Electric Machines
Chapter 2
Transformers
Dr Jalal Al Roumy
Israa University
2019/2020
Introduction:
 The transformer is an electrical device that converts AC
electrical energy at one voltage level to another voltage
level through the action of a magnetic field.
 It consists of two or more coils of wire wrapped around a
common ferromagnetic core.
 The only connection between the coils is usually the
common magnetic flux present within the core
2
Importance of Transformers:
 Electric energy is generated at voltages between 12 to 25
kV and then transformers step up voltage between 110 kV
to 1200 kV for transmission over long distances.
3
Types of Transformers:
 Transformers are classified, based on core structure,
into:
Core form:
Transformer windings are wrapped
on two legs.
Shell form:
Transformer windings are wrapped
only on centre leg.
(leakage flux is minimized)
Both types deploy thin laminations isolated from
each other in order to minimize eddy currents.
4
Types of Transformers:
5
Types of Transformers:
 The primary and secondary windings are wrapped one
on top of the other (to reduce the leakage flux) with the
low-voltage winding innermost (to simplify insulating of
the high-voltage winding from the core).
6
Types of Transformers:
 Step up (unit) transformer:
 Located at output of a generator to step up the voltage level to
transmit the power.
7
Types of Transformers:
 Step down (substation) transformer:
 Located at main distribution or secondary level transmission
substations to lower the voltage levels for distribution 1st level
purposes.
8
Types of Transformers:
 Distribution transformer:
 Located at small distribution substation. It lowers the voltage
levels for 2nd level distribution purposes.
9
Types of Transformers:
 Special purpose transformer:
 Potential transformer.
 Current transformer.
10
The Ideal Transformer:
 An ideal transformer is a lossless device with an input
winding and an output winding in which magnetic core
has an infinite permeability.
11
The Ideal Transformer:
 The relationship between the primary and the secondary
voltage is as follows (a is the turns ratio):
 The relationship between the primary and secondary
current is Np ip(t) = Ns is(t).
12
The Ideal Transformer:
 In terms of phasor quantities:
 Note:
 The turns ratio of the ideal transformer affects the magnitudes of the
voltages and currents, but not their angles.
13
Polarity:
 To specify the secondary’s polarity, transformers are to
be opened & windings examined. To avoid this
examination, transformers employ a dot convention.
Note:
1. Voltage polarities are the same with
respect to the dots on each side of the
core.
2. If the primary current of the
transformer flows into the dotted end
of the primary winding, the secondary
current will flow out of the dotted end
of the secondary winding.
14
Power in Ideal Transformer:
 The power supplied to the transformer:
 The power supplied by the transformer:
 Since voltage and current angles are unaffected by an
ideal transformer:
15
Impedance Transformation:
 Load impedance could be matched to source impedance
by picking the proper turns ratio of a transformer.
16
17
I G  I Line  I Load
I Line 
V
Z Line  Z Load
480 0
0.18  j 0.24  4  j 3
480 0
I Line 
 90.8 37.8 A
4.18  j 3.24
V Load  I Line  Z Load
I Line 
V Load  90.8 37.8   4  j 3 
V Load  454 0.9 V
2
PLoss  I Line
 R Line
PLoss  90.82  0.18  1484 W
18
'
Z eq
 a 2 Z eq
'
'
Z eq
 a 2 ( Z Line  Z Load
)
2
'
Z eq
'
Z eq
 1 

 (400.18  j 300.24)
 10 
 4.0018  j 3.0024
'
Z eq
 5.003 36.88 Ω
IG 
Z
'
Load
 a Z Load
2
'
Z Load
 102  (4  j 3)
'
Z Load
 400  j 300 Ω
V
480 0

