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EE : ELECTRICAL ENGINEERING
Electrical Machine

INDEX

Sr. Pg.
Contents Sub−Topics
No. No.
1. Transformer(Single Phase & Three Phase Transformer)
Notes Introduction 1
Transformer Construction 2
Working of Transformer 4
Transformer Phasor Diagram 7
Equivalent Circuit of a Transformer 11
Voltage Regulation 19
Efficiency 22
Three Phase Transformers 24
Three Phase Transformer Connections 26
Parallel Operation of 3φ Transformer 30
Auto Transformers 31
Three Winding Transformer 39
LMR (Last Minute Revision) 40
Assignment−1 Questions 43
Test Paper−1 Questions 46
2. D.C. Machine (D.C. Generator and Motors)
Notes Principles of Energy conversion and Types of Windings 49
D.C. Machine Principle and Construction 56
D.C. Generator 59
Generator Characteristics 63
D.C. Motor 70
D.C. Motor Characteristics 71
D.C. Motor Starting and Speed Control 75
Armature Reaction and Commutation 83
Testing of D.C. Machine 88
Assignment−2 Questions 94
Test Paper−2 Questions 98

3. Single Phase Induction Motor

Notes Classifications of A.C. Motor 101
Introduction 102
Construction and Operating Principle 102
Performance Equation 104
Phasor Diagram and Equivalent Circuit 107
Torque slip and Power Slip characteristics 109
Performance characteristics 113
Starting Method of Induction Motors 114
Speed control of Induction Motors 119
Types of Squirrel Cage Induction Motor 122
Abnormal Operation of Induction Motor 126
Single Phase Induction Motors 128
Operating Principle of 1φ I.M. 129
Starting Methods of Single−Phase Induction Motor 131
LMR(Last Minute Revision) 142
Assignment−3 Questions 143

Test Paper−3 Questions 146

4. Synchronous Machine (Generators and Motors)
Notes Synchronous Alternators 149
Method of Synchronization and Parallel Operation 157
Characteristics of Synchronous Alternators 167
Synchronous Machine Stability 170
Synchronous Condenser 174
LMR (Last Minute Revision) 176
Assignment−4 Questions 178
Test Paper−4 Questions 181
5. Special Machine
Notes Stepper Motor 183
Printed Circuit (Disc) DC Motor 189
A.C. Series Motor 190
Reluctance Motor 193
Hysteresis Motor 194
Permanent Magnet DC Motor 195
Servomotors 197
Assignment−5 Questions 200
Test Paper−5 Questions 203
Practice Problems 205

Model Solutions
Answer Key to Assignments 212
Model Solutions to Assignment − 1 213
Model Solutions to Assignment − 2 214
Model Solutions to Assignment − 3 216
Model Solutions to Assignment − 4 220
Answer Key to Test Paper 222
Model Solutions to Test Paper − 1 223
Model Solutions to Test Paper − 2 227
Model Solutions to Test Paper − 3 231
Model Solutions to Test Paper − 4 235
Model Solutions to Test Paper − 5 237
Answer Key to Practice Problems 238
Model Solutions to Practice Problems 239
 Topic 1 : Transformer
(Single Phase and Three Phase Transformer)

INTRODUCTION
• The Transformer is a static electromagnetic energy conversion device that transfers
electrical energy from one electrical circuit to another electrical circuit through the
magnetic field and without a change in the frequency. The electric circuit which
receives energy from the supply mains is called primary winding and other circuit
which delivers electric energy to the load is called secondary winding.

• Transformer is an electromagnetic energy conversion device, since the energy

received by primary is first converted to magnetic energy and it is then reconverted to
electrical energy in the other circuits. Thus primary and secondary windings of a
transformer are not connected electrically, but are coupled magnetically. This
coupling magnetic field allows the transfer of energy in either direction, from
high−voltage to low−voltage circuits or from low−voltage to high−voltage circuits. If
transfer of energy occurs at the same voltage, the purpose of transformer is merely to
isolate the two electric circuits and this use is very rare in power applications.

Fig. 1 : Two winding transformer

• If the secondary winding has more turns then the primary winding, then secondary
voltage is lower than the primary voltage and the transformer is called step−up
transformer.

• In case of secondary winding have less turns than the primary winding, then the
secondary voltage is lower than the primary voltage and the transformer is called a
step−down transformer.

• The step up transformer becomes step−down transformer in which case the

secondary of step−up transformer becomes the primary of step−down transformer.
As it is static device needs almost negligible amount of maintenance and supervision
therefore it has highest possible efficiency out of all electrical machines.

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Vidyalankar : GATE – EE

• Insulation considerations limit the generation of alternator voltages from about 11 to

22 KV. By means of transformer, this voltage is stepped up to higher economical
transmission voltage, 400 kV or even higher, in order to reduce the transmission
losses. At the distributions centre transformers are installed to step down the voltage
suitable for it utilization for motors, illumination purposes etc. Thus the transformer is
the main reason for widespread popularity of a.c. systems over d.c. systems.
In addition to power systems, transformers are widely used in other prominent areas of
electrical engineering. In communication system, input transformers connect the
microphone output to the first stage of an electronic amplifier. Interstage and output
transformer we used extensively in radio and television circuits. In electronic and control
circuits, transformers are used for impedance matching for maximum power transfer from
source to the load. Pulse transformers find wide application in radar, television and digital
computers. In power electronics, transformer are extensively used (i) for gate−pulse
triggering and (ii) for synchronizing the pulse gating signals with a.c. supply voltage given
to the main power circuits.
Functions of Transformer :
• for decreasing or increasing voltage and current levels from one circuit to another
circuit in low and high current circuits.
• for isolating d.c. while permitting the flow of a.c. between two circuits or for isolating
one circuit from another
• for matching the impedance of a source and its load for maximum power transfer in
electronic and control circuits.
Thus transformer is, an essential piece of apparatus both for high and low current circuits.

TRANSFORMER CONSTRUCTION
• There are tow general types of transformers, the core type and shell type. These two
types differs from each other by the manner in which the windings are around the
magnetic core.

• The magnetic core is a stack of thin silicon−steel laminations about 0.35 mm thick
form 50 Hz transformers. In order to reduce the eddy current losses, these
laminations are insulated from one another by thin layers of varnish. For reducing the
core losses, nearly all transformers have their magnetic core made from cold−rolled
grain−oriented sheet−steel (C.R.G.O.). This material, when magnetized in the rolling
direction, has low core loss and high permeability.

• In core−type, the windings surround a considerable part of steel core as shown in

Fig.2(a). In shell−type, the steel core surrounds a major part of the windings as shown
in Fig.2(b). For a given output and voltage rating, core type transformer requires less
iron but more conductor material as compared to a shell−type transformer. The vertical
portions of the core are usually called limbs or legs and the top and bottom portions are
called the yoke. This means that for single−phase transformers, core type has two −
legged core whereas shell−type has three−legged core.

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.2
 Transformer

Fig. 2

• In iron−core transformer, most of flux is confined to high permeability core. There is,
however, some flux that leaks through the core legs and completes its path through
surrounding the core. This flux, called leakage flux is undesirable. Consequently, an
effort is always made to reduce it. In core type transformer, this is achieved by
placing half of the low voltage (L.V.) winding over one leg and other half over the
second leg or limb. For high voltage (H.V.) winding also, half of the winding is over
one leg and other half over the second leg.

• As thickness of insulation depends upon the voltage rating hence L.V. is placed
adjacent to the steel core and H.V. winding outside.

• In shell type transformer, the L.V. and H. V. winding are wound over the central limb
and are interleaved or sandwiched as shown in Fig.2(b). The bottom and top L.V.
coils are of half the size of other L.V. coils. Shell type transformers are preferred for
low−voltage, low−power levels, whereas core type construction is used for
high−voltage, high power transformer.

Types of Winding :

• There are two types of windings employed for transformers. The concentric coils are
used for core−type transformers as shown in Fig.2(a) and interleaved (or sandwiched)
coils for shell−type transformers as shown in Fig.2(b). Hence leakage flux in shell
type is comparatively less.
Type of winding

Concentric Interleaved
(used in core type) (or sandwiched used in shell type)

Cross over Helical Disc

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Vidyalankar : GATE – EE

Cooling :

Low power transformers are air cooled whereas large power transformers are immersed
in oil for better cooling. In oil−cooled transformers, the oil serves as a coolant and also as
an insulating medium.

WORKING OF TRANSFORMER
Principle of Transformer Action :

• A transformer works on the principle of electromagnetic induction between two

(or more) coupled circuits or coils. According to this principle, an e.m.f. is induced in
the coil if it links a changing flux. Schematic diagram for two winding transformer is
shown in the Fig.3.

• The primary P is connected to

alternating voltage source, therefore,
an alternating current Ie starts flowing
through N1 turns. The alternating m.m.f.
N1Ie sets up alternating flux φ which is
confined to the high permeability iron
path. The alternating flux induces
voltage E1 in the primary P and E2 in
the secondary S. If the load is connected
Fig. 3
across the secondary, a load current
starts flowing.

In addition to the secondary winding, there may be third (or tertiary) winding on the same
iron core. The e.m.f.

• Induced in the secondary or tertiary winding is referred as the e.m.f. due to

transformer action. Thus the transformer action requires the existence of alternating
mutual flux linking the various windings on a common magnetic core.

• A transformer having primary and secondary windings is called a two−winding

transformer whereas a transformer having primary, secondary and tertiary windings
is known as a three winding transformer.

