MFM, AGRO-TECHNICAL AND TECHNOLOGY COLLEGE

DEPARTMENT OF ELECTRICAL AND ELECTRONICS TECHNOLOGY

Introduction to Electrical Machines (EET 3311)

TRANSFORMERS

Prepared by: MfM, Agro-Technical and Technology College

Hamdi Mohammed
 Outline
2

 Introduction
 Principle of operation
 Construction
 Ideal Transformer models
 EMF Equation of a Transformer
 Equivalent Circuit of Transformer
 Open and Short circuit Test
 Transformer losses and Efficiency
 Auto Transformer and parallel operation of Transformers
 3-phase Transformer
 Instrument Transformers

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 Introduction
3

 The transformer is a static piece of apparatus by means of which an electrical power is

transformed from one alternating current circuit to another with the desired change in
voltage and current, without any change in the frequency and power.

Fig. 1 Use of transformers in transmission system

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 Working Principle of Transformer
4

 The transformer works on the principle of mutual induction.

 It consists of two inductive coils which are electrically separated but linked through a
common magnetic circuit.
 The two coils have high mutual inductance. One of the two coils is connected to a source
of alternating voltage. This coil in which electrical energy is fed with the help of source is
called primary winding (P). The other winding is connected to load. The electrical energy
transformed to this winding is drawn out to the load. This winding is called secondary
winding (S).

• The primary winding has N1 number of

turns while the secondary winding has
N2 number of turns.

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 Working Principle of Transformer Cont…
5

 Symbolically the transformer is indicated in Fig. below. When primary winding is excited by
an alternating voltage, it circulates an alternating current.

 This current produces an alternating flux (Φ) which completes its path through common
magnetic core.
 Thus an alternating flux links with the secondary winding. As the flux is alternating, according
to Faraday’s law of electromagnetic induction, mutually induced e.m.f. gets developed in the
secondary winding.
 If now load is connected to the secondary winding, this e.m.f. drives a current through it. Thus
though there is no electrical contact between the two windings, an electrical energy get
transferred from primary to the secondary.

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 Can D.C. Supply be used for Transformers?
6

 The d.c. supply cannot be used for the transformers.

 The transformer works on the principle of mutual induction, for which current in one
coil must change uniformly.
 If d.c. supply is given the current will not change due to constant supply and
transformer will not work.
 Practically, winding resistance is very small. For d.c. the inductive reactance XL is zero
as d.c. has no frequency. So total impedance of winding is very low for d.c.. thus
winding will draw very high current if d.c. supply is given to it.
 This may cause the burning of windings due to extra heat generated and may cause
permanent damage to the transformer.

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 Construction
7

 There are two basic parts of a transformer i) Magnetic Core ii) Winding or coils.

 The core of the transformer is either square or rectangular in size. It is further divided into
two parts. The vertical portion on which coils are wound is called limb while the top and
bottom horizontal portion is called yoke of the core.
 Core is made up of laminations. Because of laminated type of construction, eddy current
losses get minimized. Generally high grade silicon steel laminations are used. These
laminations are insulated from each other by using insulation like varnish.
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 Construction Cont…
8

 The coils used are wound on the limbs and are insulated from each other.
 To reduce the leakage flux, windings should be very close to each other to have high
mutual inductance.
 To achieve this, the two windings are split into number of coils and are wound adjacent
to each other on the same limb.
 The classification of the transformers based on the relative arrangement or disposition
of the core and windings.
1) Core type 2) Shell type 3) Berry type

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 Core type Transformer
9

 It has a single magnetic circuit. The core is rectangular having two limbs.
 The winding encircles the core. The coils used are of cylindrical type. Both the coils
are placed on both the limbs.
 The Fig. (a) shows the schematic representation of the core type transformer while the
Fig. (b) shows the view of actual construction of the core type transformer.

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 Shell type Transformer
10

 It has double magnetic circuit. The core has three limbs. Both the windings are placed on
the central limb.
 The core encircles most part of the windings. The coils used are multilayer disc type or
sandwich coils.
 High voltage coil is in between two low voltage coils and low voltage coils are nearest
to top and bottom of the yokes.

• Generally for very high voltage

transformers, the shell type construction
is preferred. As the windings are
surrounded by the core, the natural
cooling does not exist.
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 Berry type Transformer
11

 This has distributed magnetic circuit. The number of independent magnetic circuits are
more than 2. Its core construction is like spokes of a wheel.

 The entire transformer assembly is immersed in the oil. The oil serves two functions:
i) Keeps the coils cool by circulation and
ii) Provides the transformers an additional insulation.
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12

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 E.M.F. Equation of a Transformer
13

From Faraday’s law of electromagnetic induction the average e.m.f. induced in each turn is proportional
to the average rate of change of flux.
average e.m.f. per turn = average rate of change of flux
average e.m.f. per turn = dΦ/dt
Now dΦ/dt = Change in flux
Time required for change in flux
 Consider the 1/4th cycle of the flux as shown in the Fig. complete cycle gets completed in 1/f seconds.
In 1/4th time period, the change in flux is from 0 to Φm.
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 E.M.F. Equation of a Transformer Cont…
14

dΦ/dt = Φm - 0 as dt for 1/4th

[1/4f]
= 4fΦm Wb/sec
average e.m.f. per turn = 4fΦm volts
 As Φ is sinusoidal, the induced e.m.f. in each turn of both windings is also sinusoidal in

nature. For sinusoidal quantity,

 Form Factor = R.M.S. value = 1.11
Average value
 R.M.S. value = 1.11 x Average value
 R.M.S. value of induced e.m.f. per turn = 1.11 x 4 f Φm = 4.44 f Φm

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 E.M.F. Equation of a Transformer Cont…
15