'
Z eq
5.003 36.88
I G  95.94 36.88 A
V Load  I Load  Z Load
V Load  95.94 36.88   4  j 3 
V Load  479.7 0.01 V
'
Z eq  Z Line  Z Load
2
PLoss  I Line
 R Line
Z eq  400.18  j 300.24 Ω
PLoss  9.594 2  0.18  16.7 W
19
Operation of Real Transformers:
The primary winding is connected to
an AC power source while the
secondary winding is open-circuited.
d
d
eind   N
dt
dt
20
Transformer Voltage Ratio:
d M
eP  N P
dt
d M
eS  N S
dt
e
eP
d
 M  S
NP
dt
NS
eP N P

a
eS
NS
The smaller the leakage fluxes, the closer the total voltage
ratio approximates that of the ideal transformer.
21
Transformer Current Ratio:
FP  N P i P
FS  N S i S
Fnet  N P i P  N S i S   R
Fnet  0  N P i P  N S i S
iP N S 1


iS N P a
Note:
Reluctance of a well-designed transformer core is very small.
22
Converting Real Transformer into Ideal:
 The assumptions required to convert a real transformer
into the ideal transformer are:
 The core must have no hysteresis or eddy currents.
 The magnetization curve must have this shape (for an
unsaturated core, the net magnetomotive force = 0 & NPIP=NSIS):
 The leakage flux in the core must be zero.
 The resistance of the transformer windings must be zero.
23
Example:
An ideal transformer has a 150-turn primary and 750-turn
secondary. The primary is connected to a 240-V, 50-Hz
source. The secondary winding supplies a load of 4 A at a
lagging power factor of 0.8, Determine: (a) the a-ratio, (b) the
current in the primary, and (c) the power supplied to the load.
Solution:
a  150 / 750  0.2
I2
4
I1 

 20 A
a
0.2
V 1 240
V2 

 1200 V
a
0.2
PL V 2 I 2 cos   1200  4  0.8  3840 W
24
Losses in Real Transformer:
 Copper Losses:
 The resistive heating losses in the primary and secondary windings
of the transformer.
 Eddy current losses:
 The resistive heating losses in the core of the transformer.
 Hysteresis losses:
 Due to the rearrangement of the magnetic domains in the core
during each half-cycle.
 Leakage flux:
 The fluxes which escape the core and pass through only one of the
transformer windings and produce a self-inductance in the coils.
25
Transformer Equivalent Circuit:
 To construct an equivalent circuit that takes into account
all the major imperfections in real transformers:
 Copper losses are modeled by placing a resistor Rp in the
primary circuit of the transformer and a resistor Rs in the
secondary circuit.
 Leakage flux is modeled by primary and secondary inductors.
 The magnetization current is modeled by a reactance XM
connected across the primary voltage source.
 The core-loss current is modeled by a resistance RC connected
across the primary voltage source.
26
Exact Equivalent Circuit:
27
Approximate Equivalent Circuits:
28
Approximate Equivalent Circuits:
VR 
Transformer voltage regulation:
VR 
V
S , nl
V S ,fl
V S ,fl
 100%
V P / a  V S ,fl
V S ,fl
 100%
29
Voltage Regulation:
 Output voltage of transformer varies with load due to
voltage drop on series impedance of equivalent model.
 Full load regulation parameter compares output no-load
voltage with its full load voltage:
VR 
VR 
V
S , nl
V S ,fl
V S ,fl
 100%
V P / a  V S ,fl
V S ,fl
 100%
 For ideal transformer, V.R. = 0.
30
Phasor Diagram:
 To determine the voltage regulation of a transformer, the
voltage drops should be determined.
 Transformer equivalent circuit referred to the secondary
side shown:
VP
V S  Req I S  jX eq I S
a
31
Phasor Diagram:
Lagging P.F.
Unity P.F.
Leading P.F.
32
Simplified Voltage Regulation Calculation:
 Since transformer usually operate at lagging P.F., a
simplified method is introduced.
 For lagging loads, vertical components partially cancel
each other as the angle of (VP/a) is very small.
33
Transformer Efficiency:
 Losses in Transformer:
 Copper (I²R) losses.
 Core Hysteresis losses.
 Core Eddy current losses.
 Transformer Efficiency:

Pout
x100%
Pin

Pout
x100%
Pout  Ploss
VS I S cos

x100%
PCu  Pcore  VS I S cos
34
A 23-kVA, 2300/230-V, 60-Hz, step-down transformer has the following
resistance and leakage-reactance values: R1=4Ω, R2=0.04Ω, X1=12Ω,
and X2=0.12Ω. The equivalent core-loss resistance and the magnetizing
reactance on the primary side of the transformer are 20 kΩ and 15kΩ,
respectively The transformer is operating at its rated voltage and rated
load. If the power factor of the load is 0.866 lagging, determine the
efficiency of the transformer and plot the phasor diagram.
Solution:
35
V 12 2494.932
Pc 

Rc
20000
Pc  311.2349 Watts
Pcu  I 2'  R eqP
2
V 2'  aV 2  10  230 0  2300 0 V
I 2'  10 30 A
R e 1  R1  a 2 R 2  4  102  0.04  8 
X e 1  X 1  a 2 X 2  12  102  0.12  24 
Z e 1  R e 1  X e 1  8  j 24 
V 1 V 2'  I 2' Z e 1  2300 0  10 30 (8  j 24)
V 1  2494.93 3.8575
Pcu  102  8  800 Watts
Pout  23000  0.866
Pout  19918 Watts

Pout
Pout  Ploss
19918
100
19918  311.2349  800
  94.72%

36
Autotransformer:
 On some occasions, it is desirable to change voltage
levels by only a small amount to compensate for voltage
drops that occur a long way from the generators.
 Here, it is wasteful and expensive to wind a transformer
with two full windings. Autotransformer is used instead.
37
V-I Relations in an Autotransformer :
V H V C V SE and V L V C
I L  I C  I SE and I SE  I H
VC
NC

V SE N SE
N SE
IC 
I SE
NC
VH
N SE
V C 
VC
NC
N SE
IL 
I SE  I SE
NC
VH
N SE  N C

VL
NC
N SE  N C
IL 
IH
NC
NC
VL

V H N SE  N C
I L N SE  N C

IH
NC
38
Power Rating & Advantages:
S in  S out V L I L V H I H
S in  S out  S IO
SW V C I C V L (I L  I H ) V L I L V L I H
IH  IL
NC
N SE  N C
SW V L I L V L I L
SW  V L I L
SW  S IO
NC
N SE  N C
N SE
N SE  N C
N SE
N SE  N C
S IO N SE  N C

SW
N SE
Note:
Not all power transferring
from primary to secondary
in autotransformer pass
through windings.
Hence, if a conventional
transformer is reconnected
as an autotransformer, it
can handle much more
power than its original
rating.
39
A 100-VA, 120/12-V transformer is to be connected so as to form a stepup autotransformer as shown. A primary voltage of 120 V is applied to the
transformer. (a) what is the secondary voltage of the transformer, (b) what
is the maximum volt-ampere rating in this mode of operation?, and (c)
calculate the rating advantage of this autotransformer connection over the
conventional 120/12-V operation.
Solution:
40
VH 
N SE  N C
VL
NC
12  120
120  132 V
120
S SE ,conv 100
I SE ,max 

 8.33 A
V SE
12
VH 
I SE  I S  I H
V S V H  132 V
S out V S I S V H I H
S out  S in  132  8.33  1100 VA
S IO 1100

 11 or
SW
100
S IO N SE  N C 12  120


 11
SW
N SE
12
41
Usages & Disadvantages:
 Autotransformers are used:
 When two voltages fairly close.
 As variable transformers (where L.V. tap moves up & down the
winding).
 Disadvantages:
 Due to the direct physical connection between primary &
secondary circuits, the electrical isolation of the two sides is lost.
 The series impedance (which is required to limit current flows
during power system faults) is way smaller than that in
conventional transformers.
42
Three-Phase Transformers:
 Three-phase transformers are either:
 constructed from three single-phase transformers.
 or from three sets of windings wrapped on a common core.
 The first approach had the advantage that each unit in
the bank could be replaced individually in the event of
failure.
 The construction of a single three-phase transformer is
the preferred practice today, as it is lighter, smaller,
cheaper, and more efficient.
43
Three-Phase Transformers:
Three Single-Phase Transformers
One Three-Phase Transformers
44