Ideal Two−winding Transformers :

Practically it does not exist. For a transformer to be ideal one, the various assumptions
are as follows :

1. Winding resistance is negligible

2. All the flux set up by the primary links the secondary winding.
3. The core losses are negligible.
4. The core has constant permeability, i.e. the magnetization curve for the core is linear.

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.4
 Transformer

No load working :

• Consider an ideal transformer as shown in Fig.3 whose secondary is open and

whose primary is connected to sinusoidal alternating voltage V1. This potential
difference causes an alternating current to flow in the primary. Since the primary coil
is purely inductive and there is no output (secondary being open) the primary draws
the magnetizing current Iμ only. The function of this current is merely to magnetize
the core, it is small in magnitude and lags V1 by 90°. This alternating current Iμ
produce an alternating flux φ which is at all times, proportional to the current (assuming
permeability of magnetic circuit to be constant) and is in phase with it.

• The changing flux is linked both with the primary and the secondary windings.
Therefore, it produces self−induced e.m.f. in the primary. This self−induced e.m.f.
E1 is, at every instant, equal to and in phase opposition to V1. It is also known as
counter e.m.f. or back e.m.f. of the primary.

• Similarly emf E2 is induced in the secondary which is known as mutually induced

e.m.f. This e.m.f. is antiphase with V1 and its magnitude is proportional to the rate of
change of flux and the number of secondary turns.

Fig. 4 : Ideal transformer (a) Phasor diagram and (b) time diagram

• The instantaneous values of applied voltage, induced e.m.f.s, flux and magnetizing
current are shown by sinusoidal waves in Fig.4(b) and Fig.4(a) shows phasor
representation of effective values of above quantities.

E.m.f. Equation of a Transformer :

cycle
Let
φm
N1 = No. of turns in primary
N2 = No. of turns in secondary
φm = maximum flux in core in webers T
f
= Bm × A 4
f = frequency of a.c. input in Hz. T = 1/f

Fig. 5

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Vidyalankar : GATE – EE

As shown in Fig.5, flux increases from its zero value to maximum value φm in one quarter
of the cycle. i.e. in 1/4 f second.
φ
∴ average rate of change of flux = m
1/ 4f

Rate of change of flux per turn means induced e.m.f. in volts.

Average e.m.f./turn = 4f φm volts

If flux φ varies sinusoidally, then r.m.s. value of induced e.m.f. is obtained by multiplying
the average value with form factor.

r.m.s. value
Form factor = = 1.11
average value

∴ r.m.s. value of e.m.f./turn = 1.11 × 4 × f × φm = 4.44 f Bm A volts

Now r.m.s. value of the induced e.m.f. in the whole of primary winding
= (induced e.m.f./turn) × (No. of primary turns)
E1 = 4.44 f N1 φm = 4.44 f N1 Bm A ….(i)
Similarly, r.m.s. value of the e.m.f. induced in secondary is
E2 = 4.44 f N2 φm = 4.44 f N2 Bm A ….(ii)

E1 E2
It is seen from (i) and (ii) that = = 4.44f φm
N1 N2

It means that e.m.f./turn is the same in both the primary and secondary windings.

In an ideal transformer no−load V1 = E1 and E2 = V2 where V2 is the terminal voltage.

Voltage Transformation Ratio (K) :

From equation (i) and (ii) we get,

E2 N2
= =K
E1 N1

This constant K is known as voltage transformation ratio.

(i) If N2 > N1 i.e. K > 1, the transformer is called step−up transformer

(ii) If N2 < N1 i.e. K < 1, the transformer is called step−down transformer

For an ideal transformer, input VA = output VA

V1I1 = V2I2
I2 V1 1
or = =
I1 V2 K

Hence, currents are in the inverse ratio of (voltage) transformation ratio.

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.6
 Transformer

On Load Working :

In Fig. 6 switch S is closed then secondary circuit is completed and transformer is said to
be on load.
Let I2 be current through secondary whose magnitude and phase angle depend on the
load. I2 will produce inphase flux φ2 which has demagnetizing effects on φm but φm is
constant as if φm changes V1′ = 4.44 f φm T reduces but V1′ ≈ V which is constant.

Fig. 6 : Ideal Transformer on load

Hence additional flux φ1′ is produced by primary to compensate effect of φ2.

i.e. φ1′ = φ2

∴ I1′N1 = I2N2 I1′ is called load component of primary current.

As exciting current Ie is very small 1% to 3 % of Full load current.

∴ I1 ≈ I1′
∴ I1N1 = I2N2

N1 V1
but =
N2 V2

∴ I1V1 = I2 V2

i.e. Power input to transformer = output power of transformer.

TRANSFORMER PHASOR DIAGRAM

1) No Load :

The magnetic flux φ being common to both the primary and secondary, is drawn first.
The induced e.m.f.s E1 and E2 lag φ by 90° and are shown accordingly in Fig.7.

Here V1′ = voltage drop in primary, in direction flow of primary current.

The various imperfections in a real transformer are now considered one by one.

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Vidyalankar : GATE – EE

(a) Effect of transformer core loss :

• The core loss consists of hysteresis loss and

eddy current loss. These losses are always
present in the ferromagnetic core of the
transformers. Since the transformer is an ac
operated magnetic device. The hysteresis
loss in the core is minimized by using high
grade mats such as cold − rolled − grain−
oriented steel and the eddy current loss is
minimized by suing thin laminations for the
core.

• It is seen from Fig.7 that exciting current

Ie leads the magnetic flux φ by some
time angle α. This angle of lead or lag
being dependent on the hysteresis loop,
is called the hysteretic angle.

• The no load primary current Ie is called

Fig. 7 : Phasor under No Load
exciting current of transformer and can
be resolved into two components.

• The component Im along φ is called the reactive or magnetizing current, since

its function is to provide the required magnetic flux φ. The second
component along V1′ is Ic and this component is called the core loss
component, or power component, of Ie . Since Ic when multiplied by V1′ gives
the total core loss Pc.
P
V1′Ic = Pc or Ic = c Amp
Vi
Also, Ie = Im2 + I2c

For an ideal transformer, core loss current Ic = 0 and therefore exciting

current Ie and therefore exciting current Ie = Im (magnetizing current).

(b) Effect of Transformer Resistance :

The effect of primary resistance r1 can be accounted for by adding to V1′ , a
voltage drop equal to r1Ie , as shown is in phase with Ie and is drawn parallel to Ie
in phasor diagram.

(c) Effect of leakage flux :

The existence of electric potential is essential for the establishment of current in
an electric circuit. Similarly the magnetic potential difference is necessary for the
establishment of flux in a magnetic circuit. As path is mainly through air gap
hence does not incorporate hysteresis loop hence 90° leading to Ie.

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.8
 Transformer

Transformer on load :

Fig. 8(a) : Transformer under load.

(b) (c)
Fig. 8 : Transformer phasor diagram for
(b) lagging p.f. load and (c) leading p.f. load

From Fig.8, we can write, the voltage equation for the primary circuit under load as

V1 = V1′ + I(r ′ 1 1
1 + jx1 ) = V1 + IZ ….(i)

where Z1 = primary leakage impedance of transformer

Note that the angle θ1 between V1 and I1 is the primary power factor angle under load.

Also the voltage equation for secondary circuit can be written as

E2 = V2′ + I2 (r2 + jx 2 ) = V2′ + I2 Z2

where Z2 = is the secondary leakage impedance of transformer for leading p.f. refer
Fig.8(b).
Phasor diagram is helpful only (i) when transformer is to be studied along and (ii)
when the internal behavior of the transformer is to be understood.

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Vidyalankar : GATE – EE

Impedance Transformation :

From above analysis we know,

V1 N1 I N
= and 1 = 2 ;
V2 N2 I2 N1
N1 N
V1 = V2 & I1 = I2 ⋅ 2
N2 N1
Now
V1 V
Z1′ = & Z2 = 2
I1 I2
N1
V2
V1 N2
∴ =
I1 N2
I2
N1
2
⎛N ⎞ Z
∴ Z1′ = Z2 ⎜ 1 ⎟ = 22
N
⎝ 2⎠ K

Secondary parameter refers to primary side :

Secondary voltage w.r.t secondary current w.r.t secondary impedance

to primary to primary w.r.t to primary
2
N N ⎛N ⎞
V1′ = V2 1 I1′ = I2 ⋅ 2 Z1′ = Z2 ⎜ 1 ⎟
N2 N1 ⎝ N2 ⎠

Similarly we can derive primary parameter w.r.t. secondary side

Rating of Transformer :

The rated output, the rated voltages, the rated frequency etc. are included in rating of
Transformer.
The rated output of transformer is expressed in kVA (Kilo−volt amperes) rather than in
kilowatts (kW). This is due to fact that rated transformer output is limited by heating and
hence by the losses in the transformer. The two types of losses in a transformer are core
loss and ohmic (−2r) loss. The core loss depends on transformer voltage and ohmic loss
on the transformer current. As these losses depend upon transformer voltage (V) and
current (I) and are almost unaffected by the load pf, the transformer rated output is
expressed in VA (V × I) or in kVA and not in kW.
Since the transformer operates at a very high η, losses may be ignored. Further, the
primary p.f. cosθ1, and the secondary p.f. cosθ2 are nearly equal. Therefore, the rated
kVA marked on name plate of a transformer, refers to both the windings, i.e. rated kVA of
the primary winding and the secondary winding are equal.

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.10
 Transformer

EQUIVALENT CIRCUIT OF A TRANSFORMER

If any electrical device is to be analysed and investigated further for suitable
modifications, its appropriate equivalent circuit is necessary. Equivalent circuit is simply a
circuit representation of equations describing the performance of the device.

