Transformer

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1
P
INTRODUCTION
This chapter has been divided into following topics
1) Working Principle of Transformer
2) E.M.F. Equation of transformer
3) Transformer on NO load and ON load (with phasor diagram)
4) Equivalent circuit of transformer
5) Transformer Tests (O.C. and S.C. Test)
6) Losses in transformer and its efficiency
7) All day efficiency
WORKING PRINCIPLE OF TRANSFORMER
1) The transformer is a static piece of device by means of which an electric
power is transformed from one circuit to another circuit without change in
the frequency.
2) A transformer operates on the principle of mutual inductance between two
inductively coupled coils. The winding in which electrical energy is fed is
called Primary winding and the other from which energy is drawn out is
called Secondary winding.
Single Phase Transformers

2
3) The primary winding has 1N number of turns, while the secondary winding
has N2 number of turns. The basic transformer and its symbol is shown in
fig.(1) and (2) respectively.
Fig.(1)
Fig.(2)
4) When an alternating voltage V1 is applied to primary winding, an
alternating current I1 flows in it producing alternating flux in the core.
According to Faraday’s law of electromagnetic induction an e.m.f. is
induced in the primary winding which is given by,
1 1
d
e N
dt


where N1 is the number of turns in primary winding. The induced e.m.f. in
the primary winding is nearly equal and opposite to the applied voltage V1.
5) Assuming leakage flux to be negligible, almost whole flux produced in
primary winding links with the secondary winding. Hence e.m.f. e2 is
induced in the secondary winding.
2 2
d
e N
dt


where N1 is the number of turns in primary winding. If secondary circuit is
closed through the load, a current I2 flows in the secondary winding.
6) Thus energy is transformed from primary winding to secondary winding
through magnetic field.

3
TYPES OF TRANSFORMERS
According to their construction
On the basis of construction, the transformers are classified in three types :
1) Shell Type Transformer, 2) Core Type Transformer
3) Berry Type Transformer
1. Shell Type Transformer :
1) For the construction of Shell type transformer a three limbed iron is
used. Both primary and secondary windings are placed on central limb.
2) First primary winding is wound on the central limb with required
number of turns. Then insulating paper is wrapped on this primary
winding, then secondary winding is wound over this paper.
3) It has double magnetic circuit. The core encircles most of the part of
winding. Generally it is preferred for very high voltage transformers.
4) As the windings are surrounded by core, the natural cooling does not
exist. For removing any winding for maintenance, large number of
laminations is required to be removed.
2. Core Type Transformer:-

4
1) For core type of transformer two limbed iron core is used. Primary and
secondary winding turns are split in two parts and each part is
arranged on both the limbs of core.
2) Initially first part of the primary windings is wound on one limb and
remaining part of it is wound on other limb, then insulating paper is
wrapped on each part of primary winding. Now secondary winding is
wound on this paper in two parts, each part on one limb.
3) It has a single magnetic circuit. The winding encircles the core i.e. core
is surrounded by the winding. Core is made up of large number of thin
laminations.
4) As the windings are uniformly distributed over the two limbs the
natural cooling is more effective. The coils can be easily removed by
removing the laminations of the top yoke for maintenance.
3. Berry Type Transformer:-
1) This has distributed magnetic circuit. Its core construction is like
spokes of a wheel.
2) These transformers are generally kept in tightly fitted sheet metal
tanks. These tanks are filled with the special insulating oil.
3) The entire transformer assembly is immersed in the oil. The oil serves
two functions:
(a) keeps the windings cool by circulation and
(b) provides additional insulation to transformer.

5
According to their working
On the basis of working, the transformers are classified in three types :
1) Step up Transformer
2) Step down Transformer
3) Isolation Transformer
1. Step up Transformer :
1) When the number of turns of secondary winding (N2) are greater than
number of turns of primary windings (N1), then voltage available across
secondary is greater than the applied across primary.
2) This kind of transformer is known as Step up Transformer  1K  where
1
2
N
K
N
 .
3) These type of transformers are generally used in generating power
stations or in communication circuits as EHT Transformers.
2. Step down Transformer :
1) When the number of turns of secondary winding (N2) are less than
number of
turns of primary windings (N1), then voltage available across secondary
is less than the applied across primary.
2) This kind of transformer is known as Step down Transformer (K >1)
where 1
2
.
N
K
N

3) These type of transformers are generally used in distribution power
stations and in electronics and communication circuits.
3. Isolation Transformer :
1) When the number of turns of secondary winding (N2) are equal to the
number of turns of primary windings (N1), then voltage available across
secondary is equal to voltage applied across primary.
2) This kind of transformer is known as Isolation Transformer (K = 1)
where 1
2
.
N
K
N

3) These type of transformers are mainly used to isolate the electrical or
electronic circuit under testing from the supply.