 There are N1 number of primary turns hence the R.M.S. value of induced e.m.f. of
primary denoted as E1 is,
E1 = N1 x 4.44 f Φm volts
 While as there are N2 number of secondary turns the R.M.S. value of induced e.m.f. of
secondary denoted E2 is,
E2 = N2 x 4.44 f Φm volts
 The expressions of E1 and E2 are called e.m.f. equations of a transformer.
 Thus e.m.f. equations are,
 E1 = 4.44 f Φm N1 volts
 E2 = 4.44 f Φm N2 volts

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 Ratios of a Transformer
16

Voltage ratio
 From Eqs. above, we get

This ratio of secondary induced e.m.f. to primary induced e.m.f. is

known as voltage transformation ratio denoted as K.
Thus E2 = K E1 where K = N2/ N1
• If N2 > N1 i.e. K > 1, we get E2 > E1 then the transformer is called
step-up transformer.
• If N2 < N1 i.e. K < 1, we get E2 < E1 then the transformer is called
step-down transformer.
• If N2 = N1 i.e. K = 1, we get E2 = E1 then the transformer is
called isolation transformer or 1:1 transformer.
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 Ideal Transformer
17

 A transformer is said to be ideal if it satisfies following properties:

 It has no losses
 Its winding have zero resistance.
 Leakage flux is zero i.e. 100% flux produced by primary links with the secondary.
 Permeability of core is so high that negligible current is required to establish the flux in it.

 For an ideal transformer, the primary applied voltage V1 is same as the primary induced
e.m.f. E1 as there are no voltage drops.
 Similarly the secondary induced e.m.f. E2 is also same as the terminal voltage V2 across the
load. Hence for an ideal transformer we write,
E2 V2
---- = ---- = K
E1 V1
 No transformer is ideal in practice but the value of E1 is almost equal to V1 for properly
designed transformer
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 Ratios of a Transformer
18

Current ratio
 For an ideal transformer there are no losses. Hence the product of primary voltage V1
and primary current I1, is same as the product of secondary voltage V2 and secondary
current I2.
So V1 I1 = input VA and V2 I2 = output VA

 For an ideal transformer V1 I1 = V2 I2

V2 I1
--- = ---- = K
V1 I2
 Hence currents are in the inverse ratio of the voltage transformation ratio.

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 Volt-Ampere Rating
19

 The output maximum rating is specified as the product of output voltage and output
current i.e. V2 I2.
 This always indicates that when transformer is operated under this specified rating, its
temperature rise will not be excessive.
 Actual output power available depends on cos Φ2 which is power factor of the
secondary.
 As cos Φ2 can change depending on load, the rating is not specified in watts or kilowatts
but indicated as the product of voltage and current called VA rating.
 This rating is expressed in KVA (Kilo volt amperes rating)

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 Volt-Ampere Rating Cont…
20

 If V1 and V2 are the terminal voltages of primary and secondary then from specified
KVA rating we can decide full load currents of primary and secondary, I1 and I2.
 This is the safe maximum current limit which it may carry, keeping temperature rise
below its limiting value.
KVA rating
I1 full load = -----------------------
V1

KVA rating
I2 full load = ------------------------
V2

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 Examples
21

1. The maximum flux density in the core of a 250/3000-volts, 50-Hz single-phase

transformer is 1.2 Wb/m2. If the e.m.f. per turn is 8 volt, determine
(i) primary and secondary turns
(ii) area of the core.

1. A single-phase transformer has 400 primary and 1000 secondary turns. The net
cross-sectional area of the core is 60 cm2. If the primary winding be connected to a
50-Hz supply at 520 V, calculate
(i) the peak value of flux density in the core
(ii) the voltage induced in the secondary winding.

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 Examples
22

 A 20 KVA single phase transformer has 200 turns on the primary and 40 turn on the
secondary. the primary is Connected to 1000V, 50Hz supply. Determine the secondary
voltage on open circuit and the current flowing through the two windings on the full load.

V2= 200V
I1FL= 20*1000/1000=20A
I2FL= 20*1000/200=100A

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 Ideal Transformer on No Load
23

 Consider an ideal transformer on no load as shown in the Fig. below. The supply
voltage is V1 and as it is on no load the secondary current I2 = 0.

 E1 and E2 are in phase and both opposing supply voltage V1.

 The power input to the transformer is V1 I1 cos (V1^I1) i.e. V1 Im cos (90˚) i.e. zero.
 This is because on no load output power is zero and for ideal transformer there are
no losses hence input power is also zero. Ideal no load p.f. of transformer is zero
lagging.
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 Practical Transformer on No load
24

 Actually in practical transformer iron core causes hysteresis and eddy current losses as it
subjected to alternating flux. While designing the transformer the efforts are made to keep
these losses minimum by
 Using high grade material as silicon steel to reduce hysteresis loss.
 Manufacturing core in the form of laminations or stacks of thin laminations to reduce eddy
current loss.
 Small primary copper loss present.

 The current which accounts for the above mentioned losses is denoted as I0.
 Now the no load input current I0 has two components:
 A purely reactive component Im called magnetizing component of no load current required to
produce the flux. This is also called wattles component.
 An active component Ic which supplies total losses under no loads condition called power
component of no load current.

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 Practical Transformer on No load Cont…
25

 This is also called wattful component or core loss component of Io.

 The total no load current Io is the vector addition of Im and Ic. Īo = Īm + Īc
 It can be seen that the two components of Io are,
Im = Io sin Φo
 This is magnetizing component lagging V1 exactly by 90°.
Ic = Io cos Φo
 This is core loss component which is in phase with V1.
 The magnitude of the no load current is given by,
Io = √(I2m + I2c)
 While Φo = no load primary power factor angle.
 The total power input on no load is denoted as Wo and is given by,
Wo = V1 Io cos Φo = V1 Ic
Wo = V1 Io cos Φo = Pi = iron loss
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 Transformer on Load
26

 When the transformer is loaded, the current I2 flows through the secondary winding.
The magnitude and phase of I2 is determined by the load.
 If load is inductive, I2 lags V2. If load is capacitive I2 leads V2 while for resistive load,
I2 is in phase with V2.