Exact equivalent circuit.

r1 = primary winding resistance

x1 = primary leakage reactance
xm = magnetizing reactance
n1 = number of turns of primary
n2 = number of turns of secondary
x2 = leakage reactance of secondary
r2 = secondary winding resistance
Rc = core loss resistance
E2 = secondary induced emf
V1 = input supply
Fig. 9(a) : Exact equivalent circuit V1′ = primary induced emf
V2 = secondary output voltage
I1 = primary current
I1′= load component of I1
IE = exciting current
r1′, x1′ = primary parameter w.r.t.
secondary

Fig. 9(b) : Referred to secondary

Approximate equivalent circuit

As Ie is very small 1% to 3% of full load current, therefore we can neglect Ie and hence
exact equivalent circuit can be simplified as below.

• In this Fig.9(c) effect of Ie on r1 is neglected

Fig. 9(c)

• In this Fig.9(d) effect of Ie on r2 is added.

Fig. 9(d)

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 Vidyalankar : GATE – EE

• For transformer > 100 kVA : it can be

further simplified as

Fig. 9(e)

• For transformer > 500 kVA : it can

be most simplified as Fig. 9(f)

Fig. 9(f)
As transformer has winding resistance (r1, r2), leakage reactance (x1, x2), and magnetizing
reactance Xm. Hence losses occur. The different type of losses are as follows.
Losses in Transformers :
Losses

Core loss Ohmic Loss Stray loss Dielectric loss

I2R loss occurs in Leakage flux produces This loss occur in

Transformer winding when eddy current in transformer oil
loaded. As the standard conductor, tank, and insulation.
operating temperature of Channels, bolt and this
machine is 75°C hence loss in known as stray
this loss should be loss.
calculated at 75°C.

These two losses

Eddy Current loss Hysteresis Losses are very small and
• Ph = kh f Bmx hence neglected.
• Pe = ke (f Bm)2
ke : constant : kh depends on volume
depends upon and quality of core
thickness, volume, material.
and resistance of
• Ph ∝ Vx f 1 − x
core lamination
( As V ∝ fBm) i.e. this
• Pe ∝ V2 (as V∝ fBm) loss depends on both
i.e. eddy current loss frequency and voltage.
is independent of
frequency of supply.

Here Pe and Ph are Eddy current and Hysteresis loss

f is the frequency of supply and Bm is maximum flux density of the core

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.12
 Transformer

Testing of transformers :

Many test are performed on transformer for efficiency, regulation, insulation capacity,
over load capacity, equivalent circuit parameter for analysis purpose. Scope constraint for
the explanation of the following :

Open circuit test and short circuit test :

Performed for determination of

i) parameter of equivalent circuit
ii) voltage regulations
iii) efficiency

Open Circuit Test :

• Circuit diagram for the test is shown in Fig. 10(a)

Fig. 10

• Input power is utilized to magnitise the core, as H.V. side is open circuited i.e.
transformer is at no load hence this test is also called as no load test.

• Now, no load current vary from 1 % to 5% of full load current

⎛ 1 1 ⎞ ⎛ 5 5 ⎞
∴ copper loss varies from (I2R) ⎜ × × 100% ⎟ to ⎜ 100 × 100 × 100% ⎟ of full load.
⎝ 100 100 ⎠ ⎝ ⎠
i.e. 0.01 % to 0.25 % of full load.
but constant loss consist of eddy current and hystersis loss is approximately
proportional to (voltage)2.

• Hence ohmic loss is negligible compared to core loss. Hence wattmeter reading is
approximated to core loss PC. Corresponding equivalent circuit is shown in the
Fig.10(b).

• This test can be performed either way on L.V. & H.V. side but L.V. side is preferred
so that ordinary ranges of meter and sources could be used.

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Vidyalankar : GATE – EE

e.g.
1.1 kV/220V, 11kVA transformer.

H.V. L.V. Comment

Voltage rating 1.1kV 220V 220V source is readily
available, not 1.1kV and also
corresponding voltmeter
range.
Current rating 10A 50V
No−load current 0.5A 2.5A
Comment 2.5A could be measured
by ordinary ammeter.

Short Circuit Test :

• Circuit diagram with Ammeter, voltmeter, Wattmeter connection H.V. side is shown in
Fig 11(a).

Fig. 11(a) : Connection diagram for short Fig. 11(b) : Transformer equivalent circuit
circuit test on a transformer with short circuit test on a transformer

• As primary mmf = secondary mmf . Hence rated current in primary = rated current in
secondary.

• Under short circuit condition as load on H.V. is its own winding impedance hence 5%
to 10% of normal primary voltage is enough to circulate rated current in both primary
and secondary.

• ∴ Core loss α V2 under full applied voltage for 5 % to 10 % of input voltage.

∴ core loss varies from

⎛ 5 5 ⎞ ⎛ 10 10 ⎞
⎜ 100 × 100 × 100% ⎟ to ⎜ 100 × 100 × 100% ⎟
⎝ ⎠ ⎝ ⎠
0.25 % to 1 % of actual.

Hence wattmeter reading is approximated to ohmic loss and corresponding

equivalent circuit is shown in Fig 11(b).

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.14
 Transformer

• Short circuit test is performed on L.V. side so that ordinary ranges of meter could be
used even though can be performed on H.V. side too.
e.g. 3.3 kV / 220V, 33kVA.

H.V. L.V. Comment

10% of Voltage rating 330 V 22V 22V is a very small when
compared with ordinary
source availability
Full Load Current 10A 150A
Comment 10A is well within the
ordinary ammeter ranges
available.

Calculation of equivalent circuit parameter :

Open circuit test
Determines : Core loss resistance = Rc
Magnetizing Reactance = Xm
V1 = input voltage on L.V.
Ie = no load current
Pc = core loss
Pc = V1 Ie cos θ
Pc ⎛ P ⎞
No load power factor, cos θ0 = ; θ0 = cos −1 ⎜ c ⎟
V1Ie ⎝ V1Ie ⎠
Core loss current component = Ic = Ie cos θ0
Magnetizing current = Im = Ie sin θ0

V1 V1 ⎛ V ⎞ ⎛ V1 ⎞ V12
(A) Rc = = = ⎜ 1 ⎟⎜ ⎟=
Ic Ie cos θ0 ⎝ V1 ⎠ ⎝ Ie cos θ0 ⎠ Pc
(B) Ic 2Rc = Pc
Pc Pc
∴ Rc = =
Ic 2 (Ie cos θ0 )2
V12 Pc
∴ Rc = =
(Ie cos θ0 )
2
Pc
V1 V1
Xm = =
Im Ie sin θ0
V1
∴ Xm =
Ie sin θ0
both Rc and Xm are referred to L.V. side.

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Short circuit Test :

from Fig 11(b)

Vsc
= Ze Which is equivalent leakage impedance w.r.t. H.V. side for meter
Isc
connected. (Would have been w.r.t. L.V. side if meter were at L.V. side.)
P
& equivalent resistance w.r.t. H.V. side = re = sc2
Isc
∴ equivalent leakage reactance referred to L.V. side
x e = ze 2 − re 2
For equivalent circuit,
1
H.V. side resistance = L.V. side resistance = re
2
1
H.V. side reactance = L.V. side reactance = xe
2
Hence equivalent circuit parameter are determined.

Polarity test :
Significance of this test lies when transformer needs to be operated parallel or in
polyphase circuit.

Fig. 12 :

• Polarity of secondary winding depend upon manner of winding.

• As per IS 2026, H.V. side terminal are represents by capital letter (A1, A2) and
corresponding L.V. side by small letter (a1, a2)
• If voltmeter reading = E1 − E2 = difference of voltage rating of transformer then, 1.v
side is marked a1 corresponding to A1 of H.V.

• If voltmeter reading = E1 + E2 = addition of voltage rating of both side of transformer

then L.V. side is marked a2 corresponding to A1 of H.V.
• Also it can be concluded that the voltmeter used for the polarity test should have
rating more than addition of voltage of two side of transformer.

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.16
 Transformer

Load test (back to back or Sumpner’s Test)

• determines : i) maximum temperature rise

ii) efficiency (accuracy is less compared to open circuit and short
circuit test calculation.

• Small transformer can be put on full load but for large size transformer dissipation of
energy is difficult. They are put on full load by means of this test but for 1φ it requires
two identical transformer.

For 1φ Transformer

Fig. 13 : Sumpner’s test on two identical

single phase transformer

• In this test primaries are connected in parallel and secondary are in series.

• As secondary are open circuited hence 2I0 will be taken from input and W1 will give
core loss of both transformer, i.e. 2 Pc.

• If voltmeter V2 reads approximately zero then they are in phase opposition else
terminals connection are changed. ‘c’ will be connected to ‘b’ rather then ‘d’.

• Now variable voltage is applied to series connection of secondary so that rated

current should flow in secondary and hence in primary.

• This rated current in secondary will cause rated circulating current in primary. (which
is equivalent to consider primary in short circuit state)hence W2 give copper loss of
both transformer = 2Psc and V2 will be equal to impedance drop of two secondaries.

• Hence net power supplied = ( 2Pc + 2Psc )

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.17
Vidyalankar : GATE – EE

• As explained in open circuit and short circuit, W1 will be connected to L.V.

& W2 will be connected to H.V.

• As transformer are at full load current and rated voltage (V1) they are at full load and
hence give maximum temperature size.

• From a self explanatory Fig.13 no load current and circulating current in one
transformer is in phase while in other opposition. Hence temperature rise in previous
will be more.