6
EMF EQUATION OF TRANSFORMER
1) Let N1, N2 represents number of turns of primary and secondary windings
respectively. e1 and e2 represents voltage induced across primary and
secondary in volts.
2) As primary winding is excited by a sinusoidal alternating voltage, an
alternating current flows in the winding producing a sinusoidally varying
flux  in the core.
i.e. sinm t  
3) As per Faraday’s law of electromagnetic induction an e.m.f. e1 is induced
in the primary winding.
1 1
d
e N
dt

 
1 sinm
d
N t
dt
  
1 cosmN t   
0
1 sin 90mN t   
0
12 f sin 90mN t   
4) Therefore, maximum value of induced 1e.m.f. 2 f mN 
Hence RMS value of induced e.m.f. in primary winding is given by,
max 1
1 1
2 f
4.44f
2 2
m
m
E N
E N
 
  
 1 14.44f mE N ….(1)
5) Similarly RMS value of induced e.m.f. in secondary winding is given by,
2 24.44f mE N ….(2)
6) Also, 1 2
1 2
4.44f m
E E
N N
 
Thus, e.m.f. per turns is same in primary and secondary winding and
hence equal e.m.f. is induced in each turn of primary and secondary
winding.

7
 Transformation Ratio (K):
1) From eqn. (1) and (2) we have,
1 14.44f mE N …(3)
2 24.44f mE N …(4)
Dividing (3) by (4) we get,
1 1
2 2
E N
E N
 …(5)
The ratio 1
2
N
N
, is known as Transformation ratio and is denoted by ‘K’.
2) If losses in a transformer are neglected the volt-ampere rating in the
primary and secondary are equal.
Thus, 1 1 2 2E I E I
 1 1
2 2
E I
E I
 …(6)
3) Hence, current is transformed in the reverse ratio of the voltage. If a
transformer steps up the voltage, it steps down the current and vice
versa.
For ideal transformer,
1 1 2 2andE V E V 
 1 1
2 2
E V
E V
 ….(7)
4) From eqns (5), (6) and (7)
 Rating of Transformer:
1) Rating of a transformer indicates the output power from it. But for a
transformer, load is not fixed and its power factor goes on changing.
Hence rating is expressed in terms of product of voltage and current
called as VA rating. It is generally expressed in kVA.
2) kVA rating of Transformer = 1 1 2 2
1000 1000
V I V I


8
3) We can calculate full load currents of primary and secondary windings
from kVA rating of transformer.
Full load primary current 1
1
k rating 1000VA
I
V


Full load primary current 2
2
k rating 1000VA
I
V


IDEAL TRANSFORMER
1. It is a transformer in which there are no losses i.e.
1) Copper losses in the windings are zero. (i.e. Resistance offered by the
winding is zero.)
2) Iron losses in iron core of the transformer are zero. (i.e. core material
has strong magnetic properties and provides zero reluctance and
infinite permeability.)
3) No magnetic leakage.
2. When the primary of the ideal transformer is connected to an alternating
voltage V1 with secondary open circuited, the primary draws very small
amount of current (Iμ) from supply. This current is called as ‘magnetizing
current’. This current is purely reactive current and lags the voltage V1 by
90˚.
3. This current (Iμ) set up the flux ( m ) in iron core. The flux ( m ) induces
e.m.f.’s E1 and E2 in the primary and secondary windings.
4. These e.m.f. leads flux ( m ) by 0
90 . The induced voltage
1E always opposes applied voltage 1V . The magnitude
of 2E w.r.t. 1E depends on transformation ratio (K).
5. The vector diagram for ideal transformer with  2 1N N
i.e. K<1 is shown below :

9
TRANSFORMER ON NO LOAD
1. Practically when transformer is connected to an alternating voltage V1, the
primary will draw current (Io) from the supply. This current (Io) is called as
‘NO Load current’.
2. This current has two components :
a) Magnetizing current (Iμ) :
This component sets up the magnetic flux m in the core.
b) Working Current (Iω) :
This current provides losses in iron core.
3. The NO load primary current Io is not 90˚ behind V1 but lags it by an angle
0 . The phasor diagram for No load condition is shown below :