 The m.m.f. N2I2 is called demagnetizing ampere-turns.

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 Transformer on Load Cont…
27

 As the ampere turns are balanced we can write,

N2 I2 = N1 I’2
I’2 = [N2/N1]I2 = K I2
 Thus when transformer is loaded the primary current I1 has two components:
 The no load current Io which lags V1 by angle Φo. It has two components Im and Ic.
 The load component I’2 which is in antiphase with I2 and phase of I2 is decided by the
load.
 Hence primary current I1 is vector sum of Io and I’2.
Ī1 = Īo + Ī’2

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 Examples
28

 A single-phase transformer with a ratio of 440/110-V takes a no-load current of 5A at

0.2 power factor lagging. If the secondary supplies a current of 120 A at a p.f. of 0.8
lagging, estimate the current taken by the primary.

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 Effect of winding Resistances
29

 A practical transformer windings possess some resistances.

Let R1 = Primary winding resistance in ohms
R2 = Secondary winding resistance in ohms
 Now when current I1 flows through primary there is voltage drop I1 R1 across the winding. The
supply voltage V1 has to supply this drop. Hence primary induced e.m.f. E1 is the vector
difference between V1 and I1 R1.
Ē1 = V1 – I1 R1 …..(1)
 Similarly the induced e.m.f. in secondary is E2. When load is connected current I2 flows and
there is voltage drop I2 R2. The e.m.f. E2 has to supply this drop. The vector difference
between E2 and I2 R2 is available to the load as a terminal voltage V2.
V2 = Ē2 – I2 R2 …(2)
 The drops I1 R1 and I2 R2 are purely resistive drops hence are always in phase with the
respective currents I1 and I2.
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 Equivalent resistance
30

 The resistance of two windings can be transferred to any one side either primary or
secondary without affecting the performance of the transformer. Let us see how to
transfer the resistance on any one side.
 The total copper loss due to both the resistance obtained as

Total copper loss = I21 R1 + I22 R2 = I21[R1 + ﴾I22/ I21﴿R2] = I21 [R1 + ﴾1/K2﴿R2] …(3)
where I2/I1 = 1/K neglecting no load current
2
 This means R2/K is the resistance value of R2 shifted to primary side which causes
same copper loss with I1 as R2 causes with I2. this value of resistance R2/k2 which is
the value of R2 referred to primary is called equivalent resistance of secondary
referred to primary. It is denoted as R’2.
R’2 = R2/K2

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 Equivalent resistance Cont…
31

 Hence the total resistance referred to primary is the addition of R1 and R’2 called equivalent resistance
of transformer referred to primary and denoted as R1e.
R1e = R1 + R’2 = R1 + R2/K2
 This resistance R1e causes same copper loss with I1, total copper loss due to the individual windings.
Total copper loss = I21 R1e = I21 R1 + I22 R2
 So equivalent resistance R1e simplifies the calculations as we have to calculate parameters on one side
only.
 Similarly it is possible to refer the equivalent resistance to secondary winding.
Total copper loss = I22[﴾I21/I22﴿R1 + R2] = I22 [K2 R1 + R2] ….(7)
 Thus the resistance K2 R1 is primary resistance referred to secondary denoted as R’1.
R’1 = K2 R1
R2e = R2 + R’1 = R2 + k2 R1
 Hence the total resistance referred to secondary is the addition of R2 and R’1 called equivalent
resistance of transformer referred to secondary and denoted as R2e.

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 Equivalent resistance Cont…
32

 The concept of equivalent resistance is shown in the Fig. below (a), (b), and (c).

High voltage side → Low current side → High resistance side

Low voltage side → high current side → Low resistance side
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 Effect of Leakage Reactance
33

 In practice part of primary flux as well as secondary flux completes the path through
air and links with the respective windings only. Such a flux is called leakage flux.

X1 = Leakage reactance of primary winding.

X2 = Leakage reactance of secondary winding.
 Leakage fluxes link with the respective windings only and not to both the windings. To
reduce the leakage, both the windings are placed on same limb rather than on
separate limbs.
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 Equivalent leakage reactance
34

 Similar to the resistance, the leakage reactances also transferred from primary to
secondary or vice versa. The relation through K2 remains same for the transfer of
reactances as it is studied earlier for resistances.
 Let X1 be reactance of primary and X2 be reactance of secondary.
 Then the total leakage reactance referred to primary is X1e given by,
X1e = X1 + X’2 where X’2 = X2/K2
 While the total leakage reactance referred to secondary is X2e given by,
X2e = X2 + X’1 where X’1 = K2X1
And K = N2/N1 = transformation ratio

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 Equivalent Impedance
35

 The transformer primary has resistance R1 and reactance X1, secondary has resistance
R2 and reactance X2.
 Total impedance of primary winding is Z1,
Z1 = R1 + jX1 Ω
 Total impedance of secondary winding is Z2,
Z2 = R2 + jX2 Ω

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 Equivalent Impedance Cont…
36

 The individual magnitude of Z1 and Z2 are,

Z1 = √ ﴾ R21 + X21﴿
And Z2 = √ ﴾R22 + X22﴿
 Similar to resistance and reactance, the impedance also can be referred to any one side.

