The document discusses transformer construction, principles of operation, and testing methods to determine equivalent circuit parameters. It provides an introduction to different types of transformers and their applications. Key points covered include: - Transformers transfer power from one circuit to another through electromagnetic induction without a direct electrical connection between the circuits. - Practical transformers have equivalent circuits that account for winding resistances, core losses, and leakage fluxes/inductances not present in an ideal transformer. - Open circuit and short circuit tests are used to determine the equivalent circuit parameters like magnetizing inductance, core loss resistance, leakage reactances, and winding resistances.

1. Topic 1 : Magnetic Concept and
Transformer
Dr. Mohd Hafiz Habibuddin
Adapted from Dr. Junaidi and Dr. Nor Asiah

2. INTRODUCTION

3. Introduction
Two winding transformers
Construction and principles
Equivalent circuit
Determination of equivalent circuit parameters
Voltage regulation
Efficiency
Auto transformer
3 phase transformer

4. Introduction
Different variety of transformers

5. Introduction

6. Introduction
 The word “Transformer” means an electromagnetic device
which transforms electrical power from one end to another
at different voltages and different currents keeping frequency
constant.
 Unlike motor and generator it is static machine with different
turns ratio of primary and secondary windings through which
voltage/current is changed.
 The transfer of energy takes place through the magnetic field
and all currents and voltages are AC.
 The rating of transformer is either in kVA or MVA because
load to be connected is unknown

7. Introduction
Transformers are adapted to numerous engineering
applications and may be classified in many ways:
Power level (from fraction of a volt-ampere (VA) to over a
thousand MVA),
Application (power supply, impedance matching, circuit
isolation),
Frequency range (power, audio, radio frequency (RF))
Voltage class (a few volts to about 750 kilovolts)
Cooling type (air cooled, oil filled, fan cooled, water
cooled, etc.)-ONAN, ONAF
Purpose (distribution, rectifier, arc furnace, amplifier
output, etc.).

8. Introduction
 Examples of transformer classifications:
 Power Three phase transformers (Step up) used for transmission of
power (3-phase) at a distance
 Distribution transformer (Step down) used for utilization of power 3-
pahse/1-phase
 Instrument Transformer (VT & CT) used for measurement/practical
 Auto Transformer (Single limb, electrically connected) used for
measurement, practical, supply/utilization
 Isolation Transformer (having winding ratio of 1:1) used for safety of
human and equipment for sensitive appliances or practical purpose

9. Introduction
The invention of transformer caused transmission of
heavy AC electrical power possible thus plays
important role in electrical power technology
Functions of transformer:
Raise or lower voltage or current in AC circuit
Isolate circuit from each other
Enable to transmit electrical power energy over large
distances at about >1200kV
Provides electrical power according to the utilization
needs

10. Power transmission
Transformer- Introduction

11. Introduction
 A typical power system consists of generation, transmission
and distribution.
 Power from plant/station is generated around 11-13-20-30kV
(depending upon manufacturer and demand).
 This voltage is carried out at a distance to reach for utilization
through transmission line system by step up transformer at
different voltage levels depending upon distance and losses.
 Its distribution is made through step down transformer
according to the consumer demand.
 Here again at this stage, a transformer play an important role
to reduce the voltage to suit the consumer need.

12. Introduction
Power Transmission

13. Introduction
Transformer is a device that makes use of the
magnetically coupled coils to transfer energy.
It is typically consists of one primary winding coil
and one or more secondary windings.
The primary winding and its circuit is called the
Primary Side of the transformer.
The secondary winding and its circuit is called the
Secondary Side of the transformer.
A magnetic circuit provides the link between
primary and secondary.

14. Introduction
 When an AC voltage Vp is applied to the primary winding of
the transformer, an AC current Ip will result.
 Ip sets up a time-varying magnetic flux Ф in the core.
 A voltage Vs is induced in the secondary circuit according to
the Faraday’s law.
N1 N2
Supply Load
Primary winding Secondary winding
Laminated iron core

15. CONSTRUCTION

16. Construction
 The magnetic (iron) core is made of
thin laminated steel sheet.
 to minimize the eddy current loss by
reducing thickness.
 There are two common cross
section of core
 square or (rectangular) for small
transformers
 circular (stepped) for the large and 3
phase transformers.

17. Construction
Core (U/I) Type:
Constructed from a stack of U and I shaped laminations.
The primary and secondary windings are wound on two
different legs of the core.
Shell Type:
Constructed from a stack of E and I shaped laminations.
The primary and secondary windings are wound on the
same leg of the core, as concentric windings, one on top
of the other.

