
just from the definition of star connections of passive elements one should not proceed. In
a 3 phase transformer connection, this must be connected of course, just to point it out. If
he would have connected in the same way here while connecting I think you are getting
my point. So, I will draw it here, what I am telling the secondary I am telling.
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(Refer Slide Time: 15:49)
He also connects the only the secondary’s I am drawing, I 𝑎1 , 𝑎2 , 𝑏1 , 𝑏2 and 𝑐1, 𝑐2 . Now
in the same way star connection he does in the same way like this as he has done in the
primary. Suppose he has drawn in the same way, in that case of course, this secondary
voltage is will be balanced 3 phase voltages why? Because, if you connect like that what
we have done 𝑎2 , 𝑏1 and 𝑐2 you have shorted. So, 𝑎2 and 𝑏1 you have shorted. So, bring
this phasor here; 𝑏2 𝑏1 and 𝑐2 bring 𝑐2 there, parallel to this you will get a balanced 3 phase
supply that is there. But it is left to chances whimsically if you do you do not know.
And not only that, the terminals are marked as that is why terminal markings are very
essential and we would also like to say that supply you are giving to 𝐴1 , 𝐵1, 𝐶1 , take the
supply output from 𝑎1 , 𝑏1 , 𝑐1. Here although you will get balanced 3 phase supply if he
connects like that, but there is a mix up see this is 𝑎1 , that is 𝑏2 this is 𝑐1.
So, to avoid all confusions, we must adhere to the terminal markings. And this terminal
markings business becomes much more crucial if it is delta connections ok. Star
connection its somebody connects just like that without we with any due regard to the
polarity, then this kind of situations may happen you may not get to balanced 3 phase
output voltage of although you have connected a balanced 3 phase voltage there.
And if you see and you can argue I have connected the primary in star, secondary in star,
still I am not getting balanced 3 phase voltage. It is simply because; you have not followed
the proper dot polarity anyway. So, we have discussed about star star connection, correct
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star star connection here. Now this star star connection, as you can see which I have done
here in the first slide is called, star star Y. Generally, this is how it is written Y y star star
small y for the low voltage side, capital Y for the high voltage side.
And then, there is a number written 0° besides it Y y 0. It means that, the primary and
secondary are star connected and phase displacement of the A-phase voltage, with respect
to neutral of the secondary side. And A-phase voltage with respect to neutral of the primary
side they are same, no phase displacement vertical 𝑏1 𝑏2 this is the secondary neutral 𝑛.
So, with respect to neutral B phase voltage is in phase with B phase voltage of the primary
side 𝐵1 with respect to neutral. So, sometimes it is called Y y 0. In this connection, I will
use better a new page. So, better do not do star connection not like that ok, this is it is not
at all suggested to do like that. You always carry out the connections as per.
(Refer Slide Time: 20:33)
So, let me come back to this one, 𝐴1 , 𝐴2 and I will be henceforth drawing very quickly
and you have understood me hopefully what I mean this these are very important. 𝐵1, 𝐵2
primary, its secondary 𝑏1 small 𝑏2 these are dots. And this is the third transformer I have
named it C transformer, that is why its terminals are like this ok. This is the proper polarity
connection. Now and suppose let me go like this, I have joined 𝐴2 , 𝐵2, 𝐶2 , I have connected
supply 𝐴𝑠 to 𝐴1 ; supply 𝐵𝑠 and supply 𝐶𝑠 , I have connected like that phase sequence is A
B C.
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So, that the primary applied voltage I will connect, it will be 𝐴1 which happens to be 𝐴𝑠
supply. This three points I have joined, they would must be at the same potential and that
is the neutral. And then supply 𝐵𝑠 I have connected to 𝐵1 only. So, this is capital 𝐵1, which
happens to be same as supply 𝐵𝑠 . And then your and these are 120° apart; 𝐶1 and this is
also 𝐶𝑠 generally, 𝐴𝑠 , 𝐵𝑠 , 𝐶𝑠 we may avoid. And in the secondary as I told you I have got
suppose I have not connected anything, but these things are available these voltages, 𝑎1 𝑎2
parallel to this 𝐴1 𝐴2 ; 𝑏1 𝑏2 parallel to 𝐵1 𝐵2. In this case and this 𝑐1 𝑐2 parallel to this 𝐶1 𝐶2
these are available.
Now, suppose somebody says I will connect the secondary coils in star, but this time I will
join it like this. I will short 𝑎1 , 𝑏1 , 𝑐1 ok, I will short it. If you do this and I will take output
from 𝑎2 , 𝑏2 , 𝑐2 , then what he is essentially doing, the moment you connect this phasors
are not in isolation, 𝑎1 , 𝑏1 , 𝑐1 are at same potential. So, what you do is this you have to
now move this phasors in such a way that 𝑎1 , 𝑏1 , 𝑐1 are at the same potentials.
But nonetheless 𝑎1 𝑎2 it will be parallel to capital 𝐴1 𝐴2 that is the boundary condition.
Because, single phase transformer each one of them is. Therefore, they must satisfy the
principle of transformer. So, the way if you connect these this is suppose 𝑎1 𝑎2 now I have
joined 𝑏1 with 𝑎1 therefore, what I will do is this I will move this phasor parallel and put
it here 𝑏1 𝑏2 getting the point. Similarly 𝑐1 𝑐2 I will put on a parallel and put it here 𝑐2 , 𝑐1.
Then you know it is also a correct balance 3 phase output voltage you will get across 𝑎2
𝑏2 𝑐2 . But this time this is the neutral this is the secondary neutral 𝐴2 𝐵2 𝐶2 , is the primary
neutral, then I will say and also see the supply phase sequence is ABC everything is fine.
And there is no mix up of this terminals if it is 𝑎2 then 𝑎2 𝑏2 𝑐2 . Earlier case there was a
mix up although balanced 3 phase you will be getting here, but it is 𝑎1 it is 𝑏2 , 𝑐1.
Technically correct I mean it will work, but the point is I should be very systematic in
doing that, so that things are standardized there are.
So, this is the thing here is also no mix up 𝑎2 𝑏2 𝑐2 balanced 3 phase sequence is ABC like
that. And if you apply here each one of them 200V this length the each one of them is
100V of course, scale I am not properly drawn and these are much less. Here I think you
got the idea just to explain that I am do here.
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So, anyway these are 120°. So, balanced 3 phase voltage will be available. Now this
connection diagram, how do I specify? Earlier one was Y y 0 and this one I must then say
it is Y y star connections secondary, star primary and perhaps I will write 180°. Because,
a phase voltage with respect to neutral of the secondary side is 180° A phase voltage of
the primary is this these are parallel know. So, A phase voltage 𝐴1 with respect to the
primary neutral and a phase voltage of the secondary with respect to neutral of the
secondary side, they are 180° apart the that is why it is called Y y 180° ok.
And sometimes we will soon tell you that this is also people say it is Y y 6 why this 6? I
will come slightly later, but I hope you have understood the importance of dot connections,
you cannot just have a general way of connecting star. Three resistance do not bother
connect, in anyways star star delta delta, but in case of transformer connection you should
be extremely careful. So, I will now discuss one delta connection let us see.
(Refer Slide Time: 27:23)
So, I take suppose I want to do star delta connection of the same 3 phase transformer. So,
once again follow me follow these steps, draw the coils 𝐴1 , 𝐴2 . Each secondary small 𝑎1 ,
𝑎2 second transformer B coil; 𝐵1 𝐵2 second transformer b coil 𝑏1 , 𝑏2 and third transformer
𝐶1 , 𝐶2 and its 𝑐1, 𝑐2 . Number of turns of each of these secondaries are same, each of the
primaries are same ok.
That is understood because the ratings of each of the single phase transformer, I have taken
three numbers of this transformer what is that? 200V/100V, 5 KVA 50 Hz, single phase
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transformers. Three numbers I have taken each one of them is this. Now what I say then I
will I have as I have told you I will always maintain these dots I have marked with one
and other things left of this so, this is the dot convention.
Now, suppose I connect them in star proper star 𝐴2 , 𝐵2, 𝐶2 . And here is my supply
terminals 𝐴𝑠 , 𝐵𝑠 , 𝐶𝑠 phase sequence is, A-B-C. Therefore, the primary voltages I can
immediately draw, is not? That is this one is 𝐴1 𝐴2 . And 𝐴1 is 𝐴𝑠 that I am not writing.
Similarly here it is 𝐵1 𝐵2, 𝐴2 , 𝐵2, 𝐶2 , I shorted; 𝐶1 𝐶2 this whole thing is the primary neutral.
And the length of each one is 200V because I let us presume that I have applied 200√3
here.
Now, on the secondary as I told you next step is they first keep them in isolation and you
have this three phasors available with you, whose length is half of this length. And each
one of them is parallel to their respective primary voltages. So, 𝑏1 𝑏2 parallel to this line
and this line and this line 120° apart, so angle between them is also 120°.
Similarly, here parallel to 𝑐1 𝑐2 length of each one of them are same this slightly has
become more. So, this is the thing I hope you have understood, so this is how it is there.
Now what is delta connection? Delta connection means you have to connect for three
rheostat I was telling now, let me repeat this is. This way always think now you are now
much more matured, connect them in series closed this series and from the junctions take
the output, it has become a delta connection like this like this.
Now, here you see that suppose I connect without much discussion suppose somebody
knows this and he connects it like this he joins these two points. He joins these two points
and he is planning to join these two points. And suppose he has connected a switch there
to complete the delta suppose. Now, this three phasors are in isolation the moment you
make a connection like this, they are no longer in isolation.
So, now you position these phasors, based on these connections. For example; for example,
here was 𝑎1 𝑎2, I draw that once again 𝑎1 𝑎2, but I have connected 𝑎2 with 𝑏1 . So, what I
will do, I will ship this phasor here, keeping it parallel to this and put this 𝑏1 𝑏2 here are
you getting me? Because, I have connected so 𝑎2 and 𝑏1 must be at same potential.
Earlier I was joining 𝑎2 , 𝑏2 , 𝑐2 or 𝑎1 , 𝑏1 , 𝑐1, but I have joined only this 𝑎2 and 𝑏1 and
certain they are at same potential. So, move this phasor here, then you have to connect a
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what you have done what connection you have made 𝑏2 with 𝑐1. So, move this phasor
here, such that 𝑐1 is connected here 𝑐1 is put there. And 𝑐2 is here is not, this is how it will
be connected got the point?
Now, suppose I take a voltmeter and connect across this switch, I measure the voltage
across 𝑎1 𝑎2 this all will be 100V. Here, voltage across this coil voltage across this coil V;
what will be the voltmeter reading across this coil? That is 𝑉𝑐2 𝑎1 here is a switch which is
kept open, I have connected them in series, but somehow I have I am telling you I have
not closed it yet. Tell me what is the voltmeter reading what will be the voltmeter reading?
If you can tell me, you are half way through yes, voltmeter reading is 𝑉𝑐2 𝑎1 is not 𝑎1 and
𝑐2 these two points. And what is the voltage between 𝑐2 and 𝑎1 it is 𝑉𝑐2 𝑎1 = 0. Therefore,
even if it is closed 𝑐2 will whether it is closed or not, 𝑐2 and 𝑎1 there is no potential
difference, so you can close this no problem, no circulating current. Although, each one of
the coil has become a seat of emf they are connected in series.
And you are closing those, but these are 120° apart balanced 3 phase voltages. So, resultant
will be 0, no circulating current nothing. So, this is a successful delta connection. We will
continue with this in the next class.
Thank you.
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Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 36
Varians Connection 3 Phase Transformer – II
(Refer Side Time: 00:24)
Welcome to lecture number 36 and we have been discussing about the 3 Phase
Transformer Connections. Recall that last time, I was discussing about star delta
connection. And referred to this diagram, where the primaries are connected in star HV
side applied voltage is 200√3, then I told you when this 3 coils no connections with this
red markings have been made. Then these phasors will maintain their 120° phase
difference, but they will be in isolated conditions they are separate.
But the moment you connect it in this way suppose, somebody says delta connections
means series and closed finally, then he joins 𝑎2 𝑏1 𝑏2 𝑐1 𝑐2 and before connecting 𝑐2 with
𝑎1 suppose, I have connected a shorting switch which is opened initially. And I would like
to know; what is the voltage between these two points? Because the why have connected
a shorting switch with the understanding there will be induced voltage here induced
voltage there induced voltage there and I am going to short these 3 voltage sources
connected in series will not there be any circulating current things like that perhaps
prompted me to connect a switch preventive measure. Let me try to understand, what is
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the voltage existing there? Because mind you this is the source of EMF, each one of them
has become small 𝑎1 𝑎2, 𝑏1 𝑏2, 𝑐1 𝑐2, because of what because they are primaries have been
energized with some voltages.
So, anyway there was a shorting switch; switch was open I want to examine; what is the
voltage between these two? How to do this 𝑎1 𝑎2 these things are there. So, based on this
connection 𝑎1 𝑎2 I bring it parallel 𝑎2 is connected with 𝑏1 . So, position maintained 𝑏2 𝑏1
parallel here, but 𝑏1 is connected to 𝑎2 , place it here and then 𝑏2 with 𝑐1 place it here.
Then you see 𝑐2 and 𝑎1 will coincide although I have not connected 𝑐2 𝑎1 here. It simply
tells you that there is no voltage existing there, not that I have forced it to be at the same
potential yet, but what will be the voltmeter reading with this open voltmeter reading will
be 0. Then I asked myself each one of them is a source of EMF, they are 120° apart the
resultant voltage existing between 𝑎1 and 𝑐2 that has become 0. Therefore, no EMF exists
between these two points therefore, you can close it does not matter.
So, each coil will have induced voltage the circuit is closed now and you get a successful
delta connection. And there will be no circulating current here mind you, there cannot be
then what I told where from to take the output that I will discuss now.
(Refer Side Time: 04:18)
So, this is the thing. So, here are the coils very quickly we you are also now used to it draw
it and do not forget to mark the terminals, 𝐵1 𝐵2 and 𝐶1 𝐶2 small 𝑎1 𝑎2 secondary of A,
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secondary of B and secondary of C. And with the understanding that all 1 terminals 𝐴1 𝑎1 ,
these are dots polarity 3 individual transformers and I have shorted this fine and here I
have given supply 𝐴𝑠 , 𝐵𝑠 and 𝐶𝑠 .
So, what I am telling the primary EMFs are like this 𝐴1 𝐴2 𝐵1 𝐵2 and 𝐶1 𝐶2 . 𝐴1 is 𝐴𝑠 , 𝐵1 is
𝐵𝑠 and 𝐶1 is 𝐶𝑠 on the secondary side what I am telling you have got these 3 voltages
available now, not yet connected I will play with these phasors. So, that it becomes a
meaningful delta connection and then 𝑏1 𝑏2 will be parallel 𝑐1 𝑐2 will be parallel. Why this
will be parallel? Because it is a single phase transformer, it must be parallel to its respective
primary voltage and so on. And since the input voltages are 120° apart, they will be also
120° apart.
Now, what I am telling here is your 𝑎1 𝑎2 ok. And suppose this time, I have joined 𝑎1 with
𝑏2 suppose 𝑏1 with 𝑐2 . Delta connection I will make they must be in series and these also
should be closed, I have not yet closed.
Let us first examine, what is going to happen? Since I have already joined 𝑎1 with 𝑏2 ; 𝑎1
and 𝑏2 cannot be an isolation they must be at same potential. So, what I will do is, I will
move this phasor I will now, since 𝑏2 has been connected with 𝑎1 I will move it here 𝑏2
winding and place it here ok. Anyway this I will write 𝑏2 are you getting this 𝑏2 𝑏1 , I will
put it like this. Similarly, I have joined 𝑏1 with 𝑐2 . So, 𝑐1 I will move and 𝑐2 has been
connected with 𝑏1 , it is move parallel and 𝑐2 will be connected here.
But only thing is they are of same length, it is because of my imperfection there in drawing.
It will then come here together; are you getting? 𝑐1 and this was 𝑎2 these becomes a
anyway you have got the idea.
So, what will be the voltage across 𝑎2 and 𝑐1 0 voltage. Not because I have joined 𝑎2 and
𝑐1 these three are sources of EMF voltage is 0 therefore, I will be very confident to short
this. So, this is also a successful delta connection and I will take the outputs from here
from the junctions; 𝑎1 𝑏1 𝑐1 that is the whole idea ok.
Therefore two ways; a successful delta connections will be achieved. I will show you one
case, where I make it delta the way suppose I do not care about polarity try to make a delta
connection, what will be the consequence that I will show now.
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(Refer Side Time: 09:50)
Suppose somebody says that, this is the primaries and these are the secondaries. So, that
you can appreciate that the points, here A is 𝐴1 𝐴2 𝐵1 𝐵2 and 𝐶1 𝐶2 . And he makes star
connection this way he makes. And mind you dots are known, but he does not care. These
are 𝑎1 𝑎2 𝑏1 𝑏2 and small 𝑐1 𝑐2 .
What he does, he knows delta connection means connect 3 coils in series and close this.
And suppose he says, he connects it in this way are you getting? He joins these points. So,
these two are in series, then he makes this series connection by joining these two. And he
is planning to join this two points also, now to make a delta he tells that three elements
must be connected in series and from the junction, so, you take the output. If he does like
that what is going to happen? Let us see. So, primary is this primary voltages are 𝐴1 𝐵1
and 𝐶1 , this point is 𝐴2 𝐵2 𝐶2 this is neutral fine.
On the secondary side earlier, when no connection was made these 3 voltages was
available to me; one was 𝑎1 𝑎2 another was 𝑏1 𝑏2. It is conditioned by the primary voltages,
this is 𝑐1 𝑐2 and they were in isolation, but the moment you have joined 𝑎2 with 𝑏1 this
phasor cannot remain in isolation, because 𝑎2 has been connected to 𝑏1 . So, this thing that
is I will move it such that it remains parallel to 𝑏1 𝑏2, but 𝑎2 get connected with 𝑏1 . So,
move it, it will be like this is not that will be the thing.
Then what he has done 𝑏2 he has joined with 𝑐2 is not; that means, he has moved this
phasor and 𝑏2 with 𝑐2 he has joined. So, he will connect it like this, is not? And he is
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planning to short 𝑎1 𝑐1, now there will be a problem because what is the voltage across 𝑎1
and 𝑐1. It is not 0 now it is this length this sorry this point if it is correctly done with 60°
120° it would have been this length. 𝑎1 and 𝑐1 a large voltage will exist, source voltage
will exist secondary after all is becoming a source of the voltage. And if you are planning
to short these two points, you are virtually short circuiting it. Large current will flow
transformer may be damaged and so on.
Therefore in star connection you may make mistake, but in delta connection something
terrible is going to happen, you cannot be whimsical just telling that delta connection
means all series and close it know this you should not do. It in fact, if it is 200V, this is
100V between these two points this length will be 200V.
𝑉
𝑎1 𝑐1 = 200𝑉
⏟
𝑆 𝑂𝑝𝑒𝑛
This is 100V this is 100V. Therefore, proper delta connection is either this way or the other
way. So, this is not the proper delta not proper delta connection, never do it not proper
delta connection.
So, you should be extremely careful. And you will be always be correct provided we have
followed this rule. First find out the dot, put the dots, mark the terminals accordingly do it
everything will be fine, that is why so many words I have spent. So, that you understand
the importance of dot conventions particularly in 3 phase transformer, at least making the
connection itself demands that you understand what is what. So, far as polarity of the
induced voltages are concerned. Now, after telling all these things, let us now so, this is
some star delta connections we are discussing.
(Refer Side Time: 16:20)
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And now star delta connection I will do star delta connection and see what are the variety
is possible star delta connection. So, you take this thing primaries and these are 𝐴1 𝐴2
maybe you will be feeling slightly bored, because same thing I am doing, but you practice
it so, that it will give you confidence. And these are dots, but bear with me. And suppose
I have connected correct star connection here and this side is delta corresponding
secondaries small 𝑎1 𝑎2 , small 𝑏1 𝑏2 and small 𝑐1 𝑐2 and these are also dots ok. And delta
connections only two ways it can be done correctly. So, I choose one and this way I will
connect correct delta, easy to remember either this way or other way. So, this is delta and
this side is star ok.
Now, let us come to be phasor diagram of this whole thing. Primary phasor diagram, if
you see it is like this 𝐴1 and this is 𝐴2 𝐵2 𝐶2 I am not writing 𝐴𝑠 𝐵𝑠 𝐶𝑠 it is understood and
this is 𝐵1 and this is 𝐶1 this will be the thing. Now, the secondary voltages, I will now draw
straightaway like this. Small 𝑎1 𝑎2 cannot, but be here parallel to this it will be there. 𝑏1
𝑏2 parallel to this and I have joined 𝑎2 with 𝑏1 . So, this line I will bring it here 𝑏1 with 𝑏2
and finally. So, 𝑐1 𝑐2 and 𝑐1 is connected with 𝑏2 . So, you place this phasor here, 𝑐1 and
𝑐2 and a secondary is delta primary star ok.
Now, listen to me carefully, this connection then I should write it like this. Y delta
generally Y d you can write. d stands for delta Y for star. Now in relation with this phase
angle, how to specify in the case of star I told you it is either YY 0 or YY 180°. Only these
two things are possible here, what is the thing possible? And how to specify the phase
angle business here, what should I write? That is the thing.
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Now, what I told you that to find out the phase angle, where from I am going to take output.
See it is 𝑎1 𝑏1 𝑐1 . So, I will take output from 𝑎1 the that is this will be my 1 output I will
call it 𝑎1 from the junctions. So, this will be another output secondary output 𝑏1 and 𝑐1 is
another junction, which is the output 𝑎 𝑏 𝑐 is the output which will go to the load.
Therefore, I circle them these are my although 𝑐2 is present, but it is 𝑎1 is there it is 𝑏1 is
there it is 𝑐1 is there that way I will say the output terminals are 𝑎 𝑏 𝑐.
One may say why not you are telling 𝑎2 𝑏2 𝑐2 , it could be also circle and told this is 𝑎
voltage 𝑏 phase this is 𝑐 phase no I will not do that. This was 𝐴1 I will stick to small 𝑎1 ,
that is why it will make you much more discipline. Not that 𝑎2 𝑏2 𝑐2 that way if you mark
things will not work it will work, but be consistent that is what the idea is, so, 𝑎1 𝑏1 𝑐1, 𝐴1
𝑎1 .
Now, why this 𝑎1 is stated? See on the secondary side I have no neutral available star
connection natural neutral is available. What I told you? To find out the phase difference
of the input side and the output side voltage, it is better you consider the a phase voltage
with respect to the primary neutral that is this one. This is the A phase voltage. Go to
secondary side, where is your a phase this one and what is this voltage with respect to its
neutral, but there is no neutral available.
So, what you do? You go to the artificial neutral, in delta where is the neutral? You can
always see here is the at the centroid the neutral is there. Are you getting? Or across abc
you connect three reactors in stars that point becomes neutral with respect to that this
voltage; got the point I hope. So, this neutral is not there in the windings, but artificial the
centroid of the triangle you go. Then you say with respect to the secondary neutral, the a
phase voltage will be this one. Do not forget that the these angles are 60° because of this
120° business, this angle has become 60°.
So, these angles has become 30°, this at this and this I call small n. So, I have got a phase
voltage with respect to neutral of the primary side that is 𝑎1𝑛 . I have got the a phase voltage
of the secondary side with respect to each neutral, which is just not available you have to
artificially create it or whatever it is, but it centroid it will become with respect to that this
is the a phase voltage. Is there a phase displacement between them? Yes, if you draw the
this primary voltage A phase only stick to 1 phase other phase it will be automatically
satisfied because, each one of them is 120° apart.
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So, what you do you draw a line here, parallel to this 𝐴1𝑛 suppose, this phasor this phasor
you shift it here 𝐴1 and n. And it is these angle, then because this is vertical with vertical
how much angle this is making will tell you the respective phase relationship of the
secondary a phase voltage with respect to its neutral.
What is this angle is? 30°. So, I will be rather telling that this is Y d and secondary voltage
is, lagging the primary voltage by 30° is not? Y d (-30°) I could right. Got the point?
Therefore, the this is called the vector group of transformer connection; vector group of 3
phase transformers is this Y d (-30°).
(Refer Side Time: 26:12)
And let us see another connection for example, star delta I will make, but delta this time I
will go other way around for example, this is 𝐴1 𝐴2 this is 𝐵1 𝐵2 and this is 𝐶1 𝐶2 these
are dots not these and these dot we are talking about each secondary dot that is each
secondary these are three separate transformers no question of mutual drop. So, this is dot
𝑏1 𝑏2 and this is 𝑐1 𝑐2 this is also dot.
Now, in previous case I told you there are two valid delta connection, these way I have
connected no not this way ; this way I have connected 𝑎2 with 𝑏1 , 𝑏2 with 𝑐1, 𝑐2 with 𝑎1 .
I will now go other way round 𝑎1 𝑏2 like that. So, here that is other way means, I will now
connect it in this way. That is also valid delta and these three you join and do not worry
the potential difference 𝑎2 𝑐1, even if you do not connect will be 0 that is why you are
allowed to short it that is what I want to tell. Where from and of course, this side is I have
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connected a valid star and here is your supply with phase sequence 𝐴𝑠 𝐵𝑠 and 𝐶𝑠 got the
point.
Where from should I take. So, what will be the phasor diagram first? Phasor diagram will
be 𝐴1 𝐵1 𝐶1 these are 120° apart, this is 𝐴2 𝐵2 𝐶2 and that is the neutral of this side.
On the secondary side, once again I will draw separately first the voltage. I not drawn
separately, I can now draw straight away. Because, this is the thing we got it will be like
this. So, first I will I am slightly drawing higher this is half of this length this is 𝑎1 𝑎2; 𝑏1 𝑏2
is available parallel to this capital 𝐵1 𝐵2, move it parallel and put such that 𝑏2 gets
connected with 𝑎1 ; that is it will come here 𝑏2 , 𝑏1 .
Then what I have done 𝑏1 with 𝑐2 . So, bring this c phasor 𝑐1 𝑐2 phasor such that, it gets
connected to 𝑐2 with 𝑏1 is not? And your connection is complete. So, I will write it here
star and secondary side is delta d now I have to examine what will be this phase angle. So,
what is the next step next step is this is your a phase b phase c phase.
I am going to take the outputs from this 𝑎1 𝑏1 and 𝑐1 from the junctions and I will say this
is my a phase this is my b phase this is my c phase. The reason I told you because capital
𝐴1 small 𝑎1 like that and this is also abc phase sequence. So, this terminal I will mark it as
small a small b and small c. This is the neutral of primary this is the a phase voltage here
𝑎1𝑛 where is the a phase here where is the neutral centroid.
Therefore, your a phase voltage with respect to the secondary artificial neutral is this length
and where is your primary voltage; primary voltage with respect to primary 𝐴1 and n it is
here is your 𝐴1 and n that voltage this you bring it. So, this angle is 30°. So, secondary
voltage is leading or lagging, the primary voltage, neutral voltage.
Student: Leading.
Leading. So, I will write here Yd(+30°). Got the point? Although, I have not told you
anything about when to use star; star connection, when to use star delta connection, when
to use delta star connection no nothing like that we are getting used to how to connect the
transformers correctly polarity marketing are so, important. And what is vector group?
This vector group is also specified in terms of some clock minute hand and hour hand, that
also I will discuss next class.
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But I hope you are understanding, the idea given a 3 single phase transformers without
doing polarity test. Do not go to connect the 3 phase connections, either star; star delta star
it is the it will be then left to chances. Particularly with delta connections you may face
disasters condition short circuiting the supply. So, the point is these dots must be
ascertained separately, terminals are marked be systematic and then connect correct star
and correct delta. 2 correct star connections are possible either short 𝐴1 𝐵1 𝐶1 or short 𝐴2
𝐵2 𝐶2 .
Similarly, two correct delta connections are possible, either go this way after we have mark
the polarities or you go this way. I hope you are not only understanding, but you are taking
interest that is much more important. Please go through this notes and discuss with your
friends, you will like it I hope.
Thank you.
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Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 37
Vector Group of 3 Phase Transformer
Welcome to lecture number 37 on Electrical Machines I and we have started discussing
about 3 Phase Transformer connections. And I have told you a 3 phase transformer can be
constituted by using 3 identical single phase transformers or as a single unit of 3 phase
transformers. So, to begin with I have started of taking 3 identical single phase
transformers and connected trying to connect them externally. So, that a balanced 3 phase
voltage can be transformed into another balanced 3 phase voltage in the secondary.
But while doing so, we should be knowing the a dot markings or polarities of the
transformers and we should be systematic in describing the current. In my last lecture we
just we were seeing that the there is a shift in the secondary voltage with respect to the
primary voltage. And how that shifting by what angle the voltage is shift that is called the
and that is generally written on the nameplate of a 3 phase transformer and that is called
vector grouping. So, we were up to that, but recall that; so this was the thing.
(Refer Slide Time: 02:04)
So, you have a 3 phase transformer and once again very quickly I will tell these are suppose
the this is one transformer A, it has got a primary terminals 𝐴1 𝐴2 and this is small 𝑎1 𝑎2
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is the secondary terminal. Similarly you have the second transformer B and it has got its
primary 𝐵1 𝐵2 and correspondingly it secondary and 𝐶1 𝐶2 and this is small 𝑐1 𝑐2 . And
then you have the supply 3 terminals that is 𝐴𝑠 , 𝐵𝑠 and 𝐶𝑠 .
Now, I have to connect the supply and connect this primary and secondary in a particular
fashion has desired and I told you that while connecting the transformers in star windings
in star either 𝐴2 , 𝐵2, 𝐶2 should be shorted or 𝐴1 , 𝐵1, 𝐶1 can be shorted valid star
connections. And also remember these 1 terminals are dots like polarities at any given
instant of time similarly for the transformer B and transformer C. And these how do I get?
I will test for polarity to this individual transformers put the dots and mark the terminals
accordingly. Anyway this is how the thing is.
So, suppose I say that I will connect the transformer primary in star, I would like to connect
these in star or Y. And secondary I would like to connect it in delta or in letter it is small
d and I will of course, make valid connections. So, what I do is this I make a valid star
connection 𝐴2 , 𝐵2, 𝐶2 shorted and I will connect 𝐴1 with 𝐴𝑠 , 𝐵1 with 𝐵𝑠 and 𝐶1 with 𝐶𝑠 .
And suppose the phase sequence of supply phase sequence is 𝐴𝑠 , 𝐵𝑠 , 𝐶𝑠 ABC, A stands
for supply.
Now, the easier way of connecting this I will going to tell you now we have seen the
interesting features, if somebody connects differently what is going to happen and things
like that. Now, today’s lecture is all valid connections we will be doing ok.
The moment you do this then what you say phase sequence is this and supply is balanced.
If the supply is balanced, then it is best thing to do is draw the supply voltage first. So, I
draw the supply voltage which will be equal to like this 120° apart equal lengths and this
is suppose 𝐴𝑠 , 𝐵𝑠 , 𝐶𝑠 all lengths are equal here balanced.
And note that 𝐴𝑠 is connecting to 𝐴1 , then I will write it is also 𝐴1 is not this point.
Similarly 𝐵𝑠 is connected to 𝐵1, so comma 𝐵1 and 𝐶𝑠 is connected to 𝐶1 commas 𝐶1 and
since 𝐴2 , 𝐵2, 𝐶2 are shorted. So, 𝐴2 is here, 𝐵2 is there, 𝐶2 is there they are shorted, so
their potentials are same and this. In fact, is the neutral of the side, fine?
Now, suppose the secondary side I would like to connect it in delta and there are two
possible deltas delta connect two possible delta connections are there. So, first of all I will
connect it suppose like this; like this series and closed and output I will take from 𝑎1
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wherever 𝑎1 is there 𝑏1 and 𝑐1. Let the secondary output be 𝑎, 𝑏 and small 𝑐, I am not
writing 𝑠 because it will go to load.
Now, based on these you can easily see, the secondary voltages will be since voltage
applied across the primary of transformer A is 𝐴1 𝐴2 . Therefore, this side you have the
secondary voltage small 𝑎1 𝑎2 which will be parallel to this length. So, I write it like this
𝑎1 𝑎2 got the point parallel to capital 𝐴1 𝐴2 .
Similarly, small 𝑏1 𝑏2 the induced voltage in the secondary will be parallel to this 𝐵1 𝐵2
and 𝑏2 𝑎2 are shorted therefore, how to place these? So, 𝑏1 𝑏2 and 𝑏1 is connected to 𝑎2 .
So, so 𝑏1 𝑏2 will be like this here I am just drawing 𝑏1 𝑏2 parallel to this it is available, but
I have joined 𝑎2 with 𝑏1 . So, I will shift this fellow parallelly and bring it here like this
such that it 𝑎2 and 𝑏1 are joined. So, I must write it here 𝑏1 . So, 𝑏1 𝑎2 is this point.
So, 𝑏1 𝑏2 got the point parallel to this similarly 𝑐1 𝑐2 and 𝑐1 is joined with 𝑏2 therefore, this
phasor should be parallelly shifted and it must be this that is 𝑏2 is connected to 𝑐1. So, 𝑐1 𝑐2
got the point and this is absolutely fine.
Now, output see this is a input 𝐴1 is connected to 𝐴𝑠 , 𝐵1 is connected to 𝐵𝑠 , 𝐶1 𝐶𝑠 .
Similarly, I would like to take the output from small 𝑎1 , 𝑏1 and 𝑐1 . So, I circle them just
this will be my output 𝑎1 , 𝑏1 and 𝑐1 although 𝑏2 is connected, but I will call it my 𝑎, this
one, 𝑎1 and 𝑐2 are joined from this I am taking the output I am calling this 𝑎 to be consistent
with the primary side. So, that is the thing.
Now, the question is if you connect like this I should write it as star, secondary is delta.
Now, what is the phase angle, what should I write? Now, while doing the phase angle you
can find out first where is the artificial neutral of the secondary side got the point. So, so
this is the neutral.
So, you see that this will be the. So, this triangle we have identified 𝐴1 , 𝐵1, 𝐶1 similarly
you circle the primaries 𝐴1 with respect to this will be talking. Now, I will examine with
respect to neutral on the primary side there is 𝐴1 voltage, similarly with respect to this
artificial neutral which is centroid of this equilateral triangle and with respect to that
neutral the a phase voltage is this phasor 𝑎1𝑛 .
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Then you draw these a capital 𝐴1𝑁 here suppose thing, this primary a phase voltage with
respect to its neutral is here 𝐴1 you should not make confusion that capital 𝑁 and small 𝑛
is connected no small 𝑛 is not there in fact,, but anyway that is the 𝐴1 phase vector if I that
is vertical. So, I find that the this voltage lags the a phase voltage with respect to the neutral
on the a phase side by 30° and this is the thing.
And I then write it as it is lagging, so may be Yd(-30°) I will be telling it. But one important
thing also you note that it is not only the phase voltages that 𝐴1 with respect to neutral that
maintains 30°. What about the line to line voltage? For example, primary side line to line
voltage 𝐴1 𝐵1 because phase sequence is this. So, this voltage is 𝐴1 𝐵1 which happens to be
equal to 𝐴𝑠 𝐵𝑠 this is the line to line voltage.
On the secondary side, what is the line to line voltage? It is this vertical line and this line
to line voltage two this is 𝑎1 𝑏1 if you just that is also 30° lagging. Not only the phase
voltage with respect to the neutral or artificial neutral that will lag which also means that
line to line voltage is too will be lagging by the same amount this point you must
understand because this is 𝐴1 𝐵1 line to line voltage on the primary side your small 𝑎1 𝑏1
becomes vertical 𝑎𝑏 voltage of the same line to line voltage that lags it will be true as you
can see for 𝑏𝑐 and 𝑐𝑎 phase as well. So, examine only the lag or lead with respect to a
particular set of line to line voltage that is the thing I wanted to tell. So, this is the
connection then Yd(-30°).
Now, sometimes this vector grouping this is the vector group is also expressed in terms of
a clock minute and hour hand positions angle between them that is how. So, I will first
explain that then explore more interesting connections after that. So, first clock convention
vector group which is can be also expressed instead of showing the angles by clock
conventions; clock convention.
(Refer Slide Time: 16:34)
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So, you let us let me go to next page and you see you have a clock here let me explain that
then I will go there. Suppose you have a clock of course, you have got all the markings
here 1, 3, 6 here and 9 ok; 1 3 6 9. And of course, no it is 12; 12, 3, 6, 9, then you have in
between of course, 1 and 2, 3, 4 and 5 and 7 and 8 and 9, 10 and 11 this is usual clock our
familiar clock.
And there will be two hands hour hands and minute hand; minute hand is longer. So,
minute hand is suppose in the position of 12, hour hand could be anywhere in our usual
clock I mean that will I mean minute and hour hand both can be anywhere and you get the
time. But here while following this clock convention to describe what is the vector group
of the transformer you note that the angle between two consecutive numbers on the dial of
the clock is 30° that is known that is this angle is 30° consecutive to any numbers you take
30°.
Now, what is done is this the minute hand; minute hand is shown at 12 always you will
show it and let this minute hand represents the primary side. Say A phase voltage with
respect to neutral that is this voltage 𝐴𝑠𝑁 or 𝐴1𝑁 that is the primary side neutral voltage
that voltage I will place in the minute hand and always show it in the twelfth position of
the clock this is the primary voltage.
And secondary voltage I will once again this is suppose 𝐴𝑠 , 𝐴1 or 𝐴𝑠 either of them with
respect to neutral. Similarly, show the draw or show the show the hour hand; show the
hour hand representing; representing the secondary 𝑎 or 𝑎1 voltage with respect to neutral;
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neutral of the secondary side; of the secondary side. For delta connection use artificial
neutral that is small 𝑛 this is capital 𝑁 you have got the point.
For example the previous connection; in the previous connection minute hand is a 12
vertical 𝐴1𝑁 and secondary a phase voltage this 𝑎1 is once again a which is going to load.
So, this is 𝑎1 with respect to neutral this is my b, this is my c therefore, this a phase voltage.
Now, this angle been 30°, where do you think in the clock should I show this secondary
voltage at one is not.
Because we have seen that secondary voltage is doing this. So, this minute hand which is
bigger it is capital 𝐴1𝑁 and this is small 𝑎1𝑛 I should not write this, but I am writing you
should not be under the impression that if I write like this I meaning capital 𝑁 and small
𝑛 are shorted no it is just separate two things. Because two neutrals are located at different
positions and they are not connected electrically, but just to indicate what does this mean.
Anyway this is the thing, then I will write it as primary is star connected Y, secondary is
delta connected delta. Earlier I wrote it Yd (-30°) in the previous page this is same as Y d
1 that is the time is 1 then only this connection I mean this is the connection Y d 1 it simply
means that the with respect to neutral of a particular phase primary side voltage and
secondary voltage how they are dispersed. But since hour hand I will always represent to
indicate the secondary voltage therefore, secondary voltage lags the primary voltage by
30° ok
So, this is the vector group. For example, this I will just write for example, Y we have seen
that Y; Y with different color let me write so much space is there. For example, this
connection I will not redraw you know Yy 0° we have seen. What does this mean? It means
these are the primaries you short these are the secondaries, this is 𝐴1 𝐴2 𝑎1 𝑎2, 𝐵1 𝐵2 etcetera
𝐶1 𝐶2 small 𝑐1 𝑐2 and 𝑏1 𝑏2 and suppose you make star, then this connection indicates it is
Yy0° degree means I will write it as what, then this these voltage 𝑎1𝑛 both minute and
hour hand will be together. So, I will write it as Yy 12 time is 12, then only I will say this
is Yy 0°.
Similarly, if you short these I should have written Yy 6 because secondary voltage should
have been with respect to neutral here primary voltage is there they would have been 180°
out of phase. I hope you have understood this one. Now, let us try some more connections.
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Another connection is a see previously let me draw another one, so that with delta
connections you become comfortable. Suppose there are two ways of delta connections
either this way or other ways that is a cyclically you go this or it could be connected like
this. So, the second connection let me try, what is this?
(Refer Slide Time: 27:12)
So, once again star delta connection ok. So, these are the primaries 𝐴1 𝐴2 , 𝐵1 𝐵2, 𝐶1 𝐶2 and
these are the secondaries small 𝑎1 𝑎2, small 𝑏1 𝑏2 and small 𝑐1 𝑐2 and suppose I have shorted
these.
So, I get the primary voltages as supply voltages are this is 𝐴𝑠 , this is 𝐵𝑠 easier way of
drawing I am telling 𝐶𝑠 . So, supply voltage is balanced draw 𝐴𝑠 , 𝐵𝑠 , 𝐶𝑠 first 120° apart and
this is 𝐴𝑠 , 𝐵𝑠 , 𝐶𝑠 you draw and this will become the neutral and where 𝐴2 , 𝐵2, 𝐶2 are joined
this is capital 𝑁 this is over.
Then you know you connect it suppose like this you joined other way now and mind you
these are dots do not forget to show that which I have already told you that is important.
So, you connect this closed circuit there will be no circulating current because I have been
very meticulous very careful about polarity marking and that is why I am connecting and
this is my output terminals small a, small b which will go to load 3 phase load like that.
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Now, in this case the secondary voltage I am now drawing. Secondary voltage will be
mind you this 𝐴𝑠 is nothing, but your 𝐴1 is not; is nothing, but 𝐴1 , 𝐵𝑠 is nothing, but 𝐵1
and 𝐶𝑠 is nothing, but 𝐶1 .
Now, I will sketch the secondary voltage now you see there is the voltage 𝑎1 𝑎2 which will
be parallel to 𝐴1 𝐴2 . So, I draw it here a space is there this is suppose 𝑎1 𝑎2, it must be like
that then 𝐵1 𝐵2 parallel to that the secondary voltage has been induced that I get from single
phase funda that is induced voltages in each transformers will be co phasor with the
primary voltage that we have we know from.
So, 𝑏1 𝑏2, but this time I have joined 𝑏2 with 𝑎1 . So, move this 𝑏2 with 𝑎1 , so it must be
like this. I first write this 𝑎1 𝑎2 here this side 𝑎1 𝑎2 fine with this then 𝑏1 𝑏2 I will draw like
this with a little practice you will become conversant with this, this is the thing 𝑎1 and 𝑏2
I have joined. So, there must be at some potential and then 𝑐1 𝑐2 and 𝑐1 is joined with 𝑎2 .
So, is like this 𝑐1 𝑐2 is not`
So, it is a valid delta connection and then I am telling mark this please follow the
instructions carefully. So, that you will never make mistake while working with a 3 phase
transformer in the laboratory or in the site.
So, 𝑎1 𝑏1 𝑐1 these are the thing this is 𝑎 this is my 𝑏 this is 𝑐. So, balanced 3 phase voltage
will be obtained. So, here is the artificial neutral because delta connection no neutral. So,
a phase voltage with respect to this neutral and a phase voltage of the primary is this the
vertical line therefore, it is 30°. So, in terms of clock it will become this 12, 6, 3, 9, 10, 11
etcetera 1, 2, 4, 5, 7, 8.
So, primary voltage with the minute hand I will always show, so that was this and
secondary voltage with respect to same a phase voltage with respect to neutral is now
ahead of this. So, it must be shown here hour hand this is the secondary voltage and this is
30°. So, this connection I should say it is Y d (+30°) or you show Y d 11 like that ok. We
will continue with this try to understand very interesting, but at the same time people often
make confuse confusion I mean.
Thank you.
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Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 38
Vector Group (Contd.)
(Refer Slide Time: 00:21)
Welcome to lecture number 38 and you recall that we were discussing two connections Yd
11 and previously Yd 1 we have done and I hope you have got the idea what and do not
forget that not only the phase voltages but also the line to line voltages suffer same degree
of either lag or lead on both the sides.
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(Refer Slide Time: 00:28)
(Refer Slide Time: 00:54)
And then I have discussed about the clock this thing and we found that there are two
connections possible; so, Yd 11 or Yd 1.
Mind you if somebody says the connection is Yd 3, no you cannot realize it because these
such voltages are not available you must understand this point is not. It so happens that
30° phase displacement only possible with star delta connections. Therefore, anything you
cannot get means Yd 2 somebody says I have connected the transformer this is ridiculous,
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you cannot get it for a balanced 3 phase transformer this is just I point out. So, you should
be very methodical in drawing this.
(Refer Slide Time: 01:58)
Now, let us connect the same way, but now rather quickly these connections say D y I
would like to connect. I am not telling mentioning the vector group yet suppose and, but I
know there are two correct ways of connecting delta. So, primary I have to connect delta
secondary star and two ways of connecting star. So, suppose I say that two valid delta
connections are there. So, 𝐴1 , 𝐴2 , this is the primaries 𝐵1, 𝐵2 and this is 𝐶1 , 𝐶2 and one of
the valid connection is go this way that is first let me try.
So, this is the primary connection I will do and obviously, from 𝐴1 , 𝐵1 that is the
convention I am following 𝐶1 I will take the output, and not output I will connect my
source 𝐴𝑠 𝐵𝑠 and𝐶𝑠 this is what I will do and secondary small 𝑎1 , 𝑎2 small 𝑐1, 𝑐2 . See I
have taken a given input as if 𝐴1 𝐵1 𝐶1 . So, I would like to have outputs taken from 𝑎1 𝑏1
𝑐1 on the secondary side. So, suppose I shorted star connections.
Now, what I will be doing I have to draw the primary voltage phasors, voltage is coming
across each winding, and as I told you to begin with you better draw supply voltage is
balanced and phase sequence abc balanced. So, what you do is this, you draw this vector
diagram that primary vector as this primary side is delta connected draw it like this with
this is 𝐴𝑠 line to line voltage I am drawing 𝐵𝑠 and 𝐶𝑠 is not. This is the balance 3 phase
voltage what else 𝐴𝐵 𝐵𝐶 𝐶𝐴 they are 120° apart.
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Now, 𝐴𝑠 is connected to 𝐴1 and 𝐵2. So, I write here this point is 𝐴1 , 𝐵2 either of them or
together same point 𝐴𝑠 is connected to 𝐴1 and 𝐵2 similarly 𝐵𝑠 is connected to 𝐵1 and 𝐶2 .
So, this point is same as 𝐵1 and 𝐶2 capital and 𝐶𝑠 as you can see it is connected to 𝐶1 and
𝐴2 same 𝐶𝑠 is nothing, but 𝐶1 and 𝐴2 is not? This is the thing what is the voltage applied
across the primary of the transformer A? 𝐴1 𝐴2 where is 𝐴1 𝐴2 ? This is 𝐴1 this is 𝐴2 .
So, on the secondary side this voltage will be if I use a different color it is parallel to this
line is not parallel to this line and I will write it here 𝑎1 𝑎2 got the point. Similarly 𝐵1 𝐵2
voltage applied across the primary of the B transformer is 𝐵1 𝐵2 where is 𝐵1 where is 𝐵2
oh this is 𝐵1 this is 𝐵2. So, parallel to this small 𝑏1 𝑏2 will exist and not only that I have
joined 𝑏2 with 𝑎2 . So, bring this line parallel and it will be here 𝑏2 𝑏1 not 𝑏2 𝑏1 parallel to
this. So, this angle will be 120° apart. Mind you all these angles are 60°, I hope you
understand that geometry which I am not going. And finally, voltage applied to the primary
of the C transformer is 𝐶1 𝐶2 therefore secondary induced voltage will be parallel to this
𝐶1 𝐶2 line and 𝑐2 I have joined. So, 𝑐1 𝑐2 like this all angles are 120° I think you understand
that this is the thing
Now, if you choose a connection you get it in this way and your now I have to decide what
is the vector group of this connection primary delta D secondary star y I want to know
what is the what is the vector group. So, what I will be doing? I will identify on the delta
side there is no neutral small neutral that is artificial neutral or centroid of the triangle. So,
this is the and I will mark 𝐴1 𝐵1 𝐶1 this is where I will follow the convention, I have
connected supply similarly this is my output this is of course, obvious in this case no
problem.
Now, the a phase voltage of the primary side is this vertical line with respect to neutral.
This is the secondary neutral it exists this point this is secondary neutral. Now this a phase
voltage if you draw a this a phase voltage of the primary is this vertical line. So, this is 30°
it will be 30° and therefore, I will say that it is Yd secondary voltage lagging the primary
voltage with respect to neutral by that I could write it Dy(-30°) or you could write D y
what should be the time? 12 and this is Dy 1. So, 1 got the point that is in terms of clock
convention because your this thing it is here a phase voltage it is there.
Therefore this is how you and also do not forget that line to line voltage here that way also
one can do that line to line voltage of the primary side is 𝐴1 𝐵1 what is the line to line
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voltage of the secondary side it is this one. So, this was this once again can be shown to
be equal to 30° that is what you must understand.
Therefore both the line to line voltages and phase to neutral voltages, will suffer the same
change star delta connection ok. I will just the other connection you can try delta this way
that you decide and you will soon discover that that will be Yd 1 that I am not repeating,
but what I will say to you is this one. The same connection I will do in a slightly different
fashion just to make you understand suppose see in this connection when I did, what I have
done? I just told connect Dy, I did not specify the phase angle difference this to begin with
I did not know what it is. It looks like as if I have connected first then told oh you have
connected this is this phase group.
But suppose I want to connect Dy 1 oh sorry.
(Refer Slide Time: 12:20)
Suppose I now say that connect Dy 1 the connection I know now. The point I want to make
it is interesting you see this is the 3 transformers you have 𝐴1 , 𝐴2 , this is 𝐵1, 𝐵2 and this
is 𝐶1 , 𝐶2 good and this is secondary small 𝑎1 , 𝑎2 one terminals are dot terminals which I
am not writing all the time 𝑏1 , 𝑏2 , 𝑐1, 𝑐2 fine. Then what I am telling you connect this as
delta like this for example, same thing I am doing I mean do not worry this is 𝐴𝑠 this is
supply 𝐵𝑠 and this is supply 𝐶𝑠 .
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Now, after you energize this what I am telling. This is your supply voltages because supply
voltage I know it is balanced. So, I will write 𝐴𝑠 , draw this triangle 60° equilateral triangle
𝐶𝑠 and then say that 𝐴𝑠 is nothing but 𝐴1 𝐵2 etcetera 𝐴1 𝐵2 and 𝐵𝑠 is 𝐵1 𝐶2 and 𝐶𝑠 this is 𝐶𝑠 ;
is 𝐶1 𝐴2 and suppose secondary I have not connected that is what I am telling I want to
achieve this one connection. But what I know is this that I have now in the secondary these
phasors available to me what is that? Small 𝑎1 𝑎2 this is available then this voltage is
available, 𝑐1 𝑐2 and 𝑏1 𝑏2. Why I have shown them in isolation? Because I have not
connected yet I do not know, but since you want to connect Dy 1 what you would like to
have? Your 𝐴1 voltage with respect to neutral here that is this voltage, secondary 𝑎1
voltage small 𝑎1 must be I mean that way that is in the twelve o clock if this is your primary
you have you would like to have small 𝑎1 𝑎2 here
So, this indicates that you better join 𝑎2 𝑏2 𝑐2 I mean you ponder over this think these are
very interesting. So, such that this will be this way and then say I have joined that is in
effect what I am telling, it is not that these connections are arbitrary you can I can tell you
make this connection. Then the question is whether delta should be connected this way or
delta should be connected that way, connect it and try to get this one achieve this. So, I
have practice on this I think I have made things clearer now.
So, this is the better group, similarly you can have last connection here is delta delta
connection star star connection is so, simple.
(Refer Slide Time: 17:03)
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This I will just mention delta delta connection ok. So, delta delta connection means. See I
have not yet told you which connection to use at in what application? I am simply now
getting accounted with you what are the possible things you can do 3 phase means star
delta connection it looks like ok. 3 phase transformer, 3 coils there, 3 coils there either of
them you connect I am just telling you how to get the correct connections with confidence.
So, that is the idea of this lecture.
So, here also I say this is 𝐴1 , 𝐴2 , this is 𝐵1 , 𝐵2 what are the vector group possible here that
is what I am examining with delta delta connection and these this is the secondary. Here I
will not take much time because you are now used to it small 𝑏1 , 𝑏2 and this is small 𝑐1,
𝑐2 proper delta I have to connect. So, these are dots. So, connected this is one of the proper
way of connecting and I will give supply connect supply here 𝐴𝑠 to 𝐴1 , 𝐵𝑠 and to this
junction 𝐶𝑠 ok.
Now, suppose there also there are only two possibilities here this way you connect
secondary you connect also this way same way. Let us see what vector group it results into
because of this connection proper delta connection correct.
Now, once again delta connection. So, better draw the primary voltages line to line
voltages first because there is no neutral that is why I start with line to line voltage and it
is the 3 phase it has nothing to do with transformer this is the supply voltages which are
available here you know? Line to line voltages are also balanced 𝐴𝑠 𝐵𝑠 𝐵𝑠 𝐶𝑠 voltage lags
by 120° you know this and 𝐶𝑠 𝐴𝑠 voltage for the lags by 120° ok. I am not putting the
arrows that is understood 𝐴𝑠 𝐵𝑠 𝐶𝑠 like this.
And then what is the next step oh 𝐴𝑠 you have connected to capital 𝐴1 and 𝐶2 . So, capital
𝐴1 𝐶2 𝐵𝑠 I have connected to 𝐵1 and 𝐴2 . So, 𝐵1 𝐴2 same potential because of this
connection 𝐶𝑠 is connected to 𝐶1 and 𝐵2, 𝐶1 𝐵2 is not? This will be the thing secondary
voltages 𝑎1 𝑎2 will be parallel to this line. So, 𝑎1 𝑎2 𝑏1 𝑏2 will be parallel to this line correct
or not 𝑏1 𝑏2 will be parallel. So, 𝑏1 𝑏2 I will put it here bring it down so, but 𝑏1 is connected
to 𝑎2 .