Let Z1e = total equivalent impedance referred to primary, then

Z1e = R1e + jX1e
Z1e = Z1 + Z’2 = Z1 + Z2/K2
 Similarly Z2e = total equivalent impedance referred to secondary, then
Z2e = R2e + jX2e
Z2e = Z2 + Z’1 = Z2 + K2Z1
 The magnitude of Z1e and Z2e are,

Z1e = √﴾R21e + X21e﴿

Z2e = √﴾R22e + X22e﴿
 It can be noted that,

Z2e = K2 Z1e and Z1e = Z2e/K2

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 Equivalent Impedance Cont…
37

 The concept of equivalent impedance is shown in the Fig. below

Referred to primary Referred to secondary

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 Examples
38

 A 30 kVA, 2400/120-V, 50-Hz transformer has a high voltage winding resistance of

0.1 Ω and a leakage reactance of 0.22Ω. The low voltage winding resistance is 0.035
Ω and the leakage reactance is 0.012 Ω. Find the equivalent winding resistance,
reactance and impedance referred to the
(i) high voltage side and
(ii) the low-voltage side.

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 Examples
39

 A 50-kVA, 4,400/220-V transformer has R1 = 3.45 Ω, R2 = 0.009 Ω. The values of

reactances are X1 = 5.2 Ω and X2 = 0.015 Ω. Calculate for the transformer
(i) equivalent resistance as referred to primary
(ii) equivalent resistance as referred to secondary
(iii) equivalent reactance as referred to both primary and secondary
(iv) equivalent impedance as referred to both primary and secondary
(v) total Cu loss, first using individual resistances of the two windings and secondly, using
equivalent resistances as referred to each side.

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 Equivalent circuit of Transformer
40

 The equivalent circuit of a machine means the combination of fixed and variable
resistances and reactances, which exactly simulates performance and working of the
machine.
 For a transformer, no load primary current Io has two components,
Im = Io sin Φo = magnetizing component
Ic = Io cos Φo = active component
 From the equivalent circuit we can write,
 Ro = V1/Ic , Xo = V1/Im

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 Equivalent circuit of Transformer Cont…
41

 Hence the equivalent circuit can be shown as in the Fig. below.

 But in the equivalent circuit, windings are not shown and it is further simplified by
transferring all the values to the primary or secondary. This makes the transformer
calculations much easy.
 So transferring secondary parameters to primary we get,

R’2 = R2/ K2, X’2 = X2/ K2, Z’2 = Z2/ K2

while E’2 = E2/ K, I’2 = KI2 where K = N2/N1

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 Equivalent circuit of Transformer
42

 Thus the exact equivalent circuit referred to primary can be show as in the Fig. below.

Fig. Exact equivalent circuit referred to primary

 Similarly all the primary value can be referred to secondary and we can obtain the
equivalent circuit referred to secondary.
R’1 = K2 R1, X’1 = K2 X1, Z’1 = K2 Z1
E’1 = K E1, I’1 = I1/ K, I’o = Io/ K

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 Equivalent circuit of Transformer Cont…
43

 Similarly the exciting circuit parameters also gets transferred to secondary as R’o and
X’o. The circuit is shown in Fig. below.

Fig. Exact equivalent circuit referred to secondary

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 Approximate Equivalent Circuit
44

 To get approximate equivalent circuit, shift the no load branch containing Ro and Xo to
the left of R1 and X1. By doing this we are creating an error that the drop across R1 and
X1 due to Io is neglected.
 Hence such an equivalent circuit is called approximate equivalent circuit.

Fig. Approximate equivalent circuit referred to primary

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 Approximate Equivalent Circuit Cont…
45

 In this circuit now R1 and R’2 can be combined to get equivalent resistance referred to
primary R1e. Similarly X1 and X’2 can be combined to get X1e. Equivalent circuit
simplified as show in Fig. below.
We know that R1e = R1 + R’2 = R1 + R2/ K2
X1e = X1 + X’2 = X1 + X2/ K2
Z1e = R1e + jX1e
Ro = V1/Ic and Xo = V1/Im
and Ic = Io cos Φo and Im = Io sin Φo

 In the similar fashion the approximate equivalent circuit referred to secondary can be
obtained.
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 To Be Continued…
46

THANK YOU!!!
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 Voltage Regulation of a Transformer
47

 Constant voltage is the characteristics of most domestic, commercial and industrial loads.
 It is therefore, necessary that the output voltage of a transformer must remain within narrow
limits as the load and its power factor vary.
 Because of the voltage drop across the primary and secondary impedances it is observed that
the secondary terminal voltage drops from the no load value (E2) to load value (V2) as load and
load current increases.
 This decrease in the secondary terminal voltage expressed as a fraction of the no load
secondary terminal voltage is called regulation if a transformer
 Let E2 = Secondary terminal voltage on no load
V2 = Secondary terminal voltage on given load

 The ratio ﴾E2 - V2 ﴿/E2 is called per unit regulation.

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 Voltage Regulation of a Transformer Cont…
48

Fig. Regulation Characteristics

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 Voltage Regulation of a Transformer Cont…
49

Expression for voltage regulation

+ sign for lagging power factor while
- sign for leading power factor loads.
The regulation can be further expressed in
terms of I1, E1, R1e and X1e.
E2/E1 = I1/I2 = K
E2 = K E1, I2 = I1/K
while R1e = R2e/K2, X1e = X2e/K2

Substituting in the regulation expression we get,

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 Examples
50

1. A 250/125V, 5KVA single-phase transformer has a primary resistance of 0.2 ohm and
reactance 0.75 ohm. The secondary resistance of 0.05 ohm and reactance of 0.2 ohm.
A. Determine its regulation while supplying full load on 0.8 leading p.f.

B. The secondary terminal Voltage on full load and 0.8 leading p.f.

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 Losses in a Transformer
51

 In a transformer there exists two types of losses.

 The core gets subjected to an alternating flux, causing core losses.

 The windings carry currents when transformer is loaded, causing copper losses.