18. Construction
A) Core type B) Shell type

19. Construction

20. Construction
Primary
Winding
Secondary
Winding
Multi-layer
Laminated
Iron Core
X1
X2
H1 H2
Winding
Terminals

21. Construction
Iron core
Terminals
Secondary
winding
Insulation

22. IDEAL TRANSFORMER

23. Ideal Transformer
 Winding resistances are zero, no leakage inductance and iron
loss
 Magnetization current generates a flux that induces voltage
in both windings
Current, voltages and flux in an unloaded ideal transformer
N1 N2
m
Im
V1
E1
E2
= V2
2
N
E m
1
1



f
N
N
E
m
m




2
2
2
44
.
4
2


24. Ideal Transformer
Currents and fluxes in a loaded ideal transformer
E 2
Load
V 2
I2
 2
 1
 m
Im + I 1
V 1
E 1

25. Ideal Transformer
Turn ratio
If the primary winding has N1 turns and secondary
winding has N2 turns, then:
The input and output complex powers are equal

a 
N1
N2

E1
E2

I2
I1
*
*
I
E
S
S
I
E 2
2
2
1
1
1 



26. Ideal Transformer
Functional description of a transformer:
a = 1
 Isolation Transformer
| a | < 1
 Step-Up Transformer
 Voltage is increased from Primary side to secondary side
| a | > 1
 Step-Down Transformer
 Voltage is decreased from Primary side to secondary side

27. Ideal Transformer
Transformer Rating
Practical transformers are usually rated
based on:
Voltage Ratio (V1/V2) which gives us the
turns-ratio
Power Rating, small transformers are
given in Watts (real power) and Larger
ones (Power Transformers) are given in
kVA (apparent power)

28. Ideal Transformer
Example 1
Determine the turns-ratio of a 5 kVA 2400V/120V
Power Transformer
Turns-Ratio = a = V1/V2 = 2400/120 = 20/1 = 20
This means it is a Step-Down transformer

29. Ideal Transformer
Example 2
A 480/2400 V (r.m.s) step-up ideal transformer
delivers 50 kW to a resistive load. Calculate:
the turns ratio,
(0.2)
the primary current,
(104.17A)
the secondary current.
(20.83A)

30. Ideal Transformer
Exercise 1
A 250kVA, 1100V/400v, 50Hz single-phase
transformer has 80 turns on the secondary.
Calculate:
the approximate values of the primary and secondary
currents
 (227A, 625A)
the approximate number of primary turns
 (220)
 the maximum value of the flux
 (22.5mWb)

31. Ideal Transformer
Nameplate of a transformer

32. Ideal Transformer
Equivalent circuit of an ideal transformer
1
2
2
1
2
1
I
I
E
E
N
N
a 



33. Ideal Transformer
Equivalent circuit of an ideal transformer
Transferring impedances through a transformer
Vac Zload
T
V1 V2
I1 I2
2
2
2
2
2
1
1
1
I
V
I
V
I
V
Z a
a
a









load
a Z
Z 2
1 

34. Ideal Transformer
 Equivalent circuit when
secondary impedance is
transferred to primary side
and ideal transformer
eliminated.
 Equivalent circuit when
primary source is
transferred to secondary
side and ideal transformer
eliminated.
Vac a2
Zload
V1
I1
Vac/k Zload
V2
I2

35. PRACTICAL TRANSFORMER
Equivalent Circuit

36. Practical Transformer

37. Equivalent Circuit
In a practical magnetic core having finite
permeability, a magnetizing current Im is required to
establish a flux in the core.
This effect can be represented by a magnetizing
inductance Lm.
The core loss can be represented by a resistance Rc.

38. Equivalent Circuit
 Rc :core loss component
 Xm : magnetization component
'
1
2
2
1
2
1
I
I
E
E
N
N

 '
1
0
1 I
I
I 


39. Equivalent Circuit
Phasor diagram of an unloaded transformer

40. Equivalent Circuit
Winding resistance and leakage flux
The effects of winding resistance and leakage flux are
respectively accounted for by resistance R and leakage
reactance X (2πfL).

41. Equivalent Circuit
 Rc :core loss component
 Xm : magnetization component
 R1 and R2 are resistance of the primary and secondary winding
 X1 and X2 are reactance of the primary and secondary winding

42. Equivalent Circuit
Phasor diagram of a loaded transformer (secondary)

43. Equivalent Circuit
Phasor diagram of a loaded transformer (primary)

44. APPROXIMATE EQUIVALENT
CIRCUIT

45. Approximate Equivalent Circuit
 Since no load current is very small(3-5% of full load), the
parallel circuit of Rc and Xm can be moved close to the supply
without significant error in calculation.
 Calculations becomes easier

46. Approximate Equivalent Circuit
 Calculations will be much more easy if the primary and
secondary circuit are combined.
 Transfer the secondary circuit to the primary circuit
a
I
I
aV
V
/
'
'
2
1
2
2