So, 𝑏1 must be placed here 𝑏1 𝑏2 and finally, you know you bring this 𝑐1 𝑐2 you get this in
triangle where from I will take output to be consistent from 𝑎1 this I will call 𝑎, this will
call 𝑏, this I will call 𝑐 and this will be the thing what do you think. So, primary is delta,
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secondary is also delta small letter and what is the phase angle difference; 0° because here
also no neutral exist, but anyway with respect to artificial. This is vertical line here also
this a phase A phase mind you that is also parallel. So, time is 12 that is; so, Dd 12.
(Refer Slide Time: 22:32)
Now, the same primary connection if I make say this is 𝐴1 𝐴2 this is small 𝑎1 𝑎2 I mean
same thing I am drawing you also draw so, that you get used to it I mean small 𝑐1 𝑐2 and
suppose primary I have connected like this as I have told you delta and these are from the
junctions supply 𝐴𝑠 supply 𝐵𝑠 and supply 𝐶𝑠 this is known and secondary I will now
connect not this way, but other way these are the only 2 possible delta connections. So,
suppose the this I have connected like this and I will take output from these junctions small
𝑎1 going to load from 𝑏1 going to 𝑏 and from 𝑐1 going to 𝑐 of the supplying load.
Now, here it is. So, so what should I do? Delta connection draw supply 𝐴𝑠 𝐵𝑠 𝐶𝑠 what is
𝐴𝑠 ? 𝐴1 𝐵2 we have done it 𝐴1 𝐵2 what is 𝐵𝑠 ? 𝐵1 𝐶2 and what is 𝐶𝑠 ? 𝐶𝑠 is 𝐶1 𝐴2 like that. So,
A phase voltage secondary voltage 𝑎1 𝑎2 phasor will be parallel to this line B phase voltage
phasor will be parallel to this line with this side small 𝑏1 that side 𝑏2 and C phase voltage.
So, let me start with the a phase voltage, this will be small 𝑎1 𝑎2 parallel to this is not b
phase voltage 𝑏1 𝑏2 and 𝑏2 is 𝑏1 𝑏2 is where this 𝑏1 we drawn correctly this? Wrong sorry
there is a mistake here see 𝑎1 𝑎2 is here have I made 𝑎1 𝑎2. So, 𝑎1 𝑎2 here I should draw
like this correct or not 𝑎1 𝑎2 parallel 𝑎1 𝑎2 𝑏1 𝑏2 is here. So, that is parallel to this, but 𝑏2 is
joined with 𝑎1 where is 𝑏2 . So, 𝑏2 is joined with 𝑎1 , 𝑏2 phasor is this; so, 𝑏2 and 𝑏1 .
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And finally, I think I have done the same connection is it looks like same thing. So, it will
be 𝐷 d 0 same thing I have done. So, let me change it. So, it will be once again by mistake
I have done it 𝑐2 𝑐1. So, once again 𝑎1 𝑏1 𝑐1 and this is D d 0. See there is no difference
between this connection and the previous connection like that here also these. So, that is
why no variety is obtained. I what I intended to do let me correct that. So, anyway this is
done.
So, let me do it like this. So, that you understand because this I have already done what I
wanted to do is, I will make another valid delta connection, but this way got the point. So,
it will be like this, then you say this is 𝑎, this is your 𝑏 got the point I actually did same
thing and this is your 𝑐. So, in this case then the phasor diagram will be like this, what will
be the phasor diagram now? On the secondary side voltage 𝑎1 𝑎2 here also it will be 𝑎1 𝑎2
is not this is the thing.
(Refer Slide Time: 28:48)
Now, then or I did once again same. Now 𝑏1 𝑏2 and parallel to this line, but 𝑏1 is connected
to 𝑎2 is not? So, it will be like this 𝑏1 𝑏2 and finally, 𝑐1 𝑐2 like this and 𝑐1 is connected to
𝑏2 . So, it will be like this now it is correct 𝑐1 𝑐2 got the point?
So, in this case if you do then people say that this is a Dd 0 connections sort of. Why I am
telling people say I will take output from 𝑎2 𝑏2 𝑐2 if I take it is indeed Dd 0, got the point?
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See in this particular case it is like suppose I say that you take the output from 𝑎2 𝑏2 𝑐2
then it will appear as a Dd 180 or Dd 6 connection. If output is taken from 𝑎2 𝑏2 𝑐2 instead
of this one that is you take the output from say this you say 𝑎 this you say 𝑏 this you say
𝑐 in this case you compromise a bit and not that much a 1 for example, Yy 6 star star 6
connection what you do? Y ou have to take output from 𝑎2 𝑏2 𝑐2 , 𝑎1 𝑏1 𝑐1 shorted, but
nonetheless the phase sequence is maintained and things like that. So, this is the I will
circulate with red 𝑎2 𝑏2 and where is 𝑐2 , 𝑎2 𝑏2 𝑐2 180° apart. Therefore, you see there are
various ways of connecting the windings delta star, delta delta, star delta, star star and so,
on.
Now, in my next lecture what I am going to tell you is about a transformer which is a 3
phase unit, it is built as a 3 phase transformer at least some little bit of constructional
features of course, the connections here I have mentioned I started with threes identical
transformers these connections things will be equally valid as we will see, if there is a 3
phase transformer as a single unit. So, we will do that in the next class.
Thank you.
371
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 39
Open Delta Connection
Welcome to lecture number 39 on Electrical Machines I.
(Refer Slide Time: 00:19)
And we were discussing use of 3 single phase transformers to step up or step down voltage
it depends upon the connections, what will become the line to line voltage, transformation.
And also we have fairly understood what is the phase shift of the secondary voltage with
respect to primary voltage and what are the possible connections familiar conditions like
star delta, delta star, delta-delta and so on.
Now, you recall that, this 3 are individual transformers. And this is the delta-delta
connection with D d 0, we have discussed this earlier. That is the phase voltages are in
phase. In this connection, see this connection and I am talking in terms of 3 a bank of 3
single phase transformers; 3 identical transformers you take separate 𝐴1 𝐴2 and I can
connect them in delta-delta and so on.
Now, one interesting thing about this connection is that, you can have from this you can
derive an interesting connection which is called open delta connection. Now I will put it
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in a question form, that is suppose you are given two single phase transformers of identical
ratings not three, then is it possible to change the voltage level from the source to the load
side that is the question asked. So, 2 transformers I will give, is it possible to still transform
a 3 phase balance voltage from one level to another that is a thing. In these complete 3
transformers when you are using, we have connected this transformers as delta-delta and
D d 0 is the connection ok.
Now, suppose I say that, in this connection I will remove one transformer, say transformer
A, I will remove; then what are the things will happen, that is a what I am planning to do
I will remove this transformer as if it is not there. And other connections remain as it is,
what do I mean by it is not there? It is physically not there; so it is open circuit here. A is
gone, there is no 𝐴1 𝐴2 , got the point.
Similarly, it secondary is not there; that is there is a open circuit in place of transformer A,
but nonetheless I have connected in this fashion. Then across a transformer B of course,
same voltage has been applied 𝐵1 𝐵2 because of this line voltage is balance 3 phase what
I have done I have applied a 𝐵𝑠 𝐶𝑠 across 𝐵1 𝐵2; across 𝐶1 𝐶2 I have applied a voltage 𝐶𝑠
𝐴𝑠 because this connection is there, A is not there. But 𝐴𝑠 is connected here and 𝐶𝑠 is
connected there. So, how much voltage I have applied, 𝐶𝑠 𝐴𝑠 . So, in other words what I
am telling if you 𝐴1 𝐴2 ; where is 𝐴1 𝐴2 here, remove 𝐴1 𝐴2 , then what I am telling I have
applied a voltage across this is 𝐶2 mind you.
So, across the C transformer 𝐶1 𝐶2 has been applied and across transformer B; 𝐵𝑠 𝐶𝑠 has
been applied this phasor this phasor, but across 𝐴𝑠 𝐵𝑠 it is also balance line to line voltage
that voltage phasor exist. But 𝐴𝑠 𝐵𝑠 has not been applied to any of the primary because
that transformer A is missing, are you getting. So, this is the thing, but since I have applied
a voltage here across. So, this I will first remove across 𝑐1 𝑐2 there will be induced voltage
parallel to these voltage these voltage across 𝑏1 𝑏2 there will be induced voltage parallel
to this 𝐵𝑠 𝐶𝑠 , 𝑏1 𝑏2 and 𝑏2 𝑐1 are joined. So, it will be like this only and I will take output
from this, from this and from this.
Then what I am telling that these 2 voltage has 60° apart therefore, these voltage also will
be 60° apart it is conditioned by these 2 phasors. Even if transformer A is not there, you
will still get a balanced 3 phase output voltage, is not. So, it looks like that delta is not
complete, open delta, got the point. Therefore, you can take output and once again supply
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a 3 phase load; your 3 phase load will not fill anything balanced 3 phase voltage I am
receiving. What was the line to line voltage in the previous case when all the transformers
where there, these are the line to line voltage. Here also same, same line to line voltage
you will get.
(Refer Slide Time: 08:38)
In other words in simplified diagram, what I am telling; suppose you what you have
essentially done is, which transformers I have removed. If I draw it in a simplified diagram,
I will do it like this; this is one transformer, this is another transformer, this is it secondary,
this is it secondary and the third transformer is not there. And what I have done, I have
connected 3 phase supply there 𝐴𝑠 𝐵𝑠 𝐶𝑠 , 3 phase supply.
So these voltage will be this line to line voltage whatever it is; I mean now look at this
diagram separately ok. And what will be these voltage these will be 𝐵𝑠 𝐶𝑠 parallel to this
therefore, this third voltage gets automatically decided and 60° out of phase; therefore, I
will take output from this, these and this and supply a 3 phase load.
Then you can change the level of 3 phase voltage from one value to another just by using
2 single phase transformers; that is what I want to convey to you. You do not even require
3 transformers. Remember I took the rating of each transformer, so 2 number of
transformers mind you; and you get open delta connection like this it is mentioned. Rating
of each transformer I took as 200V/100V for easy calculation 5 KVA; single phase 50 Hz
what is the thing, what is the rated current of this side:
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This is 25Amp rated current and this is 50Amp rated current. Now I would like to know
that what is the KVA rating of this type of connections; in case of 3 phase transformers in
my first lecture on 3 phase transformer I told you, if the rating of each transformer is 5
KVA, total KVA which could be handled by this 3 transformers since 15 KVA quite
logical; each one will give you 5 KVA. But in case of open delta connection which is
called open delta connection; we would like to know how much KVA that can be handled
by this open delta connection. The idea is pretty simple, mind you in this case the
secondary winding in series with the line, what is the rating of the secondary side suppose
this is 100V side, how much maximum current I will allow to flow 50Amp because
winding rated current is 50Amp.
So, I can allow 50Amp current to flow in this line in all the line this line and if these two
are 50Amp, phasor sum will give you also this is 50Amp. So, I should not exceed this
rated current. The moment it delivers 50Amp here it will automatically by principle of
transformer will bring 100Amp there, are you getting. And these currents will be once
again balanced 3 phase current, why not load is balanced; balanced load. Therefore, what
is the total kVA handled by open delta connection will be equal to how much, if you
calculate from the L V side it will be
𝑇𝑜𝑡𝑎𝑙 𝐾𝑉𝐴 ℎ𝑎𝑛𝑑𝑙𝑒𝑑 𝑏𝑦 𝑂𝑝𝑒𝑛 𝐷𝑒𝑙𝑡𝑎 𝐶𝑜𝑛𝑛𝑒𝑐𝑡𝑖𝑜𝑛
= √3𝑉𝐿𝐿 𝐼𝐿
= √3 × 100 × 50
= √3 × 200 × 25
= 8.65𝐾𝑉𝐴
So, if you apply line to line 200V here, line to line here you will get 100V. I will neglect
the no load current etcetera.
So, this much total kVA it can handle, but it may look like you are using 2 transformer
rating of each transformer is 5 KVA perhaps can you just calculate these two.
Approximately. So, this is 8.65 KVA, but you will be rather surprise to see ok, at least I
am using 2 transformers it is kVA rating should have been 10 KVA I mean I would expect
2 transformer I am using. So, without overloading any of the transformer I see that no, it
is not 10 KVA it is slightly less than that. Therefore, why it is because this is the line
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current and the winding currents are same ok. And it is a 3 phase system and the factor by
which this KVA will be reduced is then I was expecting, see I will calculate this number
with respect to whom, when all the 3 transformers we were using what is the kVA handled
by these two transformers? 10KVA
√3 × 100 × 50 √3 × 200 × 25 √3
=
=
= 0.865
10 × 1000
10 × 1000
2
Therefore the capacity will be about 86.5% of the expected capacity or common sense tells
2 transformers you are using, then 10KVA you can extract from the transformer, no it is
not going to be like that, it will be only lesser. But nonetheless this connection was popular
I mean still may be also popular because of the fact see the advantage, although I have not
told you anything about a 3 phase single unit transformer, 3 individual transformers I am
connecting.
Now, in earlier days the advantage of using 3 single phase transformers are there are at
least two biggest advantage.
(Refer Slide Time: 18:49)
One is suppose your total KVA you have to handle 15 KVA, I mean I am going other way
around suppose my 3 phase load demand is 15 KVA. So, you by three 5 KVA transformer
to achieve this and connect this 3 individual transformers star delta whatever is necessary
depending upon the voltage level; but 15 KVA you will get. So, you will buy three single
phase transformer each of 5 KVA rating, is not. Each of 5 KVA rating you buy and no
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open delta I am telling you connect this 3 phase star delta whatever way, so total KVA of
the system will be 15 KVA fine.
But the advantage of this method is that, if suppose one transformer develops a fault ok.
Let us take that previous example in this simple connection same number because then it
will be clearer, suppose it was delta-delta connected then you apply 200V here, here you
get 100V is not. Suppose the connection is same 5 KVA each, so 15 KVA you get; what
is the rated current of the LV side we have seen, 50Amp.
So, 50Amp here now, compare delta-delta full connection then line current you can apply
50√3𝐴𝑚𝑝 you can allow and then the total KVA becomes
𝑇𝑜𝑡𝑎𝑙 𝐾𝑉𝐴 = √3 × 100 × 50√3 = 15𝐾𝑉𝐴
And you are stepping down the voltage, you are, supplying 3 phase load that you should
be always in your mind that thing is playing.
Now, suppose in this case one of the transformer develops a fault, then I will say can I still
supply the load; I will say yes. If suppose this transformer develops a fault, because this
three are three separate transformers it has developed a fault, then this transformer you
remove from the circuit. So, that your connection will become something like this and I
will say look here you still will be able to supply the load, but not 15KVA not 10KVA,
but some 8.65KVA that is what I want to convey to you; but nonetheless it is better than
nothing. Suppose one of the transformer develop fault you can immediately restore supply,
but supply a load slightly less than the rated KVA, 15 KVA certainly you cannot supply,
but it is not 10 KVA as well to be slightly less, but something is better than nothing sort
of thing happens. Not only that when you use 3 single phase transformers in such cases
what people will do, each transformer rating is 5 KVA.
So, keep buy four at the very beginning and in case of fault keep one in reserve waiting.
So buy four, 5 KVA transformer, four numbers 5 KVA, one you do not connect; spare it
is the fourth one is spared transformer, spare, spare unit waiting outside. If such a situation
occurs one transformer develops fault you remove this transformer bring that fourth one
here. What is the advantage of this method? This spare unit which you are purchasing is
only of 5 KVA rating, you will see that if it is a single 3 phase unit of transformer of 15
377
kVA; if one of the winding develops fault then you have to take shutdown and correct that
faulty.
In case of 3 phase unit you will see this open delta connection cannot be used; because
after all this transformer has developed fault you will have to take some corrective action.
And there also that transformer has developed a fault of 15 KVA single unit then the spare
of that it will be of also of 15 KVA unit. So, lot of money is to be invested you know, that
is not a good solution. Therefore, in those applications where you can use take the
advantage of open delta connection, even if you do not have any spare units still you can
maintain supply till that transformer is repaired ok; then connect it if you do not have a
spare unit and like that you can maintain ok.
So, this is a very interesting connection open delta connection, I hope you have understood
what it means; but the capacity of the transformer is reduced.
(Refer Slide Time: 25:42)
Now, I will just today tell you about a 3 phase transformer we will come back to this
connection once again; a 3 phase transformer as a single unit ok. So, how it can be first let
me draw this diagram is slightly tricky to draw; what I mean to say that suppose you have
a transformer like this, very rough sketch I will do, but you will get the idea. This is the,
suppose one transformer and to understand what is going on, this is this transformer with
iron core. And what I will do here I will have 2 windings on this limb, imagine I have 3
such iron core got the point; because this diagram I cannot draw so nicely, next time I will
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show some drawn diagram. But the idea is interesting, what is done is you take another
similar unit and suppose you have put it there, got the point you will get the idea hopefully.
Another iron core, and here also you have got 2 coils; one will be primary another will be
secondary on the same limb. Bring another third unit and put it here, this time I will make
a mess of it, but any way try to get the idea. This is the third one another similar this is like
this got the idea, you have and remove this one. Here is the third unit, where also there
will be this one its core. That is 3 units there is a central common iron; one goes like this
another like this, then need not be 120° apart nothing like that you join them somehow,
the central iron. And here also you have two windings in the outer limbs 1 and 2; I think
you have got the idea.
So, 3 iron structure of single phase transformer you have put together, mechanically. Now
what I will tell you name the terminals as 𝐴1 𝐴2 and it secondary small 𝑎1 𝑎2 on the same.
Similarly this is suppose 𝐵1 𝐵2 and this is small 𝑏1 𝑏2 and this is 𝐶1 𝐶2 and small 𝑐1 𝑐2 .
Now you imagine that this 3 coils while showing the connection I have done like this 𝐴1
𝐴2 suppose star connection I have done 𝐵1 𝐵2, 𝐶1 𝐶2 suppose primary; then this
transformer will develop a flux, secondary is open suppose 𝜑𝑎 .
Similarly 𝐵1 𝐵2 will also develop a flux say 𝜑𝑏 going these way, function of time. And
third transformer will develop a flux, this diagram so badly drawn any way try to
understand; the this is 𝜑𝑐 , but this 𝜑𝑎 , 𝜑𝑏 , 𝜑𝑐 will be 120° out of phase because the applied
voltage is balanced,
1
√3
is coming.
So, how the flux path will be completed 𝜑𝑎 goes there, through the central iron it comes
then it goes there. How 𝜑𝑏 will be completing it is path, it will go like this through the
central iron it will come down and go there. And similarly 𝜑𝑐 it will go like this through
the central it will come. If you see the top view of this thing that is easier to draw; what I
am trying to tell this is one transformer iron, this one you will be able to see. This is another
transformer plan I am saying are you getting and oh sorry this is the third one like this, if
you look from the top you will see this three things.
Now, this is the common iron here this three you will be able to see like a finding out
things. Now if I ask you what is the flux in this common iron, how much it will be? It will
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be 0, because 𝜑𝑎 (𝑡) + 𝜑𝑏 (𝑡) + 𝜑𝑐 (𝑡) they are 120° apart in time, their instantaneous
values will be
𝜑𝑎 (𝑡) + 𝜑𝑏 (𝑡) + 𝜑𝑐 (𝑡) = 0
as 𝐼𝑎 + 𝐼𝑏 + 𝐼𝑐 in a 3 phase system summed up to be
𝐼𝑎 + 𝐼𝑏 + 𝐼𝑐 = 0
Balanced 3 phase system. Therefore, it looks like this central iron whether you connect it
or not it does not matter it does not carry any flux.
So, 𝜑𝑎 will get it is return path via 𝜑𝑏 this iron and this iron, as it happens in a 3 phase 3
line current system. 𝐼𝑎 + 𝐼𝑏 + 𝐼𝑐 = 0 whether you provide that neutral path or not; if 𝐼𝑎 ,
𝐼𝑏 , 𝐼𝑐 are balanced 3 phase current it does not matter. Anyway so next step is therefore,
this iron portion central that thing can be removed. And not only that we will see in our
next class, if you remove this central iron, you save some iron definitely. Because central
portion I know it will not carry any resultant flux at any point of time if it is a balanced 3
phase system; then the weight will be reduced, but nonetheless still the structure of the
transformer will be pretty awkward. See in the working place if you keep it these
projections will make a very inconvenient piece of equipment to work with.
So, what people do is this.
(Refer Slide Time: 33:58)
380
They will I will just draw today and then they make it much more simpler. They say do
not go for this projections, make it absolutely rectangular. This is the iron laminations
stack one about them and this is one limb this is in plane. So, it will look like these are this
is, got the point. I do not know whether this we will be able to see and then you make your
connections like this is 𝐴1 𝐴2 this is we will continue with this next time do not worry,
small 𝑎1 𝑎2 . Similarly for B phase, so these are the 3 limbs 𝐵1 𝐵2 and small 𝑏1 𝑏2 and
finally, this is suppose capital 𝐶1 𝐶2 and small 𝑐1 𝑐2 number of turns of capital numbered
coils are same 𝑁1 , small number coils are same that is and this is called core type
transformer, core type 3 phase transformer. We will discuss with this next time.
Thank you.
381
Electrical Machines – I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 40
3 Phase Type and Shcle Type Transformer
Welcome, to 40th lecture on Electrical Machine I course.
(Refer Slide Time: 00:16)
And, we have started discussing on three-phase transformer as a single unit and in our last
class I so, told you that you take three single phase transformer imagine that you have
taken three single phase transformer and you have joined the one of the limbs of each of
them together and then the secondary winding you put on the outer limbs. And, then the
flux in this central limb instantaneous value of the flux will be 0, because 𝜑𝑎 , 𝜑𝑏 and 𝜑𝑐
they are 120° out of phase.
So, that material can be avoided and still this structure is somewhat awkward. So, a
simplified way of making a bit of constructional features that is in fact, I am telling is that
you take plane laminations with two windows like this. So, this is the lamination in one
plane and put several of them to get the height or the width of this transformer that is in 3dimension it will look like this which sectional view will be like this.
382
So, there will be three limbs and in each of the limbs you put coils of one phase say 𝐴1 ,
𝐴2 and its secondary instead of putting it some other place we should not do it in case of
three-phase core type transformer core type transformer which is very easy to construct.
And, then you have it is secondary and this terminals will be marked as 𝑎1 , 𝑎2 .
Similar is the case with B-phase winding. You have several its primary and its secondary.
And, they are 𝐵1, 𝐵2 primary small 𝑏1 , 𝑏2 secondary and finally, the C-phase winding 𝐶1 ,
𝐶2 and small 𝑐1, 𝑐2 . And, so, basically the analysis will remain same as that of a bank of
three single phase transformers. That is the representation of these in terms of coil I will
show it similar to the previous one 𝐴1 , 𝐴2 and its secondaries with the understanding that
capital 𝐴1 and 𝑎1 they will be the dot terminals. 𝐵1, 𝐵2 and this is small 𝑏1 , 𝑏2 and this is
capital 𝐶1 capital 𝐶2 and small 𝑐1, 𝑐2 .
And, then I could connect them in star-star, delta-star, star-delta and what not with some
specific vector grouping ok. But, only thing here one must understand that suppose you
connected this three in star in star and excited with secondary open circuited secondary
also suppose star. So, this will perhaps give me know why perhaps it will give me Y y 0
connection.
But, the point I want to tell you that with secondary open circuited primary energized, the
reluctance of the flux of 𝜑𝑎 and 𝜑𝑏 will be different because the flux 𝜑𝑎 will get divided
like this into two parallel paths sorry, 𝜑𝑎 𝜑𝑏 will be like this it will divide here itself in
two paths and 𝜑𝑐 will be once again like phi a it will be like that. Therefore, the reluctance
may be different slightly. Therefore, the no load current required, magnetizing current
required will be slightly different in the phases.
Now, if you energize it with balance three-phase source then one thing is clear as I am
repeatedly telling you. Suppose, primary star connected; so, flux created by A phase 𝜑𝑎 ,
𝑑𝜑
its value gets fixed by this voltage is not 𝑁 𝑑𝑡 suppose this primaries are capital 𝑁1 . So,
what is the value of 𝜑𝑎 𝑚𝑎𝑥 for a phase? That phase voltage whatever you have applied
across capital 𝐴1 𝐴2 that divided by √2𝜋𝑓𝑁1 . Similarly, if these voltage supply voltage is
balanced so, this 𝜑𝑏 will be also like that its strength 𝜑𝑏 and 𝜑𝑐 will be also like that,
getting the point?
383
Now, this three fluxes as if we will presume they come here and meet they vanish because
𝜑𝑎 , 𝜑𝑏 𝜑𝑐 are 120° apart clear. But, nonetheless in the limbs this limb it has to be 𝜑𝑎 , in
this limb it has to be 𝜑𝑏 because KVL is to be satisfied in the primaries and this is 𝜑𝑐 . So,
𝜑𝑎 , 𝜑𝑏 𝜑𝑐 perhaps will meet here and vanish annihilate themselves because they are 120°
apart. So, things are expected to work why not ok. So, this is the thing.
Then you can proceed. So, far as connection is concerned with the understanding that these
are dots there is no difference in the making connection. So, like star-star, delta-star, stardelta and so on with this transformer also. So, no point in repeating in that but so, this is
called core type. There is another type of a three-phase transformer possible that is called
shell type I will just sketch that shell type and write it like this shell type.
Here it is much more symmetric so far as the reluctance this dash is concerned and the flux
paths are independent. What they do is this as if you have taken. I will just mention this
core type is much more popular easier to construct, shape is fine and this is the third one
this is the sectional view of the stampings. These are windows, this is the iron portion.
Here what is done the windings are placed here primary, secondary. Similarly, this is for
A-phase I am not going to mark the terminals it will become too clumsy. So, this is for Aphase say I will take this color and this is for say B-phase it is primary, it is secondary and
C-phase primary, secondary got the point? So, here you see how this flux paths will be
there this. So, you see the windings are placed on the central limb not on the outer limbs
and the flux if you energize the primary A-phase it will produce 𝜑 here, it will nicely
closed onto itself getting the points?.
Similarly, for B-phase flux paths will be like this as if they are independent three single
phase transformers. In fact, they are, but only thing the windings are on the central limb
and here also C-phase flux will be somewhere playing this, but here the fluxes do not have
that is they interact 𝜑𝑎 , 𝜑𝑏 𝜑𝑐 directly that is why the no load current will be slightly
different. So, this is called shell type oh I have written.
Anyway in this cell type some people say a shell type is that type of transformer where
iron similarly in the single phase single phase transformer also you could have a in fact,
there are single phase 3 number of shell type transformers you can make a single phase
transformer in this way. In contrast with the core type; core type single phase was what?
It was like this here is primary winding, there is secondary winding and they may be put
384
in the same limbs distributed on both the limbs primary, secondary, but effectively it is
like that.
Here the primary and secondary will be put here just get the idea that is all. So, this is
single phase shell type people say that in single phase I do not impresses the winding in
the core type coil impresses the one as you can see iron is also showing, but anyway
operation etcetera will be similar flux paths will be different. So, so this is shell type
transformer three-phase of which this core type is more very much popular and they are
use to to make say power transformers all core type.
(Refer Slide Time: 13:38)
Now, after telling this I must tell what about the nameplate rating of a I will draw a vertical
line and write like this nameplate rating of a 3 phase transformer unit. It will be written as
suppose 30 KVA say I am just giving some number say 600V/200V which is not very
practical numbers, but just I am giving you for easy computation 600V/200V, 50Hz and
before that 3 phase 3 30 KVA 600V/200V 50Hz transformer. Mind you, it is not the
collection of three single phase identical units and this is the name plate rating.
Also you will be provided with the connections for example, it may be given as Yd1; that
means, the high voltage side will be connected in star, low voltage side in delta and I know
how to connect them. So, henceforth what I will do because connections once I have
understood, I will just tell that this transformer the normal way of drawing star and delta
connection I will do the primaries are connected like this even I will not mark 𝑎1 , 𝑎2 , 𝑎3 ,
385
𝑎4 . We know that whoever has connected he has connected properly so that you achieve
this Yd1. Similarly, the secondary; secondary is delta connected. So, it will be like this is
not and connected correctly ok.
Now, for solving problems this that it is better you draw this so that easily you can interpret
the data. Now, what this 600V means? This 600V means that this is HV side, this is LV
side, this you listen carefully. This means line to line voltage, mind you. Similarly,
secondary side voltage unless otherwise specified which will never be it is like that only,
but still I am telling if nothing is specified otherwise this will be also line to line voltage.
So, this is line to line, mind you and this is also line to line voltage. This is very important
thing to note, not that this primary winding voltage rating is 600V or 200V what about the
current rating I told you in a transformer KVA and voltages are given, current values are
not explicitly specified. From KVA and voltage informations you have to find out the rated
currents of the winding. In this case, this 30KVA is the total KVA total KVA.
Therefore, from the total KVA and line to line voltage I will be able to calculate the line
currents of the respective side not the winding currents. For example, if I say the if I call
this is side 2, this is side 1 I will say this line current 𝐼𝐿2 will be given by
𝐾𝑉𝐴 = √3𝑉𝐿𝐿1 𝐼𝐿1 = √3𝑉𝐿𝐿2 𝐼𝐿2
√3 × 600 × 𝐼𝐿1 = 30𝐾𝑉𝐴
𝐼𝐿1 =
30 × 103
√3 × 600
≈ 28𝐴𝑚𝑝
√3 × 200 × 𝐼𝐿2 = 30𝐾𝑉𝐴
𝐼𝐿2 =
30 × 103
√3 × 200
≈ 84𝐴𝑚𝑝
I will be able to calculate 𝐼𝐿2 . Mind you, this from this formula you will be able to calculate
the line current of this secondary side.
Similarly, primary side line current that is 𝐼𝐿1 neglect that no load current which will be
5% of that rated; this is the rated line current of the secondary side.
386
So, this is these are the line currents I will be able to in ampere mind you calculate it.
So, this currents I will be able to calculate I will write it down. Now, then one may ask
question what is the turns ratio of the windings? How to find out turns ratio turns ratio of
HV and LV windings or turns ratio per phase. If you are asked to calculate the turns ratio
suppose the number of turns of the primary winding is 𝑁1 and secondary winding is 𝑁2
𝑁
600
2
√3
then I will say 𝑁1 will be how much the voltage across this which will be
divided by
𝑁2 line to line voltage is 200V by 200.
600⁄ )
𝑁1 (
√3 = 3 =
=
√3
200
𝑁2
√3
This will be the turns ratio.
28Amp this will be this is the HV side. So, LV side it will be higher and we will be able
to calculate. So, the rated currents of the lines are known. Therefore, rated currents this
can you calculate tell me this value? rated current of rated currents of.
So, this is around 84Amp. So, rated currents of windings, it is not this 84Amp or 28Amp
rated current that is this current
𝐼𝐿𝑉
. Rated current will be this is 84Amp, therefore,
𝑊𝑖𝑛𝑑𝑖𝑛𝑔
that divided by √3 because it is delta connected so much ampere. Similarly,
𝐼𝐻𝑉
𝑊𝑖𝑛𝑑𝑖𝑛𝑔
rated current that will be same as the line current. So, that is 28Amp, getting the idea?
𝐼𝐻𝑉
= 𝐼𝐿1 = 28𝐴𝑚𝑝
𝑊𝑖𝑛𝑑𝑖𝑛𝑔
𝐼𝐿2 84
𝐼𝐿𝑉
=
=
𝐴𝑚𝑝
𝑊𝑖𝑛𝑑𝑖𝑛𝑔 √3 √3
Therefore, you must remember on the nameplate of a three-phase transformer of a single
unit compact three-phase transformer no three separate transformers you have put together
for connecting it a to work it as a three-phase transformer not like that a single unit of
three-phase transformer, whatever KVA is given it is the total KVA whatever voltage
ratings are given at the line to line voltages of primary and secondary side.
387
So, from which directly you will not be able to say what is the turns ratio of the coils.
Unless it is same connection that is star-star connection, turns ratio will be same as line to
line voltage, got the point, any mistake? So, so this is this point must be very clearly
understood. On the other hand, when I was using three single phase transformers the rating
of each one is known and then depending upon the connections we decided what should
be the line to line voltages that I have discussed earlier you must refer to that.
But, so far as three-phase transformer is concerned this is how it will be it is rating will be
specified ok. Unless once again I am telling the same connection star-star or delta-delta
ratio of this line to line voltages will be same as ratio of the turns ratio of the transformer
ok. So, this is how it is to be done ok. Let me correct it. So, this connection this will be.
So, so given the ratings you will be able to calculate everything including the turns ratio
of the coils. [FL] Now, I will.
(Refer Slide Time: 26:39)
Tell you about another connection first, then I will discuss about when to use stars
connection, delta connection, on the high voltage LV side etcetera those some points I will
discuss, but before that another connection I will discuss. So, now I know that so far as
connections are concerned only thing you will ask for where are my primary terminals
with polarity marking 𝐴1 , 𝐴2 capital where is secondary small 𝑎1 , 𝑎2 and so on, then you
can connect it and it is same for both the kinds.
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Now, another connection I will just tell which is also very interesting connection which is
called zigzag connection in zigzag connection it is usually denoted by z; like star Y delta
D zigzag connection is denoted by z and why zigzag connection is necessary that also we
will discuss. But, first let us for academic interest let us try to see what this connection
essentially means. At the end it will turn out to be equivalent to star connections.
Let me explain. Suppose, and it does not matter whether you are using three single phase
units or a three-phase transformer as a single unit. For example, to connected zigzag
transformer you must have for a particular phase say I am drawing here suppose three
single phase transformers I have taken. So, far I was telling this is the primary winding
and there was one secondary winding 𝑎1 , 𝑎2 is not and similarly, another transformer with
𝐵1, 𝐵2 small 𝑏1 , 𝑏2 .
But, in this case to carry out zigzag connection for each primary phase you must have two
secondary coils, identical. This connection is very interesting 𝑎3 , 𝑎4 and I will say that all
odd number terminals are dots, it means this. Similarly, for B-phase and C-phase which I
am not drawing three single phase transformer. Naturally if a you apply a voltage phasor
if the voltage applied in the primary across this transformer 𝐴1 𝐴2 on the secondaries you
will have two separate, but equal length phasors or the secondary because their number of
turns are same 𝑁2 , 𝑁2 and this is 𝑁1 , is not? If you excite it with some voltage then same
voltages will be induced and the this lines will be equal.
Therefore to carry out zigzag connection we must have for each phase two identical
secondary’s. In case of three-phase transformers it mainly means that that is if I go to
previous page which I have already drawn for A-phase I have drawn only small 𝑎1 , 𝑎2 ,
but there is no space to accommodate another coil what I will do here is that this limb only
I am sketching. Suppose, I can also do like this. This is the primary winding 𝐴1 , 𝐴2 and
there will be two secondary’s like this 𝑎1 , 𝑎2 and 𝑎3 , 𝑎4 and this I will do for B-phase and
C phase as well that is each phase must have two identical secondary coils. So, through
that is the thing.
Therefore, the connections will be. So, now, I will simply draw the coils to execute
connection for example, 𝐴1 , 𝐴2 . Now, there are now two secondary coils 𝑎1 , 𝑎2 and small
𝑎3 , 𝑎4 then for B-phase 𝐵1, 𝐵2 and it will also have two coils 𝑏1 , 𝑏2 and 𝑏3 , 𝑏4 and finally,
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it will have 𝐶1 , 𝐶2 and small 𝑐1, 𝑐2 and 𝑐3 , 𝑐4 with the understanding that these are dots all
odd numbers. The arrangement is understood absolutely.
Then what I will do? I will make it suppose primary I have connected in star ok. So, I will
clean this portion now. I know what it is ok.
(Refer Slide Time: 33:53)
Now, primary I will connect in star a valid star for example,. I will connect short 𝐴2 , 𝐵2,
𝐶2 and give supply to 𝐴𝑠 , 𝐵𝑠 and 𝐶𝑠 supply and these are the primary phasors. And, this is
𝐴1 which is same as 𝐴𝑠 , this is 𝐵1 same as 𝐵𝑠 and this is 𝐶1 lengths are equal and this is
your neutral 𝐴2 , 𝐵2, 𝐶2 . Now, on the secondary coils I have not connected it anything, but
I have now these voltages available to me to play with. What I mean by this that is I will
then have on the secondary side this two voltage 𝑎1 , 𝑎2 and 𝑎3 , 𝑎4 .
Similarly, I will have parallel to 𝐵1, 𝐵2, 𝑏1 , 𝑏2 and 𝑏3 , 𝑏4 and also I will have these voltages
𝑐1, 𝑐2 and 𝑐3 , 𝑐4 , is not? Six voltage phasors will be available to me to belonging to each
phase. Now, what is zigzag connection means these two phases two secondary coils you
connect them in series ok. The rule is these two coils take this is suppose group 1, this is
group 2 you say take a phase coil and connected in series with either b or c phase coils in
the second group that is you do not connect 𝑎1 , 𝑎2 and 𝑎3 , 𝑎4 in series there will be mixing
of phases.
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So, the these voltages are this one 𝑎1 , 𝑎2 of course, you can connect them into series then
it will be normal effectively a single coil of 2 into turns see this is 𝑁1 , this is 𝑁2 and this is
also 𝑁2 all of them are 𝑁2 , 𝑁2 , 𝑁2 , 𝑁2 , 𝑁2 , 𝑁1 , 𝑁1 . So, what you will be doing in a sense
is that you will connect two coils in series and try to achieve the connection.
So, one coil from this group this is a suppose group I and this vertical coils group II. So,
take one coil here and take either b or c whatever will be relevant that we will discuss.
Similarly, B phase and other from this one. These two you go on connecting in series and
to achieve some what is called zigzag connection ok. So, the question is, what is to be
connected and how. The whole idea is that for example, I would like to connect the primary
in Y, secondary in z I would like to connect.
And, here also it is like say 1. Zigzag connection although it is not yet done, but I told you
it will come effectively a star connection that is.
Um most probably 1, let us see ok. Y z 1 effectively star connection, but two coils will
belong to two different phase groups and at the end 3 terminals will be shorted. Therefore,
whatever will be the neutral just like start this side that will be a neutral and from neutral
to one of the lines if you try to go you will encounter two coils in series belonging to two
different phases ok.
Now, if this is the thing this is 𝐴1𝑁 therefore, A-phase from the secondary if it is 1 suppose
this is the secondary neutral which one will become neutral I am not sure yet. Suppose, it
is neutral, this is the A phase of the primary and I am telling it is one means 30° lagging.
So, what I will do I will draw a line 30° lagging here. So, neutral to A-phase on the
secondary side must lag this was my 𝐴1 this vertical line primary; secondary A phase
voltage, we will lag the primary a phase voltage by 30°.
Now, the question is how this phasor can be achieved? It can be achieved see this was 𝐴1 .
So, I must this phasor this phasor can be realized by bringing this 𝑎1 , 𝑎2 here and then you
can easily see this phasor is parallel to C-phase. Suppose, I have decided in this outer coil
see it will be 𝑎1 , 𝑎2 . So, so I will write here as 𝑐4 , 𝑐3 . See this phasor is 𝑐3 , 𝑐4 plus 𝑎2 , 𝑎1
can give me this one.
So, this is one phase voltage now once you have drawn this the result is obtained B-phase
will be 120° apart from this line. So, draw a line here and here I must expect some 𝑏1 must
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come. How that 𝑏1 will come and this phasor can be broken up into this plus this. So, if it
is 𝑏1 , 𝑏1 is available this must be 𝑏2 and who will provide me this phasor? It must come
from some A-phase 𝑎1 , 𝑎2 has already been used now 𝑎3 , 𝑎4 is remaining. So, 𝑎3 , 𝑎4 , got
the point?
And, where will be my C-phase it must be horizontal after this one this is 30°, this is 90°.
So, here it must come and these voltage must belong to some 𝑐1 .
Hm. So, so, what it should be this is 𝑐1 , 𝑐2 is available fine 𝑐1 , 𝑐2 . I mean this lines are all
equal please forgive me for this one and this will be provided by whom by B-phase. Bphase one voltage I have used 𝑏1 , 𝑏2 , so, 𝑏3 , 𝑏4 . Therefore, effectively now after knowing
this I will come for connection it will be as I told you sort of star connection three terminals
must be shorted. So, and I now know which 3 are to be shorted. So, 𝑎3 , 𝑏3 , 𝑐3 are shorted
then what I see 𝑎2 must be joined with 𝑐4 .
So, this you take a piece of wire and connect here. So, so 𝑎2 is to be connected with 𝑐4 ;
connect it, and this will go to your load, a phase. Similarly, come to the. So, this is over
𝑎4 , 𝑏2 is to be shorted where is 𝑎4 ? So, 𝑎4 and 𝑏2 are to be shorted it will be like this and
this will come as b phase terminal. Similarly, 𝑐2 , 𝑏4 are to be shorted 𝑐2 and 𝑏4 are to be
shorted and that will come as your secondary c phase terminals. So, you can see this is
called zigzag connection.
So, zigzag connection is essentially a sort of star connection three term, you must required
two identical coils you cannot do anything and we have just discussed how it can be done.
Now, the big question is what for all this things that is there, but suppose I say you that a
transformer each phase has got two secondary’s, then it looks like these way also one can
get a balanced three-phase output voltage. Rule is you take one coil here I specify the rule
belonging to one phase a-phase then you see this coil is in series with c-phase coil
belonging to a other phase; b-phase coil in series with a phase coil. And, mind you, these
are the dots and c-phase coil is connected to b-phase and you get this connection. More on
this connections in next class.
Thank you.
392
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 41
Zig Zag Connection
Welcome, to lecture number 41st lecture.
(Refer Slide Time: 00:18)
And, we were discussing Zig Zag Connection of three-phase transformers this three-phase
transformers could be three individual units. So far as connection is concerned you can
implement any connections, but to implement zig zag connection of course, there is a
condition that each phase must have two secondary coils 𝐴1 , 𝐴2 this is small 𝑎1 𝑎2 and 𝐵1,
𝐵2 B phase primary it will have two coils 𝑎1 𝑎2 and 𝑎3 𝑎4 and then small 𝑏1 𝑏2 and here
it is 𝑏3 𝑏4 and finally, the C phase and this I have assumed these are dots odd numbers.
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(Refer Slide Time: 01:36)
And, last time I told you how to implement Y z 1 connection see Y z 0 is not possible you
cannot have a Y z 0 connection you think about that, but Y z 1 is possible. Now, let us try
today say Y z 11 so that you can practice it Y z 11 connection whether it is possible or not
that will also come out Y z 11. So, in this case primary I will short steps are like this and
then primary phases I will draw. It is star connected, it is like this is 𝐴1 , this is 𝐵1, this is
𝐶1 , this point is 𝐴2 , 𝐵2, 𝐶2 , this side neutral and, on the secondary the moment you do like
this you apply voltage secondary you immediately get two sets of voltage for each phase
and these two sets are identical 𝑎1 𝑎2 , and 𝑎3 𝑎4 .
Similarly, 𝑏1 𝑏2 and you draw these not 𝑏3 𝑏4 and then finally, this two parallel to c phase
𝐶1 𝐶2 small 𝑐1 𝑐2 and 𝑐3 𝑐4 . Now, in this case this is the primary voltage whether 11 is
possible, zig zag connection I know three terminals will be shorted. So, there will be a
neutral point available looks like a star connection. Therefore, whatever will be the phase
voltage staring from that neutral of the secondary side that a phase voltage must lead this
a phase voltage.
Therefore, I draw first this a phase voltage that is at 12 and then I am expecting my thing
here at 30° you know this is also straight forward; similar to the previous one, I am just
doing so that you get used to it. So, this must here from the a phase from the secondary
side must come in and I will prefer that it comes out from small 𝑎1 , that is what I prefer
and naturally and similarly your b phase with respect to whoever will become the neutral
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that I will see after completing this b phase also should come from 𝑏1 I would prefer and
similarly the this is 120° apart c phase must come out from this.
Now, as you can easily see this 𝑎1 it can be nicely placed here 𝑎1 𝑎2 ; then how to reach
from this point to this point by a b-phase voltage; should I use 𝑏1 𝑏2 ; no, better use 𝑏3 𝑏4 .
You can use 𝑏1 𝑏2 mind you, but you should be systematic. So, this is group 1, this is group
2. It does not matter whether you either of them you can use, but I will use put this
conditions. So, this will be your 𝑏3 𝑏4 ; is not?
Similarly, this 𝑏1 phasor is this 𝑏1 𝑏2 I will quickly draw and this must be come from 𝑐3
𝑐4 and I know 𝑏2 𝑐4 is to be shorted and this fellow will come from this; that is this is 𝑐1
𝑐2 and then this is only left out thing is 𝑎3 𝑎4 . Therefore, you will be joining this time 𝑎3 ,
𝑏3 , 𝑐3 I will short and then I will take out the secondariess from this, this, this then 𝑎1 𝑎2
should be joined with 𝑏4 𝑏3 coils.
So, I will take a piece of wire then 𝑎2 should be joined with 𝑏4 . So, 𝑎2 I will take and join
with 𝑏4 then 𝑏2 I will join with 𝑐4 that is this one and 𝑐2 with 𝑎4 and your connection is
over and this is Y z 11 ok. Similarly, you can try D z connection. These I leave you as an
exercise whether primary delta, secondary zig zag can be connected or not and what are
the possibilities.
So, zig zag connection essentially means that it will be two coils in series essentially star
connections, but they will belong to two different phases there are some advantages of this
connection. Now, after I have given you how to connect logically how to implement
logically different connections I give it to you that I will be able to do. Now, I will discuss
that after this discussions I am now familiar with the connections the question is when to
use what connections.
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(Refer Slide Time: 08:36)
Before I start discussing on that now I will tell you one very important things that happen
in particularly three-phase transformers, three-phase practical transformers. You know that
if the B-H curve of a core material is linear. Suppose, the B-H curve is linear of the core
material B means this is also 𝜑, where 𝜑 = 𝐵 × 𝐴𝑟𝑒𝑎.
So, if the core material and H means what? Effectively magnetizing current it means also
that you we are now familiar with this. If this informations are important to understand
what I will be talking about; what I am telling if I say in the core of a transformer I will go
fast but it will be a qualitative discussion primarily. Suppose, you have a transformer single
phase transformer. What I told you that you apply a sinusoidal voltage, a sinusoidal flux
is guaranteed nothing doing because it has to satisfy the KVL equation there and your
supply side is a sinusoidal voltage that is fine.
So, I will now examine two things: one thing is, suppose, flux in the core of the transformer
is sinusoidal, what will be the nature of the current? Nature of the current if the flux is
sinusoidal, suppose flux is varying like this, this is B-H curve. So, go to a particular B and
sketch this side is what should I say is this the thing I mean yeah this is 𝜔𝑡 suppose I mean
be it different values and here it is 𝜔𝑡.
Then if this is B or should I have being drawn here. So, for sinusoidal flux, the projection
if you want to generate this much flux the current value too will be maximum there. It will
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be also sinusoidal because projections it is a linear line nothing is everything is fine.
Sinusoidal flux you want to create you require sinusoidal current that is the conclusion.
Reverse way, if you pass sinusoidal current you are guaranteed flux will be sinusoidal in
the core ok, but we know that B-H curve of the core of a practical transformer is neither
linear not a single valued function because of the presence of the hysteresis. See, in our
derivation of hysteresis laws I did not discuss that because till this time because I now it
is now you will appreciate this point because what happens is this B-H curve is not linear
not only linear it is double valued function for a particular H for going up.
So, it will be B-H curve will be something like this, This is the B-H curve; this is for rising
value of current you have to use this for decreasing value of current you have to use this.
Of course, we must see that this area is thin. I have drawn it exaggerated it should be like
this very thin, but nonetheless there is some hysteresis in order why it should be thin
because hysteresis loss you want to minimize. Area of this curve represents the hysteresis
loss per unit cycle per unit volume or kg of the iron material that we have discussed.
So, this side is the current and this side is the B or flux I can denote it. Now, I will ask
myself that if I say the value of the flux with respect to time changes like this I will better
do not put 𝜔𝑡 like this suppose this is time axis and I say that B value is changing 𝜑 or B
value is changing with this. The question is very innocent. Suppose, this is B or 𝜑, suppose
I want to create a sinusoidal flux what should be the nature of the current? I now know it
cannot be purely sinusoidal because it is not a linear line relationship.
For example, what I am telling when B is 0 at this point and it is increasing where from
should I read the value of the current? This is 𝜔𝑡 = 0 say, B is increasing. So, from this
curve so, you drop a for example, take this point at 𝜔𝑡 = 1 at this time, this is the flux how
much current is necessary, go horizontally. Which point should I read; this point or this
point; it is a double valued thing, phi is increasing. So, 𝜑 is increasing means this one. So,
come here and read the value of the current this much current and this where should I plot?
I will plot it at 𝜔𝑡 = 1. This much I will say is the instantaneous value of the current
needed.
At 𝜔𝑡 = 0, 𝜑 = 0 but flux is increasing; should I read from this curve or this curve; from
increasing flux. So, from this curve only. So, at 𝜔𝑡 = 0 also it will have some value here;
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not 0, 𝜑 = 0 it is not that exciting current is 0 because of the presence of this one. So, in
this way I can map each value of flux and get the current wave form.
Now, this current waveform will be periodic no doubt, but will be distorted not pure
suppose it is 50 Hz 𝜑 is changing I want to create a 50 Hz flux, one frequency and why I
want to create a single frequency flux here because my induced voltage across the
secondary which is going to supply a load that will have a single frequency voltage I do
not want any other harmonic component of voltage to be induced. Only 𝜑 should be there,
that is my ideal thing I am desiring here whatever 𝜑 is there that must be 𝜑 =
𝜑𝑚𝑎𝑥 cos(𝜔𝑡). But, what I am telling this flux distribution the current distribution here
will not be linear it will be distorted but periodic in nature.
I am not sketching in the notes I have drawn nicely. So, the thing is the conclusion is
because of the hysteresis loop because B-H curve is of this type to create sinusoidally
varying flux in the core current needed is periodic but distorted. What this distorted means?
It means that since the current wave form is periodic I can do the Fourier analysis.
Periodic with same frequency 𝜔, I mean time period will be 20ms what with that supply
periodicity this because this flux is moving with some period periodic time. So, with the
same periodic time it will be changing, but the question is that can be then Fourier analyzed
that is it will be a 50 Hz only this.
So, current waveform may be it will be like this and repeating, are you getting? This is the
magnetizing current but not sinusoidal and it can be broken up into a fundamental 50 Hz
and lot of odd harmonics because it is the negative half is same as the positive half and so
on. Therefore, only odd harmonics will be present if you Fourier analyze this exciting
current and you know with higher order harmonics the amplitude of the coefficient of the
harmonic terms they get reduced generally and therefore, is periodic, but distorted with a
strong fundamental component with predominant third harmonic component; got the
point?
That is the exciting current, in this coil then I will demand do not pass pure sinusoid
current. If your goal is to pass to create a sinusoidal flux in the core and you are planning
to get one frequency voltage across the secondary of the coil of fundamental frequency,
then the exciting current pure sinusoid will not do. If you neglect the other higher order
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harmonics you will demand that there should be a 50 Hz component current as well as 150
Hz component of current must be present.
In order that flux in the core can be sinusoidal of 50 Hz one frequency flux 50 Hz to be
created and you require a 50 Hz component current and 150 Hz component current other
higher order harmonics fifth harmonic 250 Hz ok, we are neglecting for the time being.
So, this is the thing.
(Refer Slide Time: 22:30)
The reverse problem is that exciting current is suppose sinusoidal; sinusoidal of one
frequency say 50 Hz. Then, how the flux will look like means what I do not know very
good English but anyway it is there. So, the problem is like this same core, but this time
you pass sinusoidal current from a current source.
This exciting current I will pass some 𝐼𝑚𝑎𝑥 sin(𝜔𝑡), I have forced it current a current
source imagine, you need not bother. Suppose, sinusoidal current it is excited I want to
know what will be the nature of then what will be the nature of flux waves; then better
what will be then what will be the nature of B or 𝜑? That is what. So, same problem; so,
draw the B-H curve as I have drawn previously this is the B-H curve.
Now, what I am telling is this is 𝜔𝑡 and I am telling now this current waveform mind you,
this axis is either current or H this axis is B or 𝜑 for a transformer. Now, what I am telling,
I will make the current waveforms sinusoidal 50 Hz, one frequency will B the reverse way
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current is sinusoidal what is the nature of B? In the previous case, what I told, if you want
to create the B sinusoidal what is the nature of the current; current is distorted.
In the same way if you want to create pass a 50 Hz current what is this axis? 𝑖 then the flux
will be produced once again I will sketch it suppose current is 0 and increasing this point.
Current is increasing and 0, this is the B produced, not 0; is not? So, B will be some
negative value at this time and I can go for each of the point when the current has positive
some value that is this value this is corresponding to say 𝜔𝑡1 how much is the flux;
negative.
Oh flux is 0, correct at 𝜔𝑡1 flux is. So, it will be some somewhat it will now be distorted
are you getting point by point if you go because of this double valued nature of the function
of the B-H current. So, the nature of B or 𝜑 will be distorted will be distorted, but periodic
why not, but periodic. Then this B distribution or 𝜑 once again can be Fourier analyzed
and you will see that it has got a strong fundamental with lot of odd harmonics because
the positive and negative excursion of this B will be same.
Therefore, I will say that what will be the nature of B or 𝜑? The answer is it will be
distorted, but periodic after Fourier analysis I am sorry. After Fourier analysis, you will
come to the conclusion that a strong fundamental that is 50 Hz in this case with a
predominant third harmonic component.
