MAHALAKSHMI
ENGINEERING COLLEGE
TIRUCHIRAPALLI- 621213.
QUESTION BANK
SEMESTER – III
DEPARTMENT: ECE
SUBJECT CODE: EC2201
SUBJECT NAME: Electrical Engineering
UNIT 2
TRANSFORMERS
PART – A
1. What is meant by transformation ration? [NOV/DEC 2009]
Turns ratio=
N2
N1
Transformation ratio=
E2
E1
I1
I2
K
2. What is the purpose of conducting open circuit and short circuit tests in
transformers? [NOV/DEC 2009]
The open circuit test useful to find
i) No – load loss (or) core loss ii) No – load current iii) Ro and Xo
The short circuit test useful to find
i) Full load copper loss.
ii) Equivalent resistance and reactance referred to any side.
By using above two tests, we can predetermine the
i) Efficiency of the transformer ii) Regulation of the transformer
3. Why is the core of transformer is laminated? [NOV/DEC 20010]
The core of the transformer is laminated in order to minimize eddy current loss.
4. Define voltage regulation of a transformer. [NOV/DEC 2009], [NOV/DEC
2010], [NOV/DEC 2011]
Regulation of a transformer is defined as reduction in magnitude of the terminal
voltage due to the load, with respect to the no-load terminal voltage.
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% regulation=
V2 onno load
V2 whenloaded
V2 onno load
100
5. Write down the emf equation of single phase transformer. [NOV/DEC 2009]
Emf induced in primary coil E1= 4.44fФmN1 volt emf induced in secondary
Coil E2 =4.44 fФmN2.
f--freq of AC input, Ф--maximum value of flux in the core
N1, N2--Number of primary & secondary turns.
6. State condition for maximum efficiency of a transformer. [NOV/DEC 2010]
Efficiency of the transformer will be maximum when copper losses are equal to
iron losses.
Iron loss = Copper loss (or) Constant loss = Variable loss
The load current corresponding to maximum efficiency is given by
I2
Pi
R02
7. Classify the type of transformer based on core construction. [NOV/DEC 2011]
1) Core type transformer 2) Shell type transformer 3) Berry type transformer
8. Draw the equivalent circuit of transformer. [APRIL/MAY 2010]
9. What is step-up transformer? [NOV/DEC 2011]
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In a transformer, if the number of turns in the secondary winding is higher than the
primary winding is called step up transformer.
10. Mention the function of breather in transformer. [NOV/DEC 2011]
The function of breather is to prevent the entry of moisture to inside of the
transformer tank. The breather filled with some drying agent, such as calcium
chloride or silica gel. Silica gel or calcium chloride absorbs the moisture and allows
dry air to enter in to the transformer tank
11. What is the function of the Bucholz relay in a transformer? [NOV/DEC 2008]
It is a gas operated relay which mounted in between the conservator and
transformer
tank. The Bucholz relay give an alarm in case minor fault and
disconnect the transformer from the main supply in case of severe fault.
12. Draw the no – load phasor diagram of a transformer. [NOV/DEC 2008]
13. Define all – day efficiency. [NOV/DEC 2007]
The ratio of output in kwh to input in kwh of a transformer over a 24 hour period
is known a s all – day efficiency.
all day
kwh output in 24 hrs
kwh input in 24 hrs
14. Mention two different components of core loss in a transformer. [NOV/DEC
2007]
1. Hysteresis loss
2. Eddy current loss
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15. What are the different losses occurring in a transformer? [APRIL/MAY
2008]
1. Copper loss
2. Iron loss
16. Define transformer.
The transformer works on the principle of electromagnetic induction. A
transformer is an electrical device, having no moving parts, which by mutual
induction transfers electrical energy from one circuit another at the same
frequency, with changed values of voltage and current
17. Draw the phasor diagram for a transformer with inductive load.
18. What is staggering in the construction of transformers?
In transformer, the joints in alternate layers are staggered in order to avoid the
presence of narrow gaps right through the cross – section of the core.
19. Why transformer rating is expressed in terms of KVA?
Copper loss depends on current and iron loss depends upon voltage. Hence the
total loss in a transformer depends upon volt – ampere only not on the phase angle
between voltage and current i.e. it is independent of load power factor. That is why
the rating of a transformer is given in kVA and not kW.
