1. TRANSFORMER

2. Φ C
vp
ip
P
Np
Φ= Npip/S …….(1)
where S is the reluctance
A transformer consists of a core made of laminated iron
separated by insulators and a coil of Np turns wound
around the core. This coil is supplied with an a.c voltage
supply vp which then produces a current ip. Due to this
current , a flux Φ is produced which is given by an
equation

3. dt
d
Npp
ϕ
=v
)(
dt
dN
p
2
p
p iv ×
ℜ
=
Since current varied with time , Φ also varied with time.
A back electromagnetic force (e.m.f) will be produced
which is given by the equation.
Substitute Φ = Npip/S into the above equation , then
……(2)
……(3)

4. ( )
dt
2sind
Npp
ft
v m πΦ
=
Φ = Φm sin 2πft
If ip is sinusoidal, the flux produced also sinusoidal, i.e
therefore
vp = NP2πfΦmcos 2πft = NP2πfΦmsin (2πft + π/2)
The peak value = Vpm = NP2πfΦm
and vp is leading the flux by π/2.
mm
pm
p f44.4πf2707.0
2
V
V Φ=Φ×== PP NNThe rms value
……(4)
……(5)
……(6)
……(7)

5. dt
d
Nss
ϕ
=v ……(8)
ϕ C
vp
ip
vs
is
Load
P S
N
s
N
p
primary
secondary
When another coil is wound on the other side of
the core with no of turns Ns , then the fux will
induce the e.m.f vS as given by

6. From (2) and (8) we get
p
s
p
s
N
N
V
V
=
…….(9)
p
s
p
s
N
N
=
v
v
Or in rms value
With load , is will flow in the load, mmf at load will
equal to the mmf at input, then (in rms value)
NpIp = NsIs
rearrange
p
s
s
p
N
N
I
I
=
…….(10)
…….(12)
…….(11)

7. For ideal transformer, the energy transferred will be the
same as input. Thus power at primary is same power at
secondary.
Pp = Ps
or IpVp = IsVs
NP : NS
VP VS
Primary Secondary
Symbol for ideal transformer

8. A 250 kVA,11000V/400V, 50Hz single –phase transformer
has 80 turns on the secondary. Calculate
(a) The appropriate values of the primary and secondary
currents;
(b) The approximate number of primary turns;
(c) the maximum value of the flux.
(a) Full-load primary current
AIs 625
400
10250
V
P -3
s
=
×
==
A
V
P
I
p
p 7.22
11000
10250 3
=
×
==
Full-load secondary current

9. 220011000
400
80
V
V
N
N P
s
s
P =×=×=
p
s
p
s
N
N
V
V
=
mf Φ= N44.4E
mWb
f
m 5.22
508044.4
400
N4.44
E
s
s
=
××
==Φ
(b) Number of primary turns
recall
(c) Maximum flux
recall

10. EP VP VSNS
NP
IO
Ideal transformer with no load
Io is the no load current when the secondary is open
circuit. This current consists of Iom that is required to
produce the flux in the core (it is in phase) and Io1 is to
compensate the hysteresis and eddy current losses.

11. VP= emf of supply to the primary coil
and 90o
leads the flux.
EP=emf induced in the primary coil and
same phase as VP.
VS=emf induced in the secondary coil
and 90o
lags the flux.
Iom=magnetizing current to produce flux
and it is in phase with flux.
Io1=current to compensate the losses due
to hysteresis and eddy current.
Io=the no load current and given by
2
1
2
I oomo II +=
o
o
o
I
I 1
cos =φPower factor
Phasor diagram for
no load transfomer
Φ
IOm
IOI
EP
VP
IO
VS
φο

12. Transformer 1 Transformer 2
Low
voltage
generator
Low voltage
load
High voltage line
Transformer converts the energy to high electrical
voltage and transmits in the high voltage line. At the
load, the high voltage energy is converted to low
voltage. In this way, it will compensate the losses during
transferring of the voltage energy.