(Refer Slide Time: 29:22)
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So, in summary I will say that to create flux to create 50 Hz flux in the core I am not
writing this we require we require magnetizing current magnetizing current having 50 Hz
component 50 Hz fundamental plus 150 Hz component; amplitude I am not telling a little
a thing it will be required for majority will be this.
Similarly, to create similarly I will say if magnetizing current if magnetizing current is
forced to be sinusoidal to be sinusoidal core flux will have will have a strong fundamental
strong fundamental that is 50 Hz component same result opposite way and plus 150 Hz
component of flux. So, this is necessary to know and this is all because the B-H curve is
not a straight line ok, it is because of the hysteresis loop.
So, what happens is, if suppose the core flux want to create 50 Hz one frequency that is 50
Hz only sinusoidally varying, then I am telling exciting current then should have a 50 Hz
component plus a 150 Hz component. The reverse way if somebody says magnetizing
current I will somehow manage it to be 50 Hz one frequency fundamental 50 Hz
component; no distortion. Then what will be the nature of the flux; then also the core flux
will be distorted and it will have both 50 Hz and a 150 Hz predominant harmonic
component.
We will continue with this discussion in the light of transformer connections.
Thank you.
401
Electrical Machines – I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 42
Effect of 3rd Harmonic Exciting Current and Flux
Welcome, to lecture number 42nd and we were discussing about some 3rd Harmonic
Exciting Current and 3rd Harmonic Flux. And, the conclusion drawn from discussions in
my last lecture is that if the exciting current or magnetizing current is 50 Hz sinusoidal
only pure sinusoidal 50 Hz exciting current, then flux created in the core will have a
fundamental and a third harmonic component in the reverse way if you want to create
sinusoidal flux 50 Hz pure sinusoidal flux then the exciting current must have along with
a 50 Hz component, a predominant third harmonic component exciting current will be
needed.
So, this is because of the fact the B-H curve of the material has a loop double valued
function sort of thing that creates the problem although the area of the B-H curve is very
thin, but nonetheless that third harmonic thing will be produced. Now, in the light of this
knowledge it has nothing to do with I mean as such transformer magnetic material is given
whose B-H curve is like a hysteresis loop. Then I can come to these conclusions thinking
that somehow I will pass sinusoidal current, what will be the nature of the flux? I ask
myself and get the answer.
Similarly, if I want to create sinusoidal flux what should be the nature of the exciting
current this is the thing we have discussed, although qualitatively no mathematical
analysis, but which was reasonably good it can be appreciated yes, this is going to happen.
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(Refer Slide Time: 02:43)
Now, today I will discuss for example, take the star-star connection of a three-phase
transformer ok. Now, continuing with that so, consider start star connection of three-phase
transformer and this connection can be obtained by using three single phase transformers
or a single three-phase unit transformer first let us see that, but the connections are same
in both the cases. So, suppose I will simply draw like this now because we will presume
that whoever is connecting is connecting correctly.
So, primary is star connected so, draw it star; secondary is also star connected, draw it star
this is the thing and what I told you that I will connect it to supply 𝐴𝑠 , 𝐵𝑠 , 𝐶𝑠 is the supply.
Here the output is 𝑎, 𝑏 and 𝑐 star-star connection. And, the neutral point of neither of them
is grounded nothing is done, but what have done is this I have applied a three-phase voltage
source here voltage source balanced voltage source, supply may have a neutral, but that
has not been connected anywhere only the three lines I have connected.
Now, the with secondary open suppose nothing I have connected here what in general I
was discussing you have applied line to line voltage, you will get √3 times less this voltage
𝑉
𝑁
√3
1
is not 𝐿𝐿 will come across a phase that into 𝑁2 etcetera you will get phase voltage like that
we are discussing. Now, first thing I will say in this connection one thing is clear. Now, it
will draw some magnetizing current with secondary open magnetizing current it will draw
and in the core of the transformer my attempt will be to make the flux sinusoidal.
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Why? Sinusoidal with 50 Hz component alone because of this fact that I want to get 50 Hz
voltage only to be generated in the secondary coil. I do not know want that third harmonic
voltage in any way coming across the secondary and creating problem for the load I want
a 50 Hz supply for the load.
But, if the connection is like this and with our previous knowledge we immediately
conclude that this currents whatever current magnetizing current it will draw it cannot have
a third harmonic component. To create flux in the core you require a 50 Hz suppose 50 Hz
source in that term will say and to create 50 Hz flux in the core of this transformer I require
a 50 Hz plus 150 Hz component of the exciting current also higher order fifth, seventh, but
those I am neglecting predominant thing is third harmonic.
But, what I am telling is in the line; in the line in this type of connection in the line 150 Hz
component of current will not flow will not flow ruled out why because third harmonic
which is 50 Hz third harmonic currents are co-phasor because if you say
𝑖𝑎 = 𝐼𝑀3 sin(3𝜔𝑡)
Then
𝑖𝑏 = 𝐼𝑀3 sin[3(𝜔𝑡 − 120)] = 𝐼𝑀3 sin(3𝜔𝑡)
𝑖𝑐 = 𝐼𝑀3 sin[3(𝜔𝑡 + 120)] = 𝐼𝑀3 sin(3𝜔𝑡)
Fundamental component, this will be sin(𝜔𝑡), this will be sin(𝜔𝑡 − 120) and
sin(𝜔𝑡 − 120).
So, they are co-phasor meaning that if third harmonic current flows here at any instant if
this current going this way is 5 Amp, this will be also 5 Amp, this will be also 5 Amp and
they will meet here they will not vanish KCL will be violated which is not the case with
fundamental component. Fundamental component if it is at any time suppose the instant
when a-phase current is maximum positive say 10 Amp then b-phase and c-phase will be
minis 5 Amp and minus 5 Amp. So, it will return via this path.
Therefore, in the lines I am here about one thing if no neutral etcetera is connected to the
system here about this fact no one can contest. In these lines only 50 Hz component of
current will flow, no third harmonic business. So, exciting magnetizing current will be
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sinusoidal of 50 Hz only no question of other it will be. If that be the case then we have
learned you are exciting the coils with 50 Hz current can be flux in the core be sinusoidal?
Certainly not; the flux in the core then will have both 50 Hz component and 150 Hz
component.
That is the conclusion is therefore, flux in the core will have will have both 50 Hz and 150
Hz component of course, the amplitude of the harmonic flux will be much less compared
to may be 10% of the rated flux fundamental component, that is there. But nonetheless
both of them will be present. So, in the core of this transformers there exists now two
fluxes of 50 Hz and another is 150 Hz.
Therefore, induced voltage in the coils say this is 𝐴1 , 𝐴2 this is small 𝑎1 , small 𝑎2 and so
on 𝑏1 , 𝑏2 , etc. So, here these coils will then be linked with 𝜑1 fundamental say for a-phase
𝜑𝑎1 (𝑡) plus 𝜑𝑎3 (𝑡) both of them will be linking this. Therefore, both of them will induced
voltage; induced voltage remind you it is equal to
𝑅𝑀𝑆 𝐼𝑛𝑑𝑢𝑐𝑒𝑑 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑡𝑜 𝐹𝑢𝑛𝑑𝑎𝑚𝑒𝑛𝑡𝑎𝑙 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
= √2𝜋𝑓𝜑𝑚𝑎𝑥1 𝑁1
𝑅𝑀𝑆 𝐼𝑛𝑑𝑢𝑐𝑒𝑑 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑡𝑜 𝑇ℎ𝑖𝑟𝑑 𝐻𝑎𝑟𝑚𝑜𝑛𝑖𝑐 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
= √2𝜋3𝑓𝜑𝑚𝑎𝑥3 𝑁1
And, this induced voltage will be also here in this secondary because same flux links both
primary and secondary. Therefore, across the phases of both primary and secondary as if
there will be two sources of emf between 𝑎 and this small 𝑛, there will be one source 𝐸𝑎1
and another source rms value 𝐸𝑎3 of course, they are of different frequency 𝐸𝑎3 3𝑓 and
this is 𝑓 two sources of emf will be existing between 𝑎1 and 𝑎2 and, similarly for b-phase
and c-phase.
Therefore, across the phases both in primary and secondary coils, the induced voltage will
have a 50 Hz component as well as 150 Hz component and that is what I am telling is not
desirable. Now, the question is what about line to line voltage? So, the conclusion is here
the phase voltage will be corrupted with a 150 Hz component apart from that 50 Hz
component.
405
So, so phase voltages will write here phase voltage phase voltage that is this diagram
suggest will have both 50 Hz and 150 Hz. Now, what I am asking is what will happen to
line to line voltages? See once again this is the a-phase. Let me go to the next page.
(Refer Slide Time: 16:11)
So, I am telling that I am drawing the secondary induced voltages suppose. So, I told this
is 𝐸𝑎1 50 Hz in series with another induced AC another induced voltage 𝐸𝑎3 , a-phase third
harmonic 150 Hz and then this is your a terminal. Similarly, for b-phase 𝐸𝑏3 this is 150
Hz. Then 𝐸𝑏1 fundamental component 50 Hz and finally, this c-phase 𝐸𝑐3 third harmonic
component plus minus AC 150 Hz and then fundamental component 𝐸𝑐1 50 Hz is it not?
This will be the thing, this is your b terminal going towards load c terminal. Similarly, in
the primary coils, do not worry about that.
Now, so, if you take a voltmeter, connect it here across the here suppose I take a voltmeter
connect here what it is going to read? It will be reading there are two AC voltages of
different frequencies, but this voltmeter reads rms values. So, this rms values square plus
this rms value square by √2, is not? Whatever is that formula I can I will be able to
calculate that rms value. So, it will not read this fundamental component rms value, that is
what I am telling if you connect it here.
The question is what will be the if I connect a voltmeter here between the lines what will
be it is reading? See, third harmonic voltages once again are co-phasor. Therefore, if you
traverse this path third harmonic voltage will cancel out therefore, in the line to line
406
voltage. Third harmonic component will be absent and here you will get this reading will
be √3𝐸𝑎1 what I am trying to tell because 𝐸𝑎1 , 𝐸𝑏1 , 𝐸𝑐1 that is rms value.
Therefore in the line to line voltage only 50 Hz will be there, but across each phases the
third harmonic component will be there, got the point? Therefore, in star-star connection
like this with no neutral connected anywhere isolated, the supply neutral I have never
connected, but I can tell you one thing about this.
(Refer Slide Time: 20:12)
Suppose, I say that this is the primary star this is the secondary and for a change what I
will do this is 𝐴𝑠 I wrote this is 𝐵𝑠 I wrote like that. So, anyway so, this is 𝐵𝑠 I wrote is not
𝐵𝑠 and this is 𝐶𝑠 I have drawn and the supply neutral listen to this point also I have
connected to this primary neutral.
Suppose, it is this connection is slightly different from the other balance three-phase
voltage to this and also neutral to this; supply voltage is balance three-phase voltage.
Therefore, I have applied across 𝐴1 and 𝐴2 a sinusoidal voltage across 𝐴1 𝐴2 to it I have
applied a sinusoidal voltage. Earlier how much voltage I have applied here I cannot say
the voltage across the terminal of the coil is V a by root 3 where neutral was not connected
that is the let me draw on the same page.
What I am telling in this case when neutral was not connected situation was like this 𝐴𝑠 ,
𝐵𝑠 , 𝐶𝑠 supply neutral was not connected. I can only claim that between these two lines I
407
have connected a 50 Hz voltage. But, I am not sure if I did not know anything about third
harmonic voltage I would have perhaps concluded that the voltage appearing across these
is this by √3.
But, that notion is to be changed now because of the fact that we know similarly, in the
primary side there is a fundamental component voltage and there is a third harmonic read
this capital 𝐴1 , 𝐵1, 𝐶1 also same thing. So, between the lines only fundamental will come
it is matching that KVL is not disturbed so far as supply is concerned. Only thing is that
𝑉𝐿𝐿
√3
is phase voltage you will never get because of the fact in the primary coils also the
fundamental and third harmonic will appear, got the point?
But, now in this connection when you have forced this supply neutral to be connected with
the primary side neutral of the transformer supply is being balanced. So, 𝐴𝑠 𝑁𝑠 connected
across these one is a 50 Hz voltage I am sure about that are you getting here this voltage
𝑉
applied is 𝐿𝐿 that I nobody can contest 50 Hz, only pure 50 Hz because supply is balanced
√3
50 Hz. Supply I am considering it has got no harmonics.
So, it will come here, but if this voltage is 50 Hz then the flux in the core has to be
sinusoidal because KVL is to be satisfied here. Then in this connection the core flux will
be sinusoidal 50 Hz, no this way that way. So, so in this case core flux is forced to be
sinusoidal 50 Hz only 50 Hz nothing. But, we have learned core flux is sinusoidal then
exciting current must have a fundamental and third harmonic.
Is there a path for third harmonic current to flow? Yes, there is. This current drawn from
the supply third harmonic current can be drawn very easily, because of what? They are cophasor, but there is a return path for the third harmonic current each this current, this
current, this current their instantaneous values are same so far as third harmonic current is
concerned. But, they will converge here and through the neutral it will go back to this
supply.
Therefore, in this case flux is forced to be sinusoidal with 50 Hz and magnetizing current,
then must have both 50 Hz plus 150 Hz component. Is it possible? Answer is yes because
of what? Third harmonic current will find its return path by this third harmonic current
will return via this path the neutral line, got the point?
408
And, if that be the case third harmonic current can flow in this connection flux will be
sinusoidal. Therefore, in the secondaries of this transformer small 𝑎1 , 𝑎2 etcetera only 50
Hz voltage you will get both line to line as well as phase because not third harmonic flux
exist, very nice, but only objection to this connection is this neutral wire will carry 150 Hz
current and it may disturb other electrical utilities. Earlier we used to say transmission line
used to go via the telephone lines also, earlier days telephone lines where there. So, they
can interfere with communication lines 150 Hz slightly higher frequency.
Therefore, that way it is slightly disturbing, but otherwise things will work everything will
be fine. So, third harmonic current 3 times the third harmonic current will flow for A phase
third harmonic plus B phase third harmonic plus C phase third harmonic. So, you
understand. So, 3 times the third harmonic current for each phases will flow here, but
everything will be fine, got the point? We will discuss with these with other connections,
but now looking back to single phase transformer what is going to happen?
(Refer Slide Time: 28:47)
It is qualitative discussion let us do that. So, in single phase transformer what is the
implication of this knowledge third harmonic flux, third harmonic excited current etcetera?
In single phase transformer you know this is the supply you are connecting; single phase
50 Hz voltage source and this is your secondary. And, I have ensured that this voltage
applied is 50 Hz. So, what will be the core flux? If this voltage is 50 Hz core flux has to
be 50 Hz core flux also we have only one component 50 Hz.
409
How? Because you have applied sinusoidal voltage it has to satisfy and secondary voltage
you will be also sinusoidal 50 Hz. But, then we ask what about if the core flux is sinusoidal
the magnetizing current must have a third harmonic component. It cannot be 50 Hz current
alone. The question is there a path for third harmonic current? Yes, it can flow like this no
problem because no three-phase business is there return path through star nothing like that.
So, third harmonic current can flow here like this. Now, only question is where is the
source of this third harmonic voltage? People say that you imagine first that the core flux
is suppose the magnetizing current drawn is 50 Hz only suppose you start arguing like that
50 Hz current is drawn flux will be distorted fundamental and third harmonic. So, this coil
will have a fundamental voltage induced third harmonic voltage induced and this third
harmonic voltage itself will circulate that current because it is a closed path it is getting.
Therefore, it will self correct itself and the flux will be restored to 50 Hz component alone.
So, the problem in a single phase transformer so far as third harmonic flux etcetera is
concerned is not that very critical, in the sense that if you assume the current drawn is
sinusoidal only then you will say then flux must have got disturbed 50 Hz and 150 Hz. So,
here there are two sources that is the way I drew, fundamental component and third
harmonic component. But, this circuit is closed, and third harmonic component can
produce its own third harmonic magnetizing current to restore the flux to the sinusoidal
value something like that and everything will be fine.
So, further discussion on these I will do in the next class where then will try to understand
which connection to use when.
Thank you.
410
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 43
Choosing Transformer Connection
(Refer Slide Time: 00:19)
Welcome to lecture number 43 and we have been discussing about Transformer
Connections and then we told you that in transformers because the B-H curve is in fact, a
loop although very thin loop, but nonetheless it is not a straight line relationship between
B and H. So, what happens is this in our last class the conclusion was qualitative
discussions on that.
(Refer Slide Time: 00:59)
411
Remember that these are the two things that is, if magnetizing current happens to be
sinusoidal 50 Hz then core flux will have both 50 Hz and 150 Hz component ok. On the
other hand if the flux in the core is 50 Hz only then magnetizing current must have 50 Hz
component as well as 150 Hz component, then depending on situations the current may
not have a 150 Hz component.
(Refer Slide Time: 01:54)
In case of say star star connections without neutral connected anywhere no scope for any
third harmonic current to exist in the lines of this transformers and therefore, the flux must
be having a 50 Hz as well as 150 Hz component; other higher order harmonic components
are also there, but we are neglecting that because their amplitudes will become smaller.
412
But your voltage phase voltages then will be distorted, although there will be no effect of
the line to line voltage that will only have 50 Hz components ok, but phase voltage will
have that 150 Hz component.
(Refer Slide Time: 02:44)
Anyway this sort of thing happens and we must have we must give importance to these
because of this fact. Now, although I have not drawn any voltage or current waveform
showing the fundamental flux and the third harmonic flux and the associated induced
voltages.
Today I will just draw those waveforms, so that we will understand what happens if the
flux is distorted. Suppose flux in the core has both 𝜑𝑚𝑎𝑥1 50 Hz component plus 𝜑𝑚𝑎𝑥3
that is the 150 Hz component this is sin(𝜔𝑡) and this is sin(3𝜔𝑡).
𝐹𝑙𝑢𝑥 = 𝜑𝑚𝑎𝑥1 sin(𝜔𝑡) + 𝜑𝑚𝑎𝑥3 sin(3𝜔𝑡)
Then both of them are going to induce voltage. So, in essence what I have assumed is these
the exciting or magnetizing current is 50 Hz component that is why flux has come out to
be this and then I want to see that 𝐸1 and 𝐸3 third harmonic component and the fundamental
how they will look like.
𝑑𝜑
Remember each one of them will induced voltage 𝑑𝑡 if you do and the angle with which
that induced voltage will appear is 90° that is known because you remember we draw like
413
this is the flux wave 𝜑1 then I draw 𝐸1 like this 90° relationship or in books some of the
books they will write −𝐸1 , but nonetheless the angle is 90° between them.
So, let us try to sketch that flux and induced voltage waveform to understand whether this
is really detrimental because phase voltage as in case of star star connection we will have
both fundamental and third harmonic component, what happens to the with respect to time
how the resultant voltage looks like that is the goal.
So, what I do is this suppose I say that I will try to sketch the waveform. Suppose I say
that this is suppose it is necessary I divide these are suppose fundamental 60° 180°, then
another 180° ok. So, let us see suppose the fundamental flux I am sketching; so this is half
period ok. So, it will be suppose like this; this is the fundamental component one to three
and it will be like this. So, this is this 𝜑𝑚𝑎𝑥1 sin(𝜔𝑡) I draw it.
Similarly, on the same graph I will draw and this amplitude is 𝜑𝑚𝑎𝑥1 . Then I sketch the
third harmonic component in the time axis. Now, since it is 3𝜔𝑡 when it completes one
cycle the third harmonic component will have 3 cycles over the period fundamental period.
So, it will be then how, it will look like it will be like this; this; this. So, it will be the third
harmonic component if I sketch it will be like this to 3 cycles over 360° fundamental and
this amplitude is 𝜑𝑚𝑎𝑥3 .
Then the induced voltages that is 𝐸1 and 𝐸3 I will sketch with respect to time that will be
displaced by 90°, it does not matter whether you show leading or lagging to understand
what is going to happen. So, I will just sketch the induced voltage which will be; so this
angles are 60° in the fundamental understand. So, the induced voltage because of the
fundamental I will sketch it say 90° lagging I am sketching it does not matter. So, it will
start 0 here this is 90° and where it will end it will be here half cycle, 180° this one.
Here it will be, this is becoming 180°.
Here. So, the induced voltage corresponding to fundamental it is also I am drawing blue.
So, it will end over here, but I have shifted by 90°, so it will be here.
So, this and this. So, induced voltage is suppose like this because of fundamental and it
continues. So, this is the fundamental induced voltage 𝐸1 and because of 𝜑𝑚𝑎𝑥3 that
voltage also will lag this red coloured curve by 90° with respect to its angle. So, this is 90°
414
here, is not. So, it will start from here and where will be the first cycle, it will be that is it
will be where?
That is this will be one half of this cycle are you getting. So, it will start from here, it will
go there, it will come and then it will go there and here.
Here, this will be the induced voltage and let us draw this is 𝐸3 mind you 𝐸3 and then I
draw the next cycle it will be like this it will come here.
Like that. So, 𝜑3 this is 𝜑3 first one 90° lagging with respect to this 𝜑3 if this is 𝐸3 like
that it will continue. Now, if you see the resultant voltage below this I will sketch that is
here I will sketch (𝐸1 + 𝐸3 ) that is this one and this voltage, you see the peak voltage this
voltage waveform will be peaky in nature. So, it will be in one half cycle that is in this
time frame if I sketch other half will be similar. So, you see it will be like this here this
peak and this peak coincides. So, it will be somewhat decaying next half it will continue.
So, the peak of this instantaneous voltage across the coil will become this much.
Now, the question is what is the order of the third harmonic flux? It may be of the order
of 8% to 10% ok. Suppose then if this is 𝜑𝑚𝑎𝑥1 , 𝜑𝑚𝑎𝑥3 ≈ 0.1 𝜑𝑚𝑎𝑥1 suppose let us do
that, then this induced voltage third harmonic may be as high as 30%. Why? Because
number of turns is more remember the RMS voltage is
𝑅𝑀𝑆 𝑉𝑎𝑙𝑢𝑒 𝑜𝑓 𝐼𝑛𝑑𝑢𝑐𝑒𝑑 𝐸𝑀𝐹 = √2𝜋𝑓𝜑𝑚𝑎𝑥 𝑁1
For third harmonic it is
𝑅𝑀𝑆 𝑉𝑎𝑙𝑢𝑒 𝑜𝑓 𝑇ℎ𝑖𝑟𝑑 𝐻𝑎𝑟𝑚𝑜𝑛𝑖𝑐 𝐼𝑛𝑑𝑢𝑐𝑒𝑑 𝐸𝑀𝐹 = √2𝜋3𝑓𝜑𝑚𝑎𝑥3 𝑁1
Therefore, that induced voltage will be then about 30% or so. So, the induced voltage this
𝐸3 amplitude will be about 30% of the fundamental voltage therefore, the peak voltage
will be this much.
Therefore, if third harmonic flux is present your transformer insulation there will be stress
upon the insulation of the transformer. Suppose we have not thought about third harmonic
flux and we design the transformer for the normal way 220V fundamental, 220V comes
induced voltage is also 220V based on that suppose we have designed the insulation level.
415
But now I find that every 20 ms there will be 2 peaks and whose voltage will be 1.3 times
the rated voltage, so insulation will be stressed. So, that is why people try to avoid this
third harmonic flux to be created, not only that there will be on the secondary side there
will be also not 50 Hz you will get if the flux contains third harmonic component, but at
the same time the insulation level will be under stress.
That is why you try to reduce them as far as possible or if it is not 10% it is less perhaps
you can carry on with that there will be a little bit of third harmonic voltage like that. So,
assuming the third harmonic flux to be 10% this will be the picture. So, this is the thing
ok.
(Refer Slide Time: 15:44)
Now, what I will tell we have discussed star star connection, then in star star without
neutral connection that we did last time. This was the situation across phases voltages
comes, across lines no 150 Hz component because third harmonic flux voltages will be
cancelling out in the lines because they are co phasal. Then to avoid this and therefore, this
terminal this voltage what I was telling is instantaneous value may be 30% higher than the
fundamental component of this voltage that is how this stressing upon the insulation comes
in, we cannot allow this to happen in general.
So, to avoid this I told you one method you connect this to supply neutral. The moment
you do this the flux in the core cannot, but be sinusoidal 50 Hz and therefore, current must
have a third harmonic component. Can third harmonic component of current exist? Yes, it
416
can because the return path is through this neutral wire. Mind you the third harmonic
current will be contributed by all the other phases they are co phasal, 3𝐼3𝑟𝑑 will flow
through the neutral line.
And the disadvantage of this method is that it is it may interfere with nearby
communication circuit it is not a good solution, but nonetheless it will force the phase
voltage to be 50 Hz as well as line to line voltage 50 Hz that is what we desire so far as
load is concerned is there any better method of avoiding this. We will come back to this
connections star star after some time once again to tell you another interesting observation
or right now should I tell let us see. So, this point you have understood.
(Refer Slide Time: 00:55)
Another disadvantage of star star connection let me continue with that. Suppose we have
a star star transformer primary and secondary is also star star and this of course, is and
neutrals are not connected anywhere they are living their own life it is not a member of 𝑁𝑠
supply neutral. So, this is your 𝐴𝑠 suppose 𝐵𝑠 , 𝐶𝑠 and here is your low terminal 𝑎 𝑏 𝑐 say
or better you call this 𝑏 this 𝑐.
Now, this neutrals are isolated totally; one thing is if you apply a balance free phase voltage
line to line voltage is balanced, then I told you in this case across the phases there will be
third harmonic component of voltage and now we learned about some idea if third
harmonic flux is 10% these voltage instantaneous voltage of this may be 30% higher than
the rated voltage we were thinking of, so put pressure of insulation.
417
Apart from that suppose we allow that to happen, but another thing is and if you load it
will supply load line to line voltage is sinusoidal no problem, but another interesting thing
happens here. Suppose I want to put a single phase load on the secondary of a transformer
only qualitatively I will tell you.
This is suppose transformer neutral here this is the neutral on this side, suppose I plan to
connect a load like this that is I want to attempt to supply a single phase load out of this
transformer. And these are suppose dots here 𝑏, 𝑐 are open this is quite interesting and
suppose I have connected an ammeter in series here to understand what is happening. If
you close this with some 𝑧2 connected close the switch the question is how much will be
the current in this phase.
It will be tempting that open circuit voltage whatever we had that divided by 𝑧2 will come
as the current. So, suppose some current flows here suppose, but this current will. So, far
as this transformer is concerned we will have a reflected current here and this reflected
current has to flow through the other windings whose secondary is not carrying any current
therefore, balance will be disturbed. Is not?
So, what is going to happen? I am telling you will find even if you close it ammeter reading
will be very small or no current, that is you cannot connect a single phase load. But then
𝑣
the question is what happens to Ohm’s law? This 𝑧 is the current there was voltage when
2
I when the switch was opened we have connected a 𝑧2 .
𝑣
𝑣
So, 𝑧 should be the current where then that 𝑧 = 𝑖 goes that relationship, it goes like this;
2
this secondary voltages this neutral point voltage. It can only happen if I say that after
closing you find this current is very little or no current, then I must tell that if very little
current then the voltage 𝑣 a must have been very small otherwise how that is how Ohm’s
law is also satisfied.
In other words what happens is this the moment you try to put a single phase load this
neutral will oscillate, situation will be like this. If suppose I say I will put a direct short
circuit 𝑧2 = 0, then also I am telling it is 0. What does all this mean? If you imagine the
impedance is so small 0 then the potential of 𝑎𝑛 and 𝑎 are same.
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If it is 0 this point and this point that is secondary voltage, I am telling this is neutral this
is a; these are the line voltages etcetera this 𝑛 point suppose this is the point 𝑎 this is point
𝑏, 𝑐 this 𝑛 has come here. So, that the voltage applied across the secondary is the line to
line mind you here is a balanced 3 phase voltage you have applied this voltage comes here
all the time this voltage comes here between 𝑏 and 𝑐 line to line voltage exist.
So, when you try to load the secondary of a transformer by a single phase loading this
neutral point has come over here indicating that the voltage applied to the other transformer
is 𝑏𝑛 which is almost line to line voltage 𝑏𝑎 because this is shorted. So, this neutral point
will oscillate somewhere depending upon the value of 𝑧2 you have connected and single
phasing loading you do not do do not ever try to attempt to load a Y-y connected
transformer with single phasing load because of the fact, then the voltage across the other
two coils where we have not connected load will approach to line to line voltage.
Shift sift means with 𝑧2 it will get to somewhere else.
So, in extreme case I told you 𝑧2 = 0, 𝑛 will come here that is this is not a stable point of
this neutral this must be understood and this problem will be much more great if 3 single
phase transformers with different cores are used. They will not allow a single phase loading
because other cores are other cores all the cores are magnetically independent ok.
So, this is another observation this is bank of 3 phase transformer. Anyway you think about
it, but single phase loading is not possible, but what happens if you connect the neutral to
the source neutral next step, then try to load this fine absolutely no problem. Load this it
will invite current and that current will flow here 𝑏 𝑐 is not carrying any current they are
mmfs in this line also will be 0 and everything is fine.
Therefore, without neutral connection single phase loading will not be possible with star
star connection because then the secondary coils will be very much stressed. In fact, phase
voltages it will come out to be across the secondary coils which phases are not carrying
any current shifting 𝑛 to 𝑎 in the worst case of loading that is you short circuit this. Anyway
so, this is just for your information you must remember that. Now, one way of avoiding
this third harmonic current in case of star star transformer we have discussed. Therefore,
it will be always you will be feeling that, then should I avoid star star connection.
(Refer Slide Time: 27:44)
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Now, there comes the other connections for example, star delta connection or delta star
connection one of the coils are suppose delta suppose star delta connection I will just draw
like this now correct star connection you have made and correct delta connection you have
made.
Here will be your load terminals and here is your supply terminals see this is 𝐴𝑠 , 𝐵𝑠 , 𝐶𝑠 .
Once again the neutral I have not connected to the source neutral, there is another neutral
available suppose 3 phase 4 wire system, but I have not connected. Will there be any
problem in this case? Here also I find since neutral is not connected to 𝑁𝑠 can there exist
a third harmonic exciting current? No, in the lines third harmonic current cannot exist. So,
what will be the conclusion? The exciting current whatever will be magnetizing current
that will be of 50 Hz component only and pure sinusoidal.
So, if exciting current is sinusoidal I am sure the core flux will have a fundamental and
third harmonic component and that is going to induced voltage both in the primary coils
as well as in the secondary coils and those third harmonic voltages are co phrasal. So, in
this line also there is no scope for any third harmonic current to exist on the load side
because there is no return path, but what thing happens is it is a delta winding there will
be induced voltage third harmonic component only I am telling fundamental is there they
will sum up to 0 and there will be no circulating current. But so far as the third harmonic
induced voltage is concerned in this three coils they are co phasal, is not.
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And they find a closed path. So, in this delta inside this in this closed path you find there
will be 3 emfs, 𝐸3𝑟𝑑 , then 𝐸3𝑟𝑑 ; 𝐸3𝑟𝑑 . Along with E fundamental it will be more scientific
you also sketch E fundamental this is the situation 𝐸1 . Here I will clean it better draw a
nice diagram. So, 𝐸3𝑟𝑑 𝐸1 , then once again 𝐸3𝑟𝑑 𝐸1 is there and 𝐸3𝑟𝑑 𝐸1 is there and they
are closed is not and these are the output terminals.
Now, in this closed loop you see, so far as fundamental component is concerned this
voltages are 120° apart they will sum up to 0 and no circulating current will be able to
exist here because they are 120° apart, but what happens to 𝐸3𝑟𝑑 third component of
voltage they are co phasal. So, when this is minus this is plus and have some instantaneous
value similarly this fellow same voltage same polarity and they are connected in series.
So, there will circulate a third harmonic current and this current will not be reflected in the
lines because in the lines there is no path for return that is 𝐸3𝑟𝑑 will come here and find a
path here no that is not possible. So, even if you connect a load there will be fundamental
component can only exist 3 lines only cannot have any third harmonic current. But now if
you look at the operation of the transformer I will say that fundamental component of the
magnetizing current will come from 𝐴𝑠 , 𝐵𝑠 , 𝐶𝑠 and the third harmonic exciting current
which is needed the transformer will self correct it by circulating a current third harmonic
current here.
It is not important that you pass the fundamental and third harmonic component of current
through the same primary winding it does not matter because the coils are wound on the
core and net MMF decides what is the flux. So, what all you have to see is this although
here neutral is not connected I will say fundamental 50 Hz component of current is drawn
from this side and third harmonic component of magnetizing current can circulate here
therefore, flux will be reasonably sinusoid.
And your line to line voltage here available which will go to load will be more or less
sinusoidal that is how qualitatively it can be explained to. Therefore, it looks like that
whenever you connect a transformer you want to decide whether star delta delta star in
delta delta also there is no problem because there third harmonic current can circulate see
if the connection is delta delta this is delta this is delta. You please think about it
qualitatively I am not doing any analysis it can be done, but you will never forget.
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Here also if you apply balance 3 phase voltage here, can third harmonic current exist? No,
nothing it cannot exist only fundamental component of current will be drawn. If that is
there then I will say flux will not be will have a third harmonic component, but if flux has
a third harmonic component it will third harmonic voltage induced here as well as there
and both will circulate the needed third harmonic current which will not be reflected in the
lines.
Therefore, delta winding having a delta winding on a transformer connection becomes
perhaps a very good option that is transformer connection one winding is delta at least you
can get relieve of this third harmonic problems you do not have to think too much. So, we
will continue with this in the next class.
Thank you.
422
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 44
Choosing Transformer Connection (Contd.)
Welcome to 44th lecture and we have been discussing the effect of third harmonic flux
and third harmonic currents in particularly 3 phase transformers. Star star connection
without neutral always causes problem, including single phase loading is not possible
because secondary neutral becomes oscillating and so, on.
So, the, but if it is a delta connection, if exists either in primary or secondary at least one
is delta connected, then in the lines there will be there is no possibility of any third
harmonic current to exist if it is that side is delta connected because they are co phrasal
and so, on. But nonetheless there will be induced voltage then in the coils itself and this
coils are close to itself therefore, the magnitude of the third harmonic current circulating
third harmonic current will be then, they are co phrasal 3𝐸3𝑟𝑑 of that divided by the leakage
impedance of suppose the secondary side is delta.
(Refer Slide Time: 01:23)
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This 3rd harmonic current will circulate like this. So, 3𝐸3𝑟𝑑 divided by the leakage
impedance of this side, but if the at the fundamental frequency leakage impedance is 𝑥1 it
should be 𝑗3𝑥1 .
𝐼3𝑟𝑑 =
3𝐸3𝑟𝑑
𝑟2 + 𝑗3𝑥1
So, impedance also increases and there will be circulating current little bit and this may
cause extra heating no doubt because of this circulating current, there will be copper loss,
extra copper loss and so, on. But nonetheless that major problem of getting at the output a
voltage which will also have third harmonic component is avoided that is what is
important. So, it causes extra heating slight extra heating will be caused. In any case now
the question is what is the I will tell only general guidelines. For example, generally HV
side. So, it looks like one of the sides better be having delta connection.
So generally HV side is chosen as star connection. Why? Because the voltage per phase
will be less, so insulation level etcetera will be lesser. And LV site since the coils will be
delta connected this will be low voltage, this is high voltage side this voltage will be also
less because of this one and let this be delta connected that is the guideline delta.
But this is not all in fact, there are situations other reasons will override this that is may be
the low voltage side is star, primary side is delta. For example, a distribution transformer
typical rating is 6.6KV and 440V, what is done? Because this side it is a distribution
transformer you must have the 3 phases as well as neutral. So, this must be of star
connection and this must be this can be selected as delta because of the fact one of the
windings I want to be delta that is the idea.
And from this I will be able to supply A phase with neutral to a group of consumers from
B phase and neutral to another group and C phase to neutral to another group 220V supply
etcetera can be done and everything is fine. So, depending upon the situations general
guideline is high voltage side better select star, low voltage side delta as perhaps. So, one
of the windings be delta connected is always preferable and sometimes if the voltages are
high, what people do? They use star star high voltage on both the sides relatively high
voltages. So, then select star star, but with star star there are problems we have seen, but
what people will be doing they will also put another coil which will be delta connected.
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That is there will be 2 secondaries because it is always preferable to have a delta and you
suppose you require star star nothing doing high voltage this side high voltage that side, I
will take advantage of lower insulation level this that. So, it is like this and you connect it
and another additional windings therefore, for each phase then if I draw the limb of say A
phase it will be like this 𝐴1 𝐴2 and then small 𝑎1 𝑎2 and another 𝑎3 𝑎4 . Two separate
secondary’s are there for a for each primary similarly for B phase and C phase. Now, in
this case we have seen and these two turns may not be same. In case of zig zag connection
we saw this type of secondary’s present, but those two secondary’s were identical having
same number of turns you recall that. So, here it is 𝑁1 may be this is 𝑁2 this is 𝑁3 . Now,
what I am telling is this is star connection 𝐴1 𝐴2 , 𝐵1 𝐵2 like that.
And perhaps this is the main secondary, this is main primary which is star connected may
be this is 𝐴1 this is 𝐴2 𝐵2 𝐶2 etcetera and this is also star connected this is say 𝑎1 , this is
𝑏1 and this is say 𝑐1 correct way I have connected and this is the neutral 𝑎2 𝑏2 𝑐2 I have
connected. And this will go to the load and this one I have connected in delta this 𝑎3 𝑎4 ,
similarly you will have 𝑏3 𝑏4 I think you are understanding what I am telling and you will
have 𝑐3 𝑐4 like the other two limbs and these I will connect in delta got the point.
So, you connect them in delta and 3 terminals will come out, here this winding which is
connected in delta is called tertiary winding which is primarily added so, that third
harmonic current about that problem you need not worry about circulating third harmonic
current will flow. Not only that you can get then two different level of voltages and supply
it two different loads also ok. Suppose some people use primary star secondary main
secondary star and also here whatever voltage you get you supply another three phase load.
That is which two secondaries then, but sometimes no load is connected it is merely it is
present because you provide the flow of the third harmonic magnetizing current in this
winding. So, depending upon the turns ratio two levels of voltage you will be load may or
may not be connected load may or may not be supplied; may not be supplied from tertiary
winding.
I think you have got the idea. So, high voltage high voltage star star then delta, since I was
telling you it is always better you have a delta connection somewhere and all this windings
are wound on the same magnetic circuit. So, like A phase you have B phase C phase like
here. So, that you 𝐵1 𝐵2 here also small 𝑏1 𝑏2 and 𝐶1 𝐶2 small 𝑐1 𝑐2 and then 𝑐3 𝑐4 . So,
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you have two secondary coils and they need not be of same turns depending upon your
voltage requirement, you can decide what should be your 𝑁2 this is 𝑁1 , this is 𝑁1 , this is
𝑁2 , this is 𝑁2 and this is 𝑁3 is not this is the thing.
So we now understand that the things goes 3 phase transformer connection we have learned
that is if I ask you make a 3 phase transformer connection star delta, delta star without first
really questioning where should I use it that first I discussed confidently how can you do
it do polarity test do that. And then we have pretty qualitatively of course, discussed about
the third harmonic importance of third harmonic magnetizing current we present in the
transformer in order to make the flux sinusoidal, hence the load voltage will be sinusoidal.
I do not know whether I have a previous one is there no. Let me tell you one interesting
thing; for example, zig zag connection also we have done apart from star delta and this
one, let me try today to do another zig zag connection just to telling you.
(Refer Slide Time: 13:24)
See suppose I want to make a delta zig zag connection ok. Why I am coming back to zig
zag once again is because of the fact in star star zig zag connection if you recall 2 phases
of the secondary side belonging to two different groups they are connected in series ok.
And the polarity dot polarity of this one we discussed about Y z 1 for example, let me tell
that once again so, that you also have a practice and for example, suppose I want to do Y
z 1, what should I do?.
426
Primary it will have two identical secondary’s, this time it is identical not tertiary or
secondary this is 𝑁2 this is 𝑁2 this is 𝑁1 and this is 𝐴1 𝐴2 , let me do first 𝑎1 𝑎2 and 𝑎3 𝑎4
is not, then 𝐵1 𝐵2 this is small 𝑏1 𝑏2 this is 𝑏3 𝑏4 and finally, you have C phase 𝐶1 𝐶2 and
small 𝑐1 𝑐2 𝑐3 𝑐4 . And all this odd numbers are dots relative to its primary this is one set,
this is another set and this is another set and suppose I connect this two in star because
primary is to be connected in star, then the I will connect this to supply 𝐴𝑠 𝐵𝑠 and 𝐶𝑠 and
your primary phrasal diagram will be very simple like this.
This is suppose 𝐴𝑠 𝐵𝑠 because supply voltage is balanced 𝐶𝑠 and 𝐴𝑠 mean 𝐴1 only this is
𝐵1 and this is 𝐶1 and this are 𝐴2 𝐵2 𝐶2 this is the neutral of the transformer 𝑁. Now, on the
secondary you know you have got this voltages 𝑎1 𝑎2 and 𝑎3 𝑎4 I am not explaining
because we know this and this is 120° apart 𝑏1 𝑏2 and 𝑏3 𝑏4 and finally, you are also having
parallel to C phase and this is 𝑐1 𝑐2 and 𝑐3 𝑐4 is not in this way it is there. Now, suppose I
because zig zag connection is essentially star connection we know this is group 1 and this
is group 2 ok.
So, I want to get Y z 1 means this is the primary voltage with respect to the secondary
voltage will be lagging a line voltage with respect to secondary neutral will be lagging by
30°, is not? These will be the phase voltages, this we have done I will do very quickly. So,
this is the thing. Now, while getting this voltage I want to get 𝑎1 generally here. So, your
output voltages will be this and this and as you can see it is 𝑎1 𝑎2 and this is parallel to C
phase and this I will take 𝑐3 𝑐4 is not.
Similarly, this fellow can be broken up into this 2 phasors and this is 𝑏1 𝑏2 and this is 𝑎3
𝑎4 and this can be broken up into this 2 phasors this one parallel to 𝑏3 𝑏4 and this is parallel
to 𝑐3 𝑐4 , 𝑐1 𝑐2 is not then based on this I come here and connect accordingly. For example,
𝑎1 𝑎2 is connected to 𝑐4 . So, take a piece of wire connect it and which 3 are shorted 𝑎3 𝑏3
𝑐3 short them is not short them and I connect. So, ones will be the output terminals 𝑎1 𝑏1
𝑐1 these will be my output terminals then 𝑏2 is connected to 𝑎4 . So, 𝑏2 is connected to 𝑎4
and 𝑐2 is connected to 𝑏4 and your connection is over.
Now, only one point why I have redrawn once again is that only one point I want to tell.
So, it is effectively a star connection there is a neutral point solid neutral point 𝑎3 𝑏3 𝑐3
and there are 3 lines only thing is between the neutral secondary neutral that is 𝑛 to one of
the lines, if you start your journey from neutral you will come across two coils belonging
427
to two different phases that is a phase and b phase in this case if you start from 𝑛 traverse
𝑎3 𝑎4 then traverse 𝑏2 𝑏1 and get your supply 𝑏 this is supply 𝑎 load supply side this is 𝑐.
So, two coils are in series, now suppose in this connection neutral is not connected
therefore, and there is no delta connection. Therefore, I am sure the current no third
harmonic current exists here only 50 Hz component of current will flow and in the core of
the transformer third harmonic flux must exist. And that third harmonic flux is going to
induced voltage both in the primary and the secondary coils is not and the polarity of those
voltages are shown here.
Now so, far as the third harmonic voltage is concerned you see they are co phasal. If this
is plus this is minus 𝐸3𝑟𝑑 fundamental I am not showing and then this is also third harmonic
𝐸3𝑟𝑑 in all the coils third harmonic voltage will be induced and so, plus to minus and then
minus to plus and you reach the neutral. So, what is the third harmonic voltage present
across the phases? Zero. Although in each coil third harmonic voltage exists, but in the
phases no third harmonic voltage.
So, it is a superior thing compared to simple star star, but I take a zig zag coil which will
at least ensure that there will be no third harmonic voltage in the phases that is this voltage,
it will be RMS value only in the phases neutral to any of the lines and between the line of
course, it will not be there therefore, it is a better proposition that way is not. So, zig zag
connection has another interesting connection that perhaps, I will tell later or in your course
of power system you will learn very easily now after you have understood this that, it can
be used as a grounding transformer a better choice for a 3 phase transformer anyway that
I am deferring now. So, for example this connection we have understood.
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(Refer Slide Time: 23:07)
Another small connection as I was telling with D suppose I want to do a delta zig zag
which I have not told, but let me do it. Suppose the primary winding is no delta connected
zig zag winding I have done. So, that is why I am doing so, that you become more confident
𝐵1 𝐵2 and 𝐶1 𝐶2 and delta connections and it has got 2 secondaries identical 𝑁2 and 𝑁2
turns mind you it is not tertiary one of them and this is 𝑁2 and 𝑁2 turns and what is the
point of writing. These are all same turns this is also 𝑁2 , this is also 𝑁2 two secondaries
corresponding to each phase. It is essential to make a successful connection to mark the
terminals meticulously I mean you cannot do arbitrarily something 𝑏3 𝑏4 and 𝐶1 𝐶2 small
𝑐1 𝑐2 small 𝑐3 𝑐4 .
Now, suppose I have decided I will connect it like this a delta like this and I will give
supply 𝐴𝑠 here supply 𝐵𝑠 there in the B phase and supply C phase that is how I will give
supply. Now, this side is delta delta connected. So, what is the rule? Rule is pretty simple
supply voltage is balanced better draw that; that is 𝐴𝑠 𝐵𝑠 𝐶𝑠 which is balanced I know that
is why I could make a equilateral triangle and show it and 𝐴𝑠 is once again you write with
different colour say 𝐴𝑠 means what? 𝐴1 𝐶2 𝐴1 𝐶2 and 𝐵𝑠 means what? 𝐵𝑠 means 𝐵1 𝐴2 and
𝐶𝑠 means what? 𝐶1 𝐵2 𝐶1 𝐵2 is not same points.
And on the secondaries now you have these voltages available to you, what are the
voltages? 𝑎1 𝑎2 you know 𝑎1 𝑎2 similarly parallel voltages of equal length 𝑎3 𝑎4 , similarly
𝑏1 𝑏2 another voltage is available to play with 𝑏3 𝑏4 and 𝑐1 𝑐2 ; so 𝑐1 𝑐2 and 𝑐3 𝑐4 . Now, I
429
want to make some connection delta and secondary side zig zag. In this case you will find
D z 0° or D z 60° and 180° is possible or so, how it is possible?
Because with respect to neutral on the primary side, this is the artificial neutral secondary
side voltage will be also in phase with this parallel to this, a phase voltage of the secondary
side and zig zag connection means star connection. Two coils belonging to two different
phases are to be connected in series, one should be taken from group 1 other should be
taken from group 2 the same rule.
So, and I would like to have the secondary voltage because no phase displacement, perhaps
it will be I will take 𝑎1 𝑎2 and is this voltage available? Yes 𝑎1 𝑎2 . Then I have to connect
a another voltage from other group of this kind which is available in c and I will take it as
𝑐3 𝑐4 , is not? And the b phase voltage will be 120° apart and b phase is horizontal actually
do not put it like this it is horizontal 𝑏1 𝑏2 and this is 𝑏3 𝑏4 these are horizontal why?
Because 𝑏1 𝑏2 is this. So, this voltage is the secondary phase voltage and it will be it can
be obtained like this a and b.
So, it can be drawn like that I mean this no. So, here I would like to have 𝑏1 . So, I will
sketch it 𝑏1 first which will be horizontal here 𝑏1 𝑏2 and then a like this and what is this
voltage 𝑎4 𝑎3 and finally. So, you can sketch it like this, this is 𝑎1 which is this, this is 𝑏1
and this is some c phase here similarly this can be drawn like this here and you can see
this will become 𝑏3 this will be horizontal actually 𝑏3 and 𝑏4 and this point is 𝑐1 𝑐2 is this
ok 𝑏1 𝑏2 was like this. So, this is fine 𝑎3 𝑏3 𝑐3 are to be shorted you short them and then
𝑎2 𝑐4 you join 𝑎2 𝑐4 you join and then 𝑏4 𝑐2 you join; 𝑏4 𝑐2 you join. And finally, 𝑎2 𝑐4 I
have joined 𝑎4 𝑏2 you have to join 𝑎4 and this is 𝑏2 you join these two will be output
similarly you will be able to calculate this one. Only one thing I will tell in this case
suppose I want to this turns are known 𝑁1 , 𝑁1 , 𝑁1 this is 𝑁2 , 𝑁2 . Now, same flux is linking
this suppose you have applied a line to line voltage 𝑉𝐿𝐿 here try to understand what I am
doing, I know the turns ratio I have applied 𝑉𝐿𝐿 which is known. I want to know what will
be the phase voltage on the secondary side and what will be the line to line voltage in terms
of this 𝑉𝐿𝐿 , how should I proceed?
What is the voltage per turn?
𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑝𝑒𝑟 𝑡𝑢𝑟𝑛 =
430
𝑉𝐿𝐿
𝑁1
Because it is delta connected line to line voltage comes here and what will be the induced
voltage? RMS value,
𝐼𝑛𝑑𝑢𝑐𝑒𝑑 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑃ℎ𝑎𝑠𝑒𝑠 =
𝑉𝐿𝐿
𝑁
𝑁1 2
This RMS voltage will be same for all this coils because they are all having same turns.
𝑉
So, I know the length of each phasor here to be 𝑁𝐿𝐿 𝑁2 across each secondary coil, is not?
1
𝑉𝐿𝐿
That is voltage here 𝑁 𝑁2 .
1
So, the lengths of this are known, then what will be the phase voltage? Secondary phase
voltage very simple this is 30°, this is 30° and I know this length. So, each length that is
𝑉𝐿𝐿
𝑁1
𝑁2 cos 30 will give you half and twice of that, because projection of this is also same.
So, this it will come out to be
𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑃ℎ𝑎𝑠𝑒 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 = 2𝑉𝐿𝐿
𝑁1
𝑁1
𝑁1
√3
cos 30 = 2 ×
𝑉𝐿𝐿
= √3𝑉𝐿𝐿
𝑁2
𝑁2
𝑁2
2
What will be the line to line voltage available on the secondary side?
√3 times this phase voltage that is this one is the phase voltage. So, secondary line to line
voltage will be
𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝐿𝑖𝑛𝑒 − 𝐿𝑖𝑛𝑒 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 = √3 × √3𝑉𝐿𝐿
𝑁1
𝑁1
= 3𝑉𝐿𝐿
𝑁2
𝑁2
So, what are the things it comes out to be this? Similarly try your on your own that star
star star zig zag connection if the line to line voltage is known, what is going to be the
phase voltage and the line to line voltage. So, we will continue our discussion on
transformer in next lecture.
Thank you.
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Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 45
Phase Conversion Using Transformer: Scott Connection
Welcome to lecture 45 on Electrical Machines-1 and we were discussing various
connections of 3 phase transformers and give you a sort of guideline owing to select what
connections ok; star delta, delta star or zigzag.
And zigzag connection as I told you is essentially a star connections and in my last class I
told you because with delta connection, I just view once digit 0 and it was like this and I
told you if you know the input line to line voltage and the turns of each primary and each
secondary coils, then you will be able to calculate what will be the phase voltages of these
zigzag connections between the neutral. And one of the phases of the output at the this
goes to load 𝑎, 𝑏 and 𝑐 and what is the line to line voltage and phase voltage, how to
calculate you can easily relate them ok.
(Refer Slide Time: 00:51)
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(Refer Slide Time: 01:59)
Now, another interesting thing with transformers is that using transformers it will be
possible to convert a given polyphase system to another poly phase system. That is called
phase conversion. We will briefly discuss that phase conversion, of which one very
popular connection is called Scott connection; which changes a given 3 phase voltage. It
converts to a 3 phase balanced voltage of course, to a 2 phase balanced voltage. See our
supply system is all 3 phase; generation; distribution; transmission.
Then of course, you get single phase voltage you can get between one line and neutral.
Like that that is there, but sometimes 2 phase balance voltage may be needed. For example,
if you have a 2 phase induction motor ok, then you must supply the stator coils by it 2
phase supply instead of a 3 phase balanced supply, because the motor is 2 phase and 2
phase induction motors are also used in control system and other applications.
There similarly there maybe 2 phase electric furnaces, where 2 phase supply is necessary,
but the problem is first understand the problem; problem is you have a 3 phase supply
available to you 𝐴𝑠 , 𝐵𝑠 , 𝐶𝑠 may be with a neutral 𝑁𝑠 supply is available, and you have a 2
phase load this side. So, how to convert this 3-phase supply to a balanced 2 phase supply
voltage system and then feed your load that is the problem.
So, the connection is called Scott connection and it uses 2 single phase transformers. Two
numbers of single-phase transformer; that is 3 phase to 2 conversion requires 2 single
phase transformer, separate transformers and you can show that it will give you a balance
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two phase voltage; the idea is very simple ok. So, let us start what is done is this you have
one transformer, whose primary coil I will say, because transformers are to be named, in
case of 3 phase case 3 isolated transformers can be used and there were 6 coil, so proper
naming is required.
So, in this case of course, two transformers are there, and let us call these transformer to
be transformer A with terminal say 𝐴1 𝐴2 and its secondary is having small letters 𝑎1 , 𝑎2
like the 3 phase and you have another transformers which I am drawing horizontally the
reason will be clear and its secondary is this one ok. This is suppose 𝐵1, 𝐵2 and this is also
𝑏1 , 𝑏2 .