20. Name the factors on which hysteresis loss depends.
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1. Frequency
2. Volume of the core
3. Maximum flux density
21. What are the different properties of ideal transformer?
1. No winding resistance
2. No magnetic leakage flux
3. No copper loss
4. No core loss
22. What is eddy current loss?
When a magnetic core carries a time varying flux, voltages are induced in all possible
path enclosing flux. Resulting is the production of circulating flux in core. These
circulating current do no useful work are known as eddy current and have power loss
known as eddy current loss.
23. How hysteresis and eddy current losses are minimized?
Hysteresis loss can be minimized by selecting materials for core such as silicon steel
& steel alloys with low hysteresis co-efficient and electrical resistivity. Eddy current
losses are minimized by laminating the core.
24. Why the open circuit test on a transformer is conducted at rated voltage?
The open circuit on a transformer is conducted at a rated voltage because core loss
depends upon the voltage. This open circuit test gives only core loss or iron loss of the
transformer.
25. Mention the difference between core and shell type transformers?
In core type, the windings surround the core considerably and in shell type the core
surrounds the windings i.e. winding is placed inside the core.
PART – B
1. With necessary vector diagrams, discuss about transformer on no
– load and loaded conditions. [NOV/DEC 2012]
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If the primary winding is connected to an alternating voltage and
secondary winding left open, then the transformer is said to be on n –
load. Let the supply voltage be ‘V1’ volts. This causes an alternating
current flow through the primary. Since secondary is open, this
current is called no load primary current (Io). This ‘Io’ establishes the
flux ‘ ’ weber in the core. Thus Io is not at 90o behind V1, but lags it
by an angle
o
> 90o. No load input power Po=V1Iocos
o
. ‘Io’ has two
components
i)
Active or working or iron loss or wattful component (Iw), Which
is in phase with ‘V1’ and supplies the iron loss and negligible
amount of primary copper loss.
Iw = Io cos
Where, cos
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o
----- (1)
o
= No load power factor
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ii)
Reactive or magnetizing or wattles component ( I ) which is in
quadrature with V1, and its function is to sustain the flux in the
core.
I = Io sin
o
----- (2)
From the above two equations we get
Io=
I w2
I2
2. (i) Draw the equivalent circuit of a transformer with all its
notations [NOV/DEC 2012]
Under no load condition, the primary of a transformer draws no load
current Io. It is mainly used to supply the iron loss and to produce the
flux in the core. The effect of iron loss is represented by a non –
inductive resistance Ro and the magnetizing current is represented by
Xo. Both of them are connected in parallel with primary winding. This
circuit is known as exciting branch or no – load branch.
(ii) Write a note on open circuit test on transformer. [NOV/DEC
2012]
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Open-Circuit or No-Load Test:
This test is conducted to determine the iron losses (or core losses) and
parameters R0 and X0 of the transformer. In this test, the rated
voltage is applied to the primary (usually low-voltage winding) while
the secondary is left open circuited. The applied primary voltage V1
is measured by the voltmeter, the no-load current I0 by ammeter and
no-load input power W0 by wattmeter as shown in Fig. (i). As the
normal rated voltage is applied to the primary, therefore, normal iron
losses will occur in the transformer core. Hence wattmeter will record
the iron losses and small copper loss in the primary. Since no-load
current I0 is very small (usually 2-10 % of rated current). Cu losses in
the primary under no-load condition are negligible as compared with
iron losses. Hence, wattmeter reading practically gives the iron losses
in the transformer. It is reminded that iron losses are the same at all
loads. Fig. (ii) Shows the equivalent circuit of transformer on noload.
Iron losses, Pi = Wattmeter reading = W0
No load current = Ammeter reading = I0
Applied voltage = Voltmeter reading = V1
Input power, W0 = V1 I0 cos f0
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3. (a) Deduce the equivalent circuit of the transformer from the
basic principle. [NOV/DEC 2011]
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Under no load condition, the primary of a transformer draws no load
current Io. It is mainly used to supply the iron loss and to produce the
flux in the core. The effect of iron loss is represented by a non –
inductive resistance Ro and the magnetizing current is represented by
Xo. Both of them are connected in parallel with primary winding. This
circuit is known as exciting branch or no – load branch.
Equivalent circuit of a transformer referred to primary
If all the parameters are referred to primary side, we get the
equivalent circuit of transformer referred to primary. When secondary
parameters are referred to primary resistance and reactance are
divided by K2, voltage are divided by K and currents multiplied by K.
This circuit also called as exact equivalent circuit of a transformer.