13. A single-phase transformer has 480 turns on the primary and
90 turns on the secondary. The mean length of the flux path
in core is 1.8m and the joints are equivalent to the airgap of
0.1mm. The value of the magnetic field strength for 1.1 T in
the core is 400A/m, the corresponding core loss is 1.7W/kg
at 50Hz and the density of the core is 7800kg/m3
.
If the maximum value of the flux is to be 1.1T when a p.d of
2200V at 50Hz is applied to the primary, calculate:
(a) the cross-sectional area of the core;
(b) the secondary voltage on no load;
(c) the primary current and power factor on no load

14. AB×=Φ
recall mf Φ= N44.4E
Wb
f
m 0206.0
5048044.4
2200
4.44N
E
p
p
=
××
==Φ
2
0187.0
1.1
026.0
A m
B
m
==
Φ
=
recall
(a)
(b)
p
s
p
s
N
N
V
V
=recal
l
V4132200
480
90
V
V
N
N P
s
s
P =×=×=
Practically 10% more
allow for insulator

15. AC 7208.1400HC =×=
A
B
a
o
aa 5.870001.0
104
1.1
H 7
=×
×
=×= −
πµ

A5.8075.87720 =+=
A
N
H
Iom 682.1
480
5.807
===

INH =
AIomom 19.1682.1707.0707.0I =×==
∴
(c) magnetomotive force (mmf) for the core is
mmf for the airgap is
Total mmf is
Maximum magnetizing current
recal
l
Rms value

16. 3
0337.00187.08.1 mA =×=×= 
kgdensityVol 26378000337.0. =×=×=
W4477.1263 =×=
A
V
P
I
p
o 203.0
2200
447
1 ===
lagging
I
I
o
168.0
21.1
203.0
cos 01
===φ
AIII oomo 21.1203.019.1 222
1
2
=+=+=
Volume of core
mass of core
Core loss= loss rate x mass
Core-loss component of current
No load current
Power factor

17. E1 V1 V2
I1 I2
L(θ2)
Loaded transformer
L(θ2)= load with power factor of cos θ2
V1 = emf at supply
E1=induced voltage at primary
V2=emf at load
E2=induced voltage at secondary
I1= primary current
I2=secondary current
Phasor diagram
I o
Φ
V 1
, E 1
- I 2 I 1
θ2
θ1
θO
V 2
, E 2
I 2
‘

18. A single-phase transformer has 1000 turns on the primary and
200 turns on the secondary. The no load current is 3A at a
power factor 0.2 lagging when secondary current is 280A at a
power factor of 0.8 lagging. Calculate the primary current and
the power factor. Assume the voltage drop in the windings to
be negligible.
A56280
1000
200
I
N
N
I S
P
S
P =×=×=
Recall Equation 12
p
s
s
p
N
N
I
I
=
therefore

19. ooIII φφφ sinsinsin '2'211 +=
ooIII φφφ coscoscos '2'211 +=
6.0sin '2 =∴ φ
98.0sin =∴ oφ2.0c =oosφ
8.0c '2 =φos
( ) ( ) A4.452.038.056 =×+×=
'50381
o
=φ
( ) ( ) AI 3.5854.364.45
22
1 =+=
laggingos o
78.050'38cosc 1 ==φ
805.0
4.45
54.36
tan 1 ==φ
( ) ( ) A54.3698.036.056 =×+×=
I o
Φ
V 1
, E 1
- I 2 I 1
θ2
θ1
θO
V 2
, E 2
I 2
‘
Solve for horizontal and vertical components
Power factor

20. Path of leakage
V2’
R1 L1 L2 R2
RC
Lm
V2E1
I1’I1
Equivalent circuit of transformer
Flux leakage is due to secondary current that produce flux
which encounter the primary flux. Some of the flux will link to
its own windings and produce induction. This is represented by
inductance L1. Similarly with the flux in secondary and
represented by L2.

21. There are four main losses
•Dissipated power by wire resistance of the windings (I2
R)
•Power due to hysteresis
•Power due to eddy current
•Power via flux leakages.
R1= wire resistance of primary windings
L1=inductance due to leakage flux in primary windings
RC=resistance represent power loss due to in hysteresis and eddy
current
Lm= inductance due to magnetizing current Iom
L2=inductance due to leakage flux in secondary windings
R2=wire resistance of secondary windings
E2
R1 L1 L2 R2
RC
Lm
V2V1
I1’I1
Equivalent circuit of transformer
E1