Now, we required two transformer; one transformer is transformer A another transformer
is transformer B. Now, are this two transformers are identical? Really not. What happens
is this you select this two transformers in such a way that the secondary turns are equal
𝑁2 , 𝑁2 and primary turns of this transformer is B transformer is 𝑁1 . So, secondary turns
are same, primary turns of transformer B is equal to suppose 𝑁1 then, primary turns of
transformer A should be this is let me write
𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝑇𝑢𝑟𝑛𝑠 𝑜𝑓 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝐵 = 𝑁1
𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝑇𝑢𝑟𝑛𝑠 𝑜𝑓 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝐴 =
√3
𝑁
2 1
That is how the transformer turns we have to select of this two transformers and not only
that then in transformer B primary at the 50% you must have a tapping such that this turns
𝑁
𝑁
become then 21 and this turns becomes 21; so, total turns is 𝑁1 . So, at 50% there will be a
tapping ok.
Now what you will be doing is this 𝐴2 you join it here with this 50% tapping and these
three, I will connect it to supply 𝐴1 say this I connect to 𝐵1 and this I connect to 𝐵2.
Suppose I connect like that and secondary I have not connected anything it is just isolated
and leaving like that.
Now, first of all you can see that what is the voltage applied to the primary of transformer
B. The voltage applied is line to line voltage is not and total number of turns is 𝑁1 . Mind
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you this transformer has its own magnetic circuit separate; this transformer has its own
magnetic circuit. So, volt per turn of transformer B is equal to
𝑉𝑜𝑙𝑡 𝑃𝑒𝑟 𝑇𝑢𝑟𝑛 𝑜𝑓 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝐵 =
𝑉𝐿𝐿
𝑁1
Therefore, what will be the induced voltage between in the secondary of the transformer
B.
𝐼𝑛𝑑𝑢𝑐𝑒𝑑 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑜𝑓 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝐵 = |𝑉𝑏1 𝑏2 | =
=
𝑉𝐿𝐿
× 𝑁2
𝑁1
𝑁2
× 𝑉𝐿𝐿
𝑁1
Now, you see I will draw here so, that I do not have to or shall I go to next page. I will
select this one, this diagram is needed. So, select this so that I do not have to draw, copy
and I go to next page, paste it here. So, that is the thing. Now I will erase this things, so
this was the thing.
(Refer Slide Time: 12:11)
Now and we have examined what is now, suppose I want to see in phasor term so, what
are the voltage I have applied on the primary side. Since the supply voltage is balanced
like the previous one in transformer connections, I can draw this voltage triangle and
supply is phase sequenced a𝐴𝑠 s, 𝐵𝑠 , 𝐶𝑠 . So, first I will draw this, this is 𝐴𝑠 , this is 𝐵𝑠 , and
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this is 𝐶𝑠 I will draw and 𝐴𝑠 is 𝐴1 , 𝐴𝑠 and 𝐴1 same 𝐵𝑠 and 𝐵1 is same and 𝐶𝑠 with 𝐵2, it is
also same, it is like this. So, what is the voltage applied across the primary of the
transformer B? It is this one; 𝐵1 𝐵2.
Therefore, if we say that like the previous one, this one ones are dots this one one are dots
suppose I say then, on the secondary coil I will have this induced voltage, which will be
𝑏1 𝑏2 . We have done much more complicated case in case of 3 phase ok. Now, here it will
be like this what else. So, this will be the secondary this voltage phasor parallel to 𝐵1 𝐵2
of course, I have not correctly named. So, it should be I am sorry you know this is 𝐵1 is
connected to 𝐵𝑠 and 𝐶𝑠 is connected to 𝐵2. So, voltage applied across this is 𝐵1 𝐵2. So, the
secondary induced voltage 𝑏1 𝑏2 So, I should name it accordingly nah 𝐵1 is here so, it must
be 𝑏1 and 𝑏2 , now it is correct.
So, this is the available voltage, if any mistake please point out this is the thing. Like this
several times we have done in case of 3 phase. Now, the question is what is the voltage
applied and this point I will call O where that 50% tapping has been taken. Now the
question is where this point O is in this phasor diagram? It must be at the middle point. O
lies here because half of the voltage if you move from here you get O, is not this total
𝑉
𝑉
voltage is this 2𝐿𝐿 , and 2𝐿𝐿 will come here. So, this point is O.
The moment I get this O point is a nothing but also your 𝐴2 point, because you have
connected like that and what is the voltage applied and this is 𝐴1 . So, voltage applied
across the primary of transformer A is this vertical line; is not this is the voltage applied
across the primary of transformer A. If line to line voltage is 𝑉𝐿𝐿 , what is the magnitude of
the voltage applied across 𝐴1 , 𝐴2 ? It will be this length this is 60°.
So, length 𝐴1 𝑂 is nothing, but
𝐴1 𝑂 = 𝑉𝐿𝐿 sin 60 =
√3
𝑉
2 𝐿𝐿
See my goal is to get at the secondary 2 voltages which will be 90° apart. What is balanced
two phase voltage? Balanced 2 phase voltage is like this phase voltages will be equal and
they will be 90° apart. Maybe 1 phase is A, another phase is B; that is what I want to get.
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So, one voltage phasor is like this to the primary of transformer A, I have applied this
voltage whose magnitude is
√3
𝑉 therefore, induced voltage on the secondary will be also
2 𝐿𝐿
parallel to this. That is 𝑎1 𝑎2 . I want this length to be same as this length. What is the
voltage per turn? This voltage divided by number of turns of the primary. So, voltage since
secondary turns I have said to be same, I must ensure that the voltage per turn of
transformer B and voltage per turn of transformer A they must be same.
So, I write that; so,
𝑉𝑜𝑙𝑡 𝑃𝑒𝑟 𝑇𝑢𝑟𝑛 𝑜𝑓 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝐴 = 𝑉𝑜𝑙𝑡 𝑃𝑒𝑟 𝑇𝑢𝑟𝑛 𝑜𝑓 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝐵 =
𝑉𝐿𝐿
𝑁1
They must be same then only the secondary voltage magnitudes will be same. I will simply
multiplied with 𝑁2 to get the secondary voltage for this transformer. What was the voltage
per turn?
𝑉𝐿𝐿
𝑁1
√3
. Now, this voltage I have right now seen it is 2 𝑉𝐿𝐿 . Therefore, its number of
√3 𝑉
turns if you set it to capital 𝑁1 as before, it voltage per turn would have been then 2 𝑁𝐿𝐿
1
and its secondary voltage would have been
√3 𝑉𝐿𝐿
𝑁 and that will not be same as this one.
2 𝑁1 2
So, to make the voltage per turns same, that is why its number of turns of transformer A
primary is chosen to be
√3
𝑁 , then I will write you see
2 1
√3
𝑉𝐿𝐿 𝑉𝐿𝐿
𝑉𝑜𝑙𝑡 𝑃𝑒𝑟 𝑇𝑢𝑟𝑛 𝑜𝑓 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝐴 = 2
=
𝑁1
√3
2 𝑁1
Same as the transformer B therefore, the length of this phasor length of 𝑎1 , 𝑎2 and 𝑏1 , 𝑏2
𝑁
will be then 𝑁2 × 𝑉𝐿𝐿 they are will be same. So, you see at the secondary coils of this two
1
transformers there exist 2 voltages whose magnitudes are same and which are 90° apart.
So, it was a balance 3 phase voltage and on the secondary coils although I have not
connected them yet, I get a balanced 2 phase voltage, is that correct no confusion fine. If
that be the case then what I will tell is this I will now join 𝑎2 and 𝑏1 to create the neutral
of the two phase system and I will take output 3 terminals I will take one will be marked
as 𝑎 another will be marked as 𝑏 another will be neutral. So, the moment you join this 2
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phasors cannot remain in isolation what I have joined 𝑎2 with 𝑏1 , therefore, I must place
this as 𝑎1 𝑎2 and I have joined 𝑎2 with 𝑏1 so, it should be placed here.
It will be like this or what you could do is this if you want to make the a phase voltage
leads the b phase voltage you can join 𝑎2 with 𝑏2 . Are you getting me? What I will do is I
will join 𝑎2 with 𝑏2 to create the neutral. In that case what how it will look like please be
with me what I am telling this is fine balance two phase system you will get and a phase
voltage will lead the b phase voltage and neutral is here. Another way you could do connect
the secondary coils this was the thing this was 𝑎1 𝑎2 , this was 𝑏1 , 𝑏2 is not and 𝑎1 𝑎2
voltage is here.
So, I could join 𝑎2 with 𝑏2. Did I not join that? No, why I did not join; 𝑎2 with 𝑏2 I could
short. So, in this case 𝑎2 with 𝑏2 I will short and get the output from this, get in the neutral.
Therefore, let me clear this I will do like that. So, let me slightly modified this, but you get
the idea that will also work no problem.
So, what I am telling is this is 𝑏1 𝑏2 I have drawn these are the voltage phasors like this.
So, what I will do now? I will join 𝑎2 and 𝑏2 together that is and call this to be by neutral
and I will take output from this as a phase and take output from this for b phase. And if
you do like that 𝑎2 and 𝑏2 you join then your phasor diagram will look like look like 𝑎1
𝑎2 , 𝑏1 𝑏2 and I have joined 𝑎2 and 𝑏2 and this is my neutral. That is what I made. the earlier
one will also do. But any way let us draw like that.
So, that 𝑎1 𝑏1 supply comes out like that and this is your balance 2 phase voltage. In case
of balance 2 phase voltage and this I will say as the line to line voltage; line to line voltage
will be route 2 times the phase voltage. You can easily understand; so far so, good. So,
this is what is Scott connection is. So, this two transformers in terms of number of turns,
they are not identical.
Secondary turns are must be same, 𝑁2 𝑁2 , but primary turns of transformer a which is just
drawn vertically, this vertical drawing has nothing to do with really 90°, this axis that axis;
no on the table you draw. I have drawn it so, that I understand this phasors much more
easily. So, this is transformer B having 𝑁1 turns and also I will demand a 50% tapping
available from the primary of transformer B and then I will connect like this connect it to
a balanced 3 phase system and the number of turns of primary of transformer A is not 𝑁1
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3
has that of B, but it must be √2 𝑁1 and secondaries are connected like this. Is that clear? So,
that is the thing.
Now, the question is that I will connect a balance 2 phase load on the secondary side. So,
I have got a balance 2 phase voltage, the magnitude of the voltage also I know. So, I will
connect a balance 2 phase load here on the secondary side so, some balanced current will
flow. Then how to estimate the current which will be drawn from the supply 3 phase
supply? That is important; is not I must examine, because after all you see you are
connecting a balance 2 phase load here is a 3 phase system; 3 phase voltage supply system
here.
We like that it also supplies the balance 3 phase current, not a unbalanced current like that.
So, if you connect a balance 2 phase load on the secondary side, does the primary side
current which is drawn from a 3 phase supply remain as a balance 3 phase current and how
the magnitudes of this secondary current and primary current will be related to? That will
be the next thing.
Suppose you have connected a balance two phase load; so, between 𝑎 and 𝑛. So, this is
the balance 2 phase load of same magnitude of impedance ok. If it is 𝑧2 this is also 𝑧2 per
phase impedance, got the point. Now, suppose it supplies a current 𝐼𝑎 direction of the
current I have assumed. So, 𝐼𝑎 here depending upon the power factor angle of 𝑧2 it will be
positioned. So, 𝐼𝑎 maybe there, this is I secondary. What is this angle? Is the power factor
angle of the load, is not a phase current 𝐼𝑎 . But the moment it supplies a current these are
transformer so, rules of transformer must be satisfied what is that voltage ratios are in the
ratios of turns.
Similarly, if through the dot a current comes out its primary will draw additional current
through the dot whose magnitude will be
𝐼𝑎′ =
𝑎𝐴 =
So, that is how it will be reflected upon ok.
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𝐼𝑎
𝑎𝐴
√3 𝑁1
2 𝑁2
So, so, this is 𝐼𝑎 and this is 𝐼𝑎′ which is the line current of A phase happens to be. So, a
phase current will be also in phase with these, that is what we have learnt. So, primary side
a phase current if I show it will like I mean it will be in same phase. I am just drawing
here. It has it does not mean any connection etcetera current I am drawing, this is your 𝐼𝐴𝑆
which is same as which I am calling it as Ia dashed this will be thing and this angle will be
𝜃2 .
𝐼𝐴𝑆 = 𝐼𝑎′ =
𝐼𝑎
𝑎𝐴
So, the line current of the primary side I have drawn, but there are two other line currents.
Let this line current be 𝐼𝐵𝑆 , that is how in a 3 phase system I show, let this be 𝐼𝐶𝑆 ok. Now,
I must find out in terms of the secondary current what will be the value of 𝐼𝐵𝑆 and 𝐼𝐶𝑆 . So,
same rule that is MMF must be balanced. Now, in this case it will be slightly tricky; in the
sense that this current suppose the this current I say this is 𝐼𝑏 secondary line current through
the dot 𝐼𝑏 current is coming out.
Therefore, there will be reflected current here on the primary side and that reflected current
in terms of 𝐼𝐵𝑆 and 𝐼𝐶𝑆 if I write, I will balance the MMF. Here also I could do like this 𝐼𝐴𝑆
will be such that; see this is I have followed, but I could write it like this 𝐼𝐴𝑆 , the value of
3
𝐼𝐴𝑆 I could calculate the MMF produced by primary √2 𝑁1 must be equal to
𝐼𝐴𝑆 ×
√3
2
𝑁1 = 𝐼𝑎 × 𝑁2
Same thing I will get this is
𝐼𝐴𝑆 = 𝐼𝑎′ =
𝐼𝑎
𝑎𝐴
If you bring it down here the same thing you will be getting. Got the point?
𝑎𝐴 =
√3 𝑁1
2 𝑁2
Similarly, I will balance the MMF of the secondary and primary for transformer B. Now,
let me write only on this page let it be clumsy; so that we do not miss the point. What is
the MMF produced by transformer b, current coming out from transformer b is how much
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Ib. So, 𝐼𝑏 × 𝑁2 therefore, through the dots currents will go in, but you see here there are
𝑁
two half’s of this 𝑁1 having the primary total 𝑁1 turns of which 21 turns is carrying 𝐼𝐵𝑆
current entering through the dot.
Therefore, these MMF must be equal to let me write here only so, that you understand. So,
𝑁
𝑁
this must be 𝐼𝐵𝑆 × 21 that is the thing. But this portion is 21, but it is carrying current see
if this is dot this is dot. So, this plus so, current flowing into the dot of this part is −𝐼𝐶𝑆 ×
𝑁1
2
and this should be equal to
𝐼𝐵𝑆 ×
𝑁1
𝑁1
− 𝐼𝐶𝑆 ×
= 𝐼𝑏 × 𝑁2
2
2
See this is crucial see secondary MMF through the dot 𝐼𝑏 current is coming out.
I am just telling. So, through the dot current is coming out. So, through the dot if current
comes out in its primary through the dot current will be invited as here was small 𝐼𝑎 through
the dot if there was no complication only, but here these two parts carrying different
currents.
So, I have to calculate. So, through the dot current coming out is 𝐼𝑏 therefore, I must expect
through the dot what is the current flowing. 𝐼𝐵𝑆 by virtue of its direction enters through the
𝑁
𝑁
dot, but that 𝐼𝐵𝑆 last for only 21 turns. So, it is 𝐼𝐵𝑆 × 21 I have done. Through this portion
of the coil I have assumed 𝐼𝐶𝑆 flowing like this, that is what we assume for 3 phase circuit
analysis and if this is dot this is dot we have learned in auto transformer this that. So,
current flowing in this direction is −𝐼𝐶𝑆 .
Or this is equal to
𝐼𝐵𝑆 − 𝐼𝐶𝑆 = 𝐼𝑏 ×
2 × 𝑁2
𝑁1
This is crucial. This is the thing we get. See what my problem what is the statement of the
problem statement of the problem is you get a balance 2 phase voltage, connect a balance
2 phase load. Secondary currents which are balanced this will be 𝐼𝑎 where will be 𝐼𝑏 , I will
make it more clumsy on the same paper so, that you understand. 𝐼𝑏 will be here of same
magnitude lagging the b phase voltage by 𝜃2 , this too I know.
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What I am trying to do now, then for this loading what should be my 𝐼𝐴𝑆 , 𝐼𝐵𝑆 , 𝐼𝐶𝑆 ?
Directions I have assumed in accordance with the 3 phase circuit directions. Then what
should I do? I must balance the MMF. I will neglect the no load current etcetera. So, write
the MMF equation here, what is the current coming out through the dot 𝐼𝑎 ? What will be
the current coming in through the dot? It is 𝐼𝐴𝑆 , it has to be. So, these two MMF’s are same
because they will nullify the flux.
√3
So, that is the first equation, 𝐼𝐴𝑆 × 2 𝑁1 = 𝐼𝑎 × 𝑁2 over. But for this portion it is slightly
you have to be careful the current direction 𝐼𝑏 I have assumed like this, coming out from
the dot 𝐼𝑏 like this. So, 𝐼𝑏 × 𝑁2 that is the secondary side, then through the dots must
current enter and that MMF I have to calculate. Now, 𝐼𝐵𝑆 enters through the dot, so 𝐼𝐵𝑆 ,
𝑁
𝑁
but this 𝐼𝐵𝑆 is up to this point 21; so 𝐼𝐵𝑆 × 21 .
Now, what about this portion, you have assume this current to be 𝐼𝐶𝑆 so, I can always
presume −𝐼𝐶𝑆 is flowing and this is dot. So, this MMF plus this MMF must be equal to
this. So, this is one equation. 𝐼𝐴𝑆 in terms of secondary current so, I have already calculated.
𝐼𝐴𝑆 was how much, I have calculated. See 𝐼𝐴𝑆 already I have got
𝐼𝐴𝑆 =
2
√3
×
𝑁2
× 𝐼𝑎
𝑁1
So, I have to find out the 3 phase side current in terms of secondary current of which 𝐼𝐴𝑆
already obtained.
Now, I have to find out 𝐼𝐵𝑆 and 𝐼𝐶𝑆 in terms of secondary currents, that is what we are
planning to do. To do that I find another equation is missing, because I must have 2
equations in order to express 𝐼𝐵𝑆 and 𝐼𝐶𝑆 in terms of secondary currents ok. This is one
equation. So, where is the third equation? Third equation is it is a 3 phase system,
𝐼𝐴𝑆 + 𝐼𝐵𝑆 + 𝐼𝐶𝑆 = 0
𝐼𝐵𝑆 + 𝐼𝐶𝑆 = −𝐼𝐴𝑆
Right hand side is known 𝐼𝐴𝑆 has already been obtained. I will put it here then I have two
equations I will just simply solve them. We will continue this discussion in the next class.
Thank you very much.
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Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 46
Scott Connection (Contd.)
Welcome to lecture number 46 and we have been discussing Scott connection which is
used to transform a 3 phase voltage to a balanced 2 phase voltage.
(Refer Slide Time: 00:37)
And for this you required 2 transformers, one transformer is 𝐵1 𝐵2 and its corresponding
secondary terminals is also 𝑏1 𝑏2 and an another transformer which is a separate
transformer of terminal markings capital 𝐴1 𝐴2 and small 𝑎1 𝑎2 . What is the thing we
discussed that the number of turns of the secondaries are same.
And the primary number of turns of B transformer that is 𝐵1 𝐵2 turns
𝑁𝐵1 𝐵2 = 𝑁1
𝑁𝐴1 𝐴2 =
443
√3
𝑁
2 1
So, that the voltage per turn in both the transformers become same ok, and we do the
phasor diagram it is better we draw that, so that once again and the phasor diagram will be
the supply is balanced 3 phase with phase sequence 𝐴𝑠 , 𝐵𝑠 , 𝐶𝑠 .
So, this was my primary voltages 𝐴𝑠 , 𝐵𝑠 and 𝐶𝑠 , 𝐴𝑠 happens to be also 𝐴1 and 𝐵𝑠 happens
to be 𝐵1 and 𝐶𝑠 happens to be your 𝐵2. Now, the question is what will be the applied
voltage across 𝐴1 𝐴2 , mind you in this transformer B primary has a tapping at 50% and
that point I called 𝑂 from this point such that the number of turns this side that is
𝑁𝐵1 𝑂 = 𝑁𝐵2 𝑂 =
𝑁1
2
√3
Total number of turns is 𝑁1 and the number of turns of this transformer is 2 𝑁1 , and so
this is the thing. And now, the question is what is the voltage applied across 𝐵1 𝐵2, it is 𝐵𝑠
𝐶𝑠 that is this voltage since 𝐵1 𝑏1 are dot.
So, you will get on the secondary a voltage here which I should call 𝑏1 𝑏2 that is straight
forward ok. Primary side the question is what is the voltage applied across the primary of
transformer A, this point mind you is 𝑂; because it is between 𝐵𝑠 and 𝐶𝑠 at midpoint, so
this will be 𝑂.
And this point 𝑂 happens to be your 𝐴2 , 𝑂 and 𝐴2 are same ; therefore, voltage applied
across the primary of transformer A is this one which is at 90° to voltage applied to the
transformer primary B. Therefore, you will get a voltage here which will be parallel to 𝐴1
𝐴2 and if you join this 𝑎2 and 𝑏2 . Then the phasor diagram tells you that this will be 𝑎1
this will be 𝑎2 .
Suppose the line to line voltage applied is 𝑉𝐿𝐿 therefore, a voltage applied across the
primary of B transformer is 𝑉𝐿𝐿 number of turns is 𝑁1 . So,
|𝑉𝑏1𝑏2 | = 𝑉𝐿𝐿
𝑁1
𝑁2
What will be the magnitude of the voltage available across 𝑎1 𝑎2 ?
𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝑜𝑓 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝐴 =
444
√3
𝑉
2 𝐿𝐿
√3
𝑉𝐿𝐿 𝑉𝐿𝐿
𝑉𝑜𝑙𝑡 𝑃𝑒𝑟 𝑇𝑢𝑟𝑛 𝑜𝑓 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝐴 = 2
=
𝑁1
√3
𝑁
1
2
|𝑉𝑎1 𝑎2 | = 𝑉𝐿𝐿
𝑁1
𝑁2
So, magnitude of the secondary voltages will be saying and that is why you require the
number of turns of transformer A to be root
√3
𝑁 . If you take same 𝑁1 turns voltage per
2 1
turn will not be same, our goal is to get a balance 2 phase voltage at the secondary points.
So, this is the star point and that is what we did and then we will connect a load here say a
̅̅̅2 to indicate secondary load, so
balance 2 phase load balance load. Suppose 𝑍̅ and write 𝑍
this is a balance 2 phase load and these are the voltages. Then what I told you this point is
crucial that when you connect a balance 2 phase load on the secondary side there will be
balance 2 phase current. Let this current be called 𝐼𝑎 , where this 𝐼𝑎 will be; depending upon
the power factor of the load this 𝐼𝑎 will be here lagging this voltage by the power factor
angle, 𝜃2 that is the thing.
Similarly in the b phase 𝑏1 𝑏2 same load I have connected, so this current these are the line
currents 𝐼𝑏 will be also lagging this by same angle, 𝐼𝑏 phasor and this angle too will be 𝜃2 .
Obviously, since these 2 phasors are shifted by same angle 𝜃2 and angle between the
voltages where 90°, so this angle too will be 90°. So, a balance two phase current will be
flowing to the load.
Now, the question is you have been able to transform a balance 3 phase voltage to a balance
2 phase voltage. Now, if I want to examine what will happen to the currents on the 3 phase
side, will it also remain balanced or not. The answer we will soon see that it is yes and
how it is to be calculated; therefore, I know 𝐼𝑎 what will be my 𝐼𝐴𝑆 that I want to find out,
what is my 𝐼𝐵𝑆 and what is my 𝐼𝐶𝑆 . From where shall I find out secondary currents are
known I will calculate reflected currents on this side and if you see that 𝐼𝐴𝑆 , 𝐼𝐵𝑆 , 𝐼𝐶𝑆 are of
same magnitude. But 120° apart then I will conclude that the ac side current is also
balanced.
Now how do I do that, to do this you have to balance the mmf follow the rules of sort of a
ideal transformer like the mmf of the secondary through the dot current is coming out. So,
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that is 𝑁2 𝐼𝑎 this must be equal to through the dot current is drawn in, it must be equal to
√3
𝐼𝐴𝑆 2 𝑁1.
𝐼𝐴𝑆
√3
𝑁 = 𝑁2 𝐼𝑎
2 1
𝐼𝐴𝑆 =
2 𝑁2
𝐼𝑎
√3 𝑁1
So, therefore, 𝐼𝐴𝑆 how to draw this your 𝐼𝐴𝑆 will be in same phase with 𝐼𝑎 , because this is
only a scaling factor. So, your 𝐼𝐴𝑆 will be suppose in same phase with this with 𝐼𝐴𝑆 . See
we are using whatever you have learnt from single phase transformers beat the transformer
is operating as single phase, 3 phase, 2 phase individually secondary and primary coil the
relationships they must be maintained that is there will be reflected current and so on.
So, will be your 𝐼𝐴𝑆 , so 𝐼𝐴𝑆 is known. Now, the question is how to calculate 𝐼𝐵𝑆 and 𝐼𝐶𝑆 ,
this part we have to go carefully, so these are the dot terminals. So, through the dot current
𝐼𝑏 I have shown coming out, therefore, the mmf of the; I will use different colour for these
calculations. So, it will be this one 𝑁2 𝐼𝑏 , I am not drawing bar over this things.
This must be equal to the mmf produced by the primary coil and I should always take
currents going into it through the dot the same logic. But here the problem is current in
this portion 50% turns and in these portions there different, but no problem. So, what I will
𝑁
𝑁
do for this 21 terms that is 𝐵1 𝑂 turns it should be 𝐼𝐵𝑆 21. And for this part current flowing
in this direction that is 𝐼𝐶𝑆 or I can say −𝐼𝐶𝑆 is flowing like this from right to left I can say,
and this is dot means this is also dot.
𝑁
So, through the dot what is the current flowing −𝐼𝐶𝑆 21 that is the thing, which means that
therefore,
𝐼𝐵𝑆
𝑁1
2
− 𝐼𝐶𝑆
𝑁1
2
𝐼𝐵𝑆 − 𝐼𝐶𝑆 = 2
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= 𝑁2 𝐼𝑏
𝑁2
𝑁1
𝐼𝑏
So, this is another important equation, because 𝐼𝐴𝑆 already I know and I have draw it in
terms of 𝐼𝑎 . I want to know in the similar way what will be 𝐼𝐵𝑆 and 𝐼𝐶𝑆 in terms of 𝐼𝑎 or 𝐼𝑏
whatever it is secondary currents because 𝐼𝑎 and 𝐼𝑏 as known this blue coloured phasors
from that we are trying to get.
So, I required then another equation what is the another equation; another equation is in
this side if you look at it we know
𝐼𝐴𝑆 + 𝐼𝐵𝑆 + 𝐼𝐶𝑆 = 0
That is the third equation. Therefore, I get that
𝐼𝐵𝑆 + 𝐼𝐶𝑆 = −𝐼𝐴𝑆
That is what I get. So, this is the third equation, these two equation is to be solved to get
𝐼𝐵𝑆 and 𝐼𝐶𝑆 .
So, add them and divide by 2 you will get 𝐼𝐵𝑆 supply current on the 3 phase side in the B
phase 𝐼𝐵𝑆 will be equal to
𝐼𝐵𝑆 = −
𝐼𝐴𝑆 𝑁2
+
𝐼
2
𝑁1 𝑏
Similarly, if you subtract this and this you will get 𝐼𝐶𝑆 :
𝐼𝐶𝑆 = −
𝐼𝐴𝑆 𝑁2
−
𝐼
2
𝑁1 𝑏
So, finally, what I get; I will repeat that all the 3 currents; what is this colour.
447
(Refer Slide Time: 17:57)
We got 𝐼𝐴𝑆 , 𝐼𝐵𝑆 and 𝐼𝐶𝑆 as 𝐼𝐴𝑆 =
2 𝑁2
𝐼 that is this one why not I am copying and putting
√3 𝑁1 𝑎
it. So, I will get 𝐼𝐴𝑆 here ok and so this is the 𝐼𝐴𝑆 and what is 𝐼𝐵𝑆 we have got I am just I
𝐼
𝑁
𝐼
will take straight away from this; it is let me write 𝐼𝐵𝑆 = − 𝐴𝑆
+ 𝑁2 𝐼𝑏 and 𝐼𝐶𝑆 = − 𝐴𝑆
−
2
2
1
𝑁2
𝐼 .
𝑁1 𝑏
So, I take this copy it and paste it that is all. So, this is the thing I have got I will put it in
the corner, so that you can do something. So, this is the thing 𝐼𝐴𝑆 , 𝐼𝐵𝑆 , and 𝐼𝐶𝑆 , and let me
draw the phasor diagram of the secondary side voltage and primary. So, that phasor
diagram is also needed to indicate that.
448
(Refer Slide Time: 19:51)
So, it was like this, 𝑎1 𝑏1 and this is 𝑎2 𝑏2 and primary phasor diagram was like that 𝐴𝑆 ,
𝐵𝑠 and 𝐶𝑠 and this was the point 𝑂 it was like this. Now, these currents where you recall
𝐼𝑎 and this current was 𝐼𝑏 their lengths are same, because balance load I have assumed and
this is the power factor angle 𝜃2 .
Now, from this you can see that 𝐼𝐴𝑆 will be in phase with 𝐼𝑎 and if I just want to sketch
here taking some arbitrary point it need not be neutral I just superimposing this current.
Then 𝐼𝑎 should be like this with vertical 𝜃2 it will be 𝐼𝐴𝑆 in phase with 𝐼𝑎 and here I draw
𝑁
that, this is the thing. Then if you see 𝐼𝐵𝑆 it is equal to 𝐼𝑏 is like this, so 𝑁2 𝐼𝑏 . So, this part
1
suppose 𝐼𝐵𝑆 I am plotting it is this term, so erase these two; it is not there now.
𝑁
𝑁
𝑁
1
1
1
So, 𝑁2 𝐼𝑏 , 𝐼𝑏 is here multiplied by 𝑁2. So, it will be here this is equal to this part 𝑁2 𝐼𝑏 , to
𝐼
𝐼
this I have to add − 𝐴𝑆
this is 𝐼𝐴𝑆 . So, to this you add − 𝐴𝑆
and then you will get 𝐼𝐵𝑆 .
2
2
𝑁
𝑁
1
1
Similarly, 𝐼𝐶𝑆 is 𝑁2 𝐼𝑏 negative of that, so 𝑁2 𝐼𝑏 has already been drawn. So, negative of that
𝑁2
𝐼
this length is − 𝑁 𝐼𝑏 and to this I have to add − 𝐴𝑆
.
2
1
𝐼
So, this is − 𝐴𝑆
and then this current will be your 𝐼𝐶𝑆 is that clear. So, the reflected current
2
and mind you what is 𝐼𝐴𝑆 ; 𝐼𝐴𝑆 =
2 𝑁2
𝐼 and mind you,
√3 𝑁1 𝑎
|𝐼𝑎 | = |𝐼𝑏 |
449
So this is the thing and; obviously, this angle is 90° is there.
Because I have shifted 𝐼𝑎 by 𝜃2 same angle; similarly, 𝐼𝑏 is shifted by 𝜃2 , so this angle is
90°. Therefore, if I can show that this angle is 30° then I will conclude that it will be a
balanced 3 phase currents that is the thing. That is very easy to prove you calculate tan 𝛼
from this right angle triangle
𝑁
1 2
×
× (𝑁2 ) |𝐼𝑎 |
2 √3
1
1
tan 𝛼 =
=
=
𝑁2
𝑁2
√3
( ) 𝐼𝑏
( ) |𝐼𝑏 |
𝐼
( 𝐴𝑆
2)
𝑁1
𝑁1
Which indicates that 𝛼 = 30°.
Similarly from this triangle this is also 𝛼 same ratios. Therefore, we conclude that when
the secondary current is balanced 2 phase current like this when 𝑍2 are same both in
magnitude and power factor angle then secondary current will be balanced. But it will also
ensure that the current drawn from the 3 phase side will be also balanced.
So, it is a very nice connection you require 2 single phase transformers whose which are
not identical in the sense that number of turns are different. Primary side must have a total
number of turns 𝑁1 and with a 50% tapping otherwise you cannot implement that. The
second transformer primary must have it is number of turns same as this transformer, but
√3
with 2 𝑁1 turns secondary turns are same.
So, what actually people do is this they say take 2 identical transformer having same turns
this is also 𝑁1 , this is also 𝑁1 . This fellow primary one primary should have a centre tap
√3
another primary should have a taping at 2 % that is 87% or, so. You understand that; any
way this is I do not want to complicate the figure by drawing that because this is the thing
and you get balanced 2 phase system.
If I say that it is just a question if you have understood it correctly, suppose you have a
supply neutral also available ns and I ask you can this 𝑁𝑠 be connected to somewhere in
this circuit. See star connected load supplied from a 3 phase 4 wire system neutrally
straight away connected to the star point; is not and 𝑎 𝑏 𝑐 are in the other 3. Now, I am
asking you it is not a star connection although looks like star this is one this is one this is
450
𝑂. Suppose I ask you where this 𝑁𝑠 can be connected to not at 𝑂 you should not connect
𝑁𝑠 here, where it should be at the centroid of the supply triangle and it will come over here
and that I can calculate how much it will be; centroid it is.
Therefore, at this point which will be here and I can specify at what tappings we have to
take to connect that; is not. So, so this is how a scott connection will be here only one last
point is this that this diagram I have drawn for balance load but these equations are true
for balanced and unbalanced load as well, what I mean by saying that with unbalanced
load what happens we unbalanced load connected load on secondary side no problem.
Because what I will do is this; the secondary voltages are balanced 𝑎1 𝑎2 , 𝑏1 𝑏2 and this
is 𝑎2 𝑏2 neutral and suppose the load you have connected such that it is 𝐼𝑎 and in the other
phase you have connected a load 𝐼𝑏 whose magnitudes may be different and these two
angles is not 90°. So, it is a unity power factor load, primary power factor, but what I am
telling is this equation remains intact; mmf balance equation will not change. Only thing I
will not I will be deprived of this simplification that is |𝐼𝑎 | = |𝐼𝑏 | we crossed out several
times know.
So, it should be then 𝐼𝑎 it is reflected current will be 𝐼𝐴𝑆 is this 𝐼𝑎 you have to take in phase
with this I will just indicate it. Similarly,
𝑁2
𝐼 this time it will be like this getting me and
𝑁1 𝑏
𝐼
− 𝐴𝑆
if you join I will just indicate; it is your duty to study that. So, it can be done
2
unbalanced load and you will see that primary side current is also not unbalanced and also
note that the primary current will be balanced and what will be the power factor of that
balanced current drawn from the supply same as 𝜃2 ; why?
Because this is the phase voltage of the primary side and 𝐼𝐴𝑆 lags it by an angle of 𝜃2 . If
secondary current is unbalanced primary current is unbalanced, but what I am trying to tell
these equations 𝐼𝐴𝑆 , 𝐼𝐵𝑆 and 𝐼𝐶𝑆 these equations are true no matter whether balanced load
you have connected or unbalanced but if it is unbalanced load I will not be able to use this
information, and that is why this angle 𝛼 etcetera will be different and not it will not come,
so nice as
1
; why it should; got the point?
√3
So, Scott connection is a very nice connection and another thing I will ask you to explore
is that this connection that is this is this side connected, this is this side connected. Suppose
451
I plan to apply a 2 phase voltage to this side shall I get a balance 3 phase output voltage
on this side this you please explore got the point, that is what I am telling is in the reverse
way it will work.
(Refer Slide Time: 33:17)
That is this is the secondary 2 coils and primary connection is like this I will just indicate
it. I have planned 3 phase to phase vectors 3 phase supply give here balanced 2 phase you
get, but suppose I supplied this with a balanced 2 phase voltage this side. Should I get a
balance 3 phase output voltage here, draw phasor diagram and you will get the answer.
Reverse way suppose secondary side because in any transformer after all whether you can
energize primary secondary side you get something. Secondary side you energize you will
get something in the primary.
Any way scott connection is popular because of the fact that 2 phase supply is not readably
available what is available is, 3 phase supply, single phase supply these things are available
in your lab or whatever it is and if you have a 2 phase load for example, a 2 phase induction
motor or things like that or a 2 phase furnace. Then scot connection you have to use and
of course, balanced loads are preferred ok.
So, thank you for this lecture we will see what we can teach in the next lecture
452
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 47
3 Phase to 6 phase Conversion
O.C/S.T Test on 3 Phase Transformer
(Refer Slide Time: 00:20)
Welcome to 47th lecture. We have been discussing in general actually it should come
under the topic phase conversion. And we have studied 3 phase to 2 phase, and we have
completed this and this is called Scott Connection, got the point. Similarly, these
connections another connection is there 3 phase to say 6 phase conversion. 3 phase supply
is there you require a 6 phase supply.
Anyway, there are as such you know motors which are 6 phases, 6 phase balanced 3 phase
balance to 6 phase balance system. 6 phase balance system will be what? There will be 6
voltage phasors which will be 60° apart. 3 phase system 3 voltage phasors of equal
magnitude 120° apart. Now, 3 phase to 6 phase conversion were used very much at certain
point of time. For example, we have a 3 phase to DC conversion by using a diode bridge.
3 phase supply you give you will get a DC supply.
But if you increase the number of phases of that AC to DC conversion using diodes 6
phase will be always better, you will get better DC there, is not. The ripples available is
453
not 120° apart, every 60° something you will get. Anyway, I will just tell you how to
suppose 3 phase to 6 phase I want to do. What I will do? I will take two 3 phase
transformer, not two 3 phase transformer, one 3 phase transformer with two secondary
coils, ok, identical secondary coils. For example, these are the primaries we now know,
so, I will draw.
This is quite simple, I mean to understand. This is suppose dot dot etcetera. So, every time
I am drawing dot, so that you also see it is importance, it is necessary to put the dots 𝐴1 ,
𝐴2 and then this 𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 . But and these two coils are identical, ok. These are all 𝑁1
turns, these are 𝑁2 , 𝑁2 turns 𝐵1, 𝐵2, 𝐶1 , 𝐶2 . Then this is 𝑏1 , 𝑏2 , 𝑏3 , 𝑏4 and this 𝑐1, 𝑐2 , 𝑐3 ,
𝑐4 here. And what you do is this you connect this in star. So, I know the phasor diagram.
So, phasor diagram will be primary side it will be supply is 𝐴𝑠 , 𝐵𝑠 , 𝐶𝑠 . So, 𝐴𝑠 here supply,
𝐵𝑠 , 𝐶𝑠 .
And 𝐴𝑠 is nothing but 𝐴1 , 𝐵1, 𝐶1 and this is 𝐴2 , 𝐵2, 𝐶2 . Now, what you do is this you
connect these secondaries, these two groups in stars, but one connection you make Y y 0
and the other as Y y 6 and you are done. Why? Because if you join these 3 then you get Y
y 0, is not and your this voltage will be this say 𝑎1 , 𝑏1 and 𝑐1. And the other group what
you do? That also you connect in star, but this time join this dot terminals 𝑎3 , 𝑏3 , 𝑐3 then
its phasor diagram will come out to be this one Y y 6. So, this is primary was Y y 6 this
primary is y blue I will write this is Y y 0.
Then what? The naming of the terminals will be this is 𝑎4 , 𝑏4 , 𝑐4 . And then you join these
two neutrals these two you short. Then this point becomes a global neutral on the
secondary side where 𝑎3 , 𝑏3 , 𝑐3 and 𝑎2 , 𝑏2 , 𝑐2 are shorted. So, if you see this voltage
waveform of this terminal this terminal they will be 60° apart of same magnitude. So, you
have been able to generate 6 voltages on the secondary side which are of equal magnitude
and 60° phase difference. So, balance 6 phase voltage.
We will not discuss much more, but what I am telling is what happens is this in the see
diode bridge and this may be the names of this may be changed the bit. So, that what I will
tell; what I am telling is you name this output terminals 𝑎1 to be 𝑎 suppose, ok, this 𝑐4 you
say 𝑏 phase of the secondary side, this is 𝑐 phase, this is 𝑑 phase, this is 𝑒 and this is 𝑓
phase. I could rename them then I have got these 6 voltages 𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓 whose
454
magnitudes are same and so on. So, what I will do is this I will connect it and it has also
got a neutral.
So, I have a supply system now whose terminals are 𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓 and also neutral, got
the point. What you do now? You connects diodes one of the applications I am telling you
connect diodes and then connect the resistance, load, where you want to get dc and this
point connect it to neutral things will work, ok. So, anyway, so phase conversion 3 phase
to 6 phase can be done this way. Another way you can do, I hope you have now got the
idea. Can I do it by using some delta connection?
Yes, connect one group as Dd 0 another as Dd 6. I will not do that, but I will tell you the
end results. One group this you connect in delta Dd 0 and Dd 6 this group. So, the output
phasors will be like this then. Suppose, this is Dd 6, you work out that on your own and
this is black color. This black one is Dd 0. Then also these terminals of course, there is no
scope for connecting neutral now, this is delta, this is delta, where is its neutral nothing to
be joined here as I have joined here.
So, if you connect a voltmeter for example, here even when no load is connected voltmeter
will read this much, this voltmeter will read. But here if you connect the voltmeter here
circuit is not completed because two apparently other thing. So, what is to be done, you
will get balanced voltages there provided load is connected, so connect a balanced 6 phase
load clear. So, return path will be obtained. 6 rheostat you can show it in, I should not call
it delta like a polygon it will look like 1 2 I mean equivalent resistance 3 4 5 6, suppose
this is my impedance load impedance these 6 terminals you join there then the circuit will
be completed balanced 6 phase voltage you have been able to applied.
Anyway see this 3 phase transformer connection is a very nice subjects and lot of further
more connections are there using many small-small windings, doing these adding that
phasors. For example, one of such examples where traditional zigzag connection as I told
you. But it is in nutshell whatever knowledge you have acquired, I hope you if you read
those things some advanced topics on transformer winding connections, in-corporating
more coils into it you can do several other things that is there, but these are the main things,
ok. And you will be able to I think solve simple numerical problems on this.
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(Refer Slide Time: 12:39)
So, this is the phase conversion thing. 3 phase to 2 phase, then 3 phase to 6 phase and so
on. One can go on doing like this.
(Refer Slide Time: 13:08)
Now, our next topic will be the parallel operation of transformers, that is another important
topic, ok. Our next new topics is parallel operation of transformer. Before I do the parallel
operation of transformer, I am sorry let me tell you one thing. What is that thing? That is
look here if it is a single phase transformer, I must know the equivalent circuit parameters
and this can be measured by doing some simple experiments like open circuit and short
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circuit test. Now, what do I do for a 3 phase transformer? And how to find out the
equivalent circuit parameters of a 3 phase transformers? Suppose, I say the transformer
rating is known, what I, let me write some rating.
(Refer Slide Time: 14:18)
Suppose, say 30 kVA, just writing some numbers say 400V/200V, 3 phase, 50 Hz. Now,
and I say that the transformer is star star connected, it is given, star star connected. And it
is a 3 phase transformer, not a bank of 3 single phase transformers put together. So, what
do I know about these things? About these things I know that this is the total kVA, this is
line to line voltage mind you of the HV side, this is line to line voltage.
So, in the same way you carry out the, carried out the open circuit and short circuit test
here also you should do the same thing. So, what do you do? The transformer you connect
generally as I told you from the LV side you carry out the no load test. So, suppose this
side is LV, this side is HV, therefore to carry out the no load test what I will be doing is
this I will apply rated voltage, apply rated voltage. What is the rated voltage? 100V. Where
it should be applied? Between line to line; see that is what I am telling this is line to line
voltage, so it does not tell you the 200V. You applied rated voltage, and 200V is the line
to line voltage.
This 200V never tells you that this LV side voltage rating is 200V.
200
√3
is the rated voltage
of the LV side, but anyway 200V you apply. You connect 3 phase 200V and you connect
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ammeter, and also wattmeter and better connect two wattmeters because it is a 3 phase
system. So, note down 𝐼0 and 𝑊0 wattmeter reading and this one.
Secondary side open circuit, open circuit, and applied voltage of course, must be read, so
you have a voltmeter as well connected here. So, you know applied voltage. Now, then
what you do? You calculate the parameters in this way on the per phase basis. How it is
calculated? First thing is
𝑊0 = √3𝑉𝐿𝐿 𝐼0 cos 𝜃0 = √3 × 200 × 𝐼0 cos 𝜃0
And during open circuit test it is only core loss matters, is not, copper loss we neglect
because no load current is very small may be 2% to 5% the rated current. How much is the
rated current? As I told you always do this HV, so rated current of HV side this I will
come. So, no load current will be small and you calculate this and from this you will be
able to calculate cos 𝜃0 , from that you will be able to calculate sin 𝜃0 then you will be able
to, so per phase voltage if you have applied per phase equivalent circuit, so 𝑉𝑝ℎ and this
will be your no load current. Then this is 𝜃0 , just like single phase.
Then break it up into two component and that is magnetizing (𝐼𝑚 ) and 𝐼𝑐𝑙 and then calculate
𝐼𝑐𝑙 . This is 𝜃0 , this is 𝜃0 ,
𝐼𝑐𝑙 = 𝐼0 cos 𝜃0
𝐼𝑚 = 𝐼0 sin 𝜃0
And once you do that you will be able to calculate 𝑅𝑐𝑙1 , suppose you say LV side is 1. So,
𝑅𝑐𝑙1 =
𝑉
𝐼𝑐𝑙
𝑋𝑚1 =
𝑉
𝐼𝑚
Similarly, while doing short circuit test what should I do? Generally, from the HV side the
short circuit test is done HV side and LV side is kept shorted. Of course, for this rating it
does not really matter from because it is only 200V and 400V in the laboratory all the
voltages are available and hopefully the meters required will be also available there.
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So, so there what you do? This LV side you can short them. The moment you short the LV
side can I apply rated voltage 400V here? Never, because it will then draw very large
current. Current will be only limited by its leakage impedance. So, here you must have a
3 phase auto transformer to apply rather small voltage. I should call it variac, 3 phase
variac. Because you must be knowing the difference between a variac and then
autotransformer. Autotransformer means these winding you are using different section of
wires you get advantage. Anyway. So, so you measure the voltmeter reading here 𝑉𝑠𝑐 . You
record the current. You connect 2 wattmeters. See this
𝑊0 = 𝑊1 + 𝑊2
Algebraic sum. So, this is 𝑊1 + 𝑊2 .
And while using the wattmeters, it is better because during open circuit test power factor
is rather small use a low power factor wattmeter. Anyway those are details of the
experiment, but you somehow note down this total reading 𝑊0 . Here also connect better 3
phase transformer, 2 wattmeters, current coil and pressure coil. Note down ammeter
reading and call this current 𝐼𝑠𝑐 . How much voltage should I apply so that rated current
flows? What is the rated current on the HV side? 𝐼⏟
𝐻𝑉 . It will be how much
𝑅𝑎𝑡𝑒𝑑
𝐼𝑠𝑐 = 𝐼⏟
𝐻𝑉 =
𝑅𝑎𝑡𝑒𝑑
30000
√3 × 400
≈ 43𝐴𝑚𝑝
So, you have to choose an ammeter which can read 50Amp say and similarly wattmeter
current coil. But line to line voltage voltmeter rating should be very small, may be 5% to
10% of 400V. 5% of 400V is only 20V. Voltage required will be pretty small, so that
current rated current will be circulated.
Secondary, will be also carry rated current as the primary, so that copper loss will be there.
So, wattmeter once again will read this 𝑊1 , 𝑊2 , sum of these two, 𝑊1 + 𝑊2 . We will give
you the copper loss at rated current. We neglect the core loss. Why? Because rated voltage
applied is very small. Flux is only a few percentage of the rated flux, so when you apply
rated voltage rated frequency. Therefore, copper loss we assume this and then this
2
𝑊1 + 𝑊2 = 𝑊𝑠𝑐 = 𝐹𝑢𝑙𝑙 𝐿𝑜𝑎𝑑 𝐶𝑜𝑝𝑝𝑒𝑟 𝐿𝑜𝑠𝑠 = 3𝐼𝑠𝑐
𝑟𝑒2
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From which 𝑟𝑒2 is known.
Similarly, once you get 𝑟𝑒2, you have to calculate 𝑧𝑒2, 𝑧𝑒2 will be this impedance. So, what
is the impedance?
𝑧𝑒2 =
𝑉𝑠𝑐
√3𝐼𝑠𝑐
And once you get. So, from this 𝑟𝑒2 is known, from this 𝑊𝑠𝑐 . So, 𝑟𝑒2 known, then 𝑥𝑒2 can
be separated which is equal to
𝑥𝑒2 = √(𝑧𝑒2 )2 − (𝑟𝑒2 )2
But only thing you should careful that per phase equivalent circuit the moment you carry
out test, one test from LV side, another test from HV side, and it is your duty now to choose
from which side you will draw the equivalent circuit. This I told you in this earlier class.
Therefore, if I want draw the equivalent circuit refer to the LV side then you have to
transform these parameters in that sense. What is the turns ratio here? Turns ratio of the
winding if the connections are the same star to star or delta to delta, then the turns ratio
will be the line to line voltage ratios, is not. So, that is the thing.
Now, the question is what happens? So, per phase equivalent circuit you have to draw, per
phase. See worked out problems from books as well as in my notes I have included some
problems you go through that. Per phase equivalent circuit we refer to.
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(Refer Slide Time: 27:12)
Now, suppose I say that the transformer is star delta, ok. Same rating what I told 30 kVA.
Suppose, I say that 30 kVA, 400V/200V, and the connection is star delta then what do I
do? How to draw the equivalent circuit like that? Ok. You have to carry out the open circuit
and short circuit test. Because the readings we are taking are only the line readings, is not.
We are while carrying out the test I am measuring the line quantities.
In fact, you do not need that information whether inside this is star or delta connected, is
not. We have connected everything in line, ammeter, line to line voltage I measured. I
never ask where the neutral is I will measure this phase voltage, got the point. So, one type,
one way of looking at the thing is carry out the test in the same way as you have done and
imagine as if it is star star connected and tell that per phase equivalent circuit. Listen to
me carefully what I am telling, is not.
See the transformer is now star and this side is delta. I can always imagine that no it is so
far as measurement is concerned, it does not matter, oh. It you can equivalent to this side,
you can consider it to be some equivalent star and then calculate the parameters in the
same way because what you are doing, you are measuring total power drawn by that 3
phase connection which is star shown here during the no load test. Suppose, I carry out the
no load test what do I do? This is your actual transformer I will connect 2 wattmeters here.
Let me slightly clean this thing. Try to get the idea, I will not tell. It is in fact not needed
so much of pointing out, but it should be understood what we are doing.
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(Refer Slide Time: 30:00)
Ammeter you connect, rated voltage you apply, 200V while doing open circuit test this
side is open, who bothers. That is fine. But what I am telling you can always imagine this
to be star connected while calculating the parameters and leave with them. Another way
of doing it that is looks like more appropriate, it is like this. That is I will carry out the test
no doubt, I will say that look this is the no load current you are carrying. I will start
calculating per phase what I will do? Per phase voltage applied to this winding is how
much? 200V.
I will say this is
𝐼0
, if it is delta connection I know that. So, through the winding
√3
𝐼0
√3
is
going, ok. This total wattmeter reading during open circuit test will be this sum of the cold
losses of all the 3 phases, is not. Therefore, this voltage and this
(𝑊1 + 𝑊2 )
𝐼0
= 200 × ( ) cos 𝜃0
3
√3
I will do and leave here, ok. It is a single phase transformer, this is its primary, this one is
secondary, whatever it is one of the coils and power absorbed by this is this one and
whatever is the power factor angle, I will calculate. So, in that case I will do it like this.
𝐼0
√3
𝐼
I will take whatever it is that number, And, then I will calculate ( 0 ) sin 𝜃0 to get
√3
magnetizing current and this one. And, I will while drawing the equivalent circuit referred
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to this side 200V. Only thing is in that case turns ratio will be
200
(
, that you must
100
)
√3
understand.
Anyway, solve some problems on open circuit, short circuit test because it is per phase
basis the analysis is done, regulation problem, this, that, will be all similar because single
phase transformer it can be considered to be a single phase transformer 3 such units, ok.
Think about it and we will continue in my next lecture the parallel operation as I told you.
Thank you.
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Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 48
Parallel Operation of Transformers - I
Welcome to lecture 48th and here we will start a new topic on transformer only, that is
parallel operation of transformer which is an important topics, Parallel Operation of
Transformers, ok.
(Refer Slide Time: 00:27)
First of all why parallel operation of transformers are necessary that is a the topic from the
title of the topic it looks like there will be two or more transformers which will be
connected in parallel. What does that mean? Suppose, just if you have one transformer
here with its primary and secondary like this and another transformer there; what is done;
these are the primaries of two transformers say transformer A, transformer A and this is
transformer B. Now, this not A B C phase, single phase parallel operation first I will
discuss. So, operation of transformers.
So, what does that mean is this; the primaries will be connected in parallel and excited
from voltage. Secondary winding too will be connected in parallel and load will be
connected; got the point. So, two transformers, similarly if it is more than two another
primary will be coming parallel another secondary transformer, C secondary will, A B has
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nothing to do A phase B phase just naming the transformers. So, so we let us first discuss
this single phase operation.
So, this is what is meant by parallel connection. You connect the primaries in parallel, also
secondaries in parallel to supply load and primaries to be supplied with a source. Naturally,
this, there are some conditions you can just connect any two transformers in parallel. One
of the conditions you can easily make out is that the voltage rating must be same. For
example, if it is a single phase transformer, suppose 400V/200V transformers. Rating some
kVA ratings 400V/200V, single phase transformer and all are of course, 50 Hz. So,
suppose 400 V side I will connect them in parallel and it is absolutely logical. Rating of
each one of them should be rated voltage. So, they are connected in parallel and supplied
from the same source.