R '2
R2
, X '2
2
K
X2 '
, I2
K2
KI 2 , Z'L
ZL
, V2'
2
K
V2
, Ro
K
V1
, Xo
Iw
V2
Iμ
Approximate equivalent circuit
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The no – load current Io is only 1 – 3% of rated primary current. So
I '2
practically equal to I1. Due to this, equivalent circuit can be simplified
by transferring the exciting branch (Ro and Xo) to the left position of
the circuit.
R01=R1+R’2 and X01=X1+X’1
Z 01
2
2
R01
X 01
The above figure shows all the parameters referred to primary.
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2
R02
Z02
2
X 02
The above figure shows all the parameters referred to secondary
(b) A 2200/200 V transformer draws a no load primary current of
0.6A and absorbs 400 Watts. Find the magnetizing and iron loss
current. Draw also the no load phasor diagram. [NOV/DEC 2011]
Given Data:
V1=2200V, V2=200V, Io=0.6A, Po=400W
To find? Iw, I
Solution:
Iron loss component Iw =
Po
V1
400
2200
Iw = 0.18 A
Magnetizing component Iμ
I o2
I 2w
0.62
0.182
Iμ = 0.57A
No load phasor diagram
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4 . Explain the method of conducting OC and SC test on a
transformer with neat diagram. [NOV/DEC 2011]
The circuit constants, efficiency and voltage regulation of a
transformer can be determined by two simple tests (i) open-circuit test
and (ii) short-circuit lest. These tests are very convenient as they
provide the required information without actually loading the
transformer. Further, the power required to carry out these tests is
very small as compared with full-load output of the transformer.
These tests consist of measuring the input voltage, current and power
to the primary first with secondary open-circuited (open- circuit test)
and then with the secondary short-circuited (short circuit test).
Open-Circuit or No-Load Test:
This test is conducted to determine the iron losses (or core losses) and
parameters R0 and X0 of the transformer. In this test, the rated voltage
is applied to the primary (usually low-voltage winding) while the
secondary is left open circuited. The applied primary voltage V1 is
measured by the voltmeter, the no-load current I0 by ammeter and noload input power W0 by wattmeter as shown in Fig. ((i)). As the
normal rated voltage is applied to the primary, therefore, normal iron
losses will occur in the transformer core. Hence wattmeter will record
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the iron losses and small copper loss in the primary. Since no-load
current I0 is very small (usually 2-10 % of rated current). Cu losses in
the primary under no-load condition are negligible as compared with
iron losses. Hence, wattmeter reading practically gives the iron losses
in the transformer. It is reminded that iron losses are the same at all
loads. Fig. (ii)) shows the equivalent circuit of transformer on noload.
Iron losses, Pi = Wattmeter reading = W0
No load current = Ammeter reading = I0
Applied voltage = Voltmeter reading = V1
Input power, W0 = V1 I0 cos f0
Short-Circuit or Impedance Test:
This test is conducted to determine R01 (or R02), X01 (or X02) and fullload copper losses of the transformer. In this test, the secondary
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(usually low-voltage winding) is short-circuited by a thick conductor
and variable low voltage is applied to the primary as shown in Fig.
(i). The low input voltage is gradually raised till at voltage VSC, fullload current I1 flows in the primary. Then I2 in the secondary also has
full-load value since I1/I2 = N2/N1. Under such conditions, the copper
loss in the windings is the same as that on full load. There is no output
from the transformer under short-circuit conditions. Therefore, input
power is all loss and this loss is almost entirely copper loss. It is
because iron loss in the core is negligibly small since the voltage V SC
is very small. Hence, the wattmeter will practically register the fullload copper losses in the transformer windings. Fig. (ii) shows the
equivalent circuit of a transformer on short circuit as referred to
primary; the no-load current being neglected due to its smallness.
Full load Cu loss, PC = Wattmeter reading = WS
Applied voltage
= Voltmeter reading = VSC
F.L. primary current = Ammeter reading = I1
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Note: The short-circuit test will give full-load Cu loss only if the
applied voltage VSC is such so as to circulate full-load currents in the
windings. If in a short circuit test, current value is other than full-load
value, the Cu loss will be corresponding to that current value.
Advantages of Transformer Tests
The above two simple transformer tests offer the following
advantages:
(i) The power required to carry out these tests is very small as
compared to the full-load output of the transformer. In case of opencircuit lest, power required is equal to the iron loss whereas for a
short-circuit test, power required is equal to full-load copper loss.