2
2
2
2
2
2
'
'
X
a
X
R
a
R



47. Approximate Equivalent Circuit
Phasor diagram of a loaded transformer (primary)

48. Approximate Equivalent Circuit
 For convenience, the turns is usually not shown
 The resistance and reactance can be lumped together
 We can also transfer the primary circuit to the secondary circuit

49. Approximate Equivalent Circuit
Example 3
 A 100kVA transformer has 400 turns on the primary and 80
turns on the secondary. The primary and secondary
resistance are 0.3 ohm and 0.01 ohm respectively and the
corresponding leakage reactances are 1.1 ohm and 0.035
ohm respectively. The supply voltage is 2200V. Calculate:
the equivalent impedance referred to the primary circuit
(2.05 ohm)
 the equivalent impedance referred to the secondary
circuit
(0.082)

50. TRANSFORMER TEST
Determination of equivalent circuit parameters

51. Transformer Test
 The equivalent circuit model for the actual transformer can
be used to predict the behavior of the transformer.
 The parameters Rc, Xm, R1, X1, R2, X2 and N1/N2 must be
known so that the equivalent circuit model can be used.
 These parameters can be directly and more easily
determined by performing tests:
No load test (or open circuit test)
Short circuit test

52. Transformer Test
No load/Open circuit test
Provides magnetizing reactance (Xm) and core loss
resistance (Rc)
Obtain components are connected in parallel
Short circuit test
Provides combined leakage reactance and winding
resistance
Obtain components are connected in series

53. Transformer Test – Open Circuit
 Equivalent circuit for open circuit test, measurement at the primary side
 Simplified equivalent circuit
V
A
X1 R1
X m R c
X2 R2
W
V oc
I oc
P oc
V
A
Xm Rc
W
Voc
Ioc
Poc

54. Transformer Test – Open Circuit
Open circuit test evaluation
.
.
Q
V
X
I
V
Q
I
V
P
P
V
R
oc
m
oc
oc
oc
oc
oc
oc
oc
c
2
0
1
0
2
sin
cos
















55. Transformer Test – Short Circuit
Short circuit test
Secondary (normally the LV winding) is shorted, that
means there is no voltage across secondary terminals; but
a large current flows in the secondary.
Test is done at reduced voltage (about 5% of rated
voltage, with full-load current in the secondary.
 Hence the induced flux are also 5%) The core losses is negligible
since it is approximately proportional to the square of the flux.
 So, the ammeter reads the full-load current; the wattmeter reads
the winding losses, and the voltmeter reads the applied primary
voltage.

56. Transformer Test – Short Circuit
 Equivalent circuit for short circuit test, measurement at the primary side
 Simplified equivalent circuit
R2
I sc
V
A W
X1
R1 X2
P sc
Vsc
V
A W
a2R2
X1
R1 a2X2
I sc
P sc
Vsc

57. Transformer Test – Short Circuit
 Simplified circuit for calculation of series impedance
 Primary and secondary impedances are combined
 .
 .
V
A W
Xe1
Re1
I sc
Vsc
P sc
2
2
1
1 R
a
R
Re 

2
2
1
1 X
a
X
Xe 


58. Transformer Test – Short Circuit
Short circuit test evaluation
 .
 . 2
1
2
1
1
1
2
1
e
e
e
sc
sc
e
sc
sc
e
R
Z
X
I
V
Z
I
P
R





59. Transformer Test
Equivalent circuit for a real transformer resulting
from the open and short circuit tests.
Xm Rc
Xe1 Re1

60. Transformer Test
Example 4
Obtain the equivalent circuit of a 200/400V, 50Hz 1-
phase transformer from the following test data:-
O/C test : 200V, 0.7A, 70W - on L.V. side(LV data)
S/C test : 15V, 10A, 85W - on H.V. side(HV data)
(Rc =571.4 ohm, Xm=330 ohm, Re=0.21ohm,
Xe=0.31 ohm)

61. VOLTAGE REGULATION

62. Voltage Regulation
 Most loads connected to the secondary of a transformer are
designed to operate at essentially constant voltage. However,
as the current is drawn through the transformer, the load
terminal voltage changes because of voltage drop in the
internal impedance.
 To reduce the magnitude of the voltage change, the
transformer should be designed for a low value of the
internal impedance Zeq
 The voltage regulation is defined as the change in magnitude
of the secondary voltage as the load current changes from
the no-load to the loaded condition.