Similarly, the LV side rating is 200 V and we are going to connect them in parallel voltage
cannot be different. So, the voltage rating of the primaries and secondaries must be same.
For example, you will never think of connecting a transformer of rating 400V/200V in
parallel with 100V/50V, this two cannot be parallel for obvious results, ok. So, that is the
thing. So, voltage rating there are certain condition, first of all to be full filled to connect
the two transformers in parallel.
Before coming to the conditions which need to be full filled in order the two or more
transformers connected to be connected in parallel, we must ask ourselves what is the need
of connecting two or more transformers in parallel. For example, in power system
particularly in distribution system suppose you, because in distribution system you know
at the last level there is a transformer whose rating is typically 6.6 kV/440V; 3 phase, that
we discussed earlier.
Now, suppose in a locality which is to be supplied with 440V supply. 440V means one
supply line and neutral will go to every consumer 220V or 230V like that. But the problem
is suppose a locality is growing initially the demand of power is not that high. Suppose,
you select a 100 kVA transformers to give you an idea 6.6 kV/440V, 3 phase transformer
it is; it is also true for 3 phase transformer, 3 phase 50 Hz.
KVA I am not writing. Now, what I am telling; suppose, you see that presently the kVA
rating is say 100 kVA, the will be sufficient to cater the consumers because it is a growing
area, load demand is growing. So, suppose you decide, I will now buy a 150 kVA
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transformer, 50 kVA will be another margin and start giving power to the consumers. But
as you know the as the area is growing every 5 or 7 years the load becomes doubled, ok,
approximately I mean. May be after 10 years the power demand will become 300 kVA.
Now, naturally this transformer now be in sufficient because from this transformer which
is rated at 150 kVA if you want to derivate 300 kVA that means, it will be over loaded.
See, kVA means current, mind you, the voltages as are approximately same even under
with the concept of regulation there will be slide voltage difference these that that is there.
But, nonetheless the transformers will be come over loaded. You cannot pass current kVA
is synonymous with currents, voltages are almost same.
Therefore, if the load demand increases then this transformer which you installed initially
after 7 years the load demands suppose becomes 300 kVA. What is the option left; then
you have to purchase another transformer of same voltage rating, but 300 kVA is not and
do away with this transformer remove this transformer,. But that is not too economic
proposition, because these transformer is otherwise healthy and cost of a 300 KVA
transformer with same voltage ratings is going to be definitely higher.
Therefore, perhaps what other alternative is you purchase another transformer of similar
ratings and connect them in parallel. So, that the combined kVA is 300 kVA and it will
last for another 5 or 7 years, after that perhaps another transformer you. So, your cost of
investment is reduced at a time you do not have to purchase a very big transformer and
this is the very practical solutions in some areas. For example, it is now a village, gradually
growing industries may come up and things like that a big residential complex. So, demand
will go on increasing and it is therefore, needed to connect the transformers in parallel.
The advantage of connecting the transformers in parallel is another important thing. That
is, see, if you if the load depend is one 300 kVA suppose it has gone to and suppose you
are using a single transformer which is catering that 300 kVA load, but if some fault occurs
in that transformer the transformer is to be taken out for repair works or whatever it is.
And then there will be no power to the consumers. You cannot then provide any power
because the transformer is out of order. But had there been two transformer operating in
parallel, it is very unlikely that both of them has simultaneously has become faulty or needs
servicing; is not?
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Therefore, in such a situation if two or more transformers are connected in parallel catering
a certain amount of load if it so happens that one it is likely that one transformer will
become faulty, why all of them suddenly becomes faulty. So, one of them becomes faulty;
you remove that that is to be that you take out. Then the kVA you can supply will be less
no doubt one transformer less whatever its kVA rating, but nonetheless some of the
consumers maybe supply power that is a no main thing, I mean in one case if it is a single
transformer everything is blackout, if one transformer develops fault.
But if more transformers in parallel, one transformer goes out of order take it out and then
you can you can restore service back, but not to all consumers and that you can rotate
somehow to satisfy all the consumers by switching, disconnecting, some consumers at
some point of time reconnecting some other consumers. So, these are the nice points of
having at least at the distribution level some transformers operating in parallel.
The cost of each unit is small. In fact, you can, if two or more transformers are operating
in parallel, you can keep one transformer ready, I mean not in use what it is, as a backup.
So, if one transformer is faulty temporally power will be interrupted to some consumers,
but connect that new one and take this for servicing that is the whole idea very simple logic
nothing great things. But that is why the operating transformers in parallel assumes
importance, ok.
Now, we will come to the to the important thing that is if I want to operate two transformers
in parallel what are the conditions to be satisfied. One conditions I have already seen
because you will be operating in parallel. So, voltage ratings must be same, ok; is that kVA
rating should be same? Not at all; kVA rating is current ratings. So, KVA ratings may be
different. One; you may have 100 kVA this transformer, it will be operating in parallel
with another 200V, something 50 kVA transformer no problem. kVA rating may be
different. But only thing is when they will be operating in parallel; what is kVA rating;
kVA rating mean essentially current rating of a either HV or LV.
So, you must see that this transformer both of them are operating in parallel is low voltage
side should not exceed it is rated current, that is the only condition individually. For this
transformer, its low voltage current rating will be certainly to time this that should not
exceed that is the only thing.
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Then the third important point is which is rather practical problem after you have got this
two transformers which you want to operate them in parallel, you must ensure that is very
important, polarity, polarity test, polarity must be same while connecting transformers in
parallel, while connecting that must be same, otherwise you cannot do. So, voltage rating
and it should be connected in parallel with the due regard to their polarity, otherwise there
will be problems.
So, what do I mean by this? This is suppose the polarity of transformer A marking, this is
the polarity of transformer B and this is the correct way of connecting. If somebody
connects this secondaries for example, by mistake like this then what is going to happen;
dot becomes plus 200V, here dot becomes plus 200V load you have connected, but it will
find a 400V acting in the circuit creating a huge it is equivalent to short circuit. So, this
should be avoided. It is not allowed at all. So, polarity is a must condition before you
connect transformers in parallel.
After telling this there are other conditions which will be clear as we go along, ok. Now,
where from to start? It is like this see if you see the connections the connections I have
already told. Now, I will always neglect the no load current in these analysis. No point in
taking that. Instead because no load current is only 5% of the full load current of the given
transformers compared to be load current it is nothing. Therefore, to get quick and good
estimate of the current distributions in the transformer when they are operating in parallel
you better do not take no load current into account, ok.
So, transformers which will be in parallel, I will always draw the equivalent circuit,
equivalent circuit refer to load side. For example, in this case it was 400V this is suppose
the load side. So, correct I finally, put these two correct polarity, then only you can connect.
So, this is dot, this is dot. Polarity test I will do accordingly I will connect them in parallel.
Therefore, if you draw the equivalent circuit refer to the load side then what happens is
this. The primary side will have I will only show the induced voltage 𝐸1 and this is 𝐸2 the
induced voltage on the secondary side, ok; and here I will show the all the parameters in
this way 𝑟𝑒2, 𝑥𝑒2 and here I need not show 𝑟1, 𝑥1 that is what the equivalent circuit means.
So, refer to the other side I will draw the equivalent circuit of each of the transformers,
clear.
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Therefore, if you draw the equivalent circuit referred to the secondary side or load side, it
will simply becomes now like this, because of this parallel connection this is the induced
voltage of transformer A and this is the equivalent impedance of the transformer that is 𝑟𝑒𝑎
and 𝑗𝑥𝑒𝑎 . I have paralleled it with respect to the terminals available. Similarly, the second
transformer I will only consider two transformer, 𝐸𝑏 . Mind you, this is the secondary
voltage if you want to write 2 you can write, but this I will avoid, otherwise I have to carry
on that 2. So, this is this thing and this is the second transformer.
And it will have its resistance and leakage reactance, all referred to the secondary side,
suppose this is the thing and where are parallel them here getting. So, this is thing and here
is your load. So, this thing is equivalent to this thing when translated into equivalent circuit,
got the point. So, 𝑟𝑒𝑎 , 𝑥𝑒𝑎 is what? It is secondary resistance and reflected primary
resistance, that is why I have not shown any resistance here. Similarly, 𝑟𝑒𝑏 , 𝑥𝑒𝑏 will be like
that and this is the connection.
Now, in general if the voltage rating are same, so here if you apply 400V here. So, this
induced voltage will be 200V, so here also it will be 200V, here also it will be 200V and
you get the current. Now, at this point I will tell you that, see after all these two
transformers are different transformers.
So, sometimes what happens, this secondary voltages when you have applied same
primary voltages to the primary of these two transformers; there may be a slight difference
for example, this is 200V, this may be 208V because of some they are not after all identical
there may be always slide difference. In that case what happens is that if the switch is open
then there will be a circulating current, which is limited by the sum of the leakage
impedance of these two transformers.
Anyway, people say that, little bit of difference of voltages may be allowed provided it
does not cause a large circulating current. So, this although 𝐸𝑎 = 𝐸𝑏 is the most desirable
thing, but there maybe a little bit of allowance that can be allowed and maybe it is 200V,
this maybe 210V or so 5%-10% this way, that way you can do but any way; so, generally
that is way I have wrote 𝐸𝑎 and 𝐸𝑏 . I am not sure there will be most probably equal and
fine.
Now, to understand the other important conditions to be fulfilled this circuit is drawn and
then I say that what I should ask is that let us try to calculate. Suppose, and this voltage is
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what; this voltage is the terminal voltage 𝑉2, common secondary terminal voltage here,
this is 𝑉2, is not.; and this is the secondary current 𝐼2 . Now, you see the moment you
connect these two transformers in parallel, this load current is going to be shared by
secondaries of these two and accordingly there will be reflected current in their primaries.
So, any way we are looking. So, if the secondaries of the transformers are carrying rated
current as I told you primaries are also taking rated currents no problem. Therefore, only
one side I will focus on, ok. So, it is like this 𝐼2 .
Now, the question is what will be your 𝐼𝑎 and the 𝐼𝑏 . Who decides how much of this 𝐼2
will be flowing through 𝐼𝑎 and how much of the 𝐼2 will be flowing through 𝐼𝑏 ? I am not
writing two everything is with respect to load side mind you, clear. So, that that is one
important thing I must see.
(Refer Slide Time: 25:37)
For example, so this is the circuit once again I am drawing. So, this is your 𝐸𝑎 , ac this is
your 𝑟𝑒𝑎 + 𝑗𝑥𝑒𝑎 leakage reactance. Second transformer this is the thing 𝐸𝑏 , because it is
connected in parallel. This is 𝑟𝑒𝑏 + 𝑗𝑥𝑒𝑏 and then I have connected them in parallel with
due regard to the polarity means this plus plus and then I have connected a load here. So,
what is this statement of the problem, please note down.
What I am telling is suppose load current is known, 𝐼2 is known you have connected some
load and you find it is drawing this much current it is known. I want to know what is 𝐼𝑎
and 𝐼𝑏 that is the thing and if I find out 𝐼𝑎 and 𝐼𝑏 then I will perhaps get some more
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conditions which is which will be necessary for better sharing of the load because as I told
you the KVA ratings of these two transformer may be different. For example, you just help
me if necessary in calculation simple thing.
Suppose, I have this transformer, one transformer is 400V, another is 200V, ok. I first I
will do it like this. This part at least can be done whatever time is left. Now, so how do I
find out this? It is circuit problem, what else; I know 𝐸𝑎 , 𝐸𝑏 and I can find out. But I will
do it and I will assume 𝐸𝑎 , 𝐸𝑏 are not same. They will be same that is the most desired
thing, but the as I told you there maybe a little difference. So, 𝐸𝑎 , 𝐸𝑏 are there and what
will be the current shared by this.
The best way and there are several ways you can calculate it and this voltage is 𝑉. You
can apply nodal method this that to calculate the current. But the nice way of calculating
it is this since 𝐼2 is known, what you can think of in the circuit is that this impedance can
be removed and you can tell there is a current source, you must be knowing that, that in a
circuit if current in a branch is known then replace that component by this current source.
Currents in other branches is remains same, how does it matter, is not that is. So, it can be
replaced by current source in this direction.
Since, it is known in terms of 𝐼2 I want to know what is 𝐼𝑎 and 𝐼𝑏 . Now, in this circuit there
are 3 sources 𝐸𝑎 , 𝐸𝑏 and there is a current source. I will apply super position and in one
stroke I will get 𝐼𝑎 and 𝐼𝑏 , see how. This 𝐼𝑎 should be equal to, I will apply superposition,
ok. For example, I will assume 𝐸𝑎 , 𝐸𝑏 are present and this current source is not there that
means, it is open circuited, remove it, then because of 𝐸𝑎 , 𝐸𝑏 current 𝐼𝑎 in this direction
will be simply
𝐼𝑎 =
𝐸𝑎 − 𝐸𝑏
𝑧𝑒𝑎 + 𝑧𝑒𝑏
𝑧𝑒𝑎 = 𝑟𝑒𝑎 + 𝑗𝑥𝑒𝑎
𝑧𝑒𝑏 = 𝑟𝑒𝑏 + 𝑗𝑥𝑒𝑏
So, for these two sources current will be this current source is open circuited then plus I
must remove this source because these are voltage sources.
So, I will put them short circuit and I will assume only current source is present.
471
So, if you put this short circuit means what these two are in parallel and this is the total
current. So, when this term is let me draw because for me it is simple, but so only these
two sources are present means this and this and current source is open circuit then calculate
𝐸 −𝐸
this current, that is what I did, 𝑧 𝑎 +𝑧 𝑏 . Then only current source is present then it will
𝑒𝑎
𝑒𝑏
circuit will be like this 𝑧𝑒𝑎 , 𝐸𝑎 shorted, 𝑧𝑒𝑏 , 𝐸𝑏 shorted and you have only this current
source, that is the load current 𝐼2 and how the this will be shared; this rule is already known.
So, for example, this current will be because of 𝐼2 , the total current 𝐼2 into impedance of
the other branch divided by sum of the impedances is not that is all in one line we will get
this current instead of solving so many questions. So, and this equation is important
necessary. Now, if you get this, if you get this will be I a, clear?
𝐼𝑎 =
𝐸𝑎 − 𝐸𝑏
𝑧𝑒𝑏
+ 𝐼2
𝑧𝑒𝑎 + 𝑧𝑒𝑏
𝑧𝑒𝑎 + 𝑧𝑒𝑏
(Refer Slide Time: 32:37)
Now, similarly what will be 𝐼𝑏 ;
𝐼𝑏 =
𝐸𝑏 − 𝐸𝑎
𝑧𝑒𝑎
+ 𝐼2
𝑧𝑒𝑎 + 𝑧𝑒𝑏
𝑧𝑒𝑎 + 𝑧𝑒𝑏
This is the thing.
472
Therefore, for a known load current how the currents in the winding of the transformers
on the secondary sides will look like is this and in case, in case 𝐸𝑎 = 𝐸𝑏 it will be further
simplified, so this is the circulating current part. As I am telling if there is a slight mismatch
that circulating current part can be calculated. So, in case 𝐸𝑎 = 𝐸𝑏 it will be simply
𝐼𝑎 = 𝐼2
𝑧𝑒𝑏
𝑧𝑒𝑎 + 𝑧𝑒𝑏
𝐼𝑏 = 𝐼2
𝑧𝑒𝑎
𝑧𝑒𝑎 + 𝑧𝑒𝑏
See, what we what we really want. Suppose, the rated current of this transformer is known,
if you see that I just tell you one rating. Suppose, a one transformer is 4 kVA transformer
A and transformer B is 2 kVA; kVA ratings may be different and the voltage rating of this
transformer is suppose 400V/200V and voltage rating of this transformer is also same
400V/200V and I am looking things from the 200V side, is not. That is what I told, load
side. What is the rated current of this, LV side rated current?
20Amp and 10Amp. Therefore, I will see that this transformer since I have connected one
4 kVA and 2 kVA, this transformer I should expect should supply 6 kVA, 4𝐾𝑉𝐴 +
2𝐾𝑉𝐴 = 6𝐾𝑉𝐴 and maximum load current I can allow is then 30Amp, is not?
May be load; so, 30Amp current I would like to supply to the load. But here I should also
add, this 30Amp when it will be supplied I will demand that transformer A supplies 20Amp
and transformer B supplies 10Amp, then only I will be most happiest man in the world,
ok. Your load is taking 30Amp, your transformer A is capable of delivering 20Amp,
transformer B is capable of delivering 10Amp and this is the most desirable thing, is not,
it should happen,.
But the distribution of this total current is fixed by this relations; it looks like, who decides
that out of this 30Amp, 20Amp will be carried by these and 10Amp will be carried by this
𝐼
is decided by this because if you take the ratio 𝐼𝑎 . Who decides that? If you take this ratio
𝑏
from this; it will be what?
𝐼𝑎 𝑧𝑒𝑏
=
𝐼𝑏 𝑧𝑒𝑎
473
It is not that simply I have connected, I have done my job, ok, this fellow automatically
takes, no. It depends on relative values of 𝑧𝑒𝑏 and 𝑧𝑒𝑎 . So, read this portion very carefully
and think about the answer that is I should then put condition what should be the nature
related values of 𝑧𝑒𝑏 and 𝑧𝑒𝑎 such that this 30Amp should be divided according to their
KVAs.
Thank you.
474
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 49
Parallel Operation of Transformers - II
Welcome to lecture number 49th on Electrical machines 1. And we have been discussing
about Parallel Operation of Transformer which is an important topic.
(Refer Slide Time: 00:35)
And recall that we in our last class I took an example and told that suppose you have got
two transformer, one transformer is transformer A, another is transformer B, then if you
want to connect them in parallel what does it really mean is that connect the primaries of
transformer A and B parallel and energize with its rated voltage. Of course, before
paralleling, we must ensure that the voltage ratings are same, primary voltage and
secondary voltages are same, then only you can parallel otherwise not.
Similarly, the secondary coil voltages of transformer A and transformer B, they are
connected in parallel and that will be used to supply the load. Now, the necessity of a
connecting two transformers in parallel, we also discuss that. And told that in case you
have a single transformer supplying a load and if the load demand has increased, then you
must have purchased another transformer of higher KVA, instead of doing that purchase
another transformer of lower KVA, but having same voltage ratings and connect them in
475
parallel. So, that the old one even not just moving from the circuit. So, in this way as
demand increases perhaps you can connect two or more transformers in parallel and supply
the new load demand each one will contribute to the load.
As an example I told that suppose we have two transformers of voltage rating 400V/V.
And one transformer rating is in the next page perhaps I wrote ok. These two transformers
you are connected in parallel. Now, also voltage rating should be same and while
connecting them in parallel, we must ensure polarities same polarities must be connected
both on the primary and secondary side. So, this was actually that polarity, this will ensure
plus plus here ok.
In general 𝐸𝑎 and 𝐸𝑏 maybe slightly different, because after all these two transformers are
two different transformer exact matching of the secondary voltages when primary is
energized with rated voltage which are connected in parallel is very much unlikely. But
nonetheless let us assume that the secondary voltage as a general case suppose the induced
voltage is 𝐸𝑎 and 𝐸𝑏 , and four for this transformer it will be close to 200V, this will be
also close to 200V. And these two voltages will be there and there is load connected, this
is your load impedance. And all the equivalent circuit parameters are referred to the
secondary side to load side.
And so transformer A equivalent circuit is simply 𝐸𝑎 , 𝑟𝑒𝑎 , 𝑥𝑒𝑎 and so on; here also 𝐸𝑏 this
one. Then we would like to know that this is a simple circuit otherwise that the circuit is
to be solved, but we will be interested to load for a given load current 𝐼2 , suppose this is
known, this is known, then what will be 𝐼𝑎 and 𝐼𝑏 that is how this currents will be shared.
And in case of transformer, you know current rating means KVA ratings only, voltages
are maintained same almost. Therefore, whenever I say current ratings it means their KVA
ratings.
476
(Refer Slide Time: 04:59)
So, then we just found out these expressions that for a given 𝐼2 , if this is known your 𝐼𝑎
will be this much and 𝐼𝑏 will be this much. In general if 𝐸𝑎 and 𝐸𝑏 are not equal, it will be
a general expression like this. But since 𝐸𝑎 and 𝐸𝑏 will be very close, so and if they are
absolutely equal, then of course, 𝐼𝑎 will be equal to simply this and this one that is this
part, this second terms will be much higher compared to this terms. So, approximately 𝐼𝑎
will be this and 𝐼𝑏 will be this, why approximately if 𝐸𝑎 = 𝐸𝑏 , this will be the case.
Even if 𝐸𝑎 is slightly different to 𝐸𝑏 , I can calculate the exact values, but also this will give
you an idea how 𝐼𝑎 and 𝐼𝑏 will be. Mind you these are all phasors here, 𝐼𝑎 , 𝐼𝑏 , 𝐼2 is known.
𝐼
𝑧
𝑏
𝑒𝑎
Now, the ratio of 𝐼𝑎 = 𝑧𝑒𝑏 ok. So, up to this point, we have done.
477
(Refer Slide Time: 06:12)
Therefore, we have seen that in the transformer it has come like this I a by I b is equal to
z eb by z ea is it not, which also means that this is phasor relationship. This also means
that
|𝐼𝑎 | |𝑧𝑒𝑏 |
=
|𝐼𝑏 | |𝑧𝑒𝑎 |
That is how it will be shared ok. This equation is very important. From this equation we
will also put another conditions for those two transformers A and B for better load sharing,
and what is that condition that we can get from this equation that is why this equation is
important.
Now what is that? So, suppose you have two transformer, suppose transformer A is of
rating 400V/200V, 50 Hz. And transformer B has got a same voltage rating it must have
400V/200V, 50 Hz that is fine. And the KVA rating of transformer A is suppose 4 KVA
and this transformer rating is suppose 2 KVA ok. Therefore, rated currents of the
transformers A on the low voltage side, how much it will be?
This is 20Amp, this side current. And this rated current is 10Amp. So, this information I
have. Similarly rated current of this side will be 10Amp and this side high voltage side
current will be 5Amp is it not. But anyway we have referred everything to the LV side we
have connected the secondary side 200V side in parallel.
478
So, this is the thing, and this is the circuit this is one voltage source this is 𝑧𝑒𝑎 . This is
another voltage source on the secondary side, this is 𝑧𝑒𝑏 and these two are in parallel 𝑧2 .
And this is what I told your 𝐼2 load, and this is 𝐼𝑎 , and this is 𝐼𝑏 , and this is 𝐸𝑎 , this is 𝐸𝑏 .
And assume 𝐸𝑎 = 𝐸𝑏 that is both are equal to 200V ok, so that is fine.
Now, I have this relationship ok. We will see that for a given current, load current the
current shared by transformer A and that of transformer B are in the inverse ratios of their
respective KVAs that is current shared by transformer is proportional inversely to its
leakage impedance, 𝐼𝑎 ∝
1
𝑧𝑒𝑎
and 𝐼𝑏 ∝
1
𝑧𝑒𝑏
, is it not. So, it is inversely proportional to
leakage impedance.
Now, what is the important thing? See here, I would like to have our intention will be I
told you in the last class as well that when I see transformers whenever it is in operation it
is always nice to operate the transformer as full load. It will because that is why your
investment will then become meaningful, why then purchase so much KVA, so much
voltage transformer. Therefore, when this two transformers are operating in parallel, I
would like to have 30Amp is supplied to the load, and 20Amp is carried out by transformer
A; and at that time and 10Amp is carried out by transformer B that is the most I mean best
thing one can think of is it not.
So, if 𝐼2 = 30𝐴𝑚𝑝, and 𝐼𝑎 = 20𝐴𝑚𝑝 and 𝐼𝑏 = 10𝐴𝑚𝑝. If these happens that is the best
thing, all transformers will be under full load condition, KVA handled by transformer A
will be then whatever its 4 KVA, transformer B, 2 KVA and the KVAs supplied to the
load will be approximately 6 KVA, nothing is better than that ok.
|𝐼 |
|𝑧
|
𝐼
Therefore, but we then find that |𝐼𝑎 | = |𝑧𝑒𝑏 |. Therefore, you see the ratio of 𝑎 here; this
𝑒𝑎
𝑏
𝐼𝑏
demands that
𝐼𝑎 2
=
𝐼𝑏 1
𝐼
2
If this best thing to happen, then 𝐼𝑎 = 1, which implies that the impedance of transformer
𝑏
B should be twice the impedance of transformer A.
|𝐼𝑎 | |𝑧𝑒𝑏 | 2
=
=
|𝐼𝑏 | |𝑧𝑒𝑎 | 1
479
From this now therefore; if therefore,
|𝑧𝑒𝑏 | = 2|𝑧𝑒𝑎 |
Then only such a thing is possible.
If for example, it is the other way round that is suppose you find you have purchase two
|𝑧
|
1
transformers and you find that |𝑧𝑒𝑏 | = 2, just opposite leakage impedances are like this.
𝑒𝑎
Then what is the implication? Implication is see the ratio of the currents
𝐼2 : 𝐼𝑎 : 𝐼𝑏 = 3: 2: 1
That is what I want.
But if it is other way around, then the ratio of currents will be just opposite the transformer
having a higher KVA is having higher impedance is it not. Higher KVA transformer must
have lower impedance. Otherwise, what will happen, the moment you start loading the
transformer which is capable of supplying more current for a given load current, it will
always take less current and the other transformer which is lower KVA will share more
current.
So, a situation will soon occurred, the transformer be having lower KVA will carry rated
current and the transformer having higher KVA has not yet reached rated current, I think
a I got the idea. This is very important. Therefore, I should then now add another important
point that is if you want to have a meaningful parallel operations of transformers, then I
will say voltage ratings must be same, KVA ratings maybe different. Oviously, why KVA
rating should be same, KVA rating is different, but I will demand the transformer leakage
impedance should be inversely proportional to their KVAs.
So, I must have transformer leakage impedance should be inversely proportional to their
KVAs, then only you get this nice division of load currents. Now, mind you the ratio of
|𝑧
|
the currents magnitude of the currents is |𝑧𝑒𝑏 |. Now, I say that magnitude of the
𝑒𝑎
impedances, leakage impedance should be inversely proportional to the KVA ratings fine.
480
(Refer Slide Time: 18:03)
But
𝑧𝑒𝑎 = 𝑟𝑒𝑎 + 𝑗𝑥𝑒𝑎
𝑧𝑒𝑏 = 𝑟𝑒𝑏 + 𝑗𝑥𝑒𝑏
And I saw that
|𝑧𝑒𝑎 | = √(𝑟𝑒𝑎 )2 + (𝑥𝑒𝑎 )2
|𝑧𝑒𝑏 | = √(𝑟𝑒𝑏 )2 + (𝑥𝑒𝑏 )2
So, if this magnitudes are inversely proportional to the KVA that is fine, but then another
condition maybe put that about the quality of the impedances, quality of the impedances
of the leakage impedance, this angle of this impedances are also suppose same, got this
ratio.
𝑥𝑒𝑎 𝑥𝑒𝑏
=
𝑟𝑒𝑎
𝑟𝑒𝑏
So, leakage impedance angles they are same.
And magnitudes I will put them in the inverse ratio that is
481
𝑧∝
1
𝑠
𝑧𝑒𝑎 = 𝑘
1
𝑠𝑎
𝑧𝑒𝑏 = 𝑘
1
𝑠𝑏
That you have seen. Then I am saying that ok, if you just meet this condition’s, then you
may think yourself what happens this square root of this square plus this square and square
root of this square plus this square can be chosen in such a way that ratios are, the ratios
of the currents the way I told. But what about the quality, we is it going to make any
difference? The answer is yes; about the quality also you have to think a bit.
Now, what it is? Look back, I will write here. So, here the thing is very simple idea, this
is 𝐸𝑎 and this is your 𝑟𝑒𝑎 , 𝑥𝑒𝑎 . And this is your 𝐸𝑏 equivalent circuit refers to secondary
side of both the transformer, and this is your 𝐼2 . And here is our secondary load, this is 𝑧2
ok. And these are all phasors, I am not putting bar magnitude I will put that modulus sign.
And let us assume that 𝐸𝑎 = 𝐸𝑏 that is a case we are studying, these two phases are same.
Now, voltage across each of the impedances, I will calculate. These voltage is suppose 𝑉2,
then
𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑧𝑒𝑎 = 𝐸𝑎 − 𝑉2 = 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑧𝑒𝑏 = 𝐸𝑏 − 𝑉2 = 𝑉𝑎𝑏
Because same 𝑉2 and 𝐸𝑎 = 𝐸𝑏 . So, same voltage has come here. So, suppose the voltage
across 𝑧𝑒𝑎 and 𝑧𝑒𝑏 are same let that voltage be denoted by some voltage 𝑉𝑎𝑏 suppose I say.
So, I draw that first here I will just give you the idea suppose this is the voltage across this
and across this, that I have drawn.
If the qualities of impedances are same, then 𝐼𝑎 will be lagging by some angle 𝜃 not power
factor angle ok. What is 𝜃?
tan−1
𝑥𝑒𝑎
𝑥𝑒𝑏
= tan−1
= 𝜃𝑒
𝑟𝑒𝑎
𝑟𝑒𝑏
Or say 𝜃𝑒 sum equivalent impedance, it is power factor angle. It will be 𝐼𝑎 .
482
And similarly your where you will be 𝐼𝑏 ? 𝐼𝑏 will be also lagging 𝑉𝑎𝑏 by the same angle 𝜃𝑒
under this condition, therefore, you are 𝐼𝑏 will be which transformer I took more this is 2
KVA 𝐼𝑏 . So, its current will be less. So, this is your suppose 𝐼𝑏 . I will clean it. This I will
write it as 𝐼𝑏 . And this one will be your 𝐼𝑎 , it will be like this.
What is the current supplied to the load 𝐼2 = 𝐼𝑎 + 𝐼𝑏 , phasors sum of these two, but they
are co-phasor 𝐼𝑎 and 𝐼𝑏 . So, current supplied to the load will be this one, this plus this, 𝐼2 ,
is it not, this will be the thing. What is the KVA supplied by this transformer?
𝐾𝑉𝐴 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑡𝑜 𝑙𝑜𝑎𝑑 = 𝑉2 𝐼2
𝐾𝑉𝐴 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑏𝑦 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝐴 = 𝑉2 𝐼𝑎
𝐾𝑉𝐴 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑏𝑦 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝐵 = 𝑉2 𝐼𝑏
And if the quality of the leakage impedances are same, it will be the algebraic sum of these
two is this one, that is then
𝑉2 𝐼𝑎 + 𝑉2 𝐼𝑏 = 𝑉2 𝐼2
Therefore, you see I will not only demand that the leakage impedances should be inversely
proportional to their KVA ratings, but also I will say it will be very nice if the qualities of
𝑥
𝑥
𝑟𝑒𝑎
𝑟𝑒𝑏
the leakage impedances are also same, then tan−1 𝑒𝑎 = tan−1 𝑒𝑏 = 𝜃𝑒 . And we will get
the most desirable thing.
For the example, we have taken 6 KVA, 4 KVA and 2 KVA transformer we have
paralleled them. If the leakage impedances of these two are in the ratio 1:2 or 2:1 that we
have seen. And also the angles of each of this leakage impedances are same, qualities are
same, then your output KVA you will be also 6 KVA and that is the best thing. Apart from
of course, this polarity should be same, voltage ratio should be same that is the first thing.
Second if we go much deep into it, then we have come to this conclusion. We will carry
on with this in the next class.
Thank you.
483
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 50
Parallel Operation of Transformers - III
(Refer Slide Time: 00:26)
Welcome to 50th lecture on Electrical Machines – I, and we have been discussing about
Parallel Operations of Transformer. So, we have seen that in our last class the impedances
of leakage impedances of the transformer should be inversely proportional to the KVAs
and in Ω and also of the it is always better if the qualities are also same. Then, whatever
KVA is supplied by this, whatever KVA is supplied by transformer be just algebraic sum
of this two will come.
484
(Refer Slide Time: 01:07)
Now, what is going to happen if this condition is not fulfilled? Suppose,
tan−1
𝑥𝑒𝑎
𝑥𝑒𝑏
≠ tan−1
𝑟𝑒𝑎
𝑟𝑒𝑏
tan−1
𝑥𝑒𝑎
= 𝜃𝑒𝑎
𝑟𝑒𝑎
tan−1
𝑥𝑒𝑏
= 𝜃𝑒𝑏
𝑟𝑒𝑏
Suppose, they are not equal, but impedances are inversely proportional to the KVAs
𝑧𝑒 ∝
1
𝑠
𝑧𝑒𝑎 = 𝑘
1
𝑠𝑎
𝑧𝑒𝑏 = 𝑘
1
𝑠𝑏
So, one thing is ensured that is the ratio of the currents will be
|𝐼𝑎 | |𝑧𝑒𝑏 |
=
|𝐼𝑏 | |𝑧𝑒𝑎 |
485
Because KVA s are inversely proportional. So, magnitude of 𝐼𝑎 and 𝐼𝑏 will be in the inverse
ratios, but what is the implication of this then.
We have seen already that the voltage across this impedances are same and that voltage
denoted it by 𝑉𝑎𝑏 . So, let us draw this once again this phasor diagram. Suppose, this is 𝑉𝑎𝑏
which I have already defined which will be small I am drawing it very large just to
understand. So, this is this one. Now, I am telling this two angles are not same ok. So, you
draw 𝐼𝑎 and 𝐼𝑏 this lengths are decided by this ratios for a given load current, 𝐼𝑎 and 𝐼𝑏 .
Therefore, this is the suppose this angle I say it is 𝜃𝑒𝑎 and this angle I say it is 𝜃𝑒𝑏 . So, that
the this angle is 𝜃𝑒𝑎 and this angle is 𝜃𝑒𝑎 , is it not? This way it will lag. Earlier this two
angles were same 𝐼𝑎 and 𝐼𝑏 were in phase. Now, where is your 𝐼2 ?
𝐼2 = 𝐼𝑎 + 𝐼𝑏
Phasor sum. So, add them vectorially. Add this two and you will say your load current the
this is your 𝐼2 now, is not? This will be the 𝐼2 .
Now, I redraw this circuit to avoid any confusion that is over to your planning this is my
circuit, this is your 𝑧𝑒𝑎 , this is your 𝑧𝑒𝑏 . This two are parallel this is your 𝐼2 , this is your
𝐼𝑎 , this is your 𝐼𝑏 and here is your secondary load 𝑧2 this is the thing same thing. So, this
is 𝐸𝑎 , this is 𝐸𝑏 and so on. This is the thing. So, your 𝐼2 = 𝐼𝑎 + 𝐼𝑏 .
Now, in numbers for the given example I have telling if you are the this what you will see
is this you will see that this 𝐼2 = 30𝐴𝑚𝑝 suppose the output load current is 30Amp, then
the other two transformer this current will be shared as 20Amp and 10 Amp.
You imagine ammeters are connected here. It will read 30Amp, it will read 20Amp, it will
read 10Amp because of what because of the fact this thing.
So, you see that when it will carry 20Amp and when it will carry 10Amp these two sum
of this two currents cannot be 30Amp. It will be less than this. In other words what I am
telling that this two transformer will be operating at different power factors. For example,
so far as this transformer is concerned 𝑉2 it is contributing 𝐼𝑎 at this much angle it will
because to this 𝑉𝑎𝑏 which is this voltage here if you add say I want to get what is 𝐸𝑎 where
𝐸𝑎 = 𝐸𝑏 .
486
So, it will be like this 𝑉𝑎𝑏 is which these two points are at same potential we have seen.
So, potential of this point will be this voltage this I have assumed it to be 𝑉𝑎𝑏 , this is also
𝑉𝑎𝑏 . Now, 𝐸𝑎 I want to get. So, you see that 𝐼𝑎 and 𝐼𝑏 they will be displaced and this angle
will be the difference between 𝜃𝑒𝑎 and 𝜃𝑒𝑏 , that is understood. Now, I want to get this
voltage is 𝑉2. Suppose from 𝑉2 I want to get 𝐸𝑎 .
So, I will now start from 𝑉2 I will draw 𝑉2 first, but if you start drawing from 𝑉2 that this
transformer the angle between this two is 𝜃𝑒𝑎 ~𝜃𝑒𝑏 , is not that will be the angle 𝐼𝑎 and 𝐼𝑏
angle will be that one. So, suppose this angle this current is 𝐼𝑎 and the other transformer
current is 𝐼𝑏 , such that this angle is this one, are you getting? 𝜃𝑒𝑎 ~𝜃𝑒𝑏 .
Therefore, you can easily see the power factor at which this transformer and this
transformer is operating is at different values. So, power factors at which the transformers
will operate will be different and your total load current will be in between here 𝐼2
therefore, to get the 𝐸𝑎 back or 𝐸𝑏 back you have to add I will just indicate here
𝐸𝑎 = 𝑉2 + 𝐼𝑎 (𝑟𝑒𝑎 + 𝑗𝑥𝑒𝑎 )
𝐸𝑏 = 𝑉2 + 𝐼𝑏 (𝑟𝑒𝑏 + 𝑗𝑥𝑒𝑏 )
Therefore, we find that since the vectors sum of 𝐼𝑎 and 𝐼𝑏 will be less than the sum of this
two currents when they are co-phasal therefore KVA handled will be slightly less.
Therefore, to I mean summarize this what I will say that to operate a two transformers in
parallel successfully voltage ratio should be same, leakage impedance should be inversely
proportional to their KVAs and try to also see that the qualities of the leakage impedances
are same that is the best thing ok.
487
(Refer Slide Time: 13:08)
So, one important thing sometimes people say that leakage impedance that is what we have
seen leakage impedance of the transformers should be inversely proportional to their
1
KVAs, 𝑆 ok; leakage impedance in Ω. The same statement is told in this way that per unit
values of the leakage impedance should be same. This is one and the same thing because
you know
𝑃𝑒𝑟 𝑈𝑛𝑖𝑡 𝐼𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 𝑜𝑓 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝐴 =
|𝐼𝑎 ||𝑍𝑒𝑎 |
= 𝑍𝑝.𝑢. (𝑎)
𝑉2
𝑃𝑒𝑟 𝑈𝑛𝑖𝑡 𝐼𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 𝑜𝑓 𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝐵 =
|𝐼𝑏 ||𝑍𝑒𝑏 |
= 𝑍𝑝.𝑢. (𝑏)
𝑉2
We are talking about the other side this is the per unit value of leakage impedance of
transformer A and transformer B this is the thing, all referred to the secondary side. How
much is dropped in that with respect to the rated voltage are not this two are same?
𝑍𝑝.𝑢. (𝑎) = 𝑍𝑝.𝑢. (𝑏)
What is |𝐼𝑎 ||𝑍𝑒𝑎 |? Voltage across 𝑍𝑒𝑎 and voltage across 𝑍𝑒𝑏 that is this no point in
redrawing.; voltage across 𝑍𝑒𝑎 and voltage across 𝑍𝑒𝑏 in volts they are same; this voltage
minus this voltage is this voltage in case 𝐸𝑎 = 𝐸𝑏 therefore, and what is the per unit
leakage impedance? Per unit leakage impedance physically it means that when rated
current flows through the transformer there will be some voltage drop magnitude of that
488
voltage drop expressed as the rated value of the voltage of the transformer of that side.
And, per unit values are same with respective to secondary or primary for a given
transformer.
Now, since 𝐼𝑎 𝑍𝑒𝑎 = 𝐼𝑏 𝑍𝑒𝑏 here all the parameters are with respect to the secondary side,
they are same. Therefore, instead of telling that ohmic value is inversely proportional to
the respected KVAs you simply say per unit impedance of transformer A and per unit
impedance of transformer B, they are same that is all. So, we have gone through. Now, the
last thing I will tell about the parallel operation of transformer is what happens if it is a
three-phase transformer because after all in our power system it is all three-phase
transformer.
(Refer Slide Time: 18:25)
So, I write down this conditions once again. So, I write down for successful so, for
successful and effective parallel operation of transformers we list voltage ratios. Voltages
ratios of HV, LV side should be same HV and LV side from transformers various
transformers. Voltages of HV LV side should be same.
Second thing is leakage impedances. I write I am writing in short leakage impedance 𝑧𝑒 ∝
1
𝑠
this is in Ω or same thing is 𝑧𝑝𝑢 of the transformers should be same. When this condition
is satisfied what is guaranteed? The current divisions of the transformers will be in the
ratio of these two currents fine.
489
Then, the third condition is which is desirable condition that ok, the qualities of the
transformers qualities of the equivalent of the leakage impedances the leakage impedances
should be should be same. This will ensure transformers will operate at same power factor
transformer will operate at same power factor.
Now, number 4 and of course, this I have not listed with due regard to polarities you have
to connect otherwise what. Fourth is for three-phase transformer only thing is we will say
that line voltages other things are there these are line voltages because after all voltage of
HV side LV side same because line voltage three-phase transformer that is how it is
specified. We must write that the transformers must belong to must belong to same vector
group.
What is the vector group? Plus 30° or 11 or 1 which means that are you getting the line to
line voltage rating should be same. Here I have written simply voltages of HV LV side
should be same for three-phase line to line voltages should be same, voltage ratings should
be same, got the point? Line to line voltage ratings are same and then I am telling the
transformer must belong to same vector group that is +30° or -30° and so on.
This tells me suppose you have a transformer Dy1, the other transformer need not be Dy1,
but it must be Yd1 that is what I mean same vector group parallel operation means line to
line voltages are same and they must belong to same vector group. Of course, I have see
primary you have applied a voltage, it is getting shifted both phase and line to line voltage
get shifted by some angle of 30° or so. But, primaries will be connected in parallel,
secondaries will be connected in parallel and they must be shifted by same angle then only
the voltages magnitudes of course, will same and then parallel link can be done
successfully.
So, this is the thing we must add. Of course Dy1 can be parallel with Dy1 no question, but
I should not demand that one transformer is Dy1 another transformer should be also Dy1.
No, it could be Yz1 also, it can be parallel with clear that is they must belong to same
vector group that is this numbers should be same apart from the fact that the line to line
voltages have same and then for better load sharing this things are valid that is the
impedances should be inversely per phase impedances should be inversely proportional to
their KVA ratings and per unit impedance should be same of both the transformer ok.
490
So, this is a in short or whatever you call it not so short, but I hope I have been able to give
you some idea of how transformers can be operated in parallel, what is the need of making
transformers operating in parallel, and most importantly you must have appreciated this
conditions why they are to be fulfilled in order that successful and effective parallel
operation can be implemented ok. So, we will continue with some more topics on
transformer before starting DC machines.
Thank you.
491
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 51
Specific Magnetic and electric Loudings
Welcome to 51st lecture on Electrical Machines – I and so far we have analysed the
transformers that is we found out the equivalent circuit of the transformer, how to analyse
ideal transformers, then auto transformers, then you know three-phase transformers, core
type primarily and how to represent it in per phase equivalent circuit and various tests like
open circuit, short circuit and Sumpner’s test and then finally, we in our last lectures we
told you about the parallel operation of the transformer.
See, so far, I have not really told you about the construction of the transformer in details
and what are how it will look like externally if you look at a transformer big transformer
for example, 100 MVA transformers which are used in power stations and so on.
(Refer Slide Time: 01:52)
Perhaps we are used to see transformers in the laboratory which are very small ratings for
example, 5 KVA small transformers we are used to see small transformers which are power
transformers can be termed as few KVAs and voltage rating may be 200V/100V, 50 Hz
and so on. If you see a big transformer you will find and this transformers if you look at in
492
the laboratory you have seen, you will be able to see the coils and windings just like as it
is ok.
There will be LV windings here I will just show it like this, this is suppose LV windings
sectional view LV and this may be your HV side HV core type and this is the sectional
view windings are if you see from the top, this HV windings are like this around the core,
similarly for the LV side. And, also I told you to reduce the reactance of the transformer
leakage reactance, LV windings may be divided in this two limbs. So, also HV windings
and they may be connected in series respectively so that leakage flux is reduced and
leakage reactance is reduced.
So, here I have drawn that LV on this limb and on the other limb and terminals will come
out from the transformer on a terminal box in the laboratory you know LV terminals, all
terminals will be brought in here, HV terminals are brought in. So, 𝐴1 , 𝐴2 and small 𝑎1 ,
𝑎2 will be appropriately brought in for your external use. Although you will not be able to
make out the sense of the coils otherwise it is exposed and it is everything is exposed. This
is the core, you can easily identify here is core, here is LV windings, the windings you can
see and this terminals are brought out.
But, if you see a big transformer you will always see nothing is visible except that a big
tank, steel tank will be there steel tank; tank and on the top of the tanks there will be
bushings through which terminals will come out. This is may be HV terminals, single
phase I am drawing and another pair of bushings will be there whose insulation level will
be less just to give you the idea another bushings two terminals will come out these are
LV terminals.
This is suppose capital 𝐴1 , 𝐴2 this is small 𝑎1 , 𝑎2 just I am giving you the ideas. Where
from this has come out? The transformer essentially has a core and coils this whole thing
is within the tank and the tank is filled with oil ok. Sometimes you will find if you look at
distribution transformer in your city, there will be some tubes around the walls of the coil.
This is visually what you will see several these are called cooling tubes, that is in other
words this whole thing is inside the tank and tank is filled with oil something like that.
Now, the question is for low KVA transformer I do not require any oil to cool the
transformer, but for high rating I need a transformer which should be immersed in oil. Oil
will serve two purpose insulation from the tank which is steel tank the essential thing is
493
this thing is there inside for big transformers that is what I am telling may be hundreds of
KVA transformers.
So, and these are cooling tubes. I will give you a better picture later, but a just roughly you
know that these are the things you see in real life. Now, the question is what for this oil is
there and so on. For example, these voltage may be why I mean ceramic insulators will be
there, these are called bushings things like that. To understand that cooling is a must in big
transformers will be can be understood provided we go slightly deeper into the KVA
equation of transformer and we will start with that then we will end up with this thing.
I will be trying to be very lucid to explain that, so that you can appreciate this is what it
should be otherwise how. For example, I have told you that the voltage equation forget
about that leakage impedance this that
𝑉1 ≈ 𝐸1 = 4.44𝑓𝜑𝑚𝑎𝑥 𝑁1
That is the thing.
And, also I will tell you from this equation how can you relate it to the size of the
transformer and two other things – one is called specific magnetic loading and specific
electric loading. We will try to understand these two first, then I will show that the KVA
which essentially tells you about the size of the transformer, can be shown to be
proportional to specific magnetic loading, specific electric loading and the physical
dimension of the transformer that is the first thing. You know when a transformer will be
operating and hopefully it is operating at full load condition.
Both copper loss and core loss will take place and it will raise the temperature of the whole
thing. And, suppose a transformer is now at room temperature, you have switched on and
the transformer is operating at full load then you will see temperature will grow and after
certain time the temperature will attain a constant value. What is the constant value at to
which the transformer will attain to? Anything, if you go on pumping heat into a system
temperature will rise and it will attain a constant temperature; that means, at whatever rate
you are generating heat in the system at the same rate it is also dissipating heat to the
outside world.
494
When such a thing will happen, then only that fellow will attain a constant temperature
because after all copper loss and iron loss will be constantly taking place, always you are
pumping energy into the system heat energy and your from the surfaces of the transformers
heat is being dissipated out. The rate at which heat is generated inside the transformer,
when it is saying as the rate at which that energy is given out by this surfaces of this so
called transformer if it is a transformer, then only it will attain constant temperature. It is
not only true for transformer, but for rotating machine also. There will be losses switch on
the machine from cold condition that is at ambient temperature and operate it at full load
after switching on, then temperature will exponentially grow and finally, attain a steady
value.
It is this temperature rise that limits that puts the condition that is this is the current rating
because if the temperature is allowed to rise more for example, if you overload the
transformer then your windings may be spoiled, your insulations may be spoiled and things
like that will happen. Anyway, so, we will try to see all these things. These are the
interesting topics. So, first thing is first. So, this is let me try to relate the output of a
transformer KVA rating of a transformer in terms of physical dimensions and specific
magnetic and specific electric loading.
First of all specific magnetic loading is very simple; it is 𝐵𝑚𝑎𝑥 that is all in Tesla ok. Now,
specific electric loading is the current density this one, this is denoted by 𝛿 is the current
density. If you are using copper of copper or aluminium whatever you are using. Copper
is the better because aluminium is brittle large transformers are build with copper nothing
like aluminium. So, current density and its value is if I remember correctly about 2.53Amp/mm2 where copper is used to make the coils, is not?
And, I told you LV side the section of the coil is 𝑎1 . So, LV side 𝑎1 equal to cross section
of the wire which makes the coils. Similarly, 𝑎2 suppose is the cross section of wire on the
HV side whatever it is this could be also LV HV reversed, but 𝑎1 this one then obviously,
the delta if you are using copper why they should be different? No matter whether that
where is used for LV side or HV side 𝛿 will be same. So, 𝛿 is equal to same, both for LV
and the HV side is not? Copper is used.
Let 𝐼1 be the current rating of side 1 current rating rated current of side 1 that is in this case
LV side, then
495
𝐼1 = 𝛿𝑎1
Similarly rated current of the secondary coil HV side rated current will be
𝐼2 = 𝛿𝑎2
Current rating. So, this is a rated current, this will be the thing.
Then we come to this interesting thing.
(Refer Slide Time: 17:35)
Now, KVA rating of a single phase transformer suppose single phase transformer, I will
say it is equal to
𝑆 = 𝑉1 𝐼1 ≈ 4.44𝑓𝜑𝑚𝑎𝑥 𝑁1 𝐼1
𝐼1 is the rated current forget about that leakage impedance drop etcetera. So, this is the
approximately this is the KVA very close to. Now, what my plan is, to express this KVA
rating of this transformer in terms of specific magnetic loading, specific current loading
and the physical dimension of the transformer that is my goal.
So, this I can write it as
𝜑𝑚𝑎𝑥 = 𝐵𝑚𝑎𝑥 𝐴⏟𝑖
(𝑛𝑒𝑡)
496
But net iron cross sectional area. We have seen that iron area cross sectional area to realize
this iron we use stampings. So, 𝜑𝑚𝑎𝑥 = 𝐵𝑚𝑎𝑥 𝐴⏟𝑖 . Overall area will be slightly higher, is
(𝑛𝑒𝑡)
not? If you because of the plates are kept side by side there may be little bit of air in
between. So, this is
𝑆 = 𝑉1 𝐼1 ≈ 4.44𝑓𝜑𝑚𝑎𝑥 𝑁1 𝐼1 = 4.44𝑓𝐵𝑚𝑎𝑥 𝐴⏟𝑖 𝑁1 𝛿𝑎1
(𝑛𝑒𝑡)
What is 𝑎1 ? 𝑎1 is the LV side copper wire you are using side one, the cross sectional area
of the wire which makes your turn are you getting this area 𝑎1 . I could write it in terms of
secondary also
𝑆 = 4.44𝑓𝜑𝑚𝑎𝑥 𝑁2 𝐼2
But one side we will do because I know
𝑁1 𝐼1 = 𝑁2 𝐼2
So, this is the thing. So, let us understand each term. 𝐵𝑚𝑎𝑥 that happens to be the specific
magnetic loading in Weber/m2 in typical transformer the values of 𝐵𝑚𝑎𝑥 where will be
very close to 1Tesla, may be 0.9Tesla or 1.01Tesla etc and 𝛿 if you are using copper I
wrote it may be 2.5-3Amp/mm2 ok, CRGO material – 1Tesla, copper this one and you get
this.
Therefore, at least one physical dimension I have been able to brought in here that is the
with this cross sectional area has something to do with physical area. Now, this length into
this breadth this is the cross section of the core. So, this length into this breadth if you do
what you get is 𝐴𝑖 𝑔𝑟𝑜𝑠𝑠 and through which flux is passing not through air. So, it is slightly
more. So, 𝐴𝑖 𝑛𝑒𝑡 will be slightly less than this.
𝐴𝑖 𝑔𝑟𝑜𝑠𝑠 means you take a scale measure this outside dimension of the core get the area.
So, it must be multiplied by a factor 𝑠𝑓 . This factor 𝑠𝑓 is called Stack Factor. The value of
which will be 0.95% very close to 100%, but may be 0.9% like that, got the idea?
𝐴𝑖 𝑛𝑒𝑡 = 𝑠𝑓 × 𝐴𝑖 𝑔𝑟𝑜𝑠𝑠
497
So, I will because overall dimension I want to know. So, this can be written as this is
already specific magnetic loading, then this I will write
𝑆 = 4.44𝑓𝐵𝑚𝑎𝑥 𝑠𝑓 𝐴𝑖 𝑔𝑟𝑜𝑠𝑠 𝑁1 𝑎1 𝛿 × 10−3
If these are in volts etc. I hope you are getting this, this will be the KVA rating.
Now, you look at this sectional view of the transformer core and this one. I am sorry these
two are same width make it corrected ok. Now, how these windings are there? LV
windings or side one winding coils. How many sections you will see here? 𝑁1 . What is the
sectional area of each conductor? 𝑎1 . Similarly, on the other side the coils are wound like
this sectional view will appear like that. How many things you will see here? 𝑁2 and what
is the cross sectional area? 𝑎2 .
This area is called the window area in which copper will be receding. So this area internal
rectangular area is the window area; 𝐴𝑤 is the window area, this length into this height.
There is a factor like stack factor we called it window space factor window space factor.
It is usually denoted by 𝐾𝑤 . This tells you how much of this window area is utilized by
copper. So,
𝐾𝑤 =
𝐴𝑟𝑒𝑎 𝑢𝑠𝑒𝑑 𝑏𝑦 𝐶𝑜𝑝𝑝𝑒𝑟 𝑁1 𝑎1 + 𝑁2 𝑎2
=
≈ 0.35
𝐴𝑤
𝐴𝑤
Generally, the window factor is about 35% or so I mean some may be 0.35, not 100% is
utilized to fill it with copper. We will see because there will be some oil will be filled up
circulation of heat is necessary and so on. So, practical value of this is only 35% of this
window area will be utilized for that. So, this is called window space factor. But, 𝐴𝑤 will
certainly going to decide 𝐴𝑤 along with 𝐴𝑖 , physical dimension of this core is going to
decide the physical dimension of the transformer.