(ii) These tests enable us to determine the efficiency of the
transformer accurately at any load and p.f. without actually loading
the transformer
(iii) The short-circuit test enables us to determine R01 and X01 (or R02
and X02). We can thus find the total voltage drop in the transformer as
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referred to primary or secondary. This permits us to calculate voltage
regulation of the transformer.
5. (a) List the various parts of transformer. Also explain the
constructional features of a transformer [MAY/JUNE 11]
Various parts of transformer
1. Transformer tank
2. Transformer oil
3. Core
4. Winding
5. Conservator
6. Breather
CONSTRUCTION OF A TRANSFORMER
We usually design a power transformer so that it approaches the
characteristics of an ideal transformer. To achieve this, following design
features are incorporated:
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(i)
The core is made of silicon steel which has low hysteresis loss and
high permeability. Further, core is laminated in order to reduce
eddy current loss. These features considerably reduce the iron
losses and the no-load current.
(ii)
Instead of placing primary on one limb and secondary on the
other, it is a usual practice to wind one-half of each winding on
one limb. This ensures tight coupling between the two windings.
Consequently, leakage flux is considerably reduced.
(iii) The winding resistances R1 and R2 are minimized to reduce I2R
loss and resulting rise in temperature and to ensure high efficiency.
TYPES OF TRANSFORMERS:
Depending upon the manner in which the primary and secondary
are wound on the core, transformers are of two types viz.,
(i) core-type transformer and
(ii) shell-type transformer.
(I)
Core-type transformer.
In a core-type transformer, half of the primary winding and
half of the secondary winding are placed round each limb as
shown in Fig. This reduces the leakage flux. It is a usual
practice to place the low-voltage winding below the highvoltage winding for mechanical considerations.
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(II) Shell-type transformer.
This method of construction involves the use of a double magnetic
circuit. Both the windings are placed round the central limb (See Fig,
the other two limbs acting simply as a low-reluctance flux path. The
choice of type (whether core or shell) will not greatly affect the
efficiency of the transformer. The core type is generally more suitable
for high voltage and small output while the shell-type is generally
more suitable for low voltage and high output.
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6. A 100 KVA transformer has iron loss of 720 Watts and full
load copper loss of 2100 Watts. Determine its efficiency at
full load at unity power factor and 0.8 power lagging.
[MAY/JUNE 11]
Given data:
Transformer rating = 100 kVA, Iron loss Pi=720 W
Full load copper loss Pcufl = 2100 W, Power factor cos =0.8
To find:
Full load efficiency
%η
nkVA cosΦ
100
nkVA cosΦ Pi Pcufl
Here n = 1 (for full load)
%
%
1 100 103 0.8
1 100 103 0.8 720 2100
100
96.59
7. (a) Draw the phasor diagram for an ideal transformer with
resistive and inductive load. [MAY/JUNE 11]
An ideal transformer is one that has
(i) No winding resistance
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(ii) No leakage flux i.e., the same flux links both the windings
(iii) No iron losses (i.e., eddy current and hysteresis losses) in the
core
Although ideal transformer cannot be physically realized, yet
its study provides a very powerful tool in the analysis of a
practical transformer. In fact, practical transformers have
properties that approach very close to an ideal transformer.
Consider an ideal transformer on no load i.e., secondary is opencircuited. Under such conditions, the primary is simply a coil of pure
inductance. When an alternating voltage V1 is applied to the primary,
it draws a small magnetizing current Im which lags behind the applied
voltage by 90°. This alternating current Im produces an alternating
flux
which is proportional to and in phase with it. The alternating
flux
links both the windings and induces e.m.f. E1 in the primary
and e.m.f. E2 in the secondary. The primary e.m.f. E1 is, at every
instant, equal to and in opposition to V1 (Lenz’s law). Both e.m.f.s E1
and E2 lag behind flux
by 90°. However, their magnitudes depend
upon the number of primary and secondary turns. Fig. (ii) shows the
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phasor diagram of an ideal transformer on no load. Since flux
is
common to both the windings, it has been taken as the reference
phasor. The primary e.m.f. E1 and secondary e.m.f. E2 lag behind the
flux
by 90
V1 and 180° out of phase with it.
7. The emf induced per turn of a single phase 11000 V/ 440 V,
50Hz single phase transformer approximately 11 volts.