Core or Iron Losses

 Due to alternating flux set up in the magnetic core of the transformer, it undergoes a

cycle of magnetization and demagnetization. Due to hysteresis effect there is loss of

energy in this process which is called hysteresis loss. It is given by,

MfM, Agro-Technical and Technology College 23 April 2023

 Losses in a Transformer Cont…
52

 The induced e.m.f. in the core tries to set up eddy currents in the core and hence
responsible for the eddy current losses. The eddy current loss is given by,

 The flux in the core is almost constant as supply voltage V1 at rated frequency f as
always constant. Hence the flux density Bm in the core and hence both hysteresis and
eddy current losses are constants at all the loads. Therefore, the core or iron losses are
called constant losses. The iron losses are denoted as Pi.
 The iron losses are minimized by using high grade core material like silicon steel having
very low hysteresis loop and by manufacturing the core in the form of laminations.

MfM, Agro-Technical and Technology College 23 April 2023

 Losses in a Transformer Cont…
53

Copper (Ohmic) losses

 The copper losses are due to the power wasted in the form of I2R loss due to the resistances of

the primary and secondary windings.

 The copper loss depends on the magnitude of the currents flowing through the windings.

 The copper losses are denoted as Pcu. If the current through the windings is full load current, we
get copper losses at full load. If the load on the transformer is half then we get copper losses at
half load. Thus copper losses are called variable losses.
• For transformer VA rating is V1I1 or V2I2. As V1 is constant,
we can say that copper losses are proportional to the
square of the KVA rating and square of the current. So,
MfM, Agro-Technical and Technology College 23 April 2023
 Efficiency of a Transformer
54

 Due to the losses in a transformer, the output power of a transformer is less than the
input power supplied.
Power output = Power input - Total losses
Power input = Power output + Total losses
= Power output + Pi + Pcu
 The efficiency of any device is defined as the ratio of the power output to power input.
So for a transformer the efficiency expressed as,

MfM, Agro-Technical and Technology College 23 April 2023

 Efficiency of a Transformer Cont…
55

MfM, Agro-Technical and Technology College 23 April 2023

 Efficiency of a Transformer Cont…
56

 But if the transformer is subjected to fractional load then using the appropriate values of
various quantities, the efficiency can be obtained.

MfM, Agro-Technical and Technology College 23 April 2023

 Condition for Maximum Efficiency
57

 When a transformer works on a constant input voltage and frequency then efficiency
varies with the load. As load increases the efficiency increases. At a certain load current
it achieves a maximum value. If the transformer is loaded further the efficiency starts
decreasing. The graph of efficiency against load current I2 is shown in the fig.
 The load current at which the efficiency attains maximum value is denoted as I2m and
maximum efficiency is denoted as ηmax.

 Let us determine,
 Condition for maximum efficiency

 Load current at which ηmax.

 KVA supplied at maximum efficiency

MfM, Agro-Technical and Technology College 23 April 2023

 Condition for Maximum Efficiency Cont…
58

 The efficiency is a function of load i.e. load current I2 assuming cosΦ2 constant. The
secondary terminal voltage V2 is also assumed constant. So for maximum efficiency,
dη/dI2 = 0

MfM, Agro-Technical and Technology College 23 April 2023

 Condition for Maximum Efficiency Cont…
59

 Load Current I2m at Maximum Efficiency

MfM, Agro-Technical and Technology College 23 April 2023

 Condition for Maximum Efficiency Cont…
60

 KVA Supplied at Maximum Efficiency

MfM, Agro-Technical and Technology College 23 April 2023

 Examples
61

1. A 200 kVA rated transformer has a full-load copper loss of 1.5 kW and an iron loss of
1 kW. Determine the transformer efficiency at (a) full load and 0.85 power factor. (b)
a half full load and 0.85 power factor.
2. A 3300/110 V, 50 Hz, 60 KVA single-phase transformer has iron losses of 600 watts.
Primary and secondary resistances are 3.3 ohm and 0.011ohm respectively. Determine
the efficiency of the transformer on full load at 0.8 lag p.f load.
3. A 250 KVA single-phase transformer has iron loss of 1.8 KW. The full load copper loss
is 2000 W. Calculate
A. Efficiency at full load 0.8 lag p.f
B. KVA supplied at max efficiency
C. Max efficiency at 0.8 lag p.f

MfM, Agro-Technical and Technology College 23 April 2023

 Indirect loading Test on Transformers
62

 The efficiency and regulation of a transformer on any load condition and at any power
factor condition can be predetermined by indirect loading method. In this method, the
actual load is not used on transformer. But the equivalent circuit parameters of a
transformer are determined by conducting two tests on a transformer which are,
1. Open circuit test (O.C. Test)
2. Short circuit test (S.C. Test)
 The parameters calculated from these test results are effective in determining the
regulation and efficiency of a transformer at any load and power factor condition,
without actually loading the transformer.
 The advantage of this method is that without much power loss the tests can be
performed and results can be obtained.
 Complete analysis of the transformer can be carried out, once its equivalent circuit
parameters are known.
MfM, Agro-Technical and Technology College 23 April 2023
 Open Circuit (No-Load) Test
63

 In this diagram, a voltmeter, wattmeter and an ammeter are connected on the low voltage side
of the transformer.
 The high voltage side is left open circuited.

 The rated frequency, voltage applied to the primary, i.e. low voltage side, is varied with the
help of a variable ratio auto-transformer.
 When the voltmeter reading is equal to the rated voltage of the L.V. winding, all three
instrument readings are recorded.