Now, you see when the rated current will be flowing we know 𝑁1 𝐼1 = 𝑁2 𝐼2 , but current
density of copper remain same in the primary and secondary. Therefore, I will write
𝑁1 𝑎1 𝛿 = 𝑁2 𝑎2 𝛿
𝑁1 𝑎1 = 𝑁2 𝑎2
So, come back here windows space factor then will look like. It will look like
498
𝐾𝑤 =
2𝑁1 𝑎1
𝐴𝑤
So, this is the thing.
Now, we will come back to this equation and we note that from this
𝑁1 𝑎1 =
𝐾𝑤 𝐴𝑤
2
So, for this 𝑁1 𝑎1 I will put this there. So, I do that in the next page.
(Refer Slide Time: 28:24)
So, it will be equal to we have seen that output equation is what
𝑆 = 4.44𝑓𝐵𝑚𝑎𝑥 𝑠𝑓 𝐴𝑖 𝑔𝑟𝑜𝑠𝑠
𝐾𝑤 𝐴𝑤
𝛿 × 10−3
2
𝑆 = 2.22𝑓𝑠𝑓 𝐾𝑤 𝐴𝑖 𝑔𝑟𝑜𝑠𝑠 𝐴𝑤 𝐵𝑚𝑎𝑥 𝛿 × 10−3
So, these are the two factors s𝑠𝑓 about 0.95, 𝐾𝑤 is about 0.35, 0.33 like that. 𝐴𝑖 𝑔𝑟𝑜𝑠𝑠 area
is window area is this area and this is the window area, are you getting? 𝐴𝑤 .
So, this will determine the physical dimension of the transformer. Height of the
transformer into this because the moment you know 𝐴𝑖 𝑔𝑟𝑜𝑠𝑠 you know these things. So, it
is related, but what I am telling. So, these are the you decide the physical dimension of the
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transformer. It is going to decide that is all ok. So, this is the output equation of the
transformer. Similarly, for the three-phase transformer it can be also derived, that we will
do later, but for the time being look at the single phase equation and listen carefully what
I am telling.
A transformer when it will be operating at full load condition your 𝐵𝑚𝑎𝑥 will be at rated
value I told you rated voltage, similarly your current density also will be highest than
corresponding to rated current. At that time the full load losses will take place and finally,
that is going to decide what will be the temperature rise.
Now, I am telling you that suppose I increase let all the linear dimensions of the
transformer is increased by a factor of 𝑥; all the linear dimensions are increased, keeping
specific magnetic loading 𝐵𝑚𝑎𝑥 and 𝛿 constant, got the point? All the linear dimensions
are increased and 𝛿 I will not touch because that is the capacity of that iron CRGO material,
1 Tesla and 𝛿 for copper say 3 Amp/mm2 and so on.
So, if you increase all the linear dimensions by a factor of 𝑥, what do you think the size of
the transformer will be? So, size KVA of the transformer will increase by a factor of what?
Look at this equation linear dimensions have been increased by factor of 𝑥, so, areas will
be increased by a factor of 𝑥 2 . So, there are two areas, this area will increase by a factor
of 𝑥 2 , this also will increase by factor of 𝑥 2 . So, your initial KVA was 𝑆. So, it will become
𝑥 4 𝑆, is it not? So, will be increased by a factor of 𝑥 4 .
At least from this I can say that your lab transformer same CRGO you are using 𝐵𝑚𝑎𝑥
same copper you are using 𝐵𝑚𝑎𝑥 and 𝛿 is same. So, you imagine a large transformer is
nothing, but you have increased the physical dimensions by several factors because no
more copper can be accommodated now, keeping of course, stack factor same, windows
space factor same more copper you are using definitely and your size of the transformer
will increase by a factor of 𝑥 4 .
As I told you earlier, that when heat will be generated, that heat will be also dissipated
through its exposed areas and when the rate at which it is generated within the transformer
matches with the rate at which it is dissipated out to the outside world, then constant
temperature will be attained.
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Now, the question is I have increased all the I am imagining all the dimensions are
increased by factor of 𝑥, 𝐵𝑚𝑎𝑥 and 𝛿 constant, stack factor, windows space factor
remaining same, then I conclude oh transformer rating will be increased by factor of 𝑥 4 .
Now, what will be the areas increased by what factor? Areas which ever dissipating energy
area will be because linear dimensions I have increased by a factor of 𝑥.
Areas will be increased by a factor of 𝑥 2 that is area through the areas only exposed areas
only it was dissipating therefore, that area increases by a factor of 𝑥 2 . Now, we have to
examine by what factor losses will increase you know and I think I will continue next time,
but please go through this one. This topic is very interesting.
It will give you now an fair idea how a practical power transformers is going to look like
and what are the implication of 𝐵𝑚𝑎𝑥 and 𝛿 and why elaborate cooling arrangements are a
must for a large sized transformer. Otherwise what happens I may feel a small transformer
without any extra cooling arrangement, air natural cooling is sufficient and it is working
then what happens if the size of the transformer is very large that is what we are examining.
Ok you increase the physical dimensions of the transformer then you see size KVA ratings
will increase by a factor of 𝑥 4 , giving you same large KVA rating. Then at least I know
this much area will be increased by a factor of 𝑥 2 .
In our next class, I will show you that the losses in the transformer that is copper loss and
iron loss, losses will increase by factor of 𝑥 3 that will show to show and you can also think
yourself to show that losses will increase by a factor of 𝑥 3 . So, these informations are
essential to draw conclusions about whether extra cooling arrangement is to be done or
not. So, we will continue next class.
Thank you.
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Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 52
Cooling of Transformer & Fillings of Transformer
(Refer Slide Time: 00:24)
Welcome to Electrical Machine – I course and we will have been discussing about some
general topics of transformer like what is it is output equation and what are the implications
of increasing the physical dimensions of the machines.
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(Refer Slide Time: 00:52)
So, what we did is this we started very quick review is necessary. So, that you understand
that in a transformer have starting from the KVA equation we define two things one is the
specific magnetic loading and current loading. So, same volt ampere product that is the
KVA can be expressed in terms of what is called 𝛿 and 𝐵𝑚𝑎𝑥 and do we define two factors
one is called stacking factor (𝑠𝑓 ) which is 𝐴𝑖 𝑔𝑟𝑜𝑠𝑠 if you multiply with this stacking factor
which will be over 90% and above because of laminations that is the 𝐴𝑖 𝑔𝑟𝑜𝑠𝑠 and 𝛿 is the
current density of the conducting material used say for copper.
So, 𝛿 remains fixed in primary and secondary because in both the windings we are using
copper, then what our aim was to show that the KVA rating of the transformer is
proportional to the physical dimension of the transformer. Physical dimension who
decides? The core area and the window area; over all dimension gets a determined by
𝐴𝑖 𝑔𝑟𝑜𝑠𝑠 and 𝐵𝑚𝑎𝑥 .
And this two factors stacking factor and window space factors window space factors is
about 30-35% like that because all these space in the window cannot be covered with
copper because there must be space for circulation of conducting oil I mean which will
take away heat from the windings ok. There will be also space required by the insulation
which will be covering the conductors and so on. So, about 35% of the space window
physical area will be occupied by copper. So, in terms of that for a core type transformer
we got this KVA rating. Note down this factor 2.22 ok.
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Now, see if you have a transformer in your lab may be 1 KVA, 3 KVA transformer you
will find no extra cooling arrangement is necessary because of the fact the natural cooling
by air comes in contact with the coils everything is open windings you can see only
terminals come out from the LV, HV winding and heat generated in the coil and the core
will be dissipated in the air. And I told you the rate at which heat will be dissipated and
the rate at which heat will be generated, when this two will be equal transformer will attain
a constant operating temperature ok.
So, it is the temperature rise which is important ok. If temperature rise should not exceed
the desired level of temperature rise. For example, I say that materials used is such that
temperature rise above ambient should not be more than say 75° Celsius ok. So, maximum
temperature rise is fixed decided. Therefore and no extra cooling arrangement will be
necessary for very low KVA transformer.
However, what I was trying to tell here this equation is called output equation, mind you.
This is output equation of a transformer this one. Now, you imagine that this is the physical
dimensions 𝐴𝑖 𝑔𝑟𝑜𝑠𝑠 , 𝐴𝑤 etcetera that along with this specific magnetic loading and electric
loading decides the KVA. Now, imagine I will increase all the linear dimensions by factor
of 𝑥; where 𝑥 > 1 then what will happen is this the areas KVA of the transformer will
increase by a factor of 𝑥 4 .
I will keep same 𝐵𝑚𝑎𝑥 and 𝛿 because core material same I will be using, same copper I
will be using therefore, 𝐵𝑚𝑎𝑥 and 𝛿 are same stack factor and window space factor will be
same and therefore, if you increase all the linear dimensions by factor of 𝑥 area terms will
increase by a factor of 𝑥 2 . So, (𝑥 2 × 𝑥 2 ) = 𝑥 4 therefore, KVA rating of the transformer
will increase by a factor of 𝑥 4 if all linear dimensions are increased.
So, this I must have written somewhere KVA of the transformer increased by a factor of.
Now and area surface area overall surface area through which heat is dissipated into the
atmosphere that will also increase by a factor of 𝑥 2 . Then, I stopped here we went up to
this point we can easily show that losses in the transformer for this bigger transformer
whose KVA rating is now 𝑥 4 𝑆 if 𝑆 was the original KVA rating the losses will increase
by a factor of 𝑥 3 . Why?
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(Refer Slide Time: 07:17)
It is simply because of this see the total copper loss just to show
𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑝𝑝𝑒𝑟 𝐿𝑜𝑠𝑠 = 𝐼12 𝑟1 + 𝐼22 𝑟2 = 𝛿 2 𝑎12
𝜌𝑁1 𝑙𝑚1
𝜌𝑁2 𝑙𝑚2
+ 𝛿 2 𝑎22
𝑎2
𝑎1
= 𝛿 2 𝑎1 𝜌𝑁1 𝑙𝑚1 + 𝛿 2 𝑎2 𝜌𝑁2 𝑙𝑚2
𝜌 is the resistivity of copper. Length of the copper will be proportional to the number of
turns and average length or mean length of the turns because you know this is the core
suppose turns will be like this, several turns like this.
So, mean length of one turn you take that is 𝑙𝑚 and multiply with it is number of turns that
will give you the estimate of the length of the copper conductor in the primary or secondary
𝜌𝑙
side. So, 𝑎𝑚 you remember cross sectional area of the conductor therefore, this is the thing.
So, on the top what you have physical dimension wise length into area, is not? Here also
length into area. So, copper loss is proportional to the volume of copper; volume of copper.
So, copper loss will increase by a factor of 𝑥 3 , is not? Volume means three lengths
multiplied and we are increasing each linear dimension by a factor of 𝑥. So, copper loss
increases by a factor of by a factor of 𝑥 3 . Similarly, core loss for core loss it is much more
simpler because we have already seen the core loss comprises of eddy current loss and
2
hysteresis loss and eddy current loss is proportional to 𝐵𝑚𝑎𝑥
𝑥 2 𝜏 2 where 𝜏 is the thickness
of the each lamination and that is per unit volume or kg of copper.
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Similarly, hysteresis loss the area this area it gives you hysteresis loss a measure of
hysteresis loss it is also per unit volume per unit cycle. Therefore, core loss will also
increase by a factor of 𝑥 3 . Therefore, total so, to summarize if this was your original KVA
rating this is the total loss and this is the surface area through which heat is dissipated say
we write this. So, 𝑆, 𝑃𝑇𝑜𝑡𝑎𝑙 𝐿𝑜𝑠𝑠 and this thing let me write 𝑆𝑆𝑢𝑟𝑓𝑎𝑐𝑒 some effective surface
through which heat is dissipated.
Now, if you increase the dimension by linear dimensions by 𝑥 times KVA rating will
become 𝑥 4 𝑆 total loss will become 𝑥 3 𝑃𝑇𝑜𝑡𝑎𝑙 𝐿𝑜𝑠𝑠 , where 𝑃𝑇𝑜𝑡𝑎𝑙 𝐿𝑜𝑠𝑠 when the KVA rating
was this 𝑃𝑇𝑜𝑡𝑎𝑙 𝐿𝑜𝑠𝑠 and your surface area this will increase by a factor of 𝑥 2 ; 𝑥 2 𝑆𝑆𝑢𝑟𝑓𝑎𝑐𝑒 ,
got the idea? So, if you imagine linear dimensions I have increase same magnetic loading
same electric loading this is the thing. Then you can easily see the loss increases by a factor
𝑥3𝑃
𝑃
of 𝑥 3 . So, here in this case 𝑆𝑇𝑜𝑡𝑎𝑙 𝐿𝑜𝑠𝑠 this ratio and here the same ratios that is 𝑥 2 𝑆𝑇𝑜𝑡𝑎𝑙 𝐿𝑜𝑠𝑠 =
𝑆𝑢𝑟𝑓𝑎𝑐𝑒
𝑆𝑢𝑟𝑓𝑎𝑐𝑒
𝑃
𝑥 𝑆𝑇𝑜𝑡𝑎𝑙 𝐿𝑜𝑠𝑠 , got the point? And this becomes 𝑥.
𝑆𝑢𝑟𝑓𝑎𝑐𝑒
Therefore, what we note it here is that losses increases by factor of 𝑥 3 , but the surface
through which it heat will be dissipated is only increased by factor of 𝑥 3 . Therefore,
temperature rise will become more now, because available area through which heat will
be dissipated out to the atmosphere is become less loss increases by a factor of 𝑥 3 . If your
surface area would have been increased by the same factor 𝑥 3 then the available surface
to dissipate a certain amount of loss will be same if it was 𝑥 3 , but it is now more because
𝑥 > 1 you have to dissipate more power loss through a lesser surface that is what I want
to say.
If I say that in the original transformer with this specification temperature rise was within
the limited that is say 70° celsius temperature rise. In this case you will find temperature
rise will become more, same materials you are using. Therefore, you know extra cooling
arrangement is now necessary to keep the temperature rise within the same limit compared
to this transformer.
So, this is why a transformer needs extra cooling arrangement for large transformers. For
small transformer suppose you say for 5 KVA transformer temperature rise is 70° celsius
1
or what if I reduce the dimensions linear dimensions by factor of say 𝑥 = 2 say smaller
506
size or you do not require any transformer because more surface area will be available and
your total loss will be reduced if 𝑥 < 1. We are discussing about 𝑥 > 1 for large sized
transformer what happens.
(Refer Slide Time: 16:19)
Therefore, extra cooling arrangement is a must extra area available to dissipate area
available to dissipate the losses in a large transformer will be less. Hence temperature rise
will be more; will be more got the point always it will be. So, if temperature rise allowable
temperature rise is a known to you then will say that if the transformer is higher ratings I
must adapt some other means there.
That is why what is done is this transformer core transformer core for single phase I am
telling, where your windings will be here LV winding red colors are LV windings and
suppose these are HV winding. Similarly, on this side because LV windings are also
distributed in the two limbs in a practical transformer not that LV is wound on the same
limb and this is your HV coils these are all circular around these this way. So, this is LV,
half of the LV turns there and then this is HV green one is HV, got the point?
Now, this is the transformer and this whole thing can be air natural cooling earlier it was
just air natural cooling, but here you will say that maybe I will immerse these this
transformer in a tank filled with oil mineral oil refined mineral oil. Everything will be
immersed in that and so, this is steel tank and it will be immersed in oil and then the
terminals will be brought out from the top surface of this transformer.
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So, here is no space you understand in my last lecture I showed you some diagram very
roughly sketched not. So, well a diagram, but none the less this is the it is tank inside this
transformer and terminals will be brought out through bushings through the conductors
LV side, HV side. You can easily make out which one is HV if the insulation level is more
that is HV and this is the LV side LV side thicker conductors anyway. So, it is immersed
in oil and it will go.
So, this is called the oil natural cooled here. So, filled with oil. So, oil when it comes in
contact with the core and the windings they will be heated up it will go up and it will
circulate and they will carry the heat to the surfaces through which heat will be dissipated.
I told you mind you that the final temperature rise can be easily expressed in some very
easy way.
For example, the loss in the transformer is 𝑃 rate at which heat is generated is 𝑃 ok. And
this should be equal to
𝑃𝑙𝑜𝑠𝑠 = 𝑆𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝜆𝜃𝑡𝑒𝑚𝑝
What is 𝜆? Emissivity that is how much power it is dissipating. So, this is called emissivity
and this is surface, mind you not KVA. When this two are constant then only final
temperature will be attend. If loss is taking place at a much faster rate than the rate at which
it is being dissipated out to the atmosphere, temperature will go on rising like an R-L
circuit it can be shown, but we will not go to that extent.
What I am telling
𝜃𝑡𝑒𝑚𝑝 =
𝑃𝑙𝑜𝑠𝑠
𝑆𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝜆
That is why I was telling if your surface area available is less temperature rise will be more
can be easily seen from this also ok. In any case oil will take the heat, it will come to the
tank, it will dissipate like that sometimes you may require to do something extra. For
example, you can to make the surface area more what you do you connect some tubes like
this tubes here a series of tubes cooling tubes they are called and so on the surfaces.
So, oil will go up and it will have also natural convective currents it will improve the
cooling. So, heat will be coming to the cooling tubes mind you, it is not only one it is just
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one I am seeing behind this along all the surfaces you can connect. On the top surface and
bottom surface of course, no heat is assumed to be dissipated because so many fittings will
be there on the top. So, these are cooling tubes cooling tubes.
So, effectively you have to increase the surface area because you go to higher and higher
rating of transformers, your 𝑃𝑙𝑜𝑠𝑠 increases, but surface area through which heat will be
dissipated should be increased then only temperature rise you can keep to a desired limit.
Now, this is one thing for example, say any distribution transformer if you look at when
you are walking along the street pole mounted transformers you will find they will be good
enough cooling tubes are there and it is there.
(Refer Slide Time: 25:44)
Now, another thing I will tell these are tit bits, but it is better you know that this is therefore,
the tank here, is not? What are the fittings you will expect for distribution transformer say.
So, there will be cooling tubes here all along effectively increasing the surface area through
which heat will be I am not drawing dot dot dot series of cooling tubes. On the tops there
will be bushings as I told you for HV and LV to bring out the terminals, that is also fine.
So, LV and HV terminals are brought out and these are the conductors leads from the
windings. Apart from that you will find there is another fittings here which is called
conservator tank a cylindrical tank and it is connected to the through a piping here. And
this tank is filled with oil and through this piping it is also connected. When we when you
fill up fill it up with oil will also raise here it will fill the tank then it will go up here because
509
there is a connection through piping and then this is called conservator tank very
interesting conservator tank.
And so, up to a certain level this oil will be filled up and above this oil there will be air.
This is the oil-air interface. Now, what happens is this oil it is a very good dielectric
property also may be 27-28kV per cm or mm I have forgotten you just see the books the
dielectric strength. So, it will insulate the it will provide very good insulation between the
coils and the tank; tank is you know steel tank. So, it provides insulation also among the
turns as you go from one turn to the on another there is voltage differences, there exist
voltage per turn.
Therefore, a good insulating material you have provided. Air is also good insulator, but
this oil will be much better. So, oil does two things; it helps cooling, it carries the heat
from the core and winding of the transformer to the surface area heat is dissipated through
cooling tubes other things, but the quality of the oil when it comes with moisture it
deteriorates that is why it is filled with oil and interface with air takes place here.
Now, let us try to understand what will happen suppose the temperature is operating under
I transformer is operating under no load condition, loss is very less. Temperature rise will
be very less, is not? Now, you imagine and every all oil levels are here like this now you
imagine that you have connected load on the transformer only core loss was taking place
when the transformer was under no load condition.
Now, through the secondaries you have connected it is delivering full load, full KVA in
that case copper loss comes in. So, more heat will be generated and temperature will rise
if temperature rises oil is a liquid it will expand and this level of oil will go up pushing the
air whatever it was above it will be pushed up, is not? And if you do not bother go on
overloading level of oil will go up and down. So, there must be some interface between
air it should not be sealed at the top. So, what is done another interesting fitting is
connected.
Here from the say top you take out another piping here ok. These are very interesting thing
and this pipe is also filled with air. Level of oil will change as degree of loading will change
and here is a sort of small vertical vessel like a bottle; here there will be perforations and
this thing will be filled up with a material called silica gel, a chemical material silica gel.
I do not know spelling you check silica gel and there will be perforations. So, this
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transformer oil is in touch with the atmosphere via this tank air, this air this is filled with
air and this air through silica gel through this air.
Now, why this silica gel is poured? Suppose, you have not put silica gel, it is just like this
then what will happen? As degree of loading changes the air will be pushed out. Suppose,
you are increasing load, level of oil will increase air will come out and when the
transformer is operating a very light load condition oil will drops air, air will be sucked in,
is not? That is the transformer is also breathing as human being breathes depending upon
the degree of loading.
And this atmospheric air if it contains moisture that moisture will be sucked in and it will
get in touch with the oil and therefore, oil property will drastically deteriorate. It will form
sludge, thicker it will become and its dielectric property will also decrease. So, the this
particular thing is called breather.
Breather it is called that is as human being breathes similarly transformer breathes. It takes
air; it also pushes out air through this device. Now, what is this silica gel? Silica gel is a
protection. It make sure that water vapor present in the atmosphere does not come here in
the conservator tank. Why? Silica gel, the normal color of silica gel color is blue; it is in
granules you know some granules, blue. Whenever air will be sucked in the color, the
moisture will be absorbed by silica gel and its color will change to pink. When it absorbs
moisture color changes to changes to pink and of course, so no moisture is allowed here it
will go up ok.
So, this is how silica gel will prevent entry of moisture into the transformer tank and come
in contact with oil try to deteriorate it, but during rainy season and all this will soon
become, see it will be a very important job for supervisors whoever is watching this
transformer to inspect the this breather all the time whether color has become totally pink;
that means, it has become saturated with moisture. Moisture will be then going here;
therefore, he will replace this silica gel with new silica gel and once again that pink colored
silica gel you dry a bit it will become blue. It can be reused, but these are the things I
wanted to tell ok.
So, transformer has got a breather and this is the air-oil interface and another fittings will
be there on the tank of the transformer that is called explosion vent. I must tell this because.
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I am showing it here it will be properly positioned, but you will see another structure like
this and this is also a sort of vertical pipe bent like that.
And it is not open to air there will be some diaphragm connected here and on the tank
surface it is fitted. But, this tank as I told you it is made of steel, here also a not strong steel
material is used a rather little sheet of iron or steel is used while fixing it, got the idea? The
steel thickness of the tank is high I mean thick quite, but it is only little thickness ok.
Therefore, no oil neither air is allowed to be coming in because here also it is fixed. This
end is also fixed by a plate.
Therefore, no question of oil communicating with air through this then what purpose it is
given? It is called explosion vent it is like a safety valve in your pressure cooker. What
happens is this if suppose the transformer becomes overloaded some fault has occurred,
high short circuit current is flowing, heavy losses are taking place, oil is going up here, but
this place is a rather weak in it is mechanical strength. So, oil will oil pressure inside the
transformer builds and it will break this and this oil will gush out from this place.
And the direction of this explosion vent is made in a particular angle where you will keep
sand etcetera, so that fire does not breakout because the that will be very hot oil and it
causes fire got the point. Therefore, explosion so, this plate diaphragm is also weak, this
is also weak if no protection mechanism works oil will break this thing because of its own
pressure heat generated and oil will gush out and you provide a directed path on the floor
you keep some sand etcetera so that oil will be coming here. You must have read in
newspaper sometimes in some locality transformer has exploded because of overloading
or some protection mechanism did not work and it caused fire.
So, anyway this is explosion vent which acts like a safety valve and only last point I must
tell you that what happens is this between this conservator tank and this tube here a relay
is connected relay which is called Buchholz relay ok. When oil will be heated oil will push
out, density will fall, there will be some floats inside, they will move through that
mechanism you can initiate some operation or it can give you some alarm. So, that things
are not good for the transformer, oil is almost vaporized here. You know float like your
toilet float in the this one.
So, it will go up because density has fallen and that movement is used. So, we are not
discussing that, but I will just mention only there is a place in this connecting tubes there
512
will be a Buchholz relay connected and it will be operating and give you alarm or if
necessary it will disconnect the transformer you will read those things in power system
protection. So, this in nutshell will be the general fittings one is the conservator tank
bushings.
In case of three-phase transformer three-bushings will come out and on the name plate
there will be a plate KVA ratings this that will be written and there is a breather here silica
gel which is through which air will be coming in when transformer inhales because during
no load condition oil will become cooler, its level will fall air will come in and here is a
gate which will not allow moisture to enter. And the color of silica gel will change from
blue to pink and it will arrests the moisture air, but regular inspection is necessary to see
that silica gel color has become all pink, then you take out this sack in which silica gel
which are granules dry it up. So, we will continue with this next time.
Thank you.
513
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 53
Output Equation of 3 Phase Transformer
Welcome to 53rd lecture and we have been discussing some general topics on transformer,
not so much mathematical. So, we started with telling that for large and large transformers,
by large transformers I mean very large KVA transformers. Air natural cooling is not
sufficient, and you must have at least in distribution transformer hundreds of KVAs may
be 250 KVA or 500 KVA these are the fittings you will expect
(Refer Slide Time: 00:39)
Of course, this is for single phase transformer I have drawn; it could be for 3 phase
transformer as well and these are the usual fittings, cooling tubes will be there. And mind
you the maximum temperature rise is decided by 𝑃𝐿𝑜𝑠𝑠 divided by surface through which
roughly this is the thing into emissivity (𝑆
𝑃𝐿𝑜𝑠𝑠
𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝜆
of the material sum.
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), 𝜆 is called emissivity which is constant
(Refer Slide Time: 01:30)
So, the cooling’s are classified on the name plate of transformer, it may be written like
A.N. it means air natural cooling; air natural or very small transformer, no extra elaborate
arrangement is necessary. There may be A N O L cooling, air natural and oil natural. There
are lot of things combinations of this. Now there may be very big transformers ok, then
cooling tubes is also not sufficient; for example, transformers of ratings of 100 MVA
which are used in power stations, very large transformers 3 phase.
In that case what you will find, it will be there are several like oil forced, air natural this
kind of thing, but you go through books you can easily make out what do they mean. But
what I am telling, suppose this is your transformer this is the main transformer tank
bushings, etc all things are there and to increase the surface area the cooling tubes are also
not sufficient, what they then connect is called radiators.
Radiators are nothing but iron plates with tubes like this. Just I am drawing one, it is iron
plates and several things and these are rectangular type looking; iron, steel and this oil here
is connected to the radiator tubes. So, these are radiators instead of cooling tubes radiator,
large surface area and at this is connected here like this; there will be inlet air, oil it is
connected. So, oil this is filled with oil inside, this oil is connected here, oil; and oil will
fill up this plates also, these are the open space.
So, and then oil will be pumped in and it will once again come out to this connected here
to the tank. So, oil will go through this radiating tubes all along and will come here. Now
515
oil on it is own will may not go, for large transformer what they will connect a pump, oil
pump; oil will be forced to go there, go up and then return to the main tank. So, these are.
So, artificially you increase the surface area and this radiators may be at a, for a place I
mean quite at some distance the radiator plates are there.
Then it is called oil forced cooling, oil forced. So, idea is very clear, if you want to use
very large transformers you must do extra to cool the transformer, so the temperature rise
of the transformer is within the limit. Because whether it is very small or large transformers
the value of 𝐵𝑚𝑎𝑥 magnetic loading is approximately same may be 1.2 Wb/m2; if you are
using copper almost 𝛿 is same may be 3Amp/mm2, those are fixed generally.
Therefore you since you have gone for higher KVA rating, the size of the transformer has
increased and you must have elaborate arrangement. Artificially increase the surface area
of cooling and heat will be radiated not only from the usual tanks here, may be tubes also
connected natural circulation of the oil here; but also force the oil to move through the
radiators to make a closed path and the oil will be circulating. Sometimes air is also forced
for big transformer, you keep fans here; fans which will also cool the surface area of the
radiators; forced air then say.
So, these are terminologies which you will easily understand, but the basic understanding
of the whole process is that for large transformer, you have to have some elaborate cooling
arrangement for the transformer; for distribution transformer simple oil, for large
transformer it will be like this, may be hundreds of MVA transformer which is housed in
power stations ok. So, this in nutshell about the cooling; only one thing I will leave it as
an exercise to you, the what about the output equation of a transformer?
516
(Refer Slide Time: 08:20)
Output equation of a 3 phase transformer; 3 phase core type transformer in the same way
it can be found out; I leave it to you to find it out. That is you remember I leave it as an
exercise to you this is the thing. In 3 phase transformer there will be two windows like
that, there will be say LV winding on each limb and for each phase A phase B phase LV
winding and C phase LV winding; and HV winding will be here got the point. So, LV
windings LV windings, there are two windows and this two windows are equal dimensions
because old design.
So, here only thing, so you start with this thing that is KVA rating of the transformer is
𝑆 = 3𝑉𝑝ℎ 𝐼𝑝ℎ × 10−3 = 3 × 4.44𝑓𝐵𝑚𝑎𝑥 (𝑠𝑓 𝐴𝑖 𝑔𝑟𝑜𝑠𝑠 ) 𝑁1 𝛿𝑎1
You know this is how it will look like. So, 𝛿 is the electric loading, 𝐵𝑚𝑎𝑥 is the magnetic
loading. Now in this case, this once again 𝑁1 𝐼1 = 𝑁2 𝐼2 per phase mmf balance will take
place, then
𝐼1 = 𝛿𝑎1
𝐼2 = 𝛿𝑎2
So,
𝑁1 𝑎1 = 𝑁2 𝑎2
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I am telling verbally what you have to do, only thing is window space factor.
Window space factor is go to any window area covered by copper. So, suppose I say that,
it has got 𝑁1 turns and this has got 𝑁2 turns secondary so; obviously, you can see there are
two sections unlike. So, it will be what; area covered by copper
𝑘𝑤 =
2𝑁1 𝑎1 + 2𝑁2 𝑎2
𝐴𝑤
And then 𝑁1 𝑎1 = 𝑁2 𝑎2 you substitute and for 𝑁1 𝑎1 you put it there. What you will get it
is this, the total KVA will be
𝑆 = 3.33𝑓𝐵𝑚𝑎𝑥 𝛿𝑠𝑓 𝑘𝑤 𝐴𝑖 𝑔𝑟𝑜𝑠𝑠 𝐴𝑤
Note that for single phase transformer it came as 2.22, but in 3 phase it is 3.33.
So, other reasoning remaining same, same if dimensions are increased by factor of 𝑥 losses
will increase, by factor of 𝑥 3 and so on, area will increase by factor of 𝑥 2 , KVA rating will
increase by factor of 𝑥 4 and so on, so this is the thing.
(Refer Slide Time: 14:29)
Another tit bits of this transformers in terms of single phase I will just tell you, is that if
you see a leaned crossed section like this, this is the iron cross section your LV winding
will be around it; this is the LV winding and then your HV winding.
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Oh I am; so sorry, badly drawn and this is your, this blue is your HV winding, HV and this
is say LV; because nearer to the core LV winding is present mind you, is not. And this is
the top sectional view of the transformer, you now understand this is 𝐴𝑖 , this is 𝐴𝑖 hide on
area; first LV winding you wound then HV winding and so on.
Now you note one thing when the transformer will be operating at any given point of time,
we have seen that if LV winding is carrying current in this direction what will be the
direction of current in the HV winding; it will be just opposite are you getting. That is how
mmf will be balanced any, forget about that no load current, no load current is small 5%,
when secondary load is not there this current is absent only this current will circulate that
is 5%. But the moment you load it 𝐼2 then 𝐼2′ too will appear, that is if HV side is your
secondary 𝐼2 opposite current; then only this 2 mmf will balance each other and your
original flux will be restored and decided by the magnetizing current, that is what we are
telling. This will be read like this.
If this is the situation do you expect that there will be some mechanical forces acting
between these two coils? Yes, we know that if two conductors you keep side by side, they
will experience mechanical forces. What will be the nature of the forces? It is just opposite
to magnetism; in opposite current meant attractive not like that. It will be repulsive; that
means, this LV coil, now this LV coil just cannot stay alone there will be some bobbing
sort of thing over which this LV winding is put. So, when normal load current flows, but
none the less there will be force of repulsion, acting that is what will be this direction of
the force this way.
That is LV winding will be squeezed towards the core and this will try to go away. So, you
have to take proper fixing up of the coil. So, that they can withstand that mechanical forces
at rated currents at least. But what happens is this you do not know there might be short
circuit taking place and very large current is flowing, then this force will become enormous
and HV windings will become loosened, come out and this one.
That is why protection of transformer is so important, you cannot allow a very large short
circuiting current to flow apart from the fact the winding may be spoiled. But you know
to make a winding burn that large current must be there for sufficiently long time; it is not
𝑖 2 𝑟; it is 𝑖 2 𝑟𝑡 that decides how long that large current is flowing. But the mechanical forces
too will be very large, the short circuiting current may be high and transformer will not
519
only be ruined by if the fault is not clear; ruined by windings will be spoiled, mechanically
also it will be a disaster. LV winding will tries to push upon on all directions and it will
try to go out HV winding. This is just I am mentioning, you keep now after learning these
things those things can be very nicely explained. And current in the windings will be
always in opposition and this is instantaneous current deduction I am showing, when it
will be these ways, that will that way; but the deduction of force will be always like that.
Another last point I will comment I will make, sometimes what happens in transformers,
what if you load the transformer you know because of regulation voltage decreases. See
you load the transformer; connect the load we have seen that this is suppose LV side, this
is suppose HV side; you load the transformer or I will draw other way to introduce the
problem. So, this is suppose HV, this is LV side it does not matter in any way this is load,
you load the transformer; then what happens under no load condition whatever was the
voltage when you load it voltage will fall. If the load is inductive RL type and that is
usually the case is only for capacitive load may be it will increase a bit; but for RL type of
load is which is generally the case, the terminal voltage magnitude will fall.
So, compare to open circuit voltage, the moment you start drawing current because of the
internal drop inside the transformer in the equivalent resistance and leakage reactance, this
voltage will fall magnitude. And suppose you want to restore back the voltage, you cannot
do anything here; two ways you can do, one thing is you increase the supply voltage a bit;
the supply voltage whatever is coming that voltage you increase. So, that it will may
compensate that, but that is not a option left to you supply voltage whatever is coming,
coming you cannot do anything. So, what can be done is some transformers are provided
with tap changers.
Suppose primary number of turns 𝑁1 secondary turns is 𝑁2 ; idea is very nice and simple.
So, what I am telling, you suppose say that I will there will be some tapings provided here
1, 2, 3, 4 turns tapings are you getting, LV I am drawing elaborately; there are this is the
normal secondary terminal, then you are provided with extra tapings with a switch, this is
the secondary terminals.
Now a transformer with some extra turns here on the LV side and a switch connected, you
can increase the number of turns. So, what is the point here 𝑉1 number of turns is 𝑁1 and
𝑁2 number of turns is not fixed, a little change I will be able to do. In a transformer what
520
𝑉
I told, voltage per turn remains constant. So, what is voltage per turn 𝑁1 , you are simply
1
and this is 𝑁2 and this is (𝑁2 + 𝛿𝑁2 ) some few turns you increase, then your voltage will
increase on the load side, getting.
But the turns ratio that thing will not be affected too much, is not if you increase these by
𝑁
𝑁
1
few turns what 𝑁1 practically remain (𝑁 +𝛿𝑁
, got the idea. So, transformers are provided
)
2
2
2
with tap changers, have I written. And tap changers I have shown here in this example
listen to me carefully, on the LV side I have made an arrangement to increase the turn,
what should I do? Applied voltage is fixed; when there was no load this switch was open,
this voltage was fine this is my actual secondary. But when you connect a load here this
voltage falls a bit, you want to compensate you cannot do anything with this supply voltage
here. Therefore, you simply increase few turns voltage per turn is decided by
𝑉1
𝑁1
and
increase 1 or 2 turns you get more voltage. So, regulation problem can be addressed a bit
by this method.
But now the question is listen to me, should I put the tap changer on the LV side or HV
side. See the condition is this tap changer I will be operating under load condition that is I
will put, it was supplying the load. So, you find all across the load people are complaining,
voltage is small, there is a switch and those will be also oil immersed I hope and you put
it there, ok. So, that is another interesting thing how tap changers work.
But the idea I am telling, you will move it here and these are called on load tap changers.
There may be off load tap changers, what you can do, you switch off the supply first; then
bring this switch from this point to this point and once again energized, but that is not a
very good idea. So, only load tap changers are much common, I mean useful; you put it
here increase the turns a bit, increase the turns a bit in order to restore the voltage desired
voltage.
But the question I am putting, should were the tap changers should be tap changers on load
tap changers should be on which side; should I connect HV side or LV side. For example,
here while explaining the things I have shown it is on the LV side I have connected, is it a
good idea? No; because LV side current is more. So, whenever you will be moving this
switch from this point to this point, you are trying to break the current and quickly there it
will be short circuited this turn a bit. So, tap changers will be much more stressed;
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therefore, tap changers should be always on the HV side. Tap changers better connect,
because you have to make and break current which is large, load is already connected you
cannot do anything. Particularly with on load tap changer, hundreds of current is flowing
you suddenly want to move this one from this two, increase the voltage a bit.
Can I then do the same thing, if I connect the tap changers on the HV side; answer is on
HV side. I will not connect tap changers on the LV side, whether it is load or not that is
different issue, you must understand. But since I have drawn the load on this side, connect
switch; here, what will happen is this, this is the normal 𝑁1 turns I connected this whole
𝑉
thing is 𝑁1 turns, 𝑉1 is the supply voltage, voltage per turn is 𝑁1 and this is the switch,
1
under no load I was getting a voltage here. But when I connect load I find because of the
regulation voltage drops which is not desirable; then I plan I will play with number of turns
on the HV side, not with LV side because current is more.
So, what should I do, because this supply voltage is fixed I cannot do anything what should
I do? What should I do with 𝑁1 , should 𝑁1 be increased or decreased; voltage per turn
should be increased, then only secondary voltage. Secondary voltage I am not touching it
is 𝑁2 therefore, tapping should be here; this is 𝑁1 then lesser number of turns. So, your
switch should be connected from here to there, you reduce the number of turns of the HV
side same applied voltage; therefore, voltage per turn increases and it is the voltage turn
which remains constant in primary and secondary side. So, secondary voltage will
increase, N 2 I will not touch, got the idea.
Therefore it is interesting, you I will leave it to you to analyze this. So, 𝑁1 must be
decreased and you see it is nicely matching with the construction of the transformer. This
is the core LV winding, then comes your HV winding and HV windings tapings you want
to take and it is on the outer side; you easily take the taping, it is difficult nah to take
tapings from LV side from in between. So, these are very small, but very interesting thing,
you we have done mathematics we know how to calculate regulation efficiency.
But I think whatever I have told you it will give you a practical flavour of the way you
look at transformers ok; mathematics is there, but these are small, but important things.
So, there should be transformers are provided with on load tap changers also. And mind
you the number of turns you will change a little bit not more. I have not told great things
about design, but that is the first starting point you must have a broader idea and when you
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look at a transformer you may find a big transformer may be provided may not be provided
with a tap changing arrangement.
On load tap changing arrangement means that; tap changers must be on the HV side as
current is less and also it is easily accessible, taping is to be taken from the this HV. And
then it is not that you always increase the number of turns to increase the voltage, if this
side is your supply side you have to reduce the numbers of turns to increase the voltage on
the LV side. Suppose I say that this is LV side and on HV side is your load, this is 𝑁1
where you are applying a voltage 𝑉1; voltage per turn is
𝑉1
𝑁1
, LV side I am not going to
connect any tap changers. So, nothing is changing here voltage per turn is fixed; I have
connected load and find that voltage has fallen compared to the open circuit case. Then I
want to increase voltage and I am certain I have to connect the tap changer on the HV side.
In that case what should I do with 𝑁2 , I must increase it, is not. So, taping will be this way.
So, depending upon the situation you have to reduce the number of turns or increase the
number of turns. On the load side if the HV side is there, so provision for increasing and
decreasing both should be present; you do not know which side will be primary, secondary
like that. So, these is a nutshell what I want to tell about transformers; we will have a
discussion session, there will lot of problems given and I have I am going to upload several
notes whatever I have told in these lectures and we will see that so.
Thank you.
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Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 54
Introduction to D. C Machine
Welcome to our next topic in the Electrical Machines I lectures and we will start discussing
about DC Machines.
(Refer Slide Time: 00:27)
And DC machines and these are rotating machines you know rotating machines; rotating
machines, machines I will often write like this. Now, before I start discussing with DC
machines rotating, we try to understand how DC voltage can be generated? For example,
after the discovery of Faradays in 1830s or so somewhere, which says that voltage induced
𝑑𝜑
in a coil is 𝑛 𝑑𝑡 and if the coil is moved in a magnetic field which sinusoidally varies, then
there will be induced voltage which will be also alternative in nature.
This was quite obvious but, nonetheless people started developing DC machines first and
we will see that it is not so easy because, inherent voltage induced in a coil, when it is
rotated in a magnetic field will be essentially alternating in nature. So, we have to do
something extra to convert that AC voltage to DC, if this problem has been brought today
524
that ok, you want to generate DC one solution perhaps many people will give that generate
AC because, AC generation is so simple and connected rectifier get DC over.
But, in those days rectifiers were also not there at that time. So, it was DC machine people
started thinking that large power I have to somehow before that of course, people knew
about batteries DC voltages only we were used to at very small level of power batteries
were there, 1.5V connect them in series get some voltage like that but, large voltages
delivering large current hundreds of voltages that was not there, then you know DC
machines in any case was developed. So, to understand the basics of DC machine, we start
with this discussion.
You imagine that I have a coil a conductor, this is a conductor and suppose these two red
lines are rails ok, this is a conductor and these two are rails, it has wheels and it is putting.
So, that this conductor can move this way that way, these are rails and let us imagine on
the top of this arrangement, I have put a north magnet all along this line length of the rails
and below this I have put a south pole magnet all along the track, that is from in the roof
top, if it is on the plane of the of my table, on the roof top there is a North Pole and on the
floor there is a South Pole in between it is there, on the table top and there this table runs
infinitely. Now, if that be the case then the direction of 𝐵, I must show cross from top to
bottom. So, it will be say cross 𝐵 ok.
So, 𝐵 direction and it is all along this whole length and also along the breadth everywhere
it is 𝐵, flux density you know. So, this is the flux density 𝐵 from top to bottom and this is
the conductor. Now, let us imagine that this conductor is moving with a velocity say 𝑣 m/s,
this conductor is moving with a velocity 𝑣 m/s from left to right and let us assume that no
friction on the track, and also no air resistance.
Therefore, to move this conductor at a constant velocity 𝑣 m/s, on a friction less track no
air friction also, you require no force therefore, external force acting on the conductor will
be 0 that is what Newton’s told us, to move a thing in a friction less environment at a
constant velocity, you do not require any force.
If no force is acting on any body, then either it is at rest or it is moving at a constant velocity
that is what we know. So, anyway to move the conductor at a constant velocity 𝑣 m/s, I do
not require any external force to be acting, it will go on moving, under these assumptions
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ok, that is fine and suppose this side of the conductor I name it as a and this side I name it
as b, the 2 ends of the conductor. As the conductor moves, there is 𝑣 there is a velocity,
these are at right angle.
So, I can apply right hand rule like this here, they will be showing this. So, you see there
is a 𝐵 here into the paper. So, forefinger is 𝐵, there is a velocity thumb gives you velocity
and this one middle finger, the end of the middle finger will give you the polarity of the
induced voltage, that is a will be plus, b will be minus and magnitude of the induced
voltage will be 𝐵𝑙𝑣, that is equal to potential of a with respect to b, that is how we write,
a will be positive 𝐵𝑙𝑣.
So, and what is 𝑙, 𝑙 is equal to length ab, conductor length. So, this much induced voltage
will be there and it will as it moves with a constant velocity therefore, this conductor
becomes a seat of emf. So, I can say this conductor is like a battery here, a b and this
magnitude of this battery voltage is like this, it will be DC because, always the magnetic
field is into the paper I have never changed it.
And nothing will happen, although across the rail if you connect a voltmeter it will read
𝐵𝑙𝑣 because, it makes contact with the these are also conducting rails, conducting material
although no current can flow. Now, what I will do, I will connect between the rails a switch
and a load got the point. So, if you wish if you close the switch, this voltage which is a
battery across it a resistance will be connected, is it not?
And we expect a current will flow like this, if the switch is closed, current path will be this
is the source, it will go and come back. Let, the initial velocity be 𝑣0 , it was moving with
a velocity 𝑣0 say meter per second and my target is to find out when I close the switch,
how much will be the current and how long will the current flow things like that.
Now, look here the moment this conductor delivers current here, current will also flow
through this conductor, same current series. Now, we know that that a current carrying
conductor placed in a magnetic field, we will experience a force and what is the magnitude
of the force 𝐵𝑖𝑙, where is 𝐵? And you have to apply left hand rule 𝐵 is here, 𝑖 is this finger
and thumb will you give you the direction of the force.
So, 𝐵𝑖𝑙 this much force will act in this direction, opposite to the direction of motion the
moment you close the hence forth. So, shown so, long when the switch was opened as I
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told you, conductor was not experiencing any force whatsoever, it was just moving with
constant velocity as per Newton’s law but, the moment you close the switch although
friction etcetera is not there, I immediately find that it will now experience a force in the
opposite direction, is it not?
Therefore, this conductor ab is acted on by a force, if it is acted on by a force from right to
left and it had a initial velocity of 𝑣0 , what do I expect? Conductor to decelerate because,
no other force is there, velocity will go on decreasing but, if velocity goes on decreasing
this voltage will go on decreasing, this current 𝑖 will also go on decreasing.
Physically I am first telling and so in this way although there will be this resistance will
be heated up, current is flowing 𝑖 2 𝑟𝑡 but, can this how long it will sustain, it looks like that
as time passes, after we have closed the switch at 𝑡 = 0 suppose, as time passes the
magnitude of the current will decrease, magnitude of the induced voltage decrease,
velocity decreases and the time will come when, velocity will become 0 and all this game
will be over, everything will be dead I mean, there will be no velocity of the conductor
and things like that.
Now, what I am trying to tell that, when a conductor had a initial velocity 𝑣0 and without
anybody assisting its motion and it is also not needed when the switch is open, see there
1
was kinetic energy stored in the conductor, 2 𝑚𝑣02 , is not? When the switch was opened,
that kinetic energy was somehow important but, when you close the switch the energy is
getting extracted from it is kinetic energy is dissipating here and that kinetic energy was
finite therefore, finally, it will become 0, all the energy will be dissipated here.
So, these example if I mathematically analyze, we will enhance our this physical
understanding in a much better way. For example, for physical reasons I know, velocity
will become a function of time, the induced voltage 𝑒𝑎𝑏 will become a function of time
and so on. So, I must write down these two equation, equation of motion for the conductor
ab.
𝑑𝑣
What will be the equation of motion? Suppose, 𝑚 is the mass so, 𝑚 𝑑𝑡 , it is velocity is
always from left to right, must be equal to the force acting from left to right, from left to
right what is the force acting? When the switch has been closed is −𝐵𝑖𝑙; 𝐵𝑖𝑙 is acting in
the opposite direction, this is the equation. So, it is a first order differential equation.
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𝑚
𝑑𝑣
= −𝐵𝑖𝑙
𝑑𝑡
Now, the question is what is 𝑖? In a very simple circuit,
𝑖=
𝐵𝑙𝑣
𝑅
where 𝑅 is the resistance so, I put that.
𝑚
𝑑𝑣
𝐵𝑙𝑣
𝐵2𝑙2
= −𝐵𝑖𝑙 = −𝐵𝑙
=−
𝑣
𝑑𝑡
𝑅
𝑅
𝑚
𝑑𝑣 𝐵 2 𝑙 2
+
𝑣=0
𝑑𝑡
𝑅
𝑑𝑣 𝐵 2 𝑙 2
+
𝑣=0
𝑑𝑡
𝑅𝑚
This is the equation which is a first order, constant coefficient, differential equation and I
want to know what is velocity.
So, velocity the solution of this equation is well known characteristic root of this equation
𝐵2 𝑙2
is − 𝑚𝑅 . So, it will be
𝑣 = 𝐴𝑒
−
𝐵2 𝑙2
𝑚𝑅
And there is no forcing function on the right hand side. So, this is the solution and I have
to determine the 𝐴 from boundary condition, see velocity of a mass cannot change
instantaneously, when the switch was opened it is velocity was 𝑣0 , at 𝑡 = 0 + after we
have closed the switch, velocity will be still be 𝑣0 because, of it is inertia therefore, at 𝑡 =
0 +, 𝑣 = 𝑣0 only. So, this gives you 𝐴 = 𝑣0 , what is 𝑣0 ? The velocity of the conductor
when this was opened.
528
(Refer Slide Time: 18:45)
Therefore finally, what I get is this, that velocity of the conductor at any time 𝑡 is
𝑣(𝑡) = 𝑣0 𝑒
−
𝐵2 𝑙2
𝑡
𝑚𝑅
This is the thing and velocity will exponentially decay down to 0, as we were reasoning it
physically so, that is the expression. Then what is the expression and how current will
change, 𝑖(𝑡) is also very simple because,
𝐵2 𝑙2
𝐵𝑙𝑣 𝐵𝑙
−
𝑖(𝑡) =
= 𝑣0 𝑒 𝑚𝑅 𝑡
𝑅
𝑅
This will be the thing, is it not? Correct. So, this is the, it is coming correctly dimensionally
𝐵𝑙𝑣 is voltage by 𝑅. So, this is the explanation of the current.
So, both voltage and current if you sketch, they will decay down to 0, this was velocity of
the conductor, this is time and this is the current. So, current in the circuit at 𝑡 = 0, when
𝐵𝑙
you close the switch, this is coming correctly, that is 𝐵, this current value is 𝑅 𝑣0 , then it
also exponentially decay down to 0, this is time axis. So, this generator will not last long
but, we have understood several things, one is a conductor moving in a magnetic field, it
produces a unidirectional voltage, provided everything is not on the top, everything is
below is South Pole, then when it moves that voltage magnitude of course, becomes a
function of time.
529
Now, naturally the question is, if I want to make a sustainable generator not that after
sometime everything will vanish, what should I do? That means, I want to get a constant
voltage here, with some load being supplied that switch is closed and at that time I must
maintain the velocity, see if I say that my velocity will remain 𝑣0 with the switch closed,
then I must demand that there will be an opposing force coming in you must apply another
force from left to right, to compensate for that opposing force. So, that once stay again net
force will be given it will run at a constant velocity, got the idea? That is you must have a
prime mover, which will push this conductor hard, the moment you want to extract power
out of because of it is kinetic energy only, it was delivering power.
But, now the moment you demand that I want to get power in this resistance continuously,
then rather make the velocity constant by compensating this opposing force from left to
right, then nothing will be coming out from the kinetic energy to here, whoever is pushing
that conductor at that force which is equal to 𝐵𝑖𝑙, it will do that one idea. So, in general
there must be a this thing, external agency which will pushing it from left to right that is
called prime mover ok.
So, this is a simple generator without any prime mover. In general therefore, to make the
generator work, there must be a prime mover as I told and the equation of motion I will
write next class but, before that let me in the same arrangement.
(Refer Slide Time: 24:13)
530
Let me just study another thing this was my rail suppose, this was the rail and you know
this was my conductor and suppose this time what I decide is that, this conductor in the
rails was resting of length 𝑙. Now, what I will do, it was at rest ab initially at rest. Now,
what I do I connect a battery here, through the rails with a switch. So, when the switch is
open nothing happens it remains stationary and there is of course, 𝐵 as usual, I am just
drawing 2 lines but, it otherwise invading all the areas between the rails 𝐵, flux density.
If you now close the switch and the conductor was stationary, it is velocity cannot change
instantaneously, at least at 𝑡 = 0 you close and I am pretty sure that
𝑖(0 +) =
𝐸
𝑅
It has to be because, at 𝑡 = 0 + 𝑣 is still 0, conductor was not moving at 𝑡 = 0 + because
it has got mass, it is velocity cannot change. So, at 𝑡 = 0 + current will be like this and
what will be the direction of the current, it will be like this 𝑖, 𝑖 and 𝑖. So, 𝑖(0 +) is there,
the moment this is current and it is placed in a magnetic field, it will experience a force,
which I can get from left hand rule that is 𝐵 and this is 𝑖.
So, you see this is 𝐵, this is 𝑖 it is correct. So, this is 𝐵, this is direction of 𝑖. So, thumb
will give you direction of velocity force, this is called electromagnetic force experienced
by the conductor and the conductor in absence of any frictional force we will try to
accelerate, it is velocity will now start increasing, earlier it was 0 at 𝑡 = 0 you have closed
the switch, then from 𝑡 = 0 onwards, the conductor will start moving and the reduction of
the force is from left to right.
So, for 𝑡 = 0, conductor experiences conductor ab, experiences a force from left to right
left to sorry from left to right and what is the value of the force 𝐵𝑖𝑙 what else, they have
got thing starts moving I can write down the dynamic equation. What will be the dynamic
𝑑𝑣
equation? 𝑚 𝑑𝑡 is equal to the force from left to right I will velocities this way increasing.