10
4. From phasor diagram,
0 0cosI I 
And 0 0sinI I 
0I is vector sum of I and I . Hence we have,
2 2
0I I I  
No load power factor 0
0
cos
I
I

 
5. The NO load i/p power is given by,
0 1 0 0cosW V I 
6. The NO load current 0I is very small as compared to full load current 1I .
Hence copper loss is negligible and NO load input power is practically equal
to iron loss or core loss in the transformer.
 1 1 0 0Iron loss W cosV I 
TRANSFORMER ON LOAD
Fig. (1)
Fig. (2)

11
Fig. (3)
Fig. (4)
1. In fig. (1), on NO load condition the transformer primary draws a current Io
from supply which has two components I and I .
2. In fig.(2), when a load with lagging power factor is connected across the
secondary of transformer, secondary current (I2) starts flowing through the
load. This secondary current sets up flux 2 in iron core which opposes
main flux m .
3. Because of this net flux in the iron core decreases. Hence flux linking with
primary coil reduces and induced e.m.f. E1 decreases.
4. In fig.(3), Due to reduction in induced voltage across primary (E1), potential
difference is created. Thus primary will draw an extra amount of current
(I1) to overcome load connected across secondary. This current is always
out of phase with (I2).
5. This current ( 1 'I ) generates flux ( 1 ' ) which opposes flux ( 2 ). This action
cancels out flux ( 2 ) and ( 1 ' ).
6. In fig. (4), as flux ( 1 ' ) and ( 2 ) cancel out each other, the net flux in iron
core becomes m . Hence, magnetic flux in iron core of transformer remains
same whether transformer is ON load or NO load condition.

12
7. As flux in iron core of transformer remains constant,
iron losses in transformer remains constant. Hence
iron losses are called constant losses.
8. Now net current in primary is I1 which is vector
addition of 0I and 1 'I i.e.
1 0 1 'I I I 
9. The complete phasor diagram of transformer ON load
is shown below.
PRACTICAL TRANSFORMER [TRANSFORMER WITH LEAKAGE IMPEDANCE]
1. Let R1, R2  Resistance of primary and secondary.
1 2,X X  Reactance of primary and secondary.
1 2,Z Z  Impedance of primary and secondary.
Thus, 1 1 1jXZ R  and 2 2 2jXZ R 
2. Applying KVL to primary side,
1 1 1 1V I Z E   (Because E1 is opposite to V1.)
 1 1 1 1V I Z E 
 1 1 1 1 'V I Z V  (Where 1 1'V E  )
 1 1 1 1 1 1 'V I R I X V  
3. Applying KVL to Secondary side,
2 2 2 2E I Z V 
  2 2 2 2 2 2E I R I Z V  

13
Steps to draw phasor diagram (for lagging power factor)
1. Consider flux φ as reference vector.
2. Draw vector E1 lagging φ by 90˚.
3. Draw vector E2 in phase with E1. (Length of E2 depends on turns ratio)
4. Extend E1 in opposite direction and mark it as –E1.
5. Draw vector Iμ along φ and Iω along (-E1). Then draw their resultant Io
making an angle φ0 with (-E1).
6. Draw vector I2 lagging E2.
7. Draw I2R2 in phase with I2 and I2X2 perpendicular to I2R2. Draw their
resultant as I2Z2.
8. Extend I2Z2 in negative direction and draw resultant of -I2Z2 and E2. Label
this resultant as V2 and angle between V2 and I2 as φ2.
9. Extend I2 in opposite direction and get the vector I2’ such that I2’ = K I2.
10. Find the resultant of I2’ and I0. Label this resultant as I1. i.e. net primary
current.
11. Draw vectors I1R1 in phase with I1 and I1X1 perpendicular to I1R1. Draw their
resultant as I1Z1.
12. Get the resultant of I1Z1 and (-E1). Label this resultant as V1 and angle
between V1 and I1 as φ1.
PHASOR DIAGRAM FOR LAGGING POWER FACTOR

14
PHASOR DIAGRAM FOR LEADING POWER FACTOR
PHASOR DIAGRAM FOR UNITY POWER FACTOR

15
EQUIVALENT CIRCUIT OF TRANSFORMER
NO Load Condition
1) When transformer is on NO load condition it draws a current Io which has
got two components.
(a) Iμ = Io sin φo which produces flux and
(b) Iω = Io cos φo which provides iron loss.
2) Consider Two imaginary components Ro and Xo. The current Iω flows
through Ro to provide iron loss and current Iμ flows through Xo to produce
flux φ.
3) The equivalent circuit of transformer on NO load condition is shown below:
4) From circuit diagram,
1
0
V
R
I
 and 1
0
V
X
I