Circuit diagram for open-circuit test on a transformer

MfM, Agro-Technical and Technology College 23 April 2023
 Open Circuit (No-Load) Test Cont…
64

 Then the iron loss of the transformer Pi=W0

W0=V1I0 cosφ0 …………….(1)
 The no load power factor is cosφ0=W0/V1I0

 Working component IW is IW=W0/V1 ………..(2)

 Putting the value of W0 from the equation (1) in equation(2) you will get the value of
the working component as
Iw=I0 cosφ0
Magnetizing component is Im=I0 sinφ0
 No load parameters are given below:

Equivalent exciting resistance is R0=V1/IW

Equivalent exciting reactance is X0=V1/Im

MfM, Agro-Technical and Technology College 23 April 2023

 Open Circuit (No-Load) Test Cont…
65

 The phasor diagram of the transformer at no load or when an open circuit test is
performed is shown below.

 The iron losses measured by the open circuit test is used for calculating the efficiency of
the transformer.

MfM, Agro-Technical and Technology College 23 April 2023

 Open Circuit (No-Load) Test Cont…
66

 It must be kept in mind that the values of Rc and Xm, in general, refer to the side, in
which the instruments are placed (the L.V. side in the present case).
 A voltmeter is sometimes, used at the open-circuited secondary terminals, in order to
determine the turns ratio.
 Thus the open-circuit test gives the following information:
 Core loss at rated voltage and frequency,
 The shunt branch parameters of the equivalent circuit, i.e. Rc and Xm and
 Turns ratio of the transformer.

MfM, Agro-Technical and Technology College 23 April 2023

 Short-Circuit Test
67

 The low voltage-side of the transformer is short-circuited and the instruments are placed
on the high voltage side, as illustrated in Figure (a).
 The applied voltage is adjusted by auto-transformer, to circulate rated current in the
high voltage side.
 In a transformer, the primary m.m.f. is almost equal to the secondary m.m.f., therefore, a
rated current in the H.V. winding causes rated current to flow in the L.V. winding.

Figure (a) connection diagram for short circuit (b) Equivalent circuit with short-circuit on the
test on a transformer secondary side
MfM, Agro-Technical and Technology College 23 April 2023
 Short-Circuit Test Cont…
68

Thus, the short-circuit test gives the following

information:
i. ohmic loss at rated current and frequency and
ii. the equivalent resistance and equivalent
leakage reactance.
Voltage regulation of a transformer can be determined
from the data obtained from short-circuit test.

MfM, Agro-Technical and Technology College 23 April 2023

 To Be Continued…
69

THANK YOU!!!
MfM, Agro-Technical and Technology College 23 April 2023
 Examples
70

1. A 5KVA, 500/250V, 50Hz single-phase transformer gave the following readings.

O.C Test: 500V, 1A, 50W (H.V side open)
S.C Test: 25V, 10A, 60W (L.V side shorted)
Determine
i. Efficiency on full load 0.8 lag p.f
ii. Voltage regulation on full load 0.8 leading p.f
iii. Efficiency on 60% of full load 0.8 leading p.f
iv. Draw the equivalent circuit referred to primary and insert all the values in it.

MfM, Agro-Technical and Technology College 23 April 2023

 All Day Efficiency of a Transformer
71

 For a transformer, efficiency is defined as the ratio of output power to input power. This
is its power efficiency.
 But power efficiency is not the true measure of the performance of some special types
of transformers such as distribution transformers.
 Distribution transformers serve residential and commercial loads. The load on such
transformers varies considerably during the period of the day.
 For most period of the day, these transformers are working at 30 to 40% of full load
only or even less than that. But the primary of such transformers is energized at its rated
voltage for 24 hours, to provide continuous supply to the consumer.
 The core loss which depends on voltage takes place continuously for all the loads. But
copper loss depends on the load condition.
MfM, Agro-Technical and Technology College 23 April 2023
 All Day Efficiency of a Transformer Cont…
72

 For no load, copper loss is negligibly small while on full load it is at its rated value.
Hence power efficiency cannot give the measure of true efficiency of such transformers.
 In such transformers, the energy output is calculated in kilo-watt hours (kWh). Similarly,
energy spent in supplying the various losses is also determined in kilo watt hours (kWh).
 Then the ratio of total energy output to total energy input (output + losses) is calculated.
Such ratio is called Energy efficiency or All Day Efficiency of a transformer. Based on
this efficiency, the performance of various distribution transformers is compared.
 All-day efficiency is defined as,

MfM, Agro-Technical and Technology College 23 April 2023

 Examples
73

1. A 400-KVA distribution transformer has full load iron loss of 2.5 KW and copper loss
of 3.5 KW. During a day its load cycle for 24 hours is
 6 hours - 300KW at 0.8 pf
 10 hours - 200KW at 0.7 pf
 4 hours - 100 KW at 0.9 pf
 4 hours - no load

Determine its all-day efficiency.

2. A 100-KVA lighting transformer has a full-load loss of 3 KW, the losses being equally
divided between iron and copper. During a day, the transformer operates on full-load
for 3 hours, one half-load for 4 hours, the output being negligible for the remainder of
the day. Calculate the all-day efficiency.

MfM, Agro-Technical and Technology College 23 April 2023

 Three-Phase Transformers
74

 The generation of an electrical power is usually three-phase and at higher voltages like 13.2 kV,
22 kV or somewhat higher.
 Similarly transmission of an electrical power is also at very high voltages like 110 kV, 132 kV or
400 kV.
 To step up the generated voltages for transmission purposes it is necessary to have three-phase
transformers. At the time of distribution it is necessary to reduce the voltage level upto 6600 V,
440 V, 230 V etc. For which step down three-phase transformers are essential.
 Thus three-phase transformers proves to be economical in transmission and utilization of
electrical energy.
 Previously it was a common practice to employ suitably interconnected three single-phase
transformers than to have a single three-phase transformer which is popular now a days and
used widely.
 In a single three-phase transformer there is improvement in design and manufacture.