So, this must be equal to
𝑚
𝑑𝑣
= 𝐵𝑖𝑙
𝑑𝑡
531
No negative sign this time because, this is accelerating it ok, 𝐵𝑖𝑙. So, this is one equation,
what is our target? What is the velocity as a function of time is what? And 𝑖 as a function
of time is what?
Now, the this is one equation. So, I must have two equations to get 𝑣 and 𝑖, what will be
the next equation. Next equation is, what will be the expression of 𝑖? Expression of 𝑖 at
any time 𝑡, see the moment this happens it is starts moving, we have just discussed a
conductor moving in a magnetic field, will have a induced voltage 𝐵𝑙𝑣. So, for this
conductor moving from left to right with velocity 𝑣m/s must have a induced velocity and
the what will be the polarity of this induced velocity, this is what we are saying 𝐵𝑙𝑣.
Therefore, this whole circuit will now be equal to a induced voltage here between a and b,
that is 𝐵𝑙𝑣 and this thing your switch here and please I will connect also a resistance here,
otherwise there will be short circuit suppose with a resistance this source is. Or suppose
this resistance is the resistance of the conductor either way. So, there is a resistance here
and there is your battery voltage here and current direction is this therefore, I must say that
current is
𝑖=
𝐸 − 𝐵𝑙𝑣
𝑅
So, I have got two equations, if you want to find out 𝑣, eliminate I from this put it here,
you will get a one first order differential equations solve for 𝑣, then use the other equation
to get the current and this we will continue in the next class.
Thank you.
532
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 55
Single Conductor D. C Generator/Motor Operation
(Refer Slide Time: 00:21)
Now, welcome to lecture 55 and before we start lecture 55, we have quick glance what we
did in lecture 54. You remember we started DC machines, and eventually we will
discussed rotating DC machines, but before that to gate the basic ideas clear what we
considered a simple linear DC machines, where a single conductor is only present that is
green line and it is suppose moving over rate frictional rails from left to right at a velocity
𝑣0 m/s and suppose this switch is opened initially and since it is frictionless.
So you do not require any force to be applied to the conductor to make it run at a constant
velocity 𝑣0 m/s.
But nonetheless it is moving from left to right and also there is a magnetic field
perpendicular to the paper here from top to bottom that is why I have shown it crossed 𝐵.
So, we applied right hand rule and we know that there is a 𝐵, there is a 𝑣0 there at
quadrature, therefore, there will be some induced voltage across the conductor a b and the
upper side of this will become plus, and lower side will become minus.
533
So, it will become a source of emf like a battery that I have shown here, the value of this
voltage will 𝐵𝑙𝑣0 . If 𝑣0 is the velocity voltage available, there will be 𝐵𝑙𝑣0 ok. So, initial
velocity is 𝑣0 ok, and suppose the switch is open open.
Now, if this because we have generated voltage, you must utilize that voltage to deliver
power to some resistance say a lamp may be. You close the switch suppose at 𝑡 = 0, then
what is going to happen because the circuit is closed, and there will be a current 𝑖 =
𝐵𝑙𝑣
𝑅
.I
have written 𝑣, because I am not sure whether the velocity will be maintained after it has
started delivering power to the resistance.
So, after you have closed the switch at 𝑡 = 0, at any time 𝑡, this is the expression of the
𝐵𝑙𝑣
current in the circuit, and 𝑖 = 𝑅 . Now, I have let us imagine that no external agency is
pushing this conductor from left to right, because it was not necessary when a even S was
opened, because it was moving over a frictionless path.
So, without absence of any external agency pushing it from left to right, we expect that it
is delivering power to the resistance where from this power will come, it must come from
1
the kinetic energy of this conductor a b, initial energy stored was 2 𝑚𝑣02 from that it is
coming and physically we then expect that velocity will gradually decrease and so also
this emf will decrease, current will decrease, and finally everything will be quiet, no
voltage, no current, initial whatever energy stored that will be dissipated in 𝑅.
And, but in between what happens, if you want to write down, this is the dynamic equation
𝑑𝑣
applied force 𝑚 𝑑𝑡 is the applied force from left to right. But we know that when a
conductor carries current 𝑖 in this direction and if there is a magnetic field here, you have
to then this conductor is going to experience also a force and by applying left hand rule,
we find that it will experience force in the opposite direction 𝐵𝑖𝑙 left hand rule you apply,
you will get it.
Therefore, equation of motion will be
𝑚
𝑑𝑣
= −𝐵𝑖𝑙
𝑑𝑡
534
Then 𝑖 is this much. So, you eliminate all these differential equation, you get this as
expected velocity will decrease progressively, and 𝐴 = 𝑣0 .
And if you like what I will add here if I add it will remain. So, if I you can verify that the
total energy loss total energy loss in 𝑅 will be equal to 𝑖 2 𝑅𝑑𝑡, and integrate it from 0 to ∞
and I know the expression of both current and current expression is known, next page I
must have done it.
∞
𝑇𝑜𝑡𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑒𝑑 𝑖𝑛 𝑅 = ∫ 𝑖 2 𝑅𝑑𝑡
0
(Refer Slide Time: 06:09)
∞
So, expression of the current is known. So, total energy loss will be equal to ∫0 𝑖 2 𝑅𝑑𝑡 in
𝑅 in the resistance from 0 to ∞ time and if you substitute the value of 𝑖 here, and calculate
it you will end up with this quantity and that is what is expected initial kinetic energy,
anyway this was the generate generated operation of a single conductor.
∞
𝑇𝑜𝑡𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑒𝑑 𝑖𝑛 𝑅 = ∫ 𝑖 2 𝑅𝑑𝑡 =
0
535
1
𝑚𝑣02
2
(Refer Slide Time: 06:55)
Then we took up the next problem can it act as a motor, that is same structure a b there is
a magnetic field perpendicular to the board, and there is a battery with this side plus, this
side minus, and switch is initially opened no current in the circuit, no voltage here either
𝐵𝑙𝑣 is also not there but if you close the switch at 𝑡 = 0.
Since velocity cannot change instantaneously at 𝑡 = 0 + after you have closed the switch,
velocity of the conductor, previously it was 0 it will maintain 0 at 𝑡 = 0 +. Therefore,
𝐸
current in the circuit will be 𝑖(0 +) = 𝑅.
What is the resistance of the circuit? Suppose all the resistance in the series circuit I have
𝐸
shown it by 𝑅. So, the current will be 𝑖(0 +) = 𝑅 this resistance may be the resistance of
the conductor also. So,
𝐸
𝑅
will be the current at 𝑖(0 +), at 𝑡 = 0 + velocity is 0.
And then we find that the current direction through the conductor is this, therefore, I have
to apply left hand rule, and I will get 𝐵𝑖𝑙 that will be the force experienced by the conductor
a b from left to right that is a electromagnetic force I will call it, generated because of the
interaction of this 𝐵 and 𝑖 and 𝑙 is the length of the conductor a b. It will start then moving
as time passes.
536
Therefore, suppose at any time 𝑡, therefore, the conductor is expected to be accelerating
ok, because there is a force which will be acting from left to right, and conductor will
accelerate, you will get motoring action.
𝑑𝑣
Now, the question is what will be the equation of motion of the conductor now; 𝑚 𝑑𝑡 if 𝑣
is the velocity at any time 𝑡 at this must be equal to the force. This time you see force is
along the same direction of velocity so plus sign and this will be the dynamic equation on
the mechanical side ok. Now, let us come to the electrical circuit equivalent electrical
circuit here.
Now, one very interesting thing happens, the moment conductor starts moving we have
just learnt that if a conductor moves in a magnetic field with some velocity 𝑣, there has to
be an emf generated across a b across this conductor, therefore, at time 𝑡 > 0, it is not only
this external voltage, but there will appear another source of emf across ab which I have
shown it by this battery and this voltage is often called back emf.
Therefore, at any time 𝑡 therefore, the current in the circuit will be
𝑖=
𝐸 − 𝐵𝑙𝑣
𝑅
And
𝑚
𝑑𝑣
= 𝐵𝑖𝑙
𝑑𝑡
So, we got these two equation. So, we start now our lecture 55, we will start from this
point. So, these are the two equations we have come across.
537
(Refer Slide Time: 11:09)
𝑑𝑣
Therefore, it will be what we have got 𝑚 𝑑𝑡 = 𝐵𝑖𝑙 and the second equation is the current
expression of the current which was equal to 𝑖 =
𝐸−𝐵𝑙𝑣
𝑅
, is it not, these are the two equations
we have got.
So, our unknown is what is 𝑖 and what is 𝑣 as a function of time we want to find out. Before
that physically let us see what is now going to happen, it will go on accelerating from left
to right and suppose this is of infinite length not that everything finishes here, the this is
rail continues infinitely long track.
Now, as it accelerates the back emf, so this 𝐵𝑙𝑣 is going to rise is it not physically I am
trying to examine what is what will be the final set of this thing, we will need to go on
accelerating forever.
The answer is no, because of the fact this voltage will go on increasing and this voltage is
fixed, supply voltage which is fixed. Therefore, a time will come when this voltage will
become equal to this voltage.
And as you can see their polarities are such that they oppose each other, therefore, at that
time I expect current will be 0 then is it not, current is 𝑖 =
𝐸−𝐵𝑙𝑣
𝑅
.
But as it is accelerating, when this back emf were reaches 𝐵𝑙𝑣, this current will vanish ok
and then what happens, then no torque on this conductor, no force on this conductor 𝐵𝑖𝑙,
538
current vanishes, therefore conductor will attain certain velocity final velocity which will
make this voltage same as this voltage and after that no further acceleration and current
will be 0; but will the conductor go on moving? Yes, it will, because we have assumed the
frictionless track.
Therefore after you have closed the switch, it was stationary, it will accelerate, accelerate,
accelerate, and it will attain such a velocity finally, which will make 𝐵𝑙𝑣 = 𝐸, or
𝑣=
𝐸
𝐵𝑙
When that velocity will be attained current will be 0, and the conductor will be moving
with that velocity indefinitely from left to right that is that from left to right. Will it violate
any of the physical rule? No, because we know in a frictionless environment to move a
thing, you do not require any force to be applied, is that clear?
Therefore, so before I mean proceeding further, I can by physically examining the system
𝐸
can conclude the final velocity final velocity has to be some 𝑣0 = 𝐵𝑙, it has to be after a
long time and final current 𝑖𝑓𝑖𝑛𝑎𝑙 = 0, this must happen and let us see indeed this thing
happens or not. For that you have to solve this circuit, two unknowns are there.
So, suppose if I solve for velocity, so replace this 𝑖 in this equation by this expression, and
you will be getting here what that is
𝑚
𝑑𝑣
𝐸 − 𝐵𝑙𝑣
= 𝐵𝑖𝑙 = 𝐵𝑙 (
)
𝑑𝑡
𝑅
So, now, in this equation only 𝑣 is present, I can solve for 𝑣 but I have to arrange the terms
or I will write
𝑚
𝐵𝑙
𝑑𝑣 𝐵 2 𝑙 2
+
𝑣= 𝐸
𝑑𝑡
𝑅
𝑅
Then what you do you divide by 𝑚 both the sides, so that it will look now like this
𝑑𝑣 𝐵 2 𝑙 2
𝐵𝑙
+
𝑣=
𝐸
𝑑𝑡
𝑚𝑅
𝑚𝑅
539
So, it is a first order differential equation. And characteristic root, in the same way
characteristic root is equal to
𝐶ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑅𝑜𝑜𝑡 = −
𝐵2𝑙2
𝑚𝑅
Therefore, velocity at any time t will be equal to A into e to the power minus B square l
square by m R.
𝐵2 𝑙2
𝑣(𝑡) = 𝐴𝑒 − 𝑚𝑅 +
𝐵2 𝑙2
𝐵𝑙 𝑚𝑅
𝐸
−
𝑚𝑅 +
𝐸
=
𝐴𝑒
2
2
𝑚𝑅 𝐵 𝑙
𝐵𝑙
It will be expected to be the velocity.
Hopefully, it is correct. If I am not correct wrong answer I will get, let us proceed ok. What
is 𝐸? 𝐸 is the supply voltage.
Now, what is the boundary condition? Boundary condition is velocity at 𝑡 = 0 is 0,
because it was stationary, 𝑣(0 −) = 0, then 𝑣(0 +) = 0, velocity of a mass cannot change
in no time, so that was the argument. So, if you put that, then, then apply this boundary
condition, and you will get
𝐴=−
𝐸
𝐵𝑙
put this condition here.
So, I will say
𝑣(𝑡) =
𝐵2 𝑙2
𝐸
(1 − 𝑒 − 𝑚𝑅 𝑡 )
𝐵𝑙
It is something like RL circuit sort of thing. So, the this is the final solution for velocity.
Let us verify this. After a long time, that is velocity as 𝑡 → ∞ how much its value will be,
𝐵2 𝑙2
𝐸
this term will vanish 𝑒 − 𝑚𝑅 𝑡 and you will be left with 𝐵𝑙 that is what we concluded
physically, is it not, I should ignore that.
𝑣𝑡→∞ =
540
𝐸
𝐵𝑙
𝐸
So, 𝐵𝑙 this is the thing, and it looks like I mean apparently. So, this is your velocity and
how to get current, now put this value of v in this expression to get the expression for the
current.
(Refer Slide Time: 22:11)
So, so current expression will be, so current 𝑖(𝑡) will be equal to we got it here it is equal
to
𝑖(𝑡) =
𝐵2 𝑙2
𝐸 − 𝐵𝑙𝑣 𝐸 𝐵𝑙
𝐸 𝐵𝑙 𝐸
𝐸 𝐵2 𝑙2
= − 𝑣(𝑡) = − [ (1 − 𝑒 − 𝑚𝑅 𝑡 )] = 𝑒 − 𝑚𝑅 𝑡
𝑅
𝑅 𝑅
𝑅 𝑅 𝐵𝑙
𝑅
So, this is the expression of 𝑖(𝑡), is it not, that is all. So, you know this is the expression
of 𝑖(𝑡). Therefore, if you sketch the as a time how it looks like I will sketch first velocity.
If I sketch velocity was 0 till 𝑡 = 0, then it was rising exponentially like this, and finally,
𝐸
reaching this velocity. What was that velocity 𝐵𝑙.
𝐸
So, this is the final velocity 𝐵𝑙 ok. This blue curve is velocity and if you sketch, oh below
this I will sketch the current. So, current wave form will be in this same this is time, this
𝐸
is time. So, before 𝑡 = 0, current was 0. So, it was there 0. Then at 𝑡 = 0 current is , it
𝑅
𝐸
shoots up to 𝑅 and then it exponentially decreases, is not it, this is how this is the expression
of the current.
541
Therefore you see in the motoring operation when you make this generator I will just draw
a replica of that this was the thing, this was the conductor here and just in short to explain
it here, this is was my conductor a b, oh my god, this was my conductor a b which was
moving and here in the rail you have connected some resistance was there and some
battery, then current was like this.
So, you start from rest, but eventually you attain a final velocity that is this conductor will
𝐸
store some energy. If I call that final velocity to be 𝑣𝑓𝑖𝑛𝑎𝑙 = 𝐵𝑙, it will be like this. But
anyway the motor does not do any mechanical work, is it not.
Let us ask ourselves in this motoring operation that suppose you run a motor to supply is
a mechanical load to overcome friction and so on, you imagine that no track is frictionless.
Suppose, this conductor has to carry a load, of course, it is not possible to some mechanical
load mass put on this, is it not, and suppose the track has got friction.
Therefore, you imagine it is moving train sort of thing on which I can dump load that is
heavier things on this one and this track is not going to be in real life frictionless and if
you put more weight on this mass, the friction to be overcome will be much more, is it not.
And therefore, this is the direction of the electromagnetic force we have seen. This is your
direction of 𝑖 that is fine. Now, I can easily say that if suppose to this movement of this
conductor, I impose an opposing force, suppose some opposing force I impose on the
system now after it has attained constant velocity.
Let us some opposing force comes in that will be the that is that is called the load force in
rotating machine we say load torque, but load force which will be in opposite direction.
Best way to think is that there is friction suppose now appears it has at least this one, but
there is now you add friction to the system.
You know in general on the moving conductor, there will be two forces acting for the
linear machines, one is the electromagnetic force which is providing the velocity and
another is the force in the mechanical system it is electromagnetic force, and it is
mechanical force in the opposite direction.
Whenever 𝐹𝑒 > 𝐹𝑜𝑝𝑝 , conductor will accelerate, must accelerate. If at any time 𝐹𝑒 < 𝐹𝑜𝑝𝑝 ,
conductor must decelerate and if 𝐹𝑒 = 𝐹𝑜𝑝𝑝 , conductor will move with constant velocity.
542
Let us try to understand. So, it has reached a final velocity 𝑣𝑓𝑖𝑛𝑎𝑙 and then you suppose
impose after it has reached the final velocity sum the conductor faces some opposite force
to be overcome, it certain suddenly appears.
The moment it appears what was the electromagnetic force at that time when it has reached
final velocity electromagnetic force developed by the conductor was 0. Suppose that that
thing occurs at 𝑡 = 𝑇1 after long time some at some 𝑇1 I have introduced opposing force.
So, at that time electromagnetic force was 0, because current was 0. So, there was no force
from left to right, but I am telling I have imposed an opposite force to the motion of the
conductor. The moment you have imposed that at that time velocity cannot change
instantaneously mind you.
So, velocity at that time 𝑇1 + velocity will still remain 𝑣𝑓𝑖𝑛𝑎𝑙 but the opposite force has
appeared. Therefore, electromagnetic force which is 0 is less than 𝐹𝑜𝑝𝑝 , and conductor will
then decelerate got the point conductor will decelerate but the moment conductor
decelerates, this 𝐸 > 𝐵𝑙𝑣 the opposite force, it remember the equivalent circuit is this, this
is 𝐵𝑙𝑣, this is your 𝑅, and this is your the supply battery voltage which is constant.
Therefore, initially when it has reached final velocity, current was 0, this that fine. But if
you put a opposite force coming onto the conductor, then it must decelerate, if velocity
will start decreasing. If velocity stars decreasing, it will draw some current it will start
because this electrical circuit 𝐸 > 𝐵𝑙𝑣 and it will draw current and the moment it draws
current, there will be 𝐵𝑖𝑙, is it not?
So, what will be the final fate of the conductor? Finally, conductor will run at that speed
which will make this 𝐵𝑖𝑙 = 𝐹𝑜𝑝𝑝 . As the speed decreases, it invites current from the source.
If it which was till not absent will now appear and it will increase a time will come when
𝐹𝑒 = 𝐹𝑜𝑝𝑝 but the conductor will run at a speed less than 𝑣𝑓𝑖𝑛𝑎𝑙 , finally it will settle down.
So, you have understood what I am telling. Therefore, so if somebody so as to say that in
general for a practical system, I will say that you have a motor, motor mode of operation.
543
(Refer Slide Time: 36:03)
For example, in this case, this is the conductor. I will reframe the problem and leave it to
you to solve it that ok. Here is your switch, here is the resistance 𝑅, and there if some
battery you will connect to the track ok; and S, I will close at 𝑡 = 0. But this time I am
telling that the track is not frictionless, a constant friction force is acting.
So, there is a constant friction force in this direction, 𝑇𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 is this is present on the track
you close this switch at 𝑡 = 0.
Then I will once again write let me write that that at any time velocity is 𝑣(𝑡), current is
𝑖(𝑡), I want to know the solution, because there will be current at 𝑡 = 0 + which is equal
𝐸
to 𝑅. So, motion will start. What will be the force acting on the conductor at 𝑡 = 0 + current
𝐸
is 𝑖𝑡=0+ = 𝑅.
Therefore, force electromagnetic force acting on the conductor is
𝐹𝑒 = 𝐵𝑖𝑙 = 𝐵
𝐸
𝑙
𝑅
And its direction is from left to right. But now the equation of motion should be 𝐵 and as
it starts moving there will appear this back emf, therefore,
𝐹𝑒 𝑡=0+ = 𝐵𝑖𝑙 = 𝐵
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𝐸
𝑙
𝑅
In general
𝐹𝑒 = 𝐵𝑙
(𝐸 − 𝐵𝑙𝑣)
𝑅
So, this is the electromagnetic force, but your equation of motion now I should write it like
this m d 2 v d t 2.
Student: First derivative.
𝑚
𝑑𝑣
= 𝐹𝑒 − 𝐹𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛
𝑑𝑡
There is the mechanical opposite force. I am not going to solve it, but physically I am I am
telling you what is going to happen to the final velocity and current.
𝐸
Do you think the final velocity will be 0? A final velocity will be equal to still 𝑣𝑓𝑖𝑛𝑎𝑙 = 𝐵𝑙,
𝐸
in this case? The answer is no, because if it is 𝐵𝑙 then current drawn will be 0, there will
be no electromagnetic torque present.
So, what will be the final velocity? Final velocity we will be talking, but before that we
must understand if the system reaches steady state at that time at steady state
𝐹𝑒 = 𝐹𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = 𝐵𝐼𝑓𝑖𝑛𝑎𝑙 𝑙
So, for this all this dynamic equation need not be solved. One can once again solve this
and you will find that this has to be your 𝐹𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 .
𝐼𝑓𝑖𝑛𝑎𝑙 =
So, I will be telling look your 𝐼𝑓𝑖𝑛𝑎𝑙 =
𝐹𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛
𝐵𝑙
𝐹𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛
𝐵𝑙
. This will be the final steady state current,
which of course after solving this differential equations one must arrive at this you can try
it.
And what will be the final velocity, how much will be the final velocity? Final velocity
will be such final velocity final steady velocity will be such thus will be such that this that
this 𝐼𝑓𝑖𝑛𝑎𝑙 current flows in the circuit; that means,
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𝐸 − 𝐵𝑙𝑣𝑛𝑒𝑤
= 𝐼𝑓𝑖𝑛𝑎𝑙
𝑅
See to find out the steady state current and speed one need not go through all these dynamic
equation, of course, it is a suggested method of studying at least for one, so that you know
there are situations when the motor or generator will run in steady state will be mostly
interested in that some few cases of dynamic behavior of DC machines also we will
discuss.
But to find out the steady state in DC things, what you have to do it if; it is; it will be
running at constant speed, then the electromagnetic and mechanical forces which are
acting in opposition they must be same, and electromagnetic force is 𝐵𝑖𝑙; from that I will
be able to calculate I final and so on.
Anyway we will continue with this in the next class.
546
Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 56
Homopolar D.C Generator
So, I was telling about very simple DC machines and a simple generator and motor
consisting of a single conductor and placed under a magnetic field. If you excite it with an
external voltage it will act as motor and if you give a movement it may act as generator.
Only thing about the motor I have completed for the generator I will not tell much except
this information I must tell, you recall that in case of Generator mode also.
(Refer Slide Time: 01:03)
In generator mode here also I assumed one thing there is no friction like that and only thing
it is in fact, a no generator at all somehow you have dumped some kinetic energy to the
conductor initially it was running with velocity 𝑣0 and the moment you try to draw power
out of this generator the power will be delivered to be load, but not for so long, voltage
will collapse gradually to 0, current will collapse. So, long kinetic energy will be present
in the conductor that will be there, but what I am telling is that, this I will tell in a nutshell
now in case of see generator just to have for completeness sec on discussion on this topic
I will say that ok.
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(Refer Slide Time: 02:09)
Suppose you have a generator on this railway track as we have drawn earlier and very
quickly I can write down the results this is my a b conductor generator mode generator and
it is running initially with velocity. And suppose I say like this it is running with velocity
𝑣 at constant speed and then I say that the generator is supplying a resistance like this and
your b is in this direction. So, polarity of the voltage induced will be 𝐵𝑙𝑣 is it not.
Now, suppose I say that this generator is found to deliver a constant current of value I and
the conductor was found to run with a constant velocity 𝑣. You get me that is the conductor
is of length 𝑙 moving it was found at one time that moving from left to right with velocity
𝑣 and we know voltage will be 𝐵𝑙𝑣 and it is delivering a constant current which has to be
𝐵𝑙𝑣
equal to 𝐼 = 𝑅 .
It was found it is sustained, now the then I will such a situation is present that it is moving
with a constant velocity and this one, is it possible? I will say yes it is possible, because
this conductor is carrying a current 𝐼 therefore, this conductor experiences and opposite
electromagnetic force 𝐹𝑒 whose value is 𝐹𝑒 = 𝐵𝐼𝑙 it has to be because if it is delivering a
current 𝐼. Then I will demand that oh it is running at a constant velocity; that means, some
external agency must have applied a force to the right mechanical force, matching this
𝐵𝑙𝑣
electromagnetic force that is why it is running at constant velocity 𝑣 and current is 𝑅
everything is in place.
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In other words what I am telling is for a generator to continuously supply power in this
case it will not happen the velocity will come to 0. So, long there is an external fellow
providing you this mechanical force which is equal to this electromagnetic force developed
by the machine which is from right to left that is why that engine comes in to run a
generator you require some mechanical assistance whatever power you deliver to the load
will come from the mechanical system or prime mover in case of rotating machine it is
called a prime mover. So, what is that prime mover? Maybe it is a diesel engine sort of
thing which will give you this mechanical force got the point.
Therefore this is the called the prime mover which drives the generator. So, to make a
generator run you must have some external agency which will provide these much needed
mechanical force to match the electromagnetic force and make the generator run at
constant speed. If any upset between these two happens then also I know because this is
the net force acting is 𝐹𝑀𝑒𝑐ℎ . So, 𝐹𝑀𝑒𝑐ℎ must act along the direction of the velocity, if
𝐹𝑀𝑒𝑐ℎ > 𝐹𝑒 it will accelerate. If 𝐹𝑀𝑒𝑐ℎ < 𝐹𝑒 it will decelerate and if these two are same it
will run at a constant velocity.
Anyway where this is in short the basic idea of a very simple DC generator which of
course, one will not manufacture because it is so, non ideal in the sense that very
impractical long infinite long rails what for, but this problem we have tackled to bring out
several interesting points that in a generator and motor take place ok. It gives you an idea
in a moving system two forces will act on the moving conductor; one is the electromagnetic
force, another is the mechanical force. In case of motoring mode electromagnetic force
will decides the direction of motion and in case of motoring mode it is the prime mover in
which direction it is driving the generator this one that will decides the direction of motion
and so on.
And will come to these points several times when we start a real practical generator. So,
go through this exercise and the as I told you can once again start if you say mechanical
force I have set then how the final operating point will be reached and so on we will discuss
about it. Now, I will talk about another interesting generator of this kind. Now, in this type
of generator as you can see this conductor is it moves from left to right and lines of forces
are into the papers. Conductors never see a reversal of field that is it is always a we will
see north to south north through south lines of forces it will cut and give you force ok. Let
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us see another generator another’s generator to make things clear, DC can be generated
because that was so impractical.
Now, I will tell you about a very practical simple DC generator which is rotating type let
us see how it will look like very interesting. Suppose you have an aluminum disk we will
be discussing another simple type of DC machine which is more practical, I mean there is
no doubt about it and this structure or the constructional feature of this machine is you take
a aluminum disc here like this.
(Refer Slide Time: 10:21)
Aluminum disc and a circular aluminum disc and it has got a spindle about which it can
be rotated vertical axis along this line that is there. Now, on the top you keep a north pole
suppose this is a north pole above it ok.
This blue colored thing is a North Pole and below you keep a South Pole all along the
covering the area of this aluminum disc so, this is South Pole. So, there will be lines of
forces crossing this aluminum disc like this way this will be the direction of 𝐵. The black
one is the aluminum plate; aluminum plates ok and this aluminum disc can be rotated about
a vertical axis.
Now, in a simpler diagram I will put it like this that you have this disc of certain radius
and this aluminum plate I now draw on the plane of this page. So, your axis of rotation
will be here this is the axis of rotation and above this aluminum plate there is the North
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Pole, below this there is a South Pole lines of forces will be into the paper all over this I
am not making it clumsier by putting so many cross here.
And if this is a rotated suppose in this direction with some rpm say 𝑛 rps, so many rotations
per second I am rotating it. Then final result I am telling then if you connect a voltmeter
between this centre point and this ream of this aluminum disc that is the periphery, if you
connect a voltmeter you will find there is voltage generator and that will be DC voltage.
So, on the top there is a north pole here and below this aluminum plate there is a south
pole. So, magnetic field will be 𝐵 and things. Now, why it happens, how do I get the
voltage?
So, for that what I am telling, you consider any line radial line here it consists of infinite
radial lines like this one of them you choose and then what I am telling you and suppose
the radius of this aluminum disc is 𝑟 ok. So, it will be this one and what you do now at a
distance this let this point be O this spindle point at a distance 𝑥 here, you consider a small
element vanishingly small element 𝑑𝑥 here this we concentrate on this element. And this
I am drawing it in a here the 𝑑𝑥 element in a larger scale 𝑑𝑥 is small mind you. Now, this
since it is moving like this therefore, it will be it is tangential velocity will be along this
line it will be moving suppose that tangential velocity is 𝑣𝑥 and this length is 𝑑𝑥 and there
is a magnetic field 𝐵 into the paper that this elemental length is seeing.
Therefore why not there will be induced voltage because there is a magnetic field, there is
a small element 𝑑𝑥 having a velocity 𝑣𝑥 m/s, 𝑣𝑥 can be related to this end and radius we
will do that, but this is the thing and how much will be the voltage? Voltage will be 𝐵𝑑𝑥𝑣𝑥
and what will be the polarity of the voltage 𝐵 right hand rule you apply 𝐵, 𝑣𝑥 and this will
be the polarity of the voltage. So, this side of this element will become plus and this side
will become minus. So, we expect a voltage is induced this one with this side plus, this
side minus.
Now, the question is why I have considered an element 𝑑𝑥? Because of the fact if you
consider this whole length you do not know what is the velocity ok. 𝐵 is perpendicular
length of the conductor is perpendicular to 𝐵, but you cannot assign a constant tangential
velocity to this whole segment is not because tangential velocity is a function of radius.
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So, that is why I have considered a small element only anyway if you divide this in small
small small things elements then all the emfs will add up and finally, give you a voltage
between this point and that point some substantial voltage if all the small elementary
lengths they become seat of emf with this polarity only they will become and if you add
them up you will get a substantial voltage. So, magnitude of the voltage is this one. Now,
the question is they you have considered a single line there are I mean infinite number of
lines like that is not if you draw close by this line many lines are there many lines ok.
And they are touching each other this line you will find another line close by touching
each other, then each one of them this segments has become plus minus not doubt then
will they not will not there be a problem this conductors are in contact with the next
conductors no insulation between no there will be no problem. Because if you considered
a radius here if you consider a radius all the points in this circle considered they will be at
equipotential. Therefore, there is no current in this 𝜃 direction or there is no voltage
induced in this direction why because tangential velocity suppose somebody says no I will
consider the element like this here.
I will consider the elements like that at a given are this element then tangential velocity
and the length of the element are in the same direction they must be mutually perpendicular
𝐵𝑙𝑣 business so anyway so, they that will be equipotential. So, no matter you consider just
one element. So, all these elements are in parallel that is what I am telling. So, finally, a
finite voltage will be available between this point and this point and we would like to and
that voltage will be DC only because 𝐵 is constant, velocity you are running it at a constant
velocity, 𝑛 rps and so on.
So, and we found that the elemental voltage for a single element is if I say this is 𝑑𝑒𝑥 in
this element what is the voltage
𝑑𝑒𝑥 = 𝐵𝑣𝑥 𝑑𝑥
Now, what is the known input? Known input is this one 𝐵 is known and 𝑛 rps is known
ok. Now, rotation per second is same at this point, at this point rotation is same, but the
distance traverse meter per second velocity will change as you move from origin to this
way. How they are related? 𝑣 = 𝜔𝑟, see for example, here what will be the velocity,
velocity will be tangential velocity at any point here is equal to
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𝑣 = 2𝜋𝑟𝑛
It is not in one second this point moves how much meters. So, therefore so much of m/s
will be the velocity.
Therefore tangential velocity 𝑣𝑥 of this element should be
𝑣𝑥 = 2𝜋𝑥𝑛
Because rotation per second are same for all these. So, this is 𝑣𝑥 mind you this is rotation
per second yes. So, this is meter per second it will become therefore, the elemental voltage
has become
𝑑𝑒𝑥 = 𝐵𝑣𝑥 𝑑𝑥 = 2𝜋𝐵𝑛𝑥𝑑𝑥
So, this will be 𝑑𝑒𝑥 , then what will be the total voltage between point O and say point A
potential and what is the polarity this side plus minus?
So, potential voltage generated voltage between points A and O will be then
𝑟
𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑝𝑜𝑖𝑛𝑡𝑠 𝐴 𝑎𝑛𝑑 𝑂 = 2𝜋𝑛𝐵 ∫ 𝑥𝑑𝑥 = 2𝜋𝑛𝐵
0
𝑟2
2
This is the final thing is equal to the generated voltage, we will continue with this in the
next lecture ok. So, you will get a DC between the center and the perimeter and this A
point there these are all equipotential point. So, whether this point or that point you I will
tell you about how to collect this voltage and use it for practical purposes.
Thank you.
553
Electrical Machines -I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 57
Homopolar D.C Motor
Welcome to lecture number 57 and we were actually discussing some elementary DC
machines in the last couple of lectures.
(Refer Slide Time: 00:30)
And, recall that in my last lecture I told you about a simple type of DC generator, where
an aluminum disk is rotated at some given rps of certain radius 𝑟. And, there will be
magnetic field from top to bottom penetrating the disk, all over the surface of the disk.
Then, what happens is if you consider a radial line, there will be induced voltage between
the center and the perimeter of this disk. And, the value of that voltage is given by this
expression ok, this expression.
This we got last time. So, it is directly proportional to 𝐵 and radius square and rps of the
machine rotation per second. Now, and also the polarity of the voltage will be this side
plus this side minus. And mind you that this induced voltage, that is along the radial lines
the voltage exist, but along the direction of the 𝜃 along the direction of the 𝑟 at any radius,
there is no voltage difference between this point and this point that does not exist. Because,
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there will be at equipotential points if you draw a circle that circle will become
equipotential. Of course, the value of the potential increases as you travel from center
towards the perimeter of the disk.
So, this type of generator of course, it is feasible, unlike your single conductor infinite
length DC machine, this machine can be easily constructed. And, this type of machine is
given a name called DC homopolar machines. Now, how to collect this voltage? What you
have to do is this you have to take two brushes, fixed brushes, 1 is here. And another at
the center somehow on the spindle suppose you connect the brush ok, this is the disk you
recall this is the disk, this is the spindle.
So, you connect I just to give you idea this is one brush, this is another brush. These two
brushes are fixed and the disk will rotate therefore, it will always touch the perimeter the
brush fixed brush. So, one brush is here, another brush is there. These will be the negative
brush, this will be the positive brush and through this fixed brushes it will supply the load.
Therefore and the magnitude of the emf generated is given by this. So, this is how you can
generate DC voltage. In fact, this type of configurations or this type of DC generators are
used, when the magnetic field can be made very large as in a superconducting coil will
create this magnets whose strength will be very large ok. But, anyway this is for
instruction, but for very large power at room temperature this machine is not of
mechanically very strong and things like that. Nonetheless this also gives us some idea
that, how easily DC voltage can be generated? And, we should not forget also that this
type of machines can also be operated as a motor.
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(Refer Slide Time: 04:40)
So, as a generator that homopolar machine, homopolar DC machine. It can be also operated
both as generator and also as motor, when it is operated as generator, we have seen it can
be modeled as the generated voltage between the brushes. And, maybe it is some resistance
effective resistance internal and then your load resistance, this is the internal resistance
and this is the generated voltage. And, generated voltage we have seen it is proportional
to this thing that is
𝑟2
𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑖𝑛 𝐷𝐶 𝐻𝑜𝑚𝑜𝑝𝑜𝑙𝑎𝑟 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 = 2𝜋𝑛𝐵 = 𝜋𝑛𝐵𝑟 2
2
Let us use different notation here this is 𝑟1 ; this 𝑟 is the radius of the disk.
And, this 𝑟1 is the internal resistance of the source 𝜋𝑛𝐵𝑟 2 . If, you want to operate the same
machine has a motor, then what you should do as this is generator operation, for motoring
operation, what you should do is this. This is disk, these are the two brushes ok. One is
connected to the spindle these are the two brushes. Here you connect a external supply,
supply is supplied voltage. If you connect a supply voltage like this with this polarity that
current will flow like this, it will come here at the center it is connected to the disk current
will all flow radially.
That is if you look at the disk during a motor operation, it will be the way I have drawn, it
will be this and this, currents will all flow radially. It will come here then come here; many
parallel paths are there for current to flow high. So, polarity of the applied voltage is such
556
it comes suppose it was stationary initially. And, let us assume that 𝐵 is in this direction
into the paper, then you can apply the left hand rule 𝐵𝑖𝑙 and this machine will start running
in the anti clockwise direction left hand rule you have to apply.
And, it will in the anti clockwise direction it will start running. And, once again it will
finally, reach a steady speed, when the applied torque that is the electromagnetic torque
developed is balanced by the opposite or present on the system. That is friction whatever
it is, but nonetheless in this case the moment it starts rotating we must not forget, each this
radial lines become source of emf, of this magnitude 𝜋𝑛𝐵𝑟 2 . At 𝑡 = 0, when it was
beginning to start you have switched on the supply with a switch at 𝑡 = 0, before that it
was stationery, you have closed the switch at 𝑡 = 0, it was stationery now it will start
accelerating electromagnetic torque will be produced. And if it can overcome the frictional
torque present on the shaft it will start moving.
And finally, it will attain a certain speed when the opposing torque and the electromagnetic
torque will be equal ok. Now, what is the direction of this force, if you calculate it? It is
we have seen 𝐵𝑖𝑙. Now, this segment what will be the polarity of the voltage. In this case
polarity of the voltage will be if you take an element here, it is moving with this 𝑣 and this
is 𝐵.
So, 𝐵 and 𝑣 this will be the polarity, that is the tip of this finger is towards the centre and
that will become plus. So, the polarity of the induced voltage will make this is plus this is
minus, it is correct this is back emf. So, it will try to oppose the inflow of current into the
motor. The model, in this case will be there will be an back emf with this polarity 𝐸𝑏 . And,
this resistance 𝑟1 in any case present and then you have your 𝐸 supply like this, 𝐸 supply
and this is the current and polarity of the voltage once again like linear motors.
So, it will try to oppose the supply voltage. And, when these things are running steadily, I
can easily say that steadily drawing some current from the supply, I will say it is operating
as a motor and they and delivering some mechanical power. How much is the mechanical
power, two ways it can be calculated. You calculate force then multiply with velocity
whatever it is integrate or simply it will be the mechanical power will be 𝑣 × 𝑖 that will be
also.
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Power supplied here is 𝐸𝑠𝑢𝑝𝑝𝑙𝑦 𝑖, a power certain power will be lost here, remaining power
will be this one. So, you can then easily write that
𝐸𝑠𝑢𝑝𝑝𝑙𝑦 = 𝑖𝑟1 + 𝐸𝑏
Multiply with 𝑖 both sides so, you will get
𝐸𝑠𝑢𝑝𝑝𝑙𝑦 𝑖 = 𝑖 2 𝑟1 + 𝐸𝑏 𝑖
This will be the thing. So, this is power drawn from the supply, a portion is lost as copper
loss in 𝑟1 and remaining power must be the mechanical power. Anyway, so, it can operate
both as a motor and generator mode like, our single conductor linear version of DC
machine ok.
So, after this idea of DC machines and here are no complications, only you require two
fixed brushes ok. To be touched one at the perimeter fixed brush mind it, they are not
moving, fixed brush. So, this is the power balance equation of this simple DC motor, which
has no complications whatsoever. So, for it is construction is concerned.
Now, we will discuss about a formal large power DC machines ok. And, that is a DC
machine with which will have a rotor, it will have a stator and then it will be having a shaft
and it will run in a magnetic field. And, there will be two terminals available to supply the
rotor coils, there will be two terminals available to these stator coils and if rotor carries
current, rotor coils and stator coils both carry current, then they can produce some force
and you get sustained rotation, when the electromagnetic torque developed by the motor
and the opposing torque present on the shaft of the machine becomes equal.
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(Refer Slide Time: 15:54)
So, before that you it will look like that the DC machine has got I have formal DC
machines, DC machine will have a stator and a rotor ok. The rotor will be laminated iron
circular plates with slots cut in it and I will that is what I want to tell that the.
(Refer Slide Time: 16:39)
For example, this is the laminated plates of the coil where this is the armature winding of
the machines, this is called the armature. So, there will be several such circular laminated
plates, they will put one above the other and that will give you the length of the machine.
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And, these are slot and teeth as usual and there will be conductors placed on the slots in a
following a particular logic there they will be placed and they will be connected ok.
And, the field structure of this machine will be this is the rotor of the DC machine rotor,
where this is called armature. And, it has got slots and teeth and in which there will be
copper conductors of coil will be placed. The stator part of the machine will look like this
one to projected poles, one on this side and the other on this side ok. And, there will be on
the stator.
So, if I draw it in this in a small version, it will look like this you have two projected poles
on the stator; two projected poles on the stators. And, there is this rotor business which has
got slot and teeth all along as I have shown here and these will carry conductors. On the
stator this is the stator iron and there will be coils here, which is simple coil stator coil is
very simple it will be like this. And, if you pass some current through these coils cross and
dot like this and these two coils are connected in say series, these two parts of the coils
may be connected in series, and you connect it to a DC supply. So, that current flows in
the stator coils. And, then this will become a north pole this will become a south pole
produced by stator structure.
So, I f is the field current. So, what I am telling these two coils connected in series, will
ultimately give you two terminals available to you from the stator and they can be marked
as F1 and F2 called the field coil terminals; field coil terminals, got the point. And armature
is slightly involved that we will see, but the stator structure is made of iron and preferably
solid iron, unlike induction machine stator, which must be also laminated.
It is suppose a solid piece of iron and it has got a length mind you. Similarly, the rotor will
have a length like that ok. This conductor will run all along this all the conductors in these
slots. So, this is the stator structure and stator is therefore, can be represented by a simple
coil. And I will connect pass some DC current 𝐼𝑓 like this. And, these two coils are
connected in series F1 and F2 is the ultimately two terminals coming out and this is
armature ok.
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(Refer Slide Time: 21:40)
But, armature will be a bit complicated we will come to that and before that I will show
you an actual DC machine armature a small DC machine armature.
(Refer Slide Time: 21:48)
This is what these slots and teeth are slots and teeth, then there are several plates stacked
together to give you the length of the machine. And, in these slots you can see multi turn
coils passing. And, they will be connected in a particular fashion. And, this part which I
will discuss extensively, why this is necessary this is called the commutator segments ok.
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And, it is this part which will help us to change it the AC voltage generated in the coil to
convert it to DC voltage that will take up later, but the idea is clear.
So, this is rotor which will rotate on this shaft like this and it will rotate in two stator
magnetic field, one on this side north, another on this side south like this and it will have
a stator. So, stator part I have not brought. And, there will be an air gap between these this
is the idea I mean you should have a physical idea how the machine looks like, that is fine.
Now, before we start this DC machine. So, stator is will give you projected poles like this,
stator is has projected poles has projected poles. And, to be excited with DC current with
constant value of current value of current called field current ok.
Rotor will have rotor is also called armature also called armature for a DC machine,
armature has several laminated, circular plates, circular plates, insulated from each other;
insulated from each other. Like transformer from each other, by a varnish coating or
whatever it is has several laminated circular plates insulated from each other, to with
number of teeth and slot and slots ok.
Now, before so, field winding field coils, they are very simple, you simply have projected
poles around it you just suppose this is a projected pole like this. And, you make the coil
like that and there is another pole suppose here. And, you have coils like this, here also
and this is the structure of the stator poles I have shown their joint like this and it will be a
circular thing. So, it comes like this red one are the coils and this is the thickness outer
thickness of the iron of the stator, it will be like this joint, this is the stator, stator iron,
which is not laminated the reason we will see.
And these two coils what you will do is this these two coils you will connect in series as I
was telling. And you connect some and these two terminals may be marked as F1 and F2
and your stator coils are ready. Now to this stator coils, if you pass some connect some
resistance in say series, you connect some in other I show connect some resistance and
connect a DC source, it will send current like this and the direction of the current will be
like that.
So, that this will become a North Pole, this will be coming South Pole, this is how stator
poles will be created and this is. So, these two coils will be connected in general series to
create a two pole structure like that. And then in the armature, armature part is more
crucial, the armature will be here like this with slot and teeth, this is armature. And, here
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there will be coils placed in the slots following a certain grammar, certain logic. So, that
you can ultimately get DC voltage in it so, we will continue with that. So, constructional
features we are discussing of formal DC machines.
Thank you.
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Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 58
Introduction to Rotating D.C Machines
(Refer Slide Time: 00:19)
Welcome to lecture number 58 and we have been discussing about rotating DC machines
formal type of DC machines as we understand and used in practice.
(Refer Slide Time: 00:33)
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And, in my last lecture, I just told you what are the basic structure in a very rough sketch,
but it brings about all the important structural features of a DC machine. For example, this
stator will have two projected poles, if it is a two pole machines 𝑝 = 2. There may be
multiple number of projected poles, if the number of poles of the machine is higher and,
this part is almost a solid piece of iron, over which there will be coils with red colors this
is these are called field coils.
One of the simplest stator coils are no complications, yes take wires and make this coil
over the projected poles. And, these two can be suitably connected; so, that when you pass
DC current in it. This produces North Pole, South Pole, because this is the direction of the
current so, lines of force will come out from this, like this. And, it will enter here creating
the South Pole.
And, these lines of force will complete their path through the stator iron like this a sample
lines of force will be. And, the armature is a number of circular thin plates made or soft
iron and they are staged together. So, that you get the length of the machine, which in this
case will be perpendicular to the paper.
Similarly, this projected pole is also having a length perpendicular to the paper. Anyway
so, this is the basic structure, but the armature is slightly complicated. And, why it will be
complicated that we will discuss in detail do not worry about that, but essentially armature
has got slot and teeth, where there will be conductors placed, ok.
And, so; and this field will be excited by some DC source and this field current is often
denoted by 𝐼𝑓 , constant value DC current, and depending upon the strength of this field
current these are essentially electromagnetic. So, the strength of the value of flux density
in the air gap will be decided. So, lines of force will be completing their paths through the
stator iron, cross the air gap, armature iron, cross the air gap and back to square one. So,
this is how the field is created.
Now, our concentration would be about the armature coil. Now, armature coil is you know
in my electrical machines II lecture, I have discussed in the winding portion of the course
the basic terminologies used in making a winding, ok. And, I will go rather quickly here
to tell you about that ok, but you can if you like please see those videos, whose links will
be given in your website for this course.
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Now, we must understand that the rotor being a structure rotor will be like this, there will
be slot and teeth, slot and teeth, all along the rotor this is the rotor, only a portion I am
drawing. And, I will place a conductor here, which will be take a piece of wire put in one
slot at the end you turn around and through another suitable slot you bring out this end.
So, at the end you will have two terminals for this coil, ok. And, you can of course, have
a multi turn coil, that is you take this wire single wire come here, then once again repeat
these. So, a multi turn coil we will look like this one take a piece of wire, this is one slot it
goes it at the end it will be like this. And, then make several turns like this, another turn it
is a two turn coil, and two terminals would be there. And, this is a coil ok; and, this coil
has got two sides, this is coil side 1 and this is coil side 2. And, this coil physically will be
placed in the two slots I am sorry I have put it in the teeth, it will be actually in slots.
So, it goes comes back like that and comes back. So, in these two slots suppose this coil is
placed it can be placed around that. And, this can be understood in this way, that suppose
these are the slots here.
(Refer Slide Time: 07:17)
Suppose this is the slots and these are the suppose slot and teeth, I just a this simple thing
and then you have a coil.
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(Refer Slide Time: 07:41)
Anyway, this you see just this is the teeth, this is the teeth, these are the slots and, therefore,
you this is one coil; this is one coil side it goes along the length then it comes back and it
has got two terminals. Similarly, another coil, it has got this coil side it goes around and
returns and two terminals for the second coil and so on, is that clear.
(Refer Slide Time: 08:21)
Physically a coil will look like this a practical coil, it is just made two turns coil, you take
a piece of wire on a format you make this two times you go around and these are the two
terminals of the coil, these two clear. And this, what I will do, I will place it in one slot,
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this coil side and the other coil side at some appropriate slot, arbitrarily it cannot be placed
ok. This is how the coils will be placed;
One good information is all coils are identical ok, this is one coil, another coil will be also
placed in these slots, only displaced from the first coil by some space angle that is all. But
the idea is very clear so, each coil will look like this and it will be in this fashion.
Now, the question is so, each coil has got two coil side, coil side 1, coil side 2, that is what
I have written here, coil side 1 and coil side 2. This is how I have tried to and it is almost
like a diamond shaped. The effective length of the machine is this one, which will be the
under the influence of the North and South Pole and, these portions that is this portion and
this portions are called overhang of the coil, overhang. There no voltages will be used,
because it will be under the purview of the magnetic field produced by the stator when it
will be running.
Now, the big question is where this, what should be the difference in number of slots
between these coil sides. So, two coil sides make a coil remember, this two coil sides make
a coil with two very distinct terminals that is all and all coils are identical.
(Refer Slide Time: 10:59)
Now, if I ask you this question that, consider a simple two fold DC machine that is here,
this can be easily understood. Suppose field coils I am not drawing it has created North
Pole and South Pole and here you have the armature. And, there are one slot here and you
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have a coil side here, which is perpendicular and by coil side I know where it is part of a
coil one coil side. Suppose this is coil side one, what will be the induced voltage in it.
Suppose this armature is moving in this direction it is like this. So, induced voltage will be
you apply you have to apply right hand rule 𝐵, 𝑣 and 𝑙; so, this will be cross will be the
induced voltage or if you change the direction of rotation. Suppose, it is moving in this
direction, then the induced voltage will be like this 𝐵 and 𝑣 is this way this is the direction
of 𝐵, North to South lines of forces are. And, your ah this coil side I will show a dot to
indicate the induced voltage polarity, is it not 𝐵𝑙𝑣.
Now, this is a single conductor, but to make it a coil side I must; it must have a another
coil side like this here. Therefore, what you do is this you suppose there are two slots,
which are diametrically opposite. And, this is one coil side and this is the another coil side
ok, that is coil side I was telling nha where the coil side should be placed, suppose it is a
two turn coil.
So, one coil side is there and the other coil side in which slot should I place there will be
number of slots, I will place it so, that induced voltage in the coil becomes maximum
across between these two terminals. Suppose, it is in generator mode, if it is running there
will be induced voltage in this conductor cross and in this conductor it would be cross,
because velocity is this way B is this way you can once again apply right hand rule.
So, what happens is this the polarity of the induced voltage, this will be plus, this will be
minus for this coil side coil side 1, and it will be minus, it will be plus if it is coil side 2
like this, because it is on the South Pole having same velocity. Therefore, across this coil
this is the induced voltage in coil side 1 with this polarity this is the induced voltage in coil
side 2 with this polarity. So, you will get 2𝐵𝑙𝑣 as the induced voltage between these two
wires, clear.
Therefore, you must place the coil sides of a coil intelligently. So, that across the coil, if
you allow that coil to move related to a magnetic field, the induced voltage in the coil
becomes maximum, which ultimately told in a different language I will always say that if
one coil side at a given instant of time. If it is under the center of the North Pole, the other
coil side must be under the center of the South Pole at the same time.
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It will rotate after some time if it is rotating it will come here this will go there, but it is
still under South Pole may be at reduced strength, but once again the voltage will be 2𝐵𝑙𝑣
with the magnitude will not remain same, is it not. Therefore, the essential consideration
is that, if this coil side to check you have properly placed it or not imagine that one coil
side, if at a given instant of time is under the center of the North Pole, the other coil side
of the same coil must be under the center of the South Pole, this is how we tell that. If, that
is the case then you check the maximum induced voltage here.
For example, if somebody says he has put the two coils he has made a coil and he has put
it like this. This is there was a slot available here and there was a slot also available there,
he; and the it is a two Pole configuration like this North and this is South and he places
one coil side, he make a coil, he makes a coil such that one coil side is here and it is return
coil side is here, when the center; when this coil side is under the center of the North Pole
the other coil side is in the magnetic neutral zone 𝐵 = 0 here, or tangential to the velocity
if velocity is this way. Can this conductor produce voltage? No, because 𝐵 and 𝑣 are along
the same lines 𝐵 is from left to right.
Therefore, in this case of course, at this instant there will be induced voltage under this
condition, what will be the nature of the voltage, nature of the voltage will be like this.
This will have induced voltage like this 𝐵𝑙𝑣 and this will have 0 voltage this coil side. So,
you will get voltage across the coil, but only 𝐵𝑙𝑣.
So, that is why you always insist upon that for a given coil, if one coil side is under the
center of the North Pole, then other coil side must be under the center of the South Pole
that will tell you in which slots the coil should be placed. And, as I told you that will be
number of coils number of slots, but all coils are identical ok, in different slots you go on
putting the coils in a particular fashion that is the, that is all.
So, this is like this if you have understood me, then you should tell what happens, if it is
suppose the machine is 4 pole. What do I mean by 4 pole? 4 pole means on the, I will go
to next page.
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(Refer Slide Time: 19:37)
4 pole means, suppose on the stator there will be 4 projections, this is your stator. Not very
nicely drawn, but it will give you the idea and this is the stator iron. So, if you and suppose
this is the sectional view of those coils. So, there will be now 4 field coils, got the point.
So, I want to make this is not, so the currents are like this cross dot here, then the next
poles will be a South Pole. So, lines of force will go in therefore, it must have cross here
and dot there. So, that it will become South Pole. And, this one once again cross and dot
so, lines of force will enter; oh, no this I have to make this cross and this will be dots.