Calculate the number of turns in high voltage and low
voltage windings and the net cross sectional area of the core
for maximum flux density of 1.5 Tesla. [MAY/JUNE 11]
Given data:
Emf per turn = 11 V, Primary voltage = 11000 V, Secondary = 440
V, Frequency f = 50 Hz, Maximum flux density Bm = 1.5 T.
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To find:
Number turns in the HV and LV windings, Net cross sectional
area
Solution:
E1 = N1 X emf induced/turn
N1 = 11000/11 =1100 turn
N2 = E2/Emf inducedper turn =440/11
N2 = 40 turns
E1 = 4.44 f N1 Bm A
A = E1/4.44 f N1 Bm = 4.44
11000
50 1100 1.5
A = 0.030 cm2
8. Derive the emf equation of a transformer. [NOV/DEC 2010]
EMF Equation of Transformer:
Since applied voltage is alternating in nature, the flux established is
also an alternating one. It is clear that the flux is attaining its
maximum value in one quarter of the cycle i.e., T/4 sec where ‘T’ is
the time period in second.
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We know that T= 1/f
Φm
Average rate of change of flux = 1 wb/seconds
4f
Form factor =
R.M .S value
Average value
1.11
RMS value of induced emf/turn =(1.11) x (4f x
m
)
RMS value of induced emf in the entire primary winding
E1= 4.44f
m
xN1
RMS value of induced emf in the entire primary winding
E2= 4.44f
m
xN2
9. A single phase transformer is rated at 10 KVA, 50 Hz. The
secondary rated voltage is 2400 V and the turn’s ratio is 10. The
resistance and leakage reactance as referred to secondary are8.4
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and 13.7 respectively. Find voltage regulation at full – loaf and
power factors of 0.8 lagging, 0.8 leading and unity. [NOV/DEC
2010]
Given data:
Transformer rating = 10 kVA
Frequency = 50 Hz
Turns ratio (K) = 10
Resistance referred to secondary R01 = 8.4 Ω
Reactance referred to secondary X01 = 13.7 Ω
To find:
Full – load voltage regulation
At 0.8 Lagging =?
At 0.8 Leading =?
At unity =?
Solution:
Secondary current (I2) =
kVA 10000
=
V2
2400
4.166 A
Full – load voltage regulation at 0.8 lagging p.f
cos
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= 0.8
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= cos-1 (0.8) = 36.86
Sin (36.86) = 0.6
% Voltage regulation =
=
I 2 (R 02 cos Φ X 02sin )
100
V2
4.166 (8.4 0.8 13.7 0.6)
100
2400
= 2.59 %
Full – load voltage regulation at 0.8 leading p.f
% Voltage regulation =
I 2 (R 02 cos Φ X 02sin )
100
V2
4.166 (8.4 0.8 13.7 0.6)
100
=
2400
= - 0.260 %
Full – load voltage regulation at unity p.f
% Voltage regulation =
=
I 2 R02 cos
V2
100
4.167 8.4 1
2400
= 1.45 %
10. A 30 KVA, 2000/200V single phase, 50Hz transformer has a
primary resistance of 3.5 ohms and reactance of 4.5 ohms. The
secondary resistance and reactance are 0.015 ohms and 0.02 ohms
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respectively. Find the equivalent resistance, reactance and
impedance
(i) Referred to primary
(ii) Referred to secondary side [APRIL/MAY 2010]
Given data:
Transformer rating = 30 kVA,
Primary voltage V1 = 2000 V
Secondary voltage V2 = 200 V, R1 = 3.5 Ω, X1 = 4.5 Ω, R2 = 0.015 Ω,
X2 = 0.02 Ω
To find: R01, X01, Z01, R02, X02, Z02
Transformation ratio K = V2/ V1=200/2000 =0.1
R01 = R1+ R!2 = R1+R2/K2
R01 = 3.5 + 0.015/0.12
R01 = 5 Ω
X01 = X1 + X!2 = X1 + X2/K2
X01 = 4.5 + 0.02/0.12
X01 = 6.5 Ω
Z01 =
2
R01
2
X 01
2
2
Z01 = 5 6.5
Z01 =8.2 Ω
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R02 = R2+ R!1 = R2+R1 K2
R02 = 0.015 + 3.5 X 0.12
R02 = 0.05 Ω
X02 = X2 + X!1 = X2 + X1 K2
X02 = 0.02 + 4.5 X 0.12
X02 = 0.065 Ω
Z02 =
2
R02
2
X 02
Z01 =
0.052 0.0652
Z01 = 0.082 Ω
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