63. Voltage Regulation
The voltage regulation is expressed as follows:
V2NL= secondary voltage (no-load condition)
 V2L = secondary voltage (full-load condition)

Voltageregulation
V2NL V2L
V2NL
Voltage regulation
E20 V2L
E20

64. Voltage Regulation
For the equivalent circuit referred to the primary:
 V1 = no-load voltage
 V2
’ = secondary voltage referred to the primary (full-load condition)
Voltageregulation 
V1 V2
'
V1

65. Voltage Regulation
Consider the equivalent circuit referred to the
secondary,
(-) : leading power factor
(+) : lagging power factor
NL
e
e
V
X
I
R
I
regulation
Voltage
2
2
2
2
2
2
2 sin
cos 
 

I2' R1'
V2NL
X1' R2
X2
V2 Z2
I2

66. Voltage Regulation
Consider the equivalent circuit referred to the
primary,
(-) : leading power factor
(+) : lagging power factor
1
2
1
1
2
1
1 sin
cos
V
X
I
R
I
regulation
Voltage e
e 
 

I1
R1
V1
X1 R2
'
X2
'
Z’2
I1
'
V2
'

67. Voltage Regulation
Example 5
Based on Example 3 calculate the voltage regulation
and the secondary terminal voltage for full load
having a power factor of
0.8 lagging
(0.0336pu,425V)
0.8 leading
(-0.0154pu,447V)

68. Approximate Equivalent Circuit
Example 3
 A 100kVA transformer has 400 turns on the primary and 80
turns on the secondary. The primary and secondary
resistance are 0.3 ohm and 0.01 ohm respectively and the
corresponding leakage reactances are 1.1 ohm and 0.035
ohm respectively. The supply voltage is 2200V. Calculate:
the equivalent impedance referred to the primary circuit
(2.05 ohm)
 the equivalent impedance referred to the secondary
circuit
(0.082)

69. Voltage Regulation

70. EFFICIENCY
Losses in transformer

71. Efficiency
Losses in a transformer
Copper losses in primary and secondary windings
Core losses due to hysteresis and eddy current. It
depends on maximum value of flux density,
supply frequency and core dimension. It is
assumed to be constant for all loads

72. Efficiency
 Equipment is desired to operate at a high efficiency.
 Efficiency is defined as
 Since it is a static device, losses in transformers are small
 The losses in the transformer are the core loss (Pc) and
copper loss (Pcu)
 
 
losses
P
P
P
power
input
P
power
output
out
out
in
out




cu
c
out
out
P
P
P
P





73. Efficiency
 The copper loss can be determined if the winding currents
and their resistances are known:
 The copper loss is a function of the load current.
 The core loss depends on the peak flux density in the core,
which in turn depends on the voltage applied to the
transformer
 Since a transformer remains connected to an essentially constant
voltage, the core loss is almost constant
2
2
2
1
2
1
2
2
2
1
2
1
eq
eq
cu
R
I
R
I
R
I
R
I
P





74. Efficiency
If the parameters of the equivalent circuit of a
transformer are known, the efficiency of the
transformer under any operating condition may be
determined
.
Normally, load voltage remains fixed
Therefore efficiency depends on load current and load
power factor
2
2
2 
cos
I
V
Pout 
2
2
2
2
2
2
2
cos
cos
eq
c
2
2
R
I
P
I
V
I
V







75. Efficiency
Efficiency on full load
where S is the apparent power (in volt amperes)
Efficiency for any load equal to n x full load
where corresponding total loss =
sc
oc
FL
FL
P
P
S
S






cos
cos
sc
oc
FL
FL
P
n
P
S
n
S
n





 2
cos
cos



sc
oc P
n
P 
 2

76. Efficiency
Example 6
The following results were obtained on a 50 kVA
transformer:
open circuit test – primary voltage, 3300 V; secondary
voltage, 400 V; primary power, 430W.
Short circuit test – primary voltage, 124V;primary current,
15.3 A; primary power, 525W; secondary current, full load
value.
Calculate the efficiency at full load and half load for
0.7 power factor.
(97.3%, 96.9%)

77. Efficiency
Exercise 2
The primary and secondary windings of a 500kVA
transformer have resistance of 0.42 ohm and 0.0019
ohm respectively. The primary and secondary
voltages are 11000V and 400V respectively and the
core loss is 2.9kW, assuming the power factor of the
load to be 0.8. Calculate the efficiency on
Full load
half load
(98.3%, 98.1%)

78. Efficiency
For constant values of the terminal voltage V2 and
load power factor angle θ2 , the maximum efficiency
occurs when
If this condition is applied, the condition for
maximum efficiency is
 that is, core loss = copper loss.
0
2

dI
d
2
2
2 e
c R
I
P 

79. Efficiency
Exercise 3
Assuming the power factor of the load to be 0.8,
find the output power at which the efficiency of the
transformer of Exercise 2 is a maximum and
calculate its value
(346.4kW, 98.4%)