So, that once again it will become a South Pole, North Pole, that is here lines of force were
coming out, here also lines of force will here. So, alternately you have to create the poles,
north-south here it is north there I want to make it south. So, it must be cross and this
should be dot and lines of force within go in. So, lines of force will complete their path
like this, north south, north south and this is north this is also create a lines of force like
this and this will also create a lines of force like this. So, north-south, north-south this is
the thing.
So, and this field coil should be connected in series in such a fashion the current
distribution in this conductor is like this, then only this pattern will be created ok. If, that
be the case then a in this simplified diagram, now I will not show the coils, I can create a
4 pole stator as simply as this with 4 projections with conductors around it. And so,
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suppose you have created this poles, north, south, north, south, then it is called a 4 pole
decimation, full stator I am not drawing.
Now, let us come back to the armature, which is a circle with lot of slots and teeth. For
example, here is a slot; here is also a slot, ok. And, there is also a slot like that it is available
and there are many other slots here there, many slots are there, ok.
Now, I say that I will make a coil like this as usual, if this is coil side one of a coil where
do you like to have your other coil side should it be here diametrically opposite. If, you
place, if you make a coil whose span is this much that is this coil side, this coil side, then
across the conductor there will be induced voltage, but the polarity of the induced voltage
will be subtracted. And, across the coil you will get zero voltage, that is why the rule is
that is you must ensure, if one coil side is under the center of the North Pole, the other coil
side must be under the center of the South Pole to ensure that maximum voltage is induced.
So this is how the coil span, this is called coil span ok. So, this is the thing ok, then a coil
span I will always say it should be equal to 180° electrical. Now, the question is what is
180° electrical? 180°, whenever we say angle we say in mechanical terms we understand.
For example, 90° is this, is it not and 180° is this, these are all mechanical angles. Now,
what do I mean by 180° electrical, the idea is simple, it is like this.
(Refer Slide Time: 26:37)
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That is if you imagine, you have a two pole structure like this and you have a single
conductor. Suppose a single conductor and you allow it to move with certain velocity or 𝑛
rps equivalent to that. If, it moves the voltage across this conductor, in this position it will
be 0 is not, because 𝐵 and 𝑣 are along the same line. It will be having some maximum
value when it comes under South Pole, then once again it will be 0, then once again it will
be maximum, but polarity will reverse, because it is now under North Pole and once again
back to 0.
Now, listen to me carefully. If, this conductor makes a complete 360° rotation, your output
voltage across this conductor; across this conductor means across this coil side, other coil
side I have not drawn I could also draw that ok, or if you fill put another coil here other
coil side. So, voltage across the coil is 2𝐵𝑙𝑣 at any time and it will be 0 here, then either
positive maximum or negative maximum.
So, with respect to time if you sketch it is 0 here, then it attain some positive value is it not
when this fellow comes here it will have some positive value say, then when this fellow
comes here it will have once again 0 voltage. So, like that it will grow by some rule, but
anyway it will have a peak back to 0 here, then when this conductor makes a complete
rotation of 360° mechanical you will get one cycle of emf generated across the coil, AC
voltage will be generated 𝐵𝑙𝑣. But if you have a 4 pole structure, if you have a 4 pole
structure like this north, south, then once again north, once again south. And, I have learned
what, where this coil should be placed here one conductor slots I am not drawing it will
have two coil sides and you allow it to rotate.
So, if this arrangement rotates by 360° mechanical, one rotation mechanical rotation we
will ensue what two cycles of emf. Because, it will undergo north then south, then once
again north and then once again south back to north. So, one complete rotation of the rotor
will induce two cycles of emf, here that is one rotation one mechanical rotation will induce
two cycles of emf.
Here in two Pole configuration one rotation; one mechanical rotation will induce one cycle
of emf, got the point. So, here the conductors or the coil undergoes sees only two poles,
north-south one cycle of emf. In this case coil will see north south once again north south
till it makes a complete mechanical rotation.
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Therefore, this angle for electrical induced voltage is the 𝜃𝑒 , that is 360° electrical, is it not
and this is 180° electrical. In case of mechanical two polar machine we will see that 𝜃𝑒 =
𝜃𝑚𝑒𝑐ℎ no difference angle in 𝜃𝑒 and 𝜃𝑚𝑒𝑐ℎ , when you plot these sketch the voltage in terms
of 𝜃, 𝜃 = 𝜔𝑡 it will become eventually. So, it will be like this.
So, in this case I will say
𝜃𝑒 =
𝑃
𝜃
2 𝑚𝑒𝑐ℎ
That is you move by 90° means you have moved by 180° electrical. So, think about this,
this is very basic concepts that I will use while discussing about the armature winding of
DC machine and understanding of this is essential, you cannot avoid. And, better look at
the detail introductory part on the winding lectures of my electrical machine two course. I
am briefly telling, but rather quickly because no point in repeating the same stuff once
again which is already available.
So, see that ideas are simple; so, I must distinguish between electrical and mechanical
angle. Number of poles of a machine will be always even, it cannot be odd, because you
cannot have a monopole. If, there is a North Pole accompanied South Pole must be there.
So, I have indicated the basic idea of this and next time we will continue from here.
Thank you.
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Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 59
Armature Winding of D.C Machines - I
Welcome to lecture number 59 and we have been discussing about the DC machines, in
our last class the diagrams were a little bit clumsy.
(Refer Slide Time: 00:28)
But nonetheless I explained to you that in a DC machine there will be a stator structure,
which will have projected poles like this and it could be a 2 pole or multipolar stators this
north south poles will be created by coils, which will be put around it these are called field
coils and on the rotor side there will be an armature.
Only thing important thing I mentioned about this equation that, each coil has got 2 coil
sides and if at any point of time this rotor will rotate, if one coil side of a coil is under the
center of the north pole the other coil side must be under the center of the south pole so,
as to maximize the induced emf. After all induced emf is 𝐵𝑙𝑣, velocity of the conductor
will remain same and therefore, the nature of the induced voltage will be same as the nature
of the 𝐵 distribution along the air gap.
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Now, today I will try to explain that in a much more clear diagrams, the situation will be
like this diagram if you look at.
(Refer Slide Time: 01:56)
So, if you see that these are the poles stator poles where there will be coils around it to
create north south; north south and this armature it is slot and teeth slot teeth where the
conductors will be placed and this whole thing has got a length perpendicular to the plane
of the paper and this fellow will rotate and the conductors which will be placed here will
be the armature winding ok.
(Refer Slide Time: 02:46)
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So,. So, this is the thing and if you look at this diagram on a paper the same diagram I am
telling, what is to be done is that this stator poles will be created by conductors placed
around it is not and suppose this carries cross current and this carried dot current then this
becomes a north pole.
Similarly there will be coils around the other poles as well and I want to create a south
pole. So, the current distribution in this coil should be this cross this dot so, that lines of
force will come out here lines of force will entail. So, this will become north south and
then once again this will become north south, north I want to create.
So, this will be cross cross dot dot and this is the another field coil identical all the coils
are and they and I want to make it south pole therefore, this should be cross and this should
be dot.
So, these coils which are wrapped around the projected poles are called the field coils. So,
north south; north south four pole machine it will become, and the rotor is this one where
there will be slots and there will be conductors placed. Conductors means coils will be
placed and each coil has got two coil sides and this I will show like this.
So, these conductors will be placed in the slots of the rotor the this structure is rotor there
will be a shaft here shaft with a key shaft perpendicular to the paper this is made of iron
and these are called armature conductors like this ok.
Now, I will tell you how this what is this coil is all about. If you see because in this diagram
I cannot show the length of the machine. So, let me try to show it this way.
(Refer Slide Time: 05:30)
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See this is the actually this will be the rotor are you getting? This way and it is because the
rotor windings is the most crucial in DC machine stator windings are very simple there
will be four coils, if it is 4 pole machine and these coils are to be connected in series and
ultimately two terminals will come out they are called field terminals, which is to be
excited by some DC current creating north south north south poles but the armature
winding is rather interesting.
(Refer Slide Time: 06:14)
So, this is a coil which is diamond shaped like this, it is a coil like this and this coil has got
two coils sides. If at any time these coil side will be under north pole the other will be
under south pole. So, how this coil will be placed? It will be placed in its designated slot
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like this here ok. That calculation we will do and it will have two terminals that is what I
want to tell this is how the coils will be placed and these are two coil sides to make a coil
you required two coil sides and these portions are over end portion of the coil.
Then in this portion of the winding there will be no induce voltage because north and south
will be all along the length of the machine there will be a north pole, south pole like that.
Similarly this side also there will be and it is a multi turn coil. In fact, it is a two turn coil,
it could be a 10 turn coil if you take a piece of wire make it like this on a wooden format
this is to be made then it is to be placed in the designated slot you cannot place it arbitrarily
of course.
Because I know that I have to be very careful to see that the coil span this degree in degree
it should be 90°, or 180° electrical and mechanical angle last time I told which will ensure
that if one coil side is under north pole the other coil side will be under south pole.
Only one relief is there that all coils are identical. If you know one coil the other coils will
be placed in some other slots that is all with these two terminals start finish start finish like
that I will put following certain grammar, certain logic which will fill up all the slots and
we will be proceeding towards making an armature winding.
Armature winding is somewhat difficult to understand compared to AC machine windings.
The reason I will tell you later because our goal will be here to produce DC. See what
happens is this if you have a single coil like this, and this rotor is rotated under the influence
of north south north south, then we know across the coil there will be AC voltage induced
but I want to make it a make it a DC generator. So, how to make that AC voltage converted
to constant value DC voltage to be generated?
So, what I have to do something extra after getting this AC voltage in all the coils there
will be AC voltage induced each coils are identical same AC voltage will be induced only
thing is the phase angle among the induced voltages will have some phase angle difference.
So, because of whatever is happening to this coil after some time same thing is going to
happen to other coils but nonetheless the voltage will be of alternating time that is one
thing.
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Second thing is what we will do is this, we will have what is called double layer winding
positively in DC machine. So, I will be using double layer winding what does that mean?
It means that each coil for example, here let us calculate here how many slots are there;
you can easily see there are 12 slots there are 4 poles then how many slots will be there
12
per pole; it will be 4 = 3.
Therefore if I will use a different color suppose this is one coil side; coil I have already
shown this is one coil side it goes and from where it will come out; because the angle
between south and north pole is 180° electrical therefore, number of slots per pole will be
3 therefore, its return should be 1 2 3 after that its return will be that is its return will be
here.
We will calculate this, this is one coil side this is another coil side that is it is a coil this
red mark thing each one coil side will occupy the upper deck of a slot and its return coil
side will occupy the lower deck of the slot that is in each slot there are two decks double
layered winding it is called. Therefore, each slot we will house two coil sides belonging in
general to two different coils.
We will see how this will be nicely filled up that we will see later. In other words what I
am trying to tell that in terms of this diagram, this is the coil this coil when I make it I will
these two coil sides I do not make it planar in one plane what I will do is I will press this
coil side a little higher than this.
Are you getting? So, this is the coil side and its return will be a little lower there these two
coil sides will not be in same plane. Then only if you place one coil here like this its return
will be in the lower up deck of the slot and it is the starting coils side of this coil will be in
the upper deck got the point hopefully.
Let us proceed. So, in each slot so, this red mark coil is a coil this is one coil side multi
turn coil it is there are several turns and upper deck will occupy that and lower deck will
occupy that. For example, if I say ah the slots number suppose 1, 2, 3, 4 then 5, 6 oh I am
sorry this is not this is a teeth 5, 6, 7, 8, 9, 10, 11, 12; is not? Number of slots per pole
12
which will be equal to 4 = 3.
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Anyway slots per pole we will calculate. So, you please correct that, you have understood
in these slots these conductors are to be placed.
(Refer Slide Time: 15:21)
Now, what I will tell you is this that, now we will be talking about double layer windings
double layer winding ok. That is each slots conductors will be placed on slots each coil
will have two sides and we will assume coil span is equal to 180° electrical and this I have
to do in order to maximize the voltage that is what I explained in the last class.
Suppose let number of slots number of rotor slots that is we are talking about armature
winding number of rotor slots I will denote it by capital 𝑆, suppose side by side I will go
with numbers suppose it is 16 and let the number of poles number of poles is equal to 𝑃 is
16
equal to say 4 then slots per pole is equal to 4 = 4 and that will be equal to coil span.
What does that mean; that is coil span is the difference between the two coil sides of a coil
in terms of slot number in terms of slot number like this two coil sides this will be placed
in one slot and this will be placed in another slot they will be numbered take the difference
of these two number and that has to be equal to 4 in this case. So, this is coil span and this
is equivalent to 180° electrical or in this case 90° mechanical that is the thing.
Anyway this is the thing. So, I will draw the diagram in this way now, now follow me very
carefully what I am doing. Suppose each slot will have two coil sides because it is a double
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layer winding therefore, total number of coils has to be equal to number of slots that is
equal to 𝑆 = 16; is not.
Because each coil requires two coil sides how many coil sides will be there; 32 because
each slot will carry two coil sides therefore, number of coils will be half of the coil sides
that is 16 itself. So, in general whatever is the number of slots that will be equal to the total
number of coils I can put in the on the periphery of the rotor slots that is.
Now, how do I go about placing the coils like that; so, that calculations are simple if you
follow me carefully what I want to.
(Refer Slide Time: 20:20)
Now, suppose I want to there are 16 coils 16 slots are there, in this diagram I want to tell
you like this, these slots I will show like this slot and teeth in the developed diagram. This
is the rotor periphery like that develop diagram means this diagram.
(Refer Slide Time: 20:44)
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Circular thing, you have unwinded it and you are showing it, this is perpendicular to the
board like this, this diagram I have drawn hope you have understood.
Now, these slots I will number like 1 2 3 4 and dot dot dot it will be 16,and after that once
again slot number 1 is not? Developed because it is after all circular once again one this is
how slots will look like in the developed diagram.
Now, what I will do, I will take one coil. So, I will write here this is called winding table
very simple otherwise if you please follow me at least for windings and this column I will
write coils and each coil I will show like this.
Suppose I say that I will start; I will take one coil and I will in the upper deck this is the
starting coil side and its returned first of all where its returned will be? Slots per pole is 4
therefore, it will be (1 + 4) that is its return will be here. This will be the one coil one coil
means this fellow I have put it and as I told you while manufacturing this will be slightly
below and this will be slightly up so, that it will be nicely placed here.
So, this slot number is 5. So, out of 16 coils, I have put only one coil and slot number one
I have put it. So, this coil side I will say it is 1, I start and its return is at 5 what all we do
is this that in the lower deck whichever coil sides will come, I will tell them as 5’ upper
deck coil sides I will number with without any prime and lower deck. So, 1-5’ is 1 coil
whereas, it is two coil sides?
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In slot number 1 and slot number 5. In slot number one it occupies the upper position of
the slot there is a vacancy here, somebody will come here do not worry. So, everything
will be filled up and in each slots there will be two coil sides. So, this is one coil it is over.
Now, what happens is this there will be 16 coils and all the 16 coils will be connected in
series and the circuit will be closed. Now the next coil I will take and I will start like this
next coil I will start from 2 and its returned will be in 6’.
So, this is coil 1 this is over and it is multi turn coil mind you although I am showing one
section and the second coil I will take which will start from 2, plus 4 and its return will be
here that is in slot number 6. Although I have not connected the coil each coil will have 2
free terminals but this is how I will continue my waiting that is what I am telling.
So, 2 and it will be 6’. So, I am writing it with red 2-6’ similarly third coil I will now not
take much space because 16 coils I do not know whether I can show it but it can be done.
So, so this is the thing or I will rather clean it what I am doing? So, coils I will do it nicely
coils.
So, we have seen first coil symbolically, I am writing and it has got two terminals 1-5’
second coil it will start from slot number 2-6’, 3-7’ number game it is all coils will have
same span identical coils, then 4-8’, 5-9’, 6-10’, 7-11’ ok. There will not be enough space.
So, let me draw it here continue, I could draw it in this way.
Then 8-12’, 9-13’, 10-14’, 11-15’, 12-16’, 13-17’ there is no 17’ what it should be; 1’ are
you getting? 17’ I am writing, but there is no 17’ it has to be equal to 1 because with the
circular 1’.
So, I clean it you have understood whenever it will be exceeding 16, you have to write
down this one 1’, then 14-2’ and 15-3’ and this is 16-4’, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,
13 all the 16 coils I have taken. So, this is how I will fill up all these slots and in each slot
each slot will house two coil sides.
For example, slot number 1 will house coil number 1 start and coil number 13 returns like
that. So, this can be done then what I am telling this diagram is not the actual coils, I mean
actually I know how they are disposed in space that is fine but so far there is electrical to
understand how they are connected I have just drawn each coil represented like this.
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Now, what I will do is this all these coils I am telling I will connect them in series. So, all
the coils I will connect them in series; got the point? This connection now I now do because
all the coil each coil has got two terminals available to me and I will connect them in series
and not only in series all the coils and I will close it like this, that is all the coils in series
and the coils are closed on to itself.
Now, here one may raise objections that is there are 16 coils which are put like this, where
are the coils gone; this is these are the coils we can put on this and each coil is identical
they are displaced from one another this is coil 1, coil 2 will be like this, coil 3 and you
have connected with piece of wires in series this coil next coil and so, on and everything
will be rotating under north south magnetic field and each one of them will become a seat
of emf; is not.
So, and I have connected all the coils in series knowing fully well that each will become a
seat of emf these are no dead terminals there is induced emf here in this coil.
Next coil will also have induced emf and so, on if you go on adding. So, when you connect
them in series what you are doing; you are adding these voltages here, this voltage, this
voltage, this voltage and so, on. Now the big question is if that is the case and if you close
will not there be a circulating current.
The answer is no; because of the fact the polarity of the induced voltages in the coils will
be such that the addition of all the voltages AC voltages there mind you, I have not told
anything about how to convert it to DC till now but I know there will be source of AC
voltage here like that and if you add all the voltages the sum of the voltages will be 0.
The reason is very simple. If first coil voltage is this he is represented by a phasor like this,
second coil voltage will be lagging this by some angle 𝛽, third coil by this one like that if
you go and you will end up with coming here to the same coil after you have traversed all
the 16 voltages.
Therefore, resultant voltage acting in this closed coil which are closed each one of them is
seat of emf then they some of the induced voltages will become 0. So, we need not fear
there will be a circulating current although I know pretty well that each one of them has
become a seat of emf.
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Let us also try to see that I told you there are how many slots under north pole. There will
be 4 slots under south pole; there will be 4 slots; machine will be running. At a given time
if you freeze if you take a snapshot, it is expected that 4 slots will be under the influence
of the north pole and next 4 slot will be under the influence of the south pole and so on.
So, suppose this coil size 1 2 3 4 are under the influence of the north pole and the direction
of rotation is such that they become plus; these 4 coils; these sides become plus. Obviously,
the other will be under south pole and they will become minus; is not? Instantaneous
voltage I am telling.
Then the next four slots 5 6 7 8 will be under the influence of the south pole they are
starting and the polarity of this induced voltage will be reversed minus, plus, minus, plus,
minus, plus and minus plus why not; and then the next 4 will be under the influence of the
north pole and this will be plus, minus, plus, minus, plus, minus and plus minus and the
other one other four; 13, 14, 15, 16 will be under south pole and this will be once again
minus, plus, minus, plus, minus, pluss, minus, plus.
So, if you start from this point traverse all the emfs you will be seeing as many plus. So,
many emf in the opposition acting end up with a zero voltage between if you start from
here traverse all the coils come back here zero voltage although there will be induced
voltage across this. So, how did I put this plus minus; let it be very clear. Coils are identical
they will be moving at some high speed no doubt fine.
And there are 16 slots are there, suppose it is expected then under the influence of this
north pole at a given time it is expected there will be 4 slots under north pole these 4 slots
under south pole 4 slots. If you take a snapshot instantaneous snapshot under that condition
what I am telling, coil sides are there under the either under the influence of north pole or
south pole.
Now, the question is let us assume that 1 2 3 4 is under the influence of north pole and
direction of rotation is such that I apply right hand rule and come to the conclusion that
then the coil sides 1 2 3 4 they are plus then about 5’, 6’, 7’, 8’ time certain they has to be
minus they must be under south pole because coil span is 180° electrical ok.
Then the next 4 slots will be under the influence of the south pole direction of rotation
remain same. So, polarity of the induced voltage reverses and so, on. So, I stop here today
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to conclude that the armature windings I am discussing I am of a DC machine which is not
complete.
I told you there will be several coils are to be placed in slots and all the coils are first to
be connected in series and circuit is to be closed questions still remains that is fine; there
will be no circulating current but my goal is to how to get DC so, that we will try to further
explore in the next class.
Thank you.
587
Electrical Machines – I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 60
Armature Winding of D.C Machines – II
Welcome to 60th lecture and we have been discussing the most interesting part of DC
Machine that is the Armature Winding. It is slightly difficult, but at the same time, so
interesting that is the point I want to make.
(Refer Slide Time: 00:37)
And you recall that, in my last class I told there will be slots, let us assume the number of
slots 16, each slot will house two coil size belonging to two different coils. All coils are
identical and the coil span is this one ok. It is a multi turned coil I have drawn, these are
the two terminals distinct terminals, this coil has got its own identity, ok. This coil I was
marking it has got a starting; start and finish terminals. This start terminal this coil sides,
if you start with slot number 1, that coil you calls coil 1, it will be designated at start 1
finish 1. There will be a second coil, which is identical, ok, this is the coil, this is identical.
This coil is identical; only thing is its number is to be changed like this. This is second coil
start from slot number 2, I am showing it because it is displaced. It will be marked as that
is what I am telling you, s2 and f2 start finish. And this is in slot number 1, this is in slot
number 2. Of course, slot number 1 and slot number 2 will be close by, it is like this it will
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come here. But for clarity I am showing you, then what I am doing I am connecting all of
them in series. In the machine actually I am doing like this; similarly s3, f3. Where in this
simplified diagram it can be shown like this, ok. This is actually start of coil 1 and finish
of coil 1; finish of coil 1 is in slot number 5’, this is start of coil 2, finish of coil 2 and so
on. Then you get this, but this type of winding is called lap winding, ok.
(Refer Slide Time: 03:30)
The reason is clear because of the fact, the first coil which was here I will just draw the by
this simplified diagram. This is the first coil, it is in slot number 1, second coil which I will
draw with a diagram like this will go like this, is not? And these are the 2 terminals of the
second coil. So, this is in slot number 1, this is in slot number 2. Where this fellow is there
for this problem we have considered, this is slot number 5, 5’ because it is in the lower
deck, this is 6’. So, you see and these coils I am connecting in them in series.
Second coil is on the lap of the first coil, similarly third coil will be here in slot number 3,
I think you have got the idea. Is not? Start finish of the third coil that will be on the lap of
the second coil, that is why it is called a lap winding, ok; in this way I have proceeded to
complete the winding of the coils, ok. If that be the case that is fine as I told you, but we
know that if a coil is moving this stator field is stationary, nobody is rotating because DC
current you are passing to the coils and stator poles have become magnetized alternatively
as north-south, north-south like that. And their magnitude is same and they are stationary
in space, unlike say induction motor, it produces a rotating magnetic field.
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Therefore, to induce voltage in the rotor coils to make it run as a generator I must use a
prime mover to run the generator. Coils will move and I know that there will be AC voltage
induced across the terminals of each coil, depending upon its position in space relative to
that field the polarity of the induced voltage will be decided, that we have discussed. Now,
the question is I want to make a decision latter, that is this voltages which have alternating
in nature with time the polarity of this voltage across start and finish of a coil will change
reverse its polarity AC voltage you will get. So, how to do it, let us see that.
(Refer Slide Time: 06:50)
Now, I will draw another diagram like this. Look at this diagram. Here I have shown it the
coils I have represented it is abstract; abstract in the sense I know how the coils are
disposed in these slots specially. Suppose for example, s1, f1 this is the coil number 1. I
know it starts in slot number 1, returns from its finish some slot number. I am sure about
one thing, if this s1 fellow this coil side of this coil 1 is under North Pole, the other its
return will be under South Pole, is not, that is what I know. Therefore, these all these 16
360°
coils, I have shown like this. They are displaced from each other by that angle 16 = 22.5°
and so on.
And what I have done. I have connected all the coils in series ok. So, I have done it like
this, finish of the; follow me very carefully, this I have shorted because all coils have been
connected in series; these what I have done. All the 16 coils have been connected and the
circuit is closed; is not, this is what I have done. Now in the previous diagram, I drew it
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much more simpler way, but this is a somewhat better representation to my mind. Now
what I will do is this, remember this conductor will be moving in a particular direction and
I have taken a snapshots at a certain time frozen everything. Then I would expect, the four
coil sides, the four slots will be under the influence of the North Pole, next four slots will
be under the influence of South Pole and so on.
Let us assume that 1, 2 these are slots number also. 1, 2, 3, 4 these slots are such that the
induced voltage polarity are positive. And obviously, they are returned they are under
South Pole that is this side, coil sides, this must be minus, this is how? Then the next four
coil sides, that is s5, s6, s7, s8 direction of prodigious same but they will be suppose under
South Pole has to be if it is under North Pole s5, s6, s7, s8 must be under South Pole and
the polarity of the voltage will reverse. So, it will be minus plus, minus plus, minus plus
and minus plus. And similarly, then 9, 10, 11, 12 will be once again under North Pole 9,
10, other will be negative 9, 10, 11 and 12.
And this will be then once again opposite, minus plus, minus plus, minus plus and minus
plus. So, I have taken a snapshot and I know this will be the induced voltage ok. But where
is DC about it? Mind you the magnitude of the voltage instantaneous values polarities are
like this may not be same with this one. Each one of them, because the value of 𝐵 might
change, under the North Pole no doubt, but the value of 𝐵 may change, is not? In fact, 𝐵
distribution let me tell, in case of DC machine it will be like this. Somewhat trapezoidal if
this is the thing; this is the 𝐵 distribution, ok. So, this will be the trapezoidal 𝐵 distribution,
flux density distribution, 𝐵𝑙𝑣. Here the conductors are moving.
So, the value of 𝐵 here, suppose this is under South Pole, this is North Pole ok. Conductors
are moving here, relative to this field. Those who are having under South Pole, they will
have some polarity plus minus etc, but each one of them will have same polarity because,
they are under one pole. But the magnitude of the voltage of this coil will be different from
this, no doubt that is this polarity of the induced voltage, although it will be plus minus,
plus minus for this four coil sides, but their magnitudes may differ, depends upon that
position of the conductor and the relative value of the 𝐵 there ok. 𝐵 distribution, what is
this axis? Space angle. There is no compulsion that 𝐵 should be sinusoidal. In case of AC
machines we insist that 𝐵 distribution let it be sinusoidal.
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But, here my goal is to get DC, why? I unnecessarily insist upon that fact make 𝐵
distributions sinusoidal. It will be made DC in a very nice way. In fact, 𝐵 distribution has
to be like this because the pole, if you draw one pole, stator pole under this if you suppose
this is South Pole, this is the center ok. Air gap air it will be minimum here it will the
influence of the previous North Pole has ended so, your 𝐵 distribution will be somewhat
like this then flattened, ok.
Rotor structure is like this. So, here air gap is more uniform it will be flattened and it will
once again come to. So, trapezoidal; 𝐵 distribution trapezoidal, but in a single coil it goes
on the south, north what will be the induced voltage, if you see in the oscilloscope. That
will be also trapezoidal. Whatever is the nature of 𝐵 distribution that will be the induced
voltage pattern, what else, is not. So, alternating voltage will be induced, that will be
trapezoidal in nature.
Now this is the crucial part now. I want to get DC voltage out of this picture I mean what
should I do and this mind you it is rotating, time elapses. Once again another it is; right
now it is 1, 2, 3, 4, 5, 6, 7, 8 will be there after some time the 16 coil will come this side
and coil 8 will go to that side, are you getting this is rotating nah, I have taken a snapshot.
But at any given time this name is now not important. What I am telling, I would expect
there will be 8 coils on this side and 8 coils on that side. Because this fellow is rotating ok,
this coil goes to that side, but it will be filled up by this coil coming to this side; are you
getting the point, this is most crucial. That is at a given time, this numbering I have done
fine to understand what is happening, but after some time 8 will come to this side, 7 will
occupy the position of coil 8 here, 6 will advance here, if it is rotating in this way and so
on, got the point?
Therefore, it looks like that in this space, these are position of this space. Suppose I say
that I will from the junctions here, I will bring out some wires; are you getting. From all
the junctions I will bring out some wire and this is conducting wire, this is suppose some
copper strip thick copper strip, like that. This is the most nicest part of this whole thing,
whoever first imagined this way it can be done excellent, I mean ideas.
So, you think in this way; that ok, from the junctions what I will do? I will drop down
some wires here and terminate it on thick copper strips each one; are you getting? So, it
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will be brought on to same distance, I mean you understand like this. So, these are junctions
of this series connected coils, you have brought.
And then if I say that look what you do, you take a carbon brush and therefore, what will
happen? If this fellow rotates this junctions will also rotate in space is not? If this coil is
rotating this can be also numbered, this fellow will also rotate it will come after sometime
here, if the direction of rotation this way. This fellow will come here like that. But, I am
certain about one thing, what is that thing? That you see this junction that is this point in
space, whoever; whichever coil comes, it will have a plus here.
Similarly, so what I will do is this, I will connect a carbon brush here, brush 1. And I will
connect a carbon brush there, in space B2. And they will be nicely placed and they are
fixed B1, B2 are not rotating; B1 and B2 carbon brushes, fixed in space, not moving. But,
what it will do is this, whenever any coil side come in this position, it will touch that coil
end plus and no matter whether it is 5, 6, 7, 8 or 1, 2, 3, 4 has come this side, this polarity
of this voltage will always be plus minus, no alterations, ok, newer and newer coils will
come this coil will go behind the same, it will go to that side. But this will be occupied by
this coil.
The interesting point is whoever comes in this position it cannot escaped. This destiny of
polarity of the voltage plus minus, it has to be this side plus this side minus no matter
whether coil 8 has right now here or coil 6 is there or coil 14 has come here; whoever
comes he will be treated in the same way, because 𝐵𝑙𝑣 decides the polarity a magnitude
of the voltage and polarity by right hand. Therefore, I will fish out from this series
connected voltage I will identify where is plus minus in space and in the proper space, I
will connect the brushes. After some time this copper strip will come in contact, mind you
this because of my inability to it is in the same position, when this coil come, it will come
in touch with B1; this junctions are moving, this brushes carbon brushes are fixed in
position.
And I am we can conclude that no matter let the armature rotate at high speed or whatever
it is, across B1, B2 you will always get DC voltage, no AC voltage. That is we are from
the space distribution of the polarity of the emfs we are ascertaining, we are satisfying with
arguments that whoever, whichever junction comes here it will have plus. This one plus,
oh I am so sorry I would connect it here this you please forgive me, this one this brush B1
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we I will connect it here. That will be the most optimum sequence, wherever plus joined
you connect it here. This should be the position of B1, B2 is ok, minus minus is not.
Always you will get across B1, B2 DC and these fellows are moving after some time this
junction will touch, after some time this junction will come but, whenever it comes it will
a plus plus here.
So, from the junctions you have to drop out some wires terminated some copper strips and
placed your fixed brushes in appropriate positions and so on, is not. Then it looks like that
wherever plus minus, plus minus, plus minus. This is here also I could place a brush very
nice, fixed brush say B3, because minus minus as joined. So, this is the negative brush,
this will be the positive brush. Similarly this one will be B3 is there a negative brush; B3
negative it will give and where plus plus has; I will connect another brush here B4. I can
tap the voltage from these positions and then what I will do is this, this is also plus.
So, have you got the idea? This is the essence of the thing. You have AC voltage
fundamentally induced in each coil, no doubt and it will be alternating in nature because
𝐵 distribution is alternating although not sinusoidally, it does not matter. It is trapezoidaly
it is varying. Therefore, I will get AC voltage induced across each coil, here if you see the
voltage waveform between these two terminals, that will be AC in the oscilloscope and
that will be trapezoidal alternating.
See the voltage between these two points, it will be AC. Only thing this voltage and this
voltage will have a phase displacement, because they are not put they will not be in phase,
their positions are different in space. But the interesting point is, I want to get a DC voltage
out of this AC voltage environment. Then it is done like this, argument is imagine that any
it is rotating at high velocity like that fine there are suppose 16 coils.
So, at a given time I will expect four starting coil sides will be under North Pole say 1, 2,
3, 4; s1, s2 s3, s4 under the influence of North Pole. And its return coil side must be that
is f1, f2, f3, f4 finished coil sides will be under the influence of this South Pole. And
suppose direction of rotation is such, then by applying right hand rule I come across this
polarity plus minus, plus minus, plus minus for this 4 coils. If s1, s2, s3, s4 under the
influence of the North Pole, I take a snapshot at that time. Then s5, s6, s7, s8 should be
under the influence of South Pole and they are returned or finished coil sides f5, f6, f7, f8
should be under the influence of North Pole. And that way I decide the instantaneous
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polarity of the induced voltage, how in space it is distributed, because each coil has got
two terminals.
Then I argue that let the rotor rotate, but at a given time it is expected not that 1, 2, 3, 4;
s1, s2, s3, s4 will remain always under North Pole. It is moving s1, s2, s3, s4 suppose it is
moving in the clockwise direction, it will move this way, but next s terminals will come
another four. And their distribution of polarity will be exactly same, when they occupied
the positions of s1, s2, s3 and s4 which was occupied earlier a bit earlier by those 4 coil
size. Therefore, in space the polarities of the voltages it remains same. Therefore, we now
know what to do to tap a DC voltage out of this AC distribution of induced EMF is that,
that where ever plus plus joins drop a; from all the junctions you drop conductors and
terminated thick copper strips, ok.
Then place some carbon brush, in space which will not move mind you these junctions
will move because they are at these junctions this fellow is moving. But this brush I will
not allow to move with some spring this type of arrangement stator structure I will hold it
in this position always similarly, B2 negative-negative, B3 negative-negative and B4
positive-positive. Then I will say that across B1, B2 you get a DC voltage.
And I will go further, I will join this positive-positive by an external wire, these are all
stationary. And I will join this negative-negative together and this I will say ultimately as
my armature terminal A1 and this negative-negative I will say as I may get terminal A2,
ok. We will continue with this in this next time very interesting, I mean how really from
AC voltage you get this DC voltage.
Thank you.
595
Electrical Machines – I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture - 61
Armature Winding of D.C Machines – III
(Refer Slide Time: 00:39)
Welcome to lecture 61 and we are going through Armature Winding which is more
complicated than an AC machine winding. And you recall that to convert it to DC we have
to do these things that is spatial distribution of polarities of induced voltage existing across
each coil. We have taken 16 number of coils for convenience to explain the things out and
assume the stator poles to be 4 mind you 𝑃 = 4 and this way I did it. I will just tell you
that suppose we say that it is a 2 pole machine with 16 coils, how this will look like.
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(Refer Slide Time: 01:25)
So, that you understand the previous one much in a better way. So, if it is a 2 pole winding
and total number of slots are 16, then I will first calculate coil span, number of coils a
number of slots per pole this is what I have to calculate slots per pole that will then become
equal to 8, is not? And with that understanding that the start of the coil under North Pole
and all the coils are connected in series of course, so, that thing I quickly do. So, that you,
what will happen if the number of poles is equal to 2 not 4.
So, as I told you all the coils should be connected in series nothing like that. So, these coils
have connected in series, will there be any circulating current? No, although each coil will
become seat of emf, because if I show the polarities of the voltages as many plus minus
voltages we will see in that series closed thing so many minus plus we will see they will
cancel each other.
So, if it is a 2 pole machine therefore, I would expect the starting coil sides of say the first
8 coils s1, s2, s3, s4 up to s8 they will have a fixed polarity. They will be under North pole
I have taken a snapshot when this 8 coils coil sides that is s1 to s8 coil sides are under say
North Pole and direction of rotation is such that polarities of s1 s2.
If s1 is plus f1 must be minus because f1 is the other coil side must be under South pole
that is how I have made the windings plus minus and s9 to s16 this coil side must be under
South pole coil. At this given instant of time; at this given instant see coils are moving, but
on an average 8 coils will be under North pole s1 s2. So, it will be like this.
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Therefore s9 to this one it will be on opposite polarity, it will be like this. Then the next
step is series closed then from the junctions I will draw it in a much better way now what
I will do I will draw a circular my god this one I will draw. Now, from the junctions I will
bring out some wires physically connecting it to this one and this side it will be a thick
copper strip, here there will be a thick copper strips, there like that and this is how I will
complete the diagram like this and here will be another 16 coil sides.
And then what I am telling? I will put carbon brushes in space which will be fixed that will
not be rotating with this armature coils rotating this copper strips 2 will rotate in a
particular direction. Then I after drawing these take after taking a snapshot and this thing
with time I am telling you this distribution of plus minus this will remain intact whoever
comes in position of s1 f1 after sometime a 16 f16 will come here and s1 f1 will advance
to that side, but it is destined to have this polarity plus minus that is the crucial point. Then
what I will do? I will put two carbon brushes I will take which are fixed where plus plus
has come I will call it B1 and where minus minus has joined this is also this carbon brush
it is positioned in such a fashion that this copper strip will be touching this whenever it
moves fast.
And this is say B1, this is say B2 carbon brush and these two I will say this is armature
terminal 1 and armature terminal 2 and this time only two brushes will do no 4 brush is
required, because you want to maximize voltage all the emfs here, no matter which coils
which 8 coils are on the right which 8 coils are on the left polarities of the voltages is such
that across B1-B2 you will always get DC voltage, got the point. Now, this is fine so, far
as understanding is concerned this is what you have to do, but in a practical DC machine
this type of arrangement it is to be made more solid I mean constructionally ok, what is
done is this in the actual DC machine.
So, brushes are to be placed in proper positions in space that is whoever has made the
winding he will tell you put the brush here put the brush there. So, that you will get DC
voltage in this case, this A1 become plus voltage it will give it will give minus voltage
rotate the armature in that field produced by the stator coils and you will get DC voltage.
So, that is very nice. Now, this one instead of doing like this it is to be done in this fashion.
This copper strips in effect I will I must tell that these are to be called commutator
segments, each coppers thick copper strips in this representative diagram and these are
fixed brushes; fixed brush.
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So, in the armature winding of a DC machine coils are closed connected in series then you
will have 2 additional thing, one is a commutator segments as many commutator segments
has the number of slots and there will be 2 or 4 brushes or 6 brushes depending upon the
number of poles if the winding is lap winding that is what I told you. Now, let us see how
this commutator segment is to be connected see from the junctions you have to drop this
one.
(Refer Slide Time: 12:53)
Now, let us come to this diagram that is suppose this is one coil let me draw in this develop
diagram, this is suppose one coil I will henceforth show with two terminals it is a multi
turn coil ok.
Now, what is to be done is this, below this on the shaft of the machine this is the shaft I
will put another thing like a ring whose which will look like this one. These are copper
strips the one I told showed with a circle it is on the shaft of the machine I will put it like
that these are called copper strips or called commutator segments. Copper strip conducting,
but each copper strip there will be as many commutator segments as the number of slots
because number of those circular thing we have seen it is equal to the number of slots.
So, those circle thick these lines we talked previously nah these I am telling these are the
copper strips in actual machines it is done like that ok. I have not brought that armature
winding today unfortunately next time I will show you, but this is on the shaft of the
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machine it will be inserted this way it will go and this is the shaft of the machine shaft got
the point.
And each commutator segment adjacent commutator segments should be insulated from
each other they should have their own electrical identity. For example, in the previous
diagram these two are insulated by space they should not be if they are touching purpose
will go very purpose will be lost. Therefore, what happens is this with a different color I
will show. The red portions are called mica insulations.
So, that these two each segment has it is own electrical identity, it is not sharing anything
up with the adjacent commutator segment, they are insulated from each other, got the idea.
Therefore, commutator segment 1 2 3 like that 16 commutator segments will be there on
a circular cylindrical structure and each one of them is separated by a layer of mica
insulation. And from the shaft also they will be separated insulated not that otherwise shaft
will become electrically aligned and it will be in fact, shaft all the commutator segments
not because shaft is metallic. So, insulations are provided so, each one with a weight like
this then a thin mica insulation thin and then another commutator segment.
(Refer Slide Time: 17:32)
So, it will be the replica of this stator armature like this it will be the replica of this one
here and it will be shown like this that is suppose this is the shaft in this sectional diagram
what I will do, I can show it like this, 16 such 1 2 3 each one is commutator segments these
lines are mica insulations like that. So, on the shaft it will be shown over a small radius.
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So, it will be replica of the rotor to slot arrangement only, but anyway these are all
commutator segment 1 2 3 4 to 16 got the idea ok. So, this is one coil, then there is a you
know another coil like this second coil is there I will draw with a different color, second
coil will be in the next slot it will start from here and it will return from the next slots like
this is it.
So, first coil and this is the second coil, this is the s1 and it is return f1 in terms of slots
this is f1 and this is start of second coil finish of second correct this is the thing. In terms
of slot number it will be in slot number 1 if 𝑆 = 16, 𝑃 = 4 then this should be f2 should
be in terms of slot number 6’ is not.
So, it will be 1-5’ and this is 2-6’ and this coils are to be connected in series that is s1 f1,
then I will start with this join s2 f2 and this process will continue it gives you a lap winding.
Now, below these I will try to show the commutator segment like this 1 commutator
segment also I will number them, 2, 3, how many commutator segments will be required?
As many slots are there, as many coils are there, because you recall that from the junctions
I need 1 commutator segments there are 16 coils. So, for each junctions I require a
commutator segments. So, this junction I will connect it in commutator segment suppose
1 that is where is the junction, first coil and second coil the junction I will drop it to 1, then
second coil I will drop it to 2 and so on ok.
(Refer Slide Time: 21:49)
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Now, I will go straight away to winding table in it is full form winding table, 𝑆 = 16, 𝑃 =
16
4, coil span for maximum voltage is equal to 4 = 4 slots. So, here now I do like this coil
and here I have to do another thing commutator segments two things must be there. Let
me try to explain to you, suppose this is first coil 1-5’, this one I will connect to
commutator segment to 1, where coil ends are terminated. It will be in 1 that is one I will
terminate it this is the coil I will terminate it in commutator segment 1 and next one I will
it is return I will terminate it in commutator segment 2.
So, commutator segments I will also numbered 1 to 16. 1 will be terminated here 5’. Next
coil 2-6’ this junction I will terminate it in commutator segment 2 and it is other end 3 like
that, 3-7’ in this way I will carry on for let us draw the full thing 5-9’, 6-10’, these are coil
ends mind you start finish start finish.
7-11’, 8-12’, then another 8 coils are there, 9-13’. Actually what will happen is this anyway
up to this point I will just do and all these coils are connected in series and continued like
this. So, it is 1-2, 2-3, then 3-4 like that, this way I will complete my winding tables up to
16, next time I will bring the complete diagram in a PDF form so, that I do not have to
spend much time.
Then what I want to tell is this there are now several coils connected in this way, that is
this is coil one each sides are connected in commutator segment 1 and next is 2, it is return.
Then next coil from 2 I will start next coil. So, next coil first thing is commutator segment
2 this blue one and it is return will be I will just show this one, it will be terminated in
commutator segment 3.
So, from the junctions those vertical lines are coming to the commutator segments and
these commutator segments because it is connected to shaft it will also rotate. Then I have
to take carbon brushes which will be not rotating and they will touch this commutator
segments at the appropriate place that I have indicated how it is to be done after showing
the polarities that I will do next time like this will be the brushes nothing else.
So, all in this diagram in this previous diagram this will be the commutator segment your
commutator segments will be moving and you make arrangements. So, that carbon brushes
will touch this commutator segments with the help of some spring and external thing. So,
that this fellow will rotate, but carbon brush will remain stationary.
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So, a typical brush will be connected like this at appropriate place that you have to decide
where plus plus sign meets where minus minus sign meets negative brush and then from
this armature terminals say A1 say. So, I will continue this discussion next time, but go
through it very carefully it is so interesting although complicated, but the analysis of DC
machine is much easier people say construction wise AC machines are simpler analysis is
difficult.
In DC machine it is opposite construction is rather difficult to understand particularly the
armature winding field winding is absolutely fine no complications, but armature winding
there will be slots there will be armature conductors. Armature conductors are to be
terminated on commutator segment which are mounted on shaft on the shaft and there will
be fixed carbon brushes which will be stationary and touching the commutator segment
and placing up the brushes is an intelligent work I mean if you put the brushes at wrong
positions you might get 0 voltage as well. So, this is how the complication starts, but
nonetheless across the brushes you will be ensured about the DC voltage.
So, in our next class I will show a complete armature diagram lap winding I am discussing
only wave winding I will do a little later and then two fundamental equations we have to
derive. One is if you know the flux distribution, if you know the number of conductors
present in the slots of the armature, then what will be the expression of the induced voltage
if the armature is rotated at so much rpm or rps that will be the generated voltage equation.
How much DC do we expect on? What factors will it depend? And similarly we have seen
that in case of DC machine for all machines in case of DC machine as well whenever
generated will supply power it will experience some electromagnetic torque when
armature will carry current, how much torque will be developed, these fundamental emf
and torque equations once we derive then life is rather easier for DC machines.
Thank you.
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Electrical Machines - I
Prof. Tapas Kumar Bhattacharya
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 62
Generated Voltage Across the Armature
Welcome to 62nd lecture on Electrical Machines I and we are discussing about DC
machines.
(Refer Slide Time: 00:26)
And, in DC machines the armature winding is slightly difficult unlike AC machines
windings, but although the basic features of a winding is still present that is each coil will
be generally full pitched coil and they will have two coil sides and the separation between
these two coil sides will be 180° electrical. So, in the simplified diagram you recall that
earlier we for a 4 pole machine we had the simplified winding diagram.
I need not go to slot and show the exact winding how it is done provided I know how I am
presenting a coil in these slots in this diagram. That is suppose 7th coil s7 and it is finished
another coil side f7 and I if at any point of time this coil side s7 is under North pole, f7 has
to be under South pole. And after assuming some direction of rotation generated voltages
were shown and then we found that no matter which coils occupy this particular position
which is presently occupied by 7th coil, whether it is 6 or 5 whoever comes here because
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if things are rotating they cannot, but have this polarity of induced voltage and it is
magnitude will be also same.
This prompted us to think that in that case I will take from the junctions with the help of
conducting wires terminate on some commentator segment circular and then we will
placed fixed brushes in space which will be stationery, but in which position where plus
plus will be joined where minus minus will be joined negative plus here plus plus joined
positive plus and this one we explained this. So, there will be in case of this lap winding
there will be four brushes B1 and B4 they can be joined together and 4 brushes can be will
have to be connected.
Then this is not how exactly this is done last day I forgot to bring this model. So, this
portion that is this and this copper strips these are done by what is known as commutator
segment. If you look at it very carefully the in the commutator segments there are several
commutator segments and the which is mounted on this shaft and two commutator
segments this and this next commutator they are insulated by it is looking like a gap, but
there is a mica insulation ok. So, that each commutator segment has got it is own identity.
Therefore commutator segments number of commutator segments will be just replica of
the number of slots here if they are equal in number. So, we will considered the simplest
case when the commutator number of commutator segments are equal to this one, because
in each commutator segment two ends of two different coils will be joined therefore,
number of commutator segments has to be equal to number of slots present in the machine.
So, this whole thing will rotate and then we will have carbon brushes suppose this strip is
a carbon brush placed in space like this and fixed. Therefore, as it rotates this fellow does
not rotate this one this brush it will touch whoever whichever coil comes in that position
it will connect that particular commutator segment, ensuring that B1 and B4 will remain
always positive B3 and B2 will always remain negative polarity of the voltage their by
giving you a DC voltage.
And another thing you note that. So, this two terminal so, commutator segment later I
showed that it can be like this if you go there will be commutator segments here.
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(Refer Slide Time: 05:41)
And each coil suppose s1 and f1 is a coil it is 2 ends will be terminated here and it will be
terminated there. The commutator segments where the coil ends will be terminated of a
particular coil the difference this length is called commutator pitch commutator pitch
which is denoted by 𝑌𝑐 .
In case lap winding 𝑌𝑐 = ±1. In this case, 𝑌𝑐 = +1. So, this coil number 1 is connected to
adjacent commutator segment, similarly coil number 2 will be actually connected between
2 and 3. So, so from 2 you start this junctions and your 2; 2 and this side it will be
terminated on 3 and like this then s3 it will continue like this. So, this is called commutator
pitch this is what the difference between the commutator segment numbers of a particular
coil which happens to be 𝑌𝑐 = +1 in case of lap winding ok.
So, and after we complete it then we make the winding table as I have shown simplified
winding table and from here I know which at which junctions the brushes are to be placed
everything is known B1 B2 B3 B4 if it is a 4 pole lap winding which two are plus which
two are minus brushes are stationeries I will connect them and ultimately get the armature
terminal.
(Refer Slide Time: 08:26)
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Now, and this one can complete I leave it as an exercise up to 16 coils you complete then
put plus minus here they their also and the commutator segment first coil 1-5’ in terms of
slot number 1-2, 2-3 like that and then you can determine the position of the brushes much
more scientifically if time permits I will further discuss on this, but it is not necessary at
this stage.
(Refer Slide Time: 08:59)
Therefore, I now know that field winding a DC machine I will represent it schematic
diagram I will show the field winding like this F1 F2 and we you know how these two
terminals are coming and armature I will show it like this, a circle and two brushes.
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Even if it is a 4 pole machine only two brushes I will show why, because of the reason that
it is the equivalent simplified representation ok, whatever is happening under a pair of pole
in the armature conductor under the next pair of pole same thing is bound to happen it is
not. North, South whatever is the happening whoever conductor is present under the next
pair of pole same thing is going to happen. Therefore, it is the simplified representation
which are called the armature terminals A1 A2 which are coming from where if it is 4 pole
machine B1 B4 we have shorted got this and B2 B3 you have shorted you have got this.
Now, if you look carefully between this one this diagram I will refer too previous to this,
this is a two pole machine if you join like that you can easily see if you start from B1 you
traverse 4 coils and you come to minus ok. So, between positive and negative brush this is
one parallel path, it could be you could reach the negative terminal because negatives are
shorted also. So, B1 another 4 coils and you reach the negative terminals so, second
parallel path ok.
So, here is one parallel path, here is the another parallel path, similarly from this positive
terminal if you want to reach the negative you see 4 coils you reach the negative, from this
another 4 coils you reach the negative. Therefore, between the positive and negative
brushes the armature circuit all the coils will be divided into as many parallel paths as the
number of poles of the machine provided it is a lap winding at least, because I have not
discussed anything about wave winding that I will do slightly later.
So, for lap winding the conclusion is between the brushes there will be a number of parallel
paths a number of parallel paths means if the number of poles of machine is 4, number of
parallel paths between positive and negative brush will be 4, if it is a 6 pole DC machine
number of parallel paths will be 6. So, this point you please see it can be easily seen from
plus to minus if I want to reach because it is this plus thing what I am telling I will write
here A1 I have written yeah I already I have written sorry.
This is the brush A1 plus this is A2. So, coming back to now the simplified diagram this
A1 is coming from the junctions of B1 and B4 which are positive and this is suppose
positive and these are called armature terminals and this is minus. Of course this plus
minus may interchange depending upon direction of rotations depending upon the
direction of flux from left to right, but anyway. So, it is if it is an ideal machine no point
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in showing this plus minus, but only a pair of brushes. This is how it will be shown, but
what I am telling if suppose at any time the armature carry some current here.
Suppose it is carrying some current here suppose it is generator mode it is delivering a
current here connected to the load because it will become a seat of emf. We will soon find
out what will be the expression of the emf, but the point is if it delivers current what will
be current in the conductors flowing. It will not be equal to 𝐼𝑎 why because there are 4
parallel paths if it is 4 pole machine. So, this current divided by the number of parallel
paths all parallel paths are identical so, current will be equally distributed in the parallel
paths.
So, if the if number of parallel paths if I say if 𝐼𝑎 is the external armature current then
current through the armature conductors through the armature coil sides or conductors
𝐼𝑎
armature conductors will be equal to 𝑎 , where 𝑎 is the number of parallel paths parallel
paths in the armature circuit in the armature circuit that is across the brushes mind you it
is important to note this external current and the armature conductor current is their not
the same.
In case of lap winding we have seen in case of lap winding a number of parallel paths 𝑎 =
𝑝 is equal to number of poles of the machine, is that clear. So, this comes naturally, if it is
a 2 polar machine lap winding number of parallel paths will be 2 it can be easily seen
because we have seen that 2 pole winding we have drawn the physical that winding
diagram 2 pole machine you see.
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(Refer Slide Time: 17:49)
There are only 2 brushes B1 and B2 to reach from plus brush to minus brush you traverse
either this way 8 coils reach minus and another alternative path is this one 2 parallel paths
for a 2 pole machine hopefully you have understood this.
Now, today I will do one very important thing that is suppose I will say I have a DC
machine with a field winding and armature winding I will run the DC machine by a prime
waver at certain rpm or rps I know will be knowing the field current which will create a
flux along this line and I would like to know what will be the EMF available across the
brush of the brushes of the machine across the armature terminals on what factors it will
depend and how to calculate that. So, let us try to do that first.
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(Refer Slide Time: 19:19)
So, consider a DC machine it is the armature terminal calculations are simple in case of
DC machine this is the field winding and this is the 𝜑 called the flux per pole if you have
passed a current 𝐼𝑓 North South will be created the other side is generally not shown. So,
flux is created and this is often called the d axis forget about this name right now it is not
necessary, so this is the flux. Incidentally if you look at it why brushes are shown here not
here also will be apparent from this dia