ON Load Condition
1. When load ZL is connected across secondary of transformer, current I2
flows through secondary winding and causes voltage drop across R2 and
X2 of the secondary winding. Total current drawn by primary of
transformer under load is I1.
2. The equivalent circuit of transformer ON load is shown below :

16
3. But, this circuit is complicated for analysis. This can be simplified by
transforming secondary parameters to primary or vice-versa.
a) Equivalent Circuit of transformer referred by primary:-
1) When secondary parameters are transferred to primary side then
equivalent circuit so obtained is known as Equivalent circuit of
transformer referred to Primary.
2) This equivalent circuit is shown below:
3) If 1
2
N
K
N
 , then secondary parameters referred to primary are,
2
2 2'R K R 2
2 2'X K X 2
2 2'Z K Z 2
'L LZ K Z
2 2'E K E 2 2'V KV
2
2 '
I
I
K
 2
2
2
'
cos '
'
R
Z
 
b) Equivalent Circuit of transformer referred to secondary :
1) When primary parameters are transferred to secondary side then
equivalent circuit so obtained is known as Equivalent circuit of
transformer referred to Secondary.
2) This equivalent circuit is shown below :
3) If 1
2
N
K
N
 then primary parameters referred to secondary are,
1
1 2
'
R
R
K
 0
0 2
'
R
R
K
 1
1 2
'
X
X
K
 0
0 2
'
X
X
K

1
1 2
'
Z
Z
K

1
1 '
V
V
K
 1
1 '
E
E
K

1 1'I K I 0 0'I K I 'I K I  'I K I 

17
APPROXIMATE EQUIVALENT CIRCUIT OF TRANSFORMER
1) The approximate equivalent circuit of transformer can be obtained by
shifting the parallel combination of imaginary components Ro and Xo to the
left of the circuit.
2) By shifting this parallel branch we are neglecting the drops across R1 and
X1 due to Io as |Io| is very small.
Approximate Circuit referred to Primary
This circuit can be further simplified as shown below :
where,
R1e  Equivalent resistance of transformer referred to primary
X1e  Equivalent reactance of transformer referred to primary
Z1e  Equivalent impedance of transformer referred to primary
2
1 1 2 1 2'eR R R R K R    , 2
1 1 2 1 2'eX X X X K X   
1 1 2 1 1' je e eZ Z Z R X   

18
APPROXIMATE CIRCUIT REFERRED TO SECONDARY
This circuit can be further simplified as shown below:
where,
R2e  Equivalent resistance of transformer referred to secondary
X2e  Equivalent reactance of transformer referred to secondary
Z2e  Equivalent impedance of transformer referred to secondary
1
2 1 2 22
'e
R
R R R R
K
    , 1
2 1 2 22
'e
X
X X X X
K
   
2 1 2 2 2' je e eZ Z Z R X   
Voltage Regulation of Transformer
1) When a transformer is loaded, the secondary terminal voltage decreases
due to drop across secondary winding resistance and leakage reactance.
2) This change in secondary terminal voltage from NO load to full load
condition expressed as fraction of the NO load secondary voltage is called
as Regulation of Transformer.
Secondary terminal Secondary terminal voltage
voltage on NO load on FULL load condition
Regulation
Secondary terminal voltage on NO load
   
   
   
2 2
2
E V
E


2 2
2
%Regulation 100
E V
E

 

19
3) The secondary terminal voltage does not depend only on load current but
also on the power factor of the load. Consider secondary circuit as shown
below:
4) Applying KVL to secondary circuit we get,
2 2 2 2E I Z V 
Thus, phasor diagrams for different power factors are shown below:
a) Lagging p.f.
For lagging p.f. 2 2E V
 Regulation is positive for lagging p.f.
b) Leading p.f.
For leading p.f. 2 2E V
 Regulation is negative for leading p.f.
c) Unity p.f.
For unity p.f. 2 2E V
 Regulation is positive for unity p.f.
5) In general the expression for Voltage
regulation is,
2 2 2 2 2 2
2
cos sin
% Regulation 100e eI R I X
V
 
 
ve lagging and unity p.f. ve leading p.f.   