For 3φ Transformer
Method involved for this is called a dummy load test.

Fig. 14 : Dummy load test on a

three−phase transformer

• From self explanatory Fig.14

W1 + W2 = total Pc

• Secondary winding in open delta is applied with 1φ supply and adjusted to obtain
rated current in winding and based on the principal already explain for 1φ
transformers sumpler’s test
W3 = Psc = total copper loss of winding.

• Voltmeter (V) connected in open delta will read IFL × 3Zeh.

where Ze is per phase leakage impedance. (w.r.t. H.V. side)

Similar to losses and temperature rise due to winding resistance and leakage reactance
voltage drop occurs in transformers which further depends upon load power factor, load
current. Hence output voltage under load condition is different from under no load
conditions; this is the measure of voltage regulation.

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.18
 Transformer

VOLTAGE REGULATION
• Definition : Voltage regulation is defined as change in secondary voltage from no
load to full load to the no load secondary voltage.
V2 = secondary terminal voltage at any load
E2 = secondary terminal voltage at no load.

E2 − V2
• Voltage regulation = × 100%
E2
at no load, primary leakage impedance is negligible hence
⎛N ⎞
E2 = V1 ⎜ 2 ⎟
⎝ N1 ⎠
⎛N ⎞
V1 ⎜ 2 ⎟ − V2
∴ Voltage Regulation = ⎝ N1 ⎠ × 100%
⎛ N2 ⎞
V1 ⎜ ⎟
⎝ N1 ⎠
⎛N ⎞
V1 − V2 ⎜ 1 ⎟
= ⎝ N2 ⎠ × 100%
V1
Derivation : Reference to approximate equivalent circuit Fig.15(a) and corresponding
phasor diagram Fig.15(b) for lagging load is shown below.

Fig.15 : (a) Approximate equivalent circuit w.r.t. secondary

(b) Phasor diagram for lagging load

Arc of radius OD intersect the extended OA at F. OF (= E2) ≅ OC.

∴ E2 ≅ OC = OA + AB + BC
= OA + AB′ cos θ2 + DB′ sin θ2
∴ E2 − V2 = I2 re2 cos θ2 + I2 xe2 sin θ2
E2 − V2 I2re2 Ix
∴ = cos θ2 + 2 e2 sin θ2
E2 E2 E2

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.19
Vidyalankar : GATE – EE

For the rated current I2 r

I2r ⋅ re2 Voltage drop across re2 at rated current

=
E2 Rated voltage E2
(∵ Rated voltage is same as base voltage)
I2r ⋅ re2
∴ = εr = per unit equivalent resistance drop.
E2
I2r ⋅ re2 I2r 2 ⋅ re2 Ohmic loss at rated current
εr = = =
E2 E2 ⋅ I2r Rated VA
Similarly
I2r ⋅ x e2
= εx
E2
∴ voltage regulation = ( εr cos θ2 + εr sin θ2 ) × 100 %
For leading load voltage regulation = ( εr cos θ2 − εr sin θ2 ) × 100 %

voltage regulation = ( εr cos θ2 ± εr sin θ2 ) × 100 % = ( εr cos θ2 ± εr sin θ2 ) in per unit.

(+) sign for lagging load and (−) for leading load
For the accurate expression for the voltage regulation
(E2 = OF ≠ OC) if considered leads to following expression.
1
voltage regulation = ( εr cos θ2 ± εr sin θ2 ) + ( εr cos θ2 ± εr sin θ2 )
2

2
(+) sign for lagging load and (−) for leading load

As regulation depends on load power factor following conclusion can be made.

Condition for zero voltage regulation :

p.u. voltage regulation

= ( εr cos θ2 + εr sin θ2 ) = 0

(for zero voltage regulation)

εr Ir r
∴ tan θ2 = − = − 2 e2 = − e2
εx I2 x e2 x e2
E2
E2

Fig.16(a) : zero V.R.

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.20
 Transformer

• as θ2 is negative therefore zero voltage

regulation occurs for leading power
factor.

• Magnitude of load p.f.

x
cos θ2 = e2
ze2

x e2
• For leading p.f. > voltage
ze2
regulation will be negative.

Fig.16(b) : negative voltage regulation

Condition for maximum voltage regulation

p.u. voltage regulation

= ( εr cos θ2 + εr sin θ2 )

Differentiating and equating to zero will give the

condition for above.
d
∴ (p.u. regulation) = −εr sin θ2 + ε x cos θ2 = 0
dθ2
ε x x e2
∴ tan θ2 = =
εr re2

Fig.16(c) : maximum voltage regulation

• Positive value of tan θ2 indicates maximum voltage regulation occurs for lagging p.f.
Magnitude of maximum voltage regulation

re2 x Ir r Ix x re2 x
= εr ⋅ + ε x e2 = 2 e2 ⋅ e2 + 2 e2 ⋅ e2 (∵ cos θ2 = and sin θ2 = e2 )
ze2 ze2 E2 ze2 E2 ze2 ze2 ze2

I2 Iz
=
E2 ze2
( )
re2 2 + x e2 2 = 2 e2 = ze2 p.u.
E2

• The maximum voltage regulation is equal to equivalent leakage impedance (in p.u.)

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.21
Vidyalankar : GATE – EE

EFFICIENCY
Efficiency is a ratio of output power to input power.

Output power
Efficiency η =
Input power
Output power
=
Output power + constant loss + cu loss
V2I2 cos θ2
=
V2I2 cos θ2 + Pc + I2 2re2
cos θ2
ηp.u. = (dividing above by V2 I2)
P I 2r
cos θ2 + c + 2 e2
V2I2 V2I2

Condition for maximum efficiency

V2I2 cos θ2
η =
V2I2 cos θ2 + Pc + I2 2re2

• As Pc, core loss is constant therefore loss variation is due to cu loss and cu loss,
depends on the load current. Therefore, differentiating w.r.t. I2 and equating to zero,
will give the condition for maximum efficiency.
dη
i.e. when =0 Yield I2 2re2 = Pc
dI2
∴ For ηmax variable ohmic loss = constant core loss

Pc
• Load current I2 for which ηmax occurs I2 =
re2
• kVA load for ηmax
P P
I2 = c = Ifl 2 c Multiply both equation by E2/ 1000
re2 Ifl re2
E2I2 EI Pc
= 2 fl
1000 1000 Ifl2re2

Pc
• (kVA )η = ( kVA )fl
max Ifl2re2
Pc
= ( kVA )fl
Full load ohmic losses

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.22
 Transformer

Effect of load power factors on efficiency

• As Pc and re2 are unaffected by a

variation in load power factor hence
ηmax does not depends on load power
factor.

• But as reduction in load power factor

reduces the output therefore efficiency
get reduced.

Efficiency can reduced but ηmax

occurs at the same load regardless of
variation in the load power factor.
Fig. 17

Energy Efficiency :

• As power transformers always operates under full load condition, hence they are
designed for maximum efficiency near full load.
• For distribution transformers variation of the load takes place through out the day.
Hence they are designed for maximum efficiency near to half load.
• With this we can also conclude that,
i) Efforts are made to reduce copper loss in power transformers as full load current
remains same.
ii) Efforts are made to reduce core loss in distribution transformers as operating
voltage remains same irrespect to load condition.

As distribution transformer is connected to consumers terminal, hence load

varies over a wide range during 24 hour day. Hence their efficiency is defined on
the basis of energy efficiency.

• Definition : Energy efficiency is a ratio of total energy output to the total energy input
during the same period.

For the period of 24 hours energy efficiency is called as All day efficiency.

All Day Efficiency

Definition : All day efficiency is a ratio of energy output to the energy input during a 24
hour period.

Output in kWh
ηall day = ( for 24 hours )
Input in kWh

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.23
Vidyalankar : GATE – EE

THREE−PHASE TRANSFORMERS
Economic generation of a three phase system now universally three phase transmission
due to the development of three phase transformer. From generating station to the
consumers, the voltage is raised and lowered several times for economic reasons using
three−phase transformer. For each voltage transformation from one value to another,
three units of single−phase transformation or one unit of three phase transformer may be
used. When three identical units of single−phase transformers are used, the arrangement
is commonly called a bank of three transformers or a three−phase transformer bank, for
the sake of distinguishing it from a three−phase transformer. Fig.18(a) illustrates a
three−phase transformer bank with both primary and secondary windings connected in
star.

Fig. 18(a) :Three−phase transformer bank consisting

of three units of single−phase transformers.

A three−phase transformer may be of core type or shell type, as in single phase

transformers. Fig.18(b) displays three single−phase transformers, with their yokes placed
120° apart. The primary and secondary windings are wound over the limbs I, II, III and
three unwound limbs (or legs) are brought in close contact. Three phase currents in three
primaries produce three−phase fluxes mutually time−displaced by 120°. These fluxes
flow through their respective yokes and then through the central limbs placed together.
The resultant flux in the three central limbs must be zero, since the phasor sum of three
equal fluxes time displaced by 120° is zero, just as the phasor sum of three−phase
balanced currents meeting at a point is zero. Alternatively, if we look Fig.18(b) from top,
then the three yokes appear to be in star and the resultant flux at the star point, made up
by the three central limbs, is zero. Since the central leg carries no flux, it can be
eliminated without disturbing any of the existing conditions. As shown in Fig.18(c). The
flux produced in any one leg, would return through the other tow legs or limbs; similar to a
3−phase system, where current in one wire at any instant returns through the other two
wires, The construction of a transformer with such type of core would be very expensive
and it is, therefore, more practicable to construct a transformer core as depicted in
Fig.18(d).