MfM, Agro-Technical and Technology College 23 April 2023

 Advantages of 3-Phase Transformers
75

 As compared to a bank of single-phase transformers, the main advantages of a 3-phase

transformer are that it occupies less floor space for same rating, less weight, cost is less and only
one unit is required to be handled which makes it easy for the operator.
 One main drawback in a 3-phase transformer is that if any one phase becomes disabled, then
the whole transformer has to be ordinarily removed from service for repairs (the shell type may
be operated open ∆, but this is not always feasible).

• However, in the case of a 3-phase bank of

single-phase transformers, if one transformer
goes out of order, the system can still be run
open-∆ at reduced capacity or the faulty
transformer can be readily replaced by a
single spare.

MfM, Agro-Technical and Technology College 23 April 2023

 Construction of Three-Phase Transformers
76

 Like single-phase transformers, the three-phase transformers are also of the core type
or shell type. The basic principle of a 3-phase transformer is illustrated in previous Fig.

• The three cores are 120° apart and their empty

legs are shown in contact with each other.
• The center leg, formed by these three, carries the
flux produced by the 3-phase currents IR, IY and
IB.
• As at any instant IR + IY + IB = 0, hence the sum of
three fluxes is also zero.

MfM, Agro-Technical and Technology College 23 April 2023

 Construction of Three-Phase Transformers Cont…
77

 Three single-phase shell-type transformers can be combined together to form a 3-phase

shell-type unit as shown in Fig. (b).

MfM, Agro-Technical and Technology College 23 April 2023

 Three-Phase Transformer Connections
78

 The primary and secondary windings of three-phase transformers as three-phase

windings can be connected in different ways such as in star or in delta.
 With suitable connection, the voltage can be raised or lowered.
 In this section, some commonly use connections for three-phase transformers are
discussed. The most useful connections are,
I. Star-Star (Y - Y ) connection
II. Delta-Delta (∆ - ∆ ) connection
III. Star-Delta (Y - ∆ ) connection
IV. Delta-Star (∆ - Y ) connection
V. Open-delta or V - V (Reading Assignment)
VI. Scott connection or T - T connection (Reading Assignment)

MfM, Agro-Technical and Technology College 23 April 2023

 Star-Star Connection
79

 In this type of connection, both the primary and secondary windings are connected in
star as shown in the Fig.

MfM, Agro-Technical and Technology College 23 April 2023

 Star-Star Connection Cont…
80

 This particular connection proves to be economical for small high-voltage transformers as

phase voltage is 1/√3 times that of line voltage.
 The number of turns per phase and the quantity of insulation required is minimum.
 The ratio of line voltages on the primary and secondary sides is the same as the
transformation ratio of each transformer.
 It can be noted that there is a phase shift of 300 between the phase voltages and line
voltages on both primary and secondary sides.
 The line voltages on both sides and the primary voltages are in phase with each other.

MfM, Agro-Technical and Technology College 23 April 2023

 Star-Star Connection Cont…
81

MfM, Agro-Technical and Technology College 23 April 2023

 Star-Star Connection Cont…
82

Advantages
 less number of turns are required. Also the stress on insulation is less.

 Phase current is same as line current. Hence windings have to carry high currents. This
makes cross section of the windings high.
 There is no phase shift between the primary and secondary voltages.

 As neutral is available, it is suitable for three-phase, four-wire system.

Disadvantages
 If the load on the secondary side is unbalanced then the performance of this connection is
not satisfactory then the shifting of neutral point is possible.
 Third harmonic present in the secondary causes voltage distortion.

 Due to the disadvantages, this connection is rare in practice.

MfM, Agro-Technical and Technology College 23 April 2023

 Delta-Delta Connection
83

 In this type of connection, both the three-phase primary and secondary windings are
connected in delta as shown in the Fig.

MfM, Agro-Technical and Technology College 23 April 2023

 Delta-Delta Connection Cont…
84

MfM, Agro-Technical and Technology College 23 April 2023

 Delta-Delta Connection Cont…
85

Advantages
1. The delta connection provides a closed path for circulation of third harmonic component
of current. The flux remains sinusoidal which results in sinusoidal voltages.
2. It allows unbalanced loading.
3. If there is bank of single phase transformers connected in delta-delta fashion and if
one of the transformers is disabled then the supply can be continued with remaining
two transformers with reduced efficiency.
4. There is no distortion in the secondary voltages.
5. Phase current is 1/√3 times the line current. Hence the cross section of the windings is
very less. This makes the connection economical for low voltage transformers.
Disadvantages
 Due to absence of neutral point it is not suitable for three-phase four wire system.

MfM, Agro-Technical and Technology College 23 April 2023

 Star-Delta Connection
86

 In this type of connection, the primary is connected in star fashion while the secondary is
connected in delta fashion as shown in the Fig.

MfM, Agro-Technical and Technology College 23 April 2023

 Star-Delta Connection Cont…
87

MfM, Agro-Technical and Technology College 23 April 2023

 Star-Delta Connection Cont…
88

Advantages
1. The primary side is star connected. Hence fewer number of turns are required.
2. The neutral available on the primary can be earthed to avoid distortion.
3. Large unbalanced loads can be handled satisfactorily.
Disadvantage
 The secondary voltage is not in phase with the primary. Hence it is not possible to

operate this connection in parallel with star-star or delta-delta connected transformer.

MfM, Agro-Technical and Technology College 23 April 2023

 Delta-Star Connection
89

• In this type of connection, the primary is connected in delta fashion while the secondary is
connected in star fashion as shown in the Fig.

MfM, Agro-Technical and Technology College 23 April 2023

 Delta-Star Connection Cont…
90

MfM, Agro-Technical and Technology College 23 April 2023

 Delta-Star Connection Cont…
91

Advantages
1. On primary side due to delta connection winding cross-section required is less.
2. On secondary side, neutral is available, due to which it can be used for 3-phase, 4
wire supply system.
3. There is no distortion due to third harmonic components.
4. The windings connected in star makes it economical due to saving in cost of insulation.
5. Large unbalanced loads can be handled without any difficulty.
Disadvantage
 Phase shift between primary and secondary voltages.