20
LOSSES IN TRANSFORMER AND EFFICIENCY
Losses in Transformer
a) Copper Losses:-
1) This loss is due to the resistance of primary and secondary windings.
2) It depends on load current and is proportional to square of the load
current.
2
1 1Primary copper loss WattsI R
2
2 2Secondary copper loss WattsI R
 2 2
1 1 2 2Totol copper loss I R I R 
b) Iron Losses (or Core Losses):-
1) This loss includes hysteresis and eddy current loss. Hysteresis loss
occurs due to setting of alternating flux in the core. Eddy current loss
occurs due to setting of eddy currents in the core which is due to
induced e.m.f. in the core.
2) These losses are also known as constant losses. Hysteresis and eddy
current loss is given by,
1.67
f VH H mP K B and 2 2 2
fE E mP K B t
 Iron loss Constant lossi H EP P P   
Efficiency of Transformer
1) We know that efficiency is the ratio of output power to input power.
Power Output
i.e.
Power Input
 
2) But for any machine, Power Input = Power Output + Losses
Power Output
Power Output + Losses
 
Power Output
% 100
Power Output + Losses
  

21
Condition for Maximum Efficiency of Transformer
1) Considering secondary side of transformer,
2 2 2
2
2 2 2 2 2
cos
cos i e
V I
V I W I R




 
Differentiating both the sides w.r.t. 2I ,
2
2 2 2 2 2 2 2 2 2 2 2 2 2 2
2
2 2 2 2 2 2
cos cos cos cos 2
cos
i e e
i e
V I W I R V V I V I Rd
dl V I W I R
   

   

 
But for maximum efficiency,
2
0
d
dl


 2
2 2 2 2 2 2 2 2 2 2 2 2 2 2cos cos cos cos 2i e eV I W I R V V I V I R      
 2 2
2 2 2 2 2 2 2 2 2 2cos cos 2i e eV I W I R V I I R    
 2
2 2i eW I R
Similarly on primary side, 2
1 1i eW I R
Thus, Efficiency of Transformer is maximum when,
Copper Loss = Iron Loss
All Day Efficiency of Transformer
1) The transformers used for distribution are energized for 24 hours of the
day. Thus, constant losses occur in the transformer for the whole day.
2) These transformers normally operate on different loads during 24 hours a
day. Because of this copper losses are different during different periods of
day.
3) Hence, efficiency of such transformers should be measured on the energy
basis.

Output in kWh
All Day efficiency
Input in kWh

Output in kWh
Output in kWh + Iron loss in kWh + Copper loss in kWh


22
TRANSFORMER TESTS
1. The performance of transformer can be studied from its equivalent circuit.
The parameters of transformer can be calculated by conducting two tests
(a) Open Circuit Test and
(b) Short Circuit Test.
2. These tests are economical and convenient, because we can find
transformer parameters without actually loading it.
(a) Open Circuit (O.C.) Test:-
1) The purpose of this test is to determine
(1) Iron loss Wi (2) Magnetizing resistance Ro and
(3) Magnetizing reactance Xo
2) The above fig. shows the circuit diagram for conducting O.C. test on
the transformer. In this test one winding is left open and other
winding is connected to a supply.
3) Ammeter, Voltmeter and wattmeter are connected on the side where
supply is connected.
4) Ammeter indicates NO load current drawn by the transformer. As
the NO load current is usually 3 to 5% of full load current, copper
losses will be negligible and wattmeter will indicate iron loss.
5) If meters are connected on primary side,
Wattmeter reading = Wi
Voltmeter reading = V1
Ammeter reading = Io
 1 0 0cosiW V I 
 0
1
cos i
o
W
V I
 
Also, coso oI I 
sino oI I 
And 1
o
V
R
I
 , 1
o
V
X
I


23
(b) Short Circuit (S.C.) Test:-
1) The purpose of this test is to determine
(1) full load copper loss,
(2) equivalent resistance R1e or R2e and
(3) equivalent reactance X1e or X2e
2) The fig. shows the circuit diagram for conducting S.C. test on
transformer. In this test one winding is short circuited, while a low
voltage is applied to the other winding.
3) The voltage is slowly increased until full load current flows in this
winding.
4) Normally, the applied voltage is 5 to 10% of rated voltage of this
winding. Hence flux produced in the core will be small and so the
iron losses are very small. Thus wattmeter indicates full load copper
loss.
5) If meter is connected on primary side,
Wattmeter reading = Wsc
Voltmeter reading = Vsc
Ammeter reading = Isc
 sc
sc
sc
V
Z
I
 and 2
sc
sc
sc
W
R
I

2 2
sc sc scX Z R 

Transformer

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