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.24
 Transformer

(b) (c) (d)

Fig. 18 : Pertaining to the evolution of three−phase core−type
transformer from three single−phase units.
A three−phase shell type transformer is obtained if three single−phase shell type cores
are placed side by side as shown in Fig.18(e) :
It can be concluded that
• Area 1 and 4 carry φ/2 flux, hence has the area half of central core area.
3
• Area 2 and 3 carry φ flux, hence has the area 86.6% of central core area.
2
• If winding B is reversed as shown in Fig.18(f) then area 1, 2, 3 & 4 will become same
and equal to half of central core area.

Fig. 18 (e) :Three−phase shell type transformer

Fig. 18(f) : Three−phase shell type transformer

with B winding reversed

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.25
Vidyalankar : GATE – EE

THREE−PHASE TRANSFORMER CONNECTIONS

All three−phase transformers connection can be grouped in followings :
1. Zero phase displacement (Yy0, Dd0, Dz0)
2. 180° phase displacement (Yy6, Dd6, Dz6)
3. 30° lag phase displacement (Dy1, Yd1, Yz1)
4. 30° lead phase displacement (Dy11, Yd11, Yz11)

• Economical for H.V. transformer as

VL = 3Vph insulation is min.

• Third Harmonic voltage is absent on line provided

no fourth wire is connected. Neutral is unstable
due to third harmonic component in exciting current.
• Suitable for large L.V. transformer
• Closed mesh serves to damp out third
harmonic voltages.
• Can tolerate large unbalanced in load.

• Not very common in use

• Has all advantages of Yy 0

• Phase shift between primary and secondary is
180°.

• Has all advantage of Dd 0

• Phase shift between primary and secondary is
180°.

• Not very common in use

• Used for supply of six−ring synchronous

converters.

• The most common connection for Yyy is start /

double − star connection.

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.26
 Transformer

• Very common for distribution transformer to

facilitate 1φ for consumer application. As a
secondary/star connection provides neutral.

• Is very common for power-supply

transformer. As has star point for mixed
loading and a delta provide close mess for
third harmonic current and which stabilizes
star point potential.
• It facilitate unbalanced loading and
reduction in third harmonic voltage but
zigzag is confined to L.V. winding.
• 15% more turns are required compared to
normal phase connection in a zigzag mode.
• Normally it is employed where delta
connection are mechanically week (due to
large number of turns and small copper
section.

• Refer Yd1.
• Phase difference between delta and star is
30°. Star leading delta by 30°
• Refer Yd1.
• Phase difference between delta and star is
30°, L.V. leading H.V. by 30°.

• Refer Yz1.
• Zigzag side leading the star side by 30°.

• Use for convertion of 3φ to 2φ and

viseversa.

• Details will be explained later.

Open−Delta or V−V connection :

• In transformer bank connected in Δ /Δ, when one phase winding is removed resulting
arrangement is called as Open delta.

• It is employed :

1. when the three−phase load is too small to warrant the installation of full
three−phase transformer bank.

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.27
Vidyalankar : GATE – EE

2. when one of the transformers in a ^ − ^

bank is disabled, so that service is
continued although at reduced capacity, till
the faulty transformer is repaired or anew
one is substituted.
3. when it is anticipated that in future the load
will increase necessitating the closing the
open delta.

Fig. 19(a)

(b) (c) (d)

Fig. 19 : (b) Transformer connected in open delta

(c) Phasor diagram for Primary line voltage
(d) Phasor diagram for Secondary line voltage

• In open delta Vline = Vphase

∴ VA rating of open delta transformer = 3 ⋅ VL ⋅ VL = 3 ⋅ VphIph

• Therefore kVA rating of open delta equals to 57.7 percent of the closed delta.

Actual kVA rating

• Utilization or rating factor is
Sum of kVA rating of transformer installed
3VphIph
= = 0.866
2VphIph
i.e. Utilization rating of open delta is 0.866 while for closed delta it is unity.

The disadvantages of this connection are :

1. the average power factor at which the V−bank operates is less than that of the
load. This power factor is actually 86.6 % of the balanced load power factor.
Another significant point to note is that, except for a balanced unity power factor
load, the two transformers in the V − V bank operate at different power factors.
2. secondary terminal voltages tend to become unbalanced to a great extent when
the load is increased, this happens even when the load is perfectly balanced.

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.28
 Transformer

• For unity power factor load one transformer operates at 0.866 lead and the other at
0.866 lag.

• For inductive load of 30° lag one transformer operates at unity power factor and carry
double the power of other which operates at 0.5 lead.

• For 0.5 lag one transformer shares zero load while other shares total load.

Scott Connection :
Use for conversion of 3φ to 2φ and vice versa.
2φ system voltages has phase displacement − of 90°. It has application in (i) single
phase are furnace. (ii) Electric tranction. (iii) 2φ control motor.

(b)

(a) Fig. 20
• Fig.20(a) shows schematic diagram of two
1Q transformer for Scott connection and
Fig.20(b) give its schematic connection
diagram.

• Here B2 is centre tap of main transformer.

• Vab = VAB = VAD + VDB

∴ VAD = VAB − VDB = VAB + VBD Fig. 20(c)

Let VBC = VL ∠0
1 1
∴ VBD = VBC = VL ∠0
2 2
VAB = VL ∠120, VCA = VL ∠ − 120
as shown in Fig. 20(c)
1
∴ VAD = VL ∠120 + VL ∠0
2
Fig. 20 (d)
VAD = 0.866VL ∠90°

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.29
Vidyalankar : GATE – EE

• Hence, VAD is 0.866.

∴ VAD = 0.866VBC ∠90°

• If T is number of turn of primary main transformer than for Vab = Vbc , number of turn
of primary teaser = 0.866 T hence it should have tapping at 0.866 of the winding.

VL
• Vph = = 0.578VL < 0.866VL
3
neutral is on teaser transformer. Therefore the neutral point is 0.866 VL − 0.578VL =
0.288VL from b.

• VAD = 0.866 VL, VAN = 0.578 VL and VDN = 0.288VL Therefore it can be concluded that
Neutral divides the teaser primary in ratio = 2 : 1.

PARALLEL OPERATION OF 3φ TRANSFORMER

Need for the parallel operation

Need for the parallel operation arises as generation, transmission and distribution of
power is three phase.
• Increases reliability
• For under load condition some transformers are switched off which reduces the
transformer losses and system becomes more economical and efficient.
• Cost of spare unit in 1φ is less compared to 3φ.

Condition for parallel operation

• The transformers must have same voltage ratio, same phase sequence and same
relative phase displacement.
• Transformers should have equal per unit leakage impedance.
• Ratio of equivalent resistance should be same.

• The transformer should have same polarity.

Volation of fourth condition will cause large circulating current within the transformer
which will damage the winding.

No load operation

• If there will be difference between secondary voltage due to any disbalance then
circulating current will flow to increase the secondary voltage of 1 transformer and
reduce the other.
• Practically such circulating current should be less than 10% of the rated current.

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.30
 Transformer

On load operation
(a) Equal voltage ratio :
i) Equal impedance but different reactance to resistance ratio :
• Full load kVA output is less than the kVA rating of individual transformer.
• Transformer with the greater leakage impedance operates at poor power
factor.
ii) Difference impedance but same reactance for resistance ratio
• kVA shared by transformers is inversely proportional to its leakage impedance.
i.e. transformer with different kVA rating can be operated in parallel provide
their equivalent leakage impedance is inversely proportional to their kVA
rating.
(b) Unequal voltage ratio :
• Will cause circulating current which reduce the power factor and increase the
load share by the transformer having larger emf.

AUTO TRANSFORMERS
• A transformer, in which a part of the winding is common to both the primary and
secondary circuits, is called an auto−transformer. In a two winding transformer,
primary and secondary windings are electrically isolated, but in an auto−transformer
the two windings are not electrically isolated.

Fig. 21 : Single phase step−down auto−transformer.

• A simple arrangement of a step−down auto−transformer is depicted in Fig.21, where
N1 and N2 are the number of turns between winding AC and winding BC
respectively. When voltage V1 is applied to winding ABC, an exciting current starts
flowing through the full winding AC. If the internal impedance drop is neglected, then
V
the voltage per turn in winding AC is 1 and, therefore the voltage across BC
N1
⎛V ⎞
is ⎜ 1 ⎟ N2 .
⎝ N1 ⎠

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.31
Vidyalankar : GATE – EE

Difference between Autotransformer and Potential Divider :

• An auto−transformers appears to be similar to a resistance potential divider. But this

is not so, as described below. A resistive potential divider can’t step up the voltage,
whereas it is possible in an auto−transformer; the potential divider has more losses
and is, therefore, less efficient. In a potential divider, almost entire power to load
flows by conduction, whereas in auto−transformer, a part of the power is conducted
and the rest is transferred to load by transformer action. In potential divider, the input
current, must always be more than the output current, this is not so in an
auto−transformer. If the output voltage in auto−transformer is less than the input
voltage, the load current is more than the input current.