MfM, Agro-Technical and Technology College 23 April 2023

 Voltage and Current Relationships for Different Types of Connections
92

 In the three-phase transformers, the phase voltage ratio is same as the turns ratio. But
due to the different types of connections the ratio of line voltages is different. The
relationships between voltages and currents for various types of connections is given in
the table. While deriving these relationships, the following assumptions have been made.
1. The primary line voltage is VL volts.

2. The primary line current is IL amperes.

3. Phase Transformation Ratio is K.

4. Loads are balanced.

5. Loads are purely resistive i.e. having unity power factor.
6. There are no losses and the transformers are ideal.

MfM, Agro-Technical and Technology College 23 April 2023

 Voltage and Current Relationships for Different Types of Connections
93

MfM, Agro-Technical and Technology College 23 April 2023

 Auto-transformer
94

 It is a transformer with one winding only, part of this being common to both primary and
secondary. Obviously, in this transformer the primary and secondary are not electrically
isolated from each other.
 But its theory and operation are similar to those of a two-winding transformer. Because of

one winding, it uses less copper and hence is cheaper.

• Fig. shows both step-down (fig a) and
step-up auto-transformers (fig b).
• As shown in Fig. (a), AB is primary winding
with N1 turns and BC is secondary winding
with N2 turns.
• Neglecting iron losses and no-load current.

MfM, Agro-Technical and Technology College 23 April 2023

 Auto-transformer Cont…
95

Auto-transformers are used when K is nearly equal to unity and where there is no objection
to electrical connection between primary and secondary. Hence, such transformers are
used:
 To give small boost to a distribution cable to correct the voltage drop.
 As auto-starter transformers to give up to 50 to 60 % of full voltage to an induction
motor during starting.
 As interconnecting power systems of different voltage levels.
 It can be used as a variac in the laboratory where variable output is required.

MfM, Agro-Technical and Technology College 23 April 2023

 Conversion of 2-Winding Transformer into Auto-transformer
96

 Any two-winding transformer can be converted into an auto-transformer either step-

down or step-up.
 Fig. (a) shows such a transformer with its polarity markings. Suppose it is a 20-kVA,
2400/240 V transformer. If we employ additive polarity between the high-voltage and
low-voltage sides, we get a step-up auto-transformer. If, however, we use the subtractive
polarity, we get a step-down auto-transformer.

MfM, Agro-Technical and Technology College 23 April 2023

 Parallel Operation of Single-phase Transformers
97

 For supplying a load in excess of the rating of an existing transformer, a second

transformer may be connected in parallel with it as shown in Fig.
 It is seen that primary windings are connected to the supply bus bars and secondary
windings are connected to the load bus-bars. In connecting two or more than two
transformers in parallel, it is essential that their terminals of similar polarities are joined
to the same bus-bars as in Fig.

MfM, Agro-Technical and Technology College 23 April 2023

 Parallel Operation of Single-phase Transformers Cont…
98

 If this is not done, the two e.m.fs. Induced in the secondaries which are paralleled with
incorrect polarities, will act together in the local secondary circuit even when supplying
no load and will hence produce the equivalent of a dead short-circuit.
 There are certain definite conditions which must be satisfied in order to avoid any local
circulating currents and to ensure that the transformers share the common load in
proportion to their KVA ratings. The conditions are:
1. Primary windings of the transformers should be suitable for the supply system voltage
and frequency.
2. The transformers should be properly connected with regard to polarity.
3. The voltage ratings of both primaries and secondaries should be identical. In other
words, the transformers should have the same turn ratio i.e. transformation ratio.

MfM, Agro-Technical and Technology College 23 April 2023

 Parallel Operation of Single-phase Transformers Cont…
99

4. The percentage impedances should be equal in magnitude and have the same X/R
ratio in order to avoid circulating currents and operation at different power factors.
5. With transformers having different KVA ratings, the equivalent impedances should be
inversely proportional to the individual KVA rating if circulating currents are to be
avoided.

MfM, Agro-Technical and Technology College 23 April 2023

 Instrument Transformers
100

 In d.c. circuit when large currents are to be measured, it is usual to use low-range
ammeters with suitable shunts. For measuring high voltages, low-range voltmeters are
used with a high resistance connected in series with them.
 But it is not convenient to use this method with alternating current and voltage
instruments.
 For this purpose, specially constructed accurate ratio instrument transformers are
employed in conjunction with standard low-range a.c. instruments.
 These instrument transformers are of two kinds :
1. Current transformers for measuring large alternating currents and

2. Potential transformers for measuring high alternating voltages.

MfM, Agro-Technical and Technology College 23 April 2023

 Current Transformers
101

 These transformers are used with low-range ammeters to measure currents in high-
voltage alternating-current circuits where it is not practicable to connect instruments and
meters directly to the lines.
 In addition to insulating the instrument from the high voltage line, they step down the
current in a known ratio.
• The current (or series) transformer has a primary coil of one
or more turns of thick wire connected in series with the line
whose current is to be measured as shown in Fig.
• The secondary consists of a large number of turns of fine wire
and is connected across the ammeter terminals.
• As regards voltage, the transformer is of step-up variety but it
is obvious that the current will be stepped down.

MfM, Agro-Technical and Technology College 23 April 2023

 Potential Transformers
102

 These transformers are extremely accurate-ratio step-down transformers and are used
in conjunction with standard low-range voltmeters (usually 150-V) whose deflection when
divided by voltage transformation ratio, gives the true voltage on the high voltage side.

MfM, Agro-Technical and Technology College 23 April 2023

 READ MORE!
103

THANK YOU!!!
MfM, Agro-Technical and Technology College 23 April 2023