• When the switch S is closed, a current I2 starts flowing through the load and current
I1 is taken from the source. Neglecting losses,

Input power = output power

or V1I1 cos θ1 = V2I2 cos θ2

If internal (or leakage) impedance drops and losses are neglected, then
cos θ1 = cos θ2

Hence V1I1 = V2I2

V2 I1 N2
or = = =k
V1 I2 N1
Here k is less than unity.
Let the terminal A, Fig.21, be positive with respect to terminal C, at the instant shown.
Then at no load, the exciting current flows from A to C and it establishes a working
m.m.f. directed vertically downward in the core. When switch S is closed, the current
in winding BC must flow from C to B, in order to create an m.m.f opposing the
exciting or working m.m.f., as per Lenz’s law. Since the working m.m.f. in a
transformer remains substantially constant at its no−load value, the primary must
draw additional current I1 from the source, in order to neutralize the effect of current
ICB. In winding AC, I1 flows from A to C while in winding BC, I2 flows from C to B.
Therefore, the current in winding BC is I1 from B to C and I2 from C to B. Here the
current I2 is greater than I1 (because V2 < V1) and their m.m.f.s are opposing each
other at every instant, therefore,
ICB = I2 − I1

m.m.f of winding AB = I1(N1 − N2 )

= (I2 − I1 )N2 [ ∴ I1N1 = I2 N2]
= ICBN2 = m.m.f. of winding CB.

• It is therefore, seen that the transformer action takes place between winding AB and
winding BC. In other words, the volt−amperes across winding AB are transferred by
transformer action to the laod connected across winding BC.

∴ Transformed VA = VAB IAB = (V1 − V2) I1

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.32
 Transformer

Total input VA to transformer = V1 I1 = output VA

Transformed VA (V1 − V2 )I1 V

∴ = = 1 − 2 = (1 − k)
Input VA V1I1 V1

Out of the input volt−amperes V1I1 , only VAB IAB = (V1 − V2) I1 are transformed to the
output by transformer action. The rest of the volt−amperes required for the output,
are conducted directly from the input.

∴ Conducted VA = Total input VA − transformed VA

= V1I1 − (V1 − V2 )I1 = V2I1

Conducted VA V2I1
∴ = = k.
Input VA V1I1

Neglecting internal impedance drops and losses, equations (1) and (2a) become
Transformed power
= (1 − k)
Input power

Conducted VA
and =k
Input VA

Inductively transferred power High voltage − Low voltage

∴ =
Total power High voltage

Advantages of Auto−transformer over two winding transformer :

(i) The weight of conductor (copper or aluminium) for any winding depends upon the
cross−sectional area and length of the conductor. Now the conductor area is
proportional to the current carried by it, whereas the length of conductor in a winding
is proportional to the number of turns in the winding. Thus the weight of conductor in
a winding is proportional to the current and number of turns in the winding.

For an auto transformer of Fig.21, winding AB carries a current of I1 and has (N1 − N2)
turns.

∴ Weight of conductor for winding AB ∝ (N1 − N2 )I1

Winding BC carries a current of (I2 − I1 ) and has N2 turns.

∴ Weight of conductor for winding BC ∝ (I2 − I1 )N2

Hence, total weight of conductor in an auto−transformer is

∝ I1(N1 − N2 ) + (I2 − I1 )N2
∝ 2(I1N1 − I1N2 )
∝ 2(N1 − N2 )I1

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.33
Vidyalankar : GATE – EE

If the weight of conductor in an auto transformer and an ordinary two winding

transformer is to be compared, the two types must have the same input (V1I1 ) , same
output (V2I2 ) and same voltage ratio V1 / V2 . Therefore, an ordinary transformer must
have voltage V1, V2 and currents I1, I2 for its primary and secondary windings
respectively. If peak flux density and core area are assumed equal, then e.m.f. per
turn Et (= 2π f Bm A i ) in the two types of transformer, is the same. Thus a two
⎛ V ⎞ ⎛ V ⎞
winding transformer must have N1 ⎜ = 1 ⎟ and N2 ⎜ = 2 ⎟ turns in its primary and
⎝ E t ⎠ ⎝ Et ⎠
secondary windings respectively.

∴ Total weight of conductor in a 2−winding transformer

∝ I1N1 + I2N2
∝ 2I1N1

Weight of conductor in an auto − transformer

∴
Weight of conductor in a 2 − winding transformer

2(N1 − N2 )I1 N
= − 1− 2 = 1− k
2N1I1 N1

or Weight of conductor in auto−transformer

= (1−k) (Weight of conductor in 2−winding transformer)

or Conductor weight in 2−winding transformer − conductor weight in auto−transformer

= (k) conductor weight in 2−winding transformer.

Saving of condcutor material ⎤ ⎡Conductor weight in

Thus, ⎥ =k× ⎢
if auto − transformer is used ⎦ ⎣2 − winding transformer

If k = 0.1, saving of conductor material is only 10% and for k = 0.9, saving of
conductor material is 90%. Hence the use of auto transformer is more economical
only when the voltage ratio k is more nearer to unity.

Other advantages of an auto−transformer, over a two−winding transformer are given

below :
(ii) Owing to the reduction in conductor and core materials, the ohmic losses in
conductor and the core loss are lowered. Therefore, an auto−transformer has higher
efficiency than a two−winding transformer of the same output.
(iii) Reduction in the conductor material means lower value of ohmic resistance. A part of
the winding being common, leakage flux and, therefore, leakage reactance is less. In
other words, an auto−transformer has lower value of leakage impedance and has
superior voltage regulation than a two−winding transformer of the same output.

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.34
 Transformer

Disadvantages :
(i) If the ratio of transformation k differs far from unity, the economic advantages of
auto− transformer over two−winding transformer decrease.

(ii) The main disadvantage of an auto−transformer is due to the direct electrical

connection between the low−tension and high−tension sides. If primary is supplied at
high voltage, then an open circuit in the common winding BC, would result in the
appearance of dangerously high voltage on the L.V. side. This high voltages may be
detrimental to the load and the persons working there. Thus a suitable protection
must be provided against such an occurrence.

(iii)The short−circuit current in an auto−transformer is higher than that in a corresponding

two−winding transformer.

I. Equivalent circuit and phasor diagram of Auto−transformer :

(a)

(b)
(c)
Fig.22 : (a) Equivalent two−winding transformer for an auto−transformer of Fig.21
(b) Equivalent circuit for an auto−transformer and
(c) Phasor diagram for an auto−transformer

From the above diagram

V2 V ⎡ ⎛ 1− k ⎞
2
⎤
V1 = Vab + V2 = + I1(re1 + jx e1 ) = 2 + I1 ⎢z1 + ⎜ ⎟ z2 ⎥
k k ⎣⎢ ⎝ k ⎠ ⎦⎥
2 2
⎛ N − N2 ⎞ ⎛ 1− k ⎞
where re1 = r1 + r2 ⎜ 1 ⎟ = r1 + r2 ⎜ ⎟
⎝ N 2 ⎠ ⎝ k ⎠

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.35
Vidyalankar : GATE – EE

2 2
⎛ N − N2 ⎞ ⎛ 1− k ⎞
x e1 = x1 + x 2 ⎜ 1 ⎟ = x1 + x 2 ⎜ ⎟
⎝ N2 ⎠ ⎝ k ⎠
z1 = r1 + jx1 = primary leakage impedance
and z2 = r2 + jx 2 = secondary leakage impedance

Equation above gives equivalent circuit for an auto−transformer as shown in

Fig.22(b) above. Here Rc, the core−loss resistance and Xm, the magnetizing
reactance are also shown to account for exciting current.

Phasor Diagram :
As in a two−winding transformer, the voltage equation for the secondary load circuit of an
auto−transformer is
E2 = V2 + ( I2 − I1 )(r2 + jx 2 )
where E2 = e.m.f. induced in N2 turns
V2 = load voltage
and (I2 − I1) = current in secondary winding CB.
Phasor sum of V2 and (I2 − I1)(r2 + jx2) is shown in Fig.22(c) where I2 is assumed to lag
behind V2 by load power angle θ2. E.m.f. E2 is induced by mutual flux φ which is shown
leading E2 by 90°. Induced e.m.f. E1 in N1 turns as shown equal to E2/k. As before,
E2
V1 = −E1 = − or V1′ =| E1 |
k
For the primary circuit, applied voltage V1 has to balance the induced e.m.f. E1, voltage
rise from B to A (against the direction of I1) and voltage drop from C to B (in the direction
of current (I2 − I1) ).
∴ V1 = E1 + I1(r1 + jx1 ) − (I2 − I1 )(r2 + jx 2 )
E1
or V1 = + I1(r1 + jx1 ) − (I2 − I1 )(r2 + jx 2 )
k
First I1(r1 + jx1 ) is added to | E1 |= V1′ and then (I2 − I1 )(r2 + jx 2 ) is subtracted as shown in
Fig.22(c) to get the applied voltage V1 for a single−phase auto−transformer.

Step−up Auto Transformer :

Figure shows a 1−phase step−up

auto−transformer. The number of turns in
winding AC is N2 and in winding BC is N1
so that winding AB has (N2 − N1) turns.
Here V2 > V1 and I1 > I2. The distribution
of currents in the windings is shown in
Fig.23.

Fig. 23 :Single−phase step−up auto−transformer.

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.36
 Transformer

As before, assuming ideal conditions,

V1 I2 N1
= = = k (less than one)
V2 I1 N2
M.m.f. of winding AB = I2 (N2 − N1)
= I1N1 − I2N1 = (I1 − I2 )N1
= m.m.f. of winding BC

This shows that transformer action occurs between winding BC and AB as in a

step−down auto−transformer.
Output of Step − up auto − transformer
Output of equivalnet two-winding transformer
V2I2 1 1
= = =
(V2 − V1 )I2 V 1− k
1− 1
V2
Thus, the advantage of enhanced power rating of step−up auto−transformer is also of the
same order as in case of step−down auto−transformer.

Comparison of characteristics of auto−transformers and two−winding transformers :

A two−winding transformer can be used as auto−transformer by connecting its two

windings in series electrically.
(i) Rating :

It is seen from Fig.21 that winding AB acts as the primary and widning BC acts as the
secondary of a 2−winding transformer. As per equation
kVA rating as an auto − transformer
kVA rating as a 2-winding transformer

Primary input voltage V1 × primary input currentI1

=
Pr imary voltage across winding AB × primary current in AB

V1I1 1 1
= = =
(V1 − V2 )I1 1 − (V2 / V1 ) 1 − k

kVA rating as an auto − transformer V2I2 1 1

Also = = =
kVA rating as a 2-winding transformer V2 (I2 − I1 ) 1 − (I1 / I2 ) 1 − k

(ii) Losses :

When a 2−winding transformer is connected as an auto−transformer, the current in

different sections and voltages across them remain unchanged. Therefore, losses
when working as an auto−transformer are the same as the losses in a 2−winding
transformer. Per unit losses, however, differ

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.37
Vidyalankar : GATE – EE

Per unit full − load losses as auto − transformer

Per unit full − load losses as 2 − winding transformer

Full − load losses kVA rating as 2 − winding transformer

= × = (1 − k)
kVA rating as auto − transformer Full − load losses

(iii) Impedance drop :

When a 2−winding transformer is used as an auto−transformer, both L.V. and H.V.

windings are utilized completely. In addition, current and voltage ratings of ach
winding section remain unaltered. Therefore, impedance drop at full load is the same
in both the transformers. Their pre unit values are, however, different. When referred
to H.V. side, per unit impedance drop as an auto−transformer is with respect to
voltage V1 and for a2−winding transformer, it is with respect to (V1 − V2) = V1 (1−k)
Per unit impedance drop as an auto − transformer
Per unit impedance drop as 2 − winding transformer

I1z1 / V1 V − V2
= = 1 = (1 − k)
I1z1(V1 − V2 ) V1

(iv) Voltage Regulation :

Regulation in transformers is proportional to per unit impedance drop.

Regulation as an auto − transformer
= (1 − k)
Regulation as a2 − winding transformer

(v) Short−circuit current :

Per unit short−circuit current is the reciprocal of the per unit impedance drop.
Per unit short − circuit current as an auto − transformer 1
∴ =
Per unit short − circuit current as 2 − winding transformer 1 − k

The value of k used in the above relations is less than one. In general, for using
these relations, the value of k for step−down or step−up auto−transformers is
k = L.V./H.V.

Uses :

Single−phase and three−phase auto−transformers are mainly employed;

i) for interconnecting power systems having voltage ratios, not differing far from unity
and
ii) for obtaining variable output voltages. When used as variable ratio auto transformers,
there are known by their trade names, such as variac, dimmerstat, autostat etc.

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.38
 Transformer

A variable ratio auto−transformer (or

variac) has a toroidal core and
toroidal winding. A sliding contact
with the winding is made by carbon
brush, Fig.24. The position of the
sliding contact can be varied by a
handwheel and with this the output
voltage gets changed. A variac used
for starting induction motors and
synchronous motors at reduced
voltages, is commonly called an Fig. 24 : Single−phase variable ratio
auto−transformer starter or a starting auto−transformer.
compensator.

Schematic diagrams for single−phase and three−phase variable ratio auto−transformers

are illustrated in Fig.25.

(a) (b)
Fig. 25: Schematic diagram for (a) single−phase variac and (b) three−phase variac.
When auto transformer is used for supplying L.V. system, then its common point, such as
C in Fig.25(b) in case of single−phase system or the neutral in case of 3−phase system,
must be earthed otherwise there is a risk of serious shock.

THREE WINDING TRANSFORMER

When the secondary of a
transformer has more than 2 V
separate winding on the
common core it is called as 2V
multiwinding transformer.
V
When a secondary has only two
separate winding it is called as
Three Winding Transformer.
V V

Fig.26
GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.39
Vidyalankar : GATE – EE

• Let z be the leakage impedance of each section ab and ac.

• Fig.26(a) shows transformer secondary with two winding. For the connection shown
in 26(b), ab and cd form a series circuit of leakage impedance 2z to supply a load
voltage 2V.

• If a connection shown in Fig.26(c) ab and cd form parallel circuit of leakage

impedance of z/2 to supply load voltage V.

• In Fig.26(d) winding ab and cd can supply 2 different load and can carry current
different in magnitude and phase.

• Hence Fig.26(b) and (c) represents two winding transformer but Fig.26(d) shows
three winding transformer.

• Practically three winding transformer are designed to supply load at different

voltages.

• Large three winding transformer are constructed for high, medium and low windings.
kVA rating of three windings are usually unequal.

• In practice kVA rating of a three winding transformer is considered equal to largest

kVA rating among the three winding.

Uses :
• It can supply two loads independently requiring different voltage and is more cheaper
and efficient than equivalent two winding transformer.

• It is used extensively in Electronic circuits. One supplying to plate and other to

cathode.

• It is used in interconnection of transmission line at different voltages.

• Used for Power factor correction and voltage regulation by connecting one of the
winding to static capacitor.

• Used for converting three phase supply to six or more phases.

LMR (LAST MINUTE REVISION)

1
• Phase voltage on star side is times delta side.
3

• Line current of star side is 3 times the delta.

• Tap changer are of two type : No load and On load type used on H.V. side for
regulation purpose.

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.40
 Transformer

• Primary and secondary voltage of transformer are 180° out of phase.

• Effect of polarity marking : In Yy0 if a2 a1 is reverse from the actual position then the
line voltage on L.V. will be unbalanced.

• Can be conclude that

i) Va1b2 = Vc2 a1 and time displaced by 60°.

ii) Line voltage Vb2 c3 is 3 time the phase voltage.

• Yd11 (group 4) can be successfully operated in parallel with by 1(group 3)

• Both in delta and star (isolated neutral) third harmonic voltage and current are
absent.
• Yy without neutral; VL = 3VPH does not hold good and effect of third harmonic
voltage is to oscillate neutral at 3f.

• Oscillating neutral cause the fluctuation in line to start point voltage.

• Yy with neutral grounded the IIIrd harmonic current flow in phase and line and cause
considerable interference with the communication circuit running parallel to
transmission line.

• Sinusoidal magnetizing flux need magnetizing current of peak nature in transformer.

This occur in Yy transformer with neutral grounded.

• If current taken by transformer is sinusoidal in nature then the flux will be flat top. And
in this case the core lose will be less than the previous case.

GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.41
Vidyalankar : GATE – EE

• Effect of current harmonics

i) Increase in ohmic loss, therefore temperature rises and efficiency decreases.
ii) Electromagnetic interference with communication line running parallel and
inductive interference may also cause the faulty operation of the protective.

• Effect of voltage harmonics :

i) Voltage strain on insulation : In Yy transformer without neutral phase voltage are
more than rated voltages which cause increase in dielectric loss which reduces
the transform efficiency and shorten the life of insulation.

ii) Electrostatic interference with communication circuits :

In case shunt capacitive reactance between transmission line and communication
line becomes equal to inductive reactance at some harmonic frequency, the series
resonance occur which increases interference.

iii) Large resonant voltages :

In Yy transformer with neutral grounded when connected to a long transmission

line, transformer inductance and the transmission line capacitances form a series
circuit. The magnetizing reactance of Transformer at 3f = line to ground capacitive
reactance at 3f then series resonant occur at third harmonic e.m.f. between line to
ground circulates third harmonic exciting current. Such a amplified harmonic
current cause third harmonic voltage between line to neutral to become
dangerously high.

• The dielectric strength of transformer oil is expected to be 33 kV.

• The permissible flux density for C.R.G.O. is around 1.7 Wb/m2.

• No load current in a transformer lags the voltage be 75°.

• For the transformer above 500 kVA Buchholz relay is used. Buchholz relay indicates
unprepared electric field.

• For the transformer upto 3000 kVA Natural oil cooling.

• In high frequency transformer air core is used.
• If supply frequency is increased by "n" then same size transformer could be used
only for
(kVA )f
(kVA )nf =
n
• Laminations : Transformer core are laminated to reduce the eddy current loss. The
steel core is assembled in such a manner that the butt joints in adjacent layers are
staggered which avoids continuous air gap and therefore the reluctance of magnetic
circuit is not increased. As continuous air gap would reduce the mechanical strength
of core.



GATE/EE/SLP/Module_1/Ch.1_Notes /Pg.42
 Transformer

ASSIGNMENT − 1
1. Primary, or secondary winding of a 3−phase transformer in delta helps to
suppress
(A) third harmonic current (B) third harmonic voltage
(C) third harmonic flux (D) all of the above

2. Scott−connected transformers, can convert

(A) 3 to 2 phases (B) 2 to 3 phases
(C) both (A) and (B) (D) 2 to 4 phases

3. The function of using delta connected, tertiary winding, in a star−star transformer is

(i) To permit the flow of sufficient fault current for the operation of protective
devices
(ii) To permit the single phase loading
(iii) To provide low−reactance path for zero−sequence currents
(iv) None of these
From above, the correct answer is
(A) (i), (ii), (iii) (B) (i), (ii)
(C) (ii), (iii) (D) (iv)

4. For a 3−phase transformer, out of positive, negative & zero sequence

impedance, the minimum impedance is
(A) zero sequence impedance
(B) positive sequence impedance
(C) negative sequence impedance
(D) depend upon the type of application.

5. The magnetizing inrush current in a transformers is rich in

rd
(A) 3 harmonics (B) 5th harmonics
th
(C) 7 harmonics (D) 2nd harmonics

6. In P.U. system, the base values chosen first are

(A) voltampere, voltage (B) voltampere, current
(C) voltage, current (D) current, impedance

7. Three 30kW, 400V single phase Induction Motor are to be supplied by

transformer connected in open−delta from an 11 kV line. At full load, each motor
has an η of 90% & operates at a p.f. of 0.866 lagging. The KVA rating of each of
the two transformers & their turns ratio is
(A) 36.67 kVA, 27.5 (B) 66.67 kVA, 27.5
(C) 60 kVA, 27.5 (D) 56.67kVA, 27.5

8. For the above question the p.f. of each transformer is

(A) 1, 0.8 (B) 0.8, 1
(C) 0.8, 0.8 (D) 1, 0.5

GATE/EE/SLP/Module_1/Ch.1_Assign /Pg.43
Vidyalankar : GATE – EE

9. For the above question if a 3rd transformer of the same rating is used to form
closed delta the available capacity will be
(A) 200 KVA (B) 220 kVA
(C) 240 kVA (D) 210 kVA

10. For the above question, the real power supplied by the transformers are
(A) 36.67 kW, 13.75 kW (B) 66.67 kW, 33.34 kW
(C) 60 kW, 13.75 kW (D) 56.67 kW, 13.75kW

11. A 30 KVA, 3500 / 400V, 1 − ph transformer has the following parameters :

H.V. winding L.V. winding
r1 = 8Ω r2 = 0.2Ω
x1 = 18 Ω x2 = 0.9 Ω

then voltage regulation at full load for a p.f. of 0.8 lagging is

(primary voltage is held constant)
(A) 11% (B) 10%
(C) 13% (D) 15%

12. In an ideal transformer, the impedance can be transformed from one side to
other in
(A) direct proportion to square of voltage − ratio
(B) direct proportion to turns−ratio
(C) inverse proportion to square of turns ratio
(D) direct proportion to square of current ratio.

13. CRGO laminations in a transformer are used to minimize

(A) eddy current loss (B) Hysteresis loss
(C) hysteresis loss & stary loss. (D) (A) & (B) both

14. The leakage flux in a transformer depends on.

(A) Applied voltage (B) the frequency
(C) the load current (D) mutual flux

15. If in a transformer PC = core loss & PSC = full load ohmic loss, then maximum kVA
delivered to the load at maximum η is equal to rated kVA multiplied by
2
PC ⎛ PC ⎞
(A) (B) ⎜ ⎟
PSC ⎝ PSC ⎠
PC PSC
(C) (D)
PSC PC

GATE/EE/SLP/Module_1/Ch.1_Assign /Pg.44
 Transformer

16. Frequency of supply voltage to a transformer at no load is increased but the

supply voltage is held fixed. With this
(i) Eddy current loss remains constant but hysteresis loss increases.
(ii) Eddy current loss remains constant but hysteresis loss decreases
(iii) magnetizing current increases but core loss current decreases
(iv) both magnetizing & core loss currents decrease
The correct − statements are
(A) (ii), (iii) (B) (ii), (iv)
(C) (i), (iii) (D) (i), (iv)

17. As the load on a transformer is increased, the core losses

(A) decrease slightly
(B) increase slightly
(C) remain constant
(D) may decrease or increase slightly depending upon the nature of load

18. When compared with power transformer, a distribution transformer has

2
(A) low % age impedance & high I R loss to core loss ratio
(B) high % age impedance & high I2R loss to core loss ratio
(C) high % age impedance & low I2R loss to core loss ratio
(D) low % age impedance & low I2R loss to core loss ratio

19. A multimeter, for measuring resistance, is connected to one terminal of primary &
other terminal of secondary. The multimeter will read
(A) zero
(B) ∞
(C) zero or infinity
(D) equal to resistance of winding

20. In a tap changer are provided on

(A) HV of power transformer
(B) LV of φ transformer
(C) Both side of transformer
(D) None of the above



GATE/EE/SLP/Module_1/Ch.1_Assign /Pg.45
Vidyalankar : GATE – EE

TEST PAPER − 1
Duration : 30 Min. Max. Marks : 25
Q1 to Q5 carry one mark each

1. Match the Type of Transformer with Applications of Transformers :

Type Applications
a Power Transformer 1. Thyristor Firing circuit
b Distribution Transformer 2. Impedance matching
c Pulse Transformer 3. At different city locations
d Audio−frequency transformer 4. At generating stations

(A) a − 3, b − 4, c − 1, d − 2 (B) a − 3, b − 4, c − 2, d − 1
(C) a − 4, b − 3, c − 1, d − 2 (D) a − 4, b − 3, c − 2, d − 1

2. Voltage Regulation of a large transformer is mainly influenced by

(A) no load current & load p.f.
(B) winding resistances & load p.f.
(C) leakage fluxes & load p.f.
(D) winding resistances & core loss
3. In transformer, the noise or hum is produced primarily due to
(A) Improper tightening of core laminations
(B) magnetostriction
(C) tank walls
(D) hysteresis loss

4. The efficiency of a 100 kVA transformer is 0.98 at full as well as half load. For
this transformer at full load, the ohmic loss
(A) is less than core loss (B) is equal to core loss
(C) is more than core loss (D) none of the above
5. Two transformers to be operated in parallel have their secondary no load emfs Ea
for transformer A & Eb for transformer B. As Ea is somewhat more than Eb, a
circulating current IC is established at no load which tends to
Ea − Eb
(A) boost both Ea and Eb with IC =
Zea + Zeb
Ea − Eb
(B) boost Ea and buck Eb with IC =
Zea + Zeb
Ea − Eb
(C) buck Ea and boost Eb with IC =
Zea + Zeb
Ea − Eb
(D) buck both Ea & Eb with IC =
Zea + Zeb

GATE/EE/SLP/Module_1/Ch.1_Test /Pg.46
 Transformer

Q6 to Q15 carry two marks each

6. Two coils with self − inductance of 1H & 4H have mutual inductance of 1H. Their
coupling factors & the coefficient of coupling is
(A) 0.5 (B) 1
(C) 0.25 (D) 0.75

7. Two Three phase transformer have the following per phase parameters referred
to secondary
re1 = 0.004Ω, x e1 = 0.018Ω
re2 = 0.002Ω, x e2 = 0.012Ω

Subscripts 1 & 2 refer to 500 kVA & 1000 kVA transformers. These two
transformers are connected for parallel operation. They share a load of 1500 kVA
at 0.8 p.f. lagging, then load shared by transformer 1 is given by
(A) 388.11 kW (B) 400.02 kW
(C) 488.11 kW (D) 712.55 kW

8. A three phase step down transformer is energized from 11 kV, 50 Hz source. If it

takes a line current of 20A from the supply mains & per phase turns ratio is 44,
then output kVA for star / star connection is
(A) 340 kVA (B) 381.05 kVA
(C) 394.02 kVA (D) Data insufficient

9. Two transformers A & B are joined in parallel to the same load. Open circuit emf
6600 V for A & 6400 V for B. Equivalent leakage impedance in terms of the
secondary = 0.3 + j3 for A & 0.2 + j1 for B. The load impedance is 8 + j6. The
current delivered by transformer B is
(A) 195 (B) 400 A
(C) 359 A (D) Data insufficient

10. A 600 kVA, Single phase transformers has an efficiency of 90% both at full load
& half load at unity power factor. Its efficiency at 65% of full load at 0.8 power
factor lag is
(A) 89.09% (B) 75%
(C) 85.03% (D) 95%

11. A 50 kVA, 1-Phase transformer has an iron loss of 400W & full copper loss of
800 W. The load at which maximum η is achieved at unity p.f. is
(A) 91% (B) 99%
(C) 95% (D) 97.79%

12. A transformer has copper loss of 1.6% and reactance drop of 4.5% when tested
at full load. It full load regulation at 0.8 p.f. leading is
(A) 1.35% (B) 1.42 %
(C) 1.5% (C) 1.25%

GATE/EE/SLP/Module_1/Ch.1_Test /Pg.47
Vidyalankar : GATE – EE

13. A short circuit test when performed on the high voltage side of a 20 kVA,
2000/400 V 1−phase transformer, gave the following data 60V, 4A, 100W
If the low voltage side is delivering full load current at 0.8 pf lag & at 400V, the
voltage applied to high voltage side approximately is
(A) 130 V (B) 139 V
(C) 141 V (D) 132 V
14. Below figure shows two 1- phase ideal transformer T1 & T2 of turns ratio 5 : 1 &
4 : 1 connected with their primaries in parallel across a 100 V, 50 Hz source.
Their secondaries are connected in series as shown. The dotpoint indicate
terminals of the corresponding polarity on the primary & secondary sides. For
R = 20Ω, the current taken from the source will be
I1 I11 T1
+
V21
I21
−
90V 5:1 V2
4:1 R = 20Ω
50Hz
source −
V22
+
T2
(A) 0.11 A (B) 0.12 A
(C) 0.125 A (D) 0.1125 A
15. The transformer secondary in star supplies power to a purely resistive load, with
one ammeter & wattmeter connected as shown in figure. The readings on the
ammeter and voltmeter are 27 A and 648 W respectively. The line to line voltage
are balanced. Considering upto third harmonic the value of the resistor R will be

W
R

A
R
R

(A) 2Ω (B) 3Ω
(C) 2.98Ω (D) 2.67Ω


GATE/EE/SLP/Module_1/Ch.1_Test/Pg.48