22. Phasor diagram for a transformer on load
V1
E1
I2
-I2’
I1X1
V2
E2
I1
I0
I1R1
I1Z1
I2X2
I2Z2
I2R2
φ1
φ2

23. 2
2
1
2
2
2
1
2'2 





≈





=
V
V
X
N
N
XX
2
2
1
2
2
1
2
2'2 





≈





=
V
V
R
I
I
RR
2
2
2'2
2
1 RIRI =
We can replace R2 by inserting R2’ in the primary thus the
equivalent resistance is
Giving us
Similarly

24. 2
2
1
21'21 





+=+=
V
V
RRRRRe
22
eee XRZ +=
e
e
R
X
tan =eφ
eee ZX φsin=
eee ZR φcos=
then
and
where
2
2
1
21'21 





+=+=
V
V
XXXXXe
(b)
Transformer simplified circuit
Ze
V1 E1=V2’
I1 I2
To
load
E2=V2

25. Phasor diagram of simplified equivalent circuit of transformer
φ 2
φ 2 I 1
R e
I 1
Z e
I 1
X e φ e
Magnified Ze portion
Φ
I 1
X e
I 1
E 1
=V 2'
V 1
I 1
Z e
I 1 R e
E 2 , V 2
I 2
φ 2
φ 1
φ e
φ 2
Complete
φo-φ2

26. voltageload-no
voltageload-fullvoltageload-no
regulationVoltage
−
=






=
1
2
12
N
N
VV
p
s
p
s
N
N
V
V
=recall
Secondary voltage on no-load
V2 is a secondary terminal voltage on full load






−





=
1
2
1
2
1
2
1
regulationVoltage
N
N
V
V
N
N
V
Substitute we have

27. ( )
1
221 sincos
V
XRI
regulationvoltageunitPer ee φφ +
=−
centper
V
N
N
VV
100
1
2
1
21
×






−
=
unitper
V
N
N
VV
1
2
1
21
regulationVoltage






−
=Or
From phasor diagram can be proved that
( )
1
21 cos
V
ZI
regulationvoltageunitPer ee φφ −
=−
Or

28. A 100kVA transformer has 400 turns on the primary and 800
turns on the secondary. The primary and secondary
resistances are 0.3Ω and 0.01Ω respectively, and the
corresponding leakage reactances are 1.1Ω and 0.035Ω
respectively. The supply voltage is 2200V. Calculate:
(a)The equivalent impedance referred to the primary circuit;
(b)The voltage regulation and the secondary terminal voltage for
full load having a power factor of (i) 0.8 lagging and (ii) 0.8
leading.
(c)The percentage resistance and leakage reactance drops of the
transformer

29. A
V
P
currentprimaryloadFull 45.45
2200
10100 3
=
×
==−
( ) unitper0336.0
2200
6.0975.18.055.045.45
=
×+×
=
Ω=





+=





+= 975.1
80
400
035.01.1
22
2
1
21
V
V
XXXe
( ) ( ) Ω=+=+= 05.2975.155.0
2222
eee XRZ
Ω=





+=





+= 55.0
80
400
01.03.0
22
2
1
21
V
V
RRRe
(a)
(b) (i)
( )
1
221 sincos
V
XRI
regulationvoltageunitPer ee φφ +
=−
centper36.3=

30. V78.44678.6440 =+=
V
N
N
V
P
S
P 440
400
80
2200 =×=×=
centper54.1−=
(b) (ii) power factor 0.8 leading
( ) unitperregulationVoltage 0154.0
2200
6.0975.18.055.045.45
−=
×−×
=
V8.140336.0440 =×=
V78.60154.0440 =×=
Sec. terminal voltage on no-load
The decreasing of full-load voltage is
V2.4258.14440 =−=Therefore the secondary full-load voltage
The increasing of full-load voltage is
Therefore the secondary full-load voltage

31. 1
e1
V
RI
=
2200
55.045.45 ×
= 0.0114=
0.0114%=
Per unit
ltageprimary vo
primarytoreferred
resistanceequivalent
currentprimary
load-full






×





=unitperdropResistance
load-noonvoltagesecondary
secondarytoreferred
resistanceequivalent
currentsecondary
load-full






×





=unitperdropResistanceOr
dropResistance∴

32. Alternative
Full load secondary current
Recall Equation 12
p
s
s
p
N
N
I
I
=
Equivalent resistance
referred to secondary
AS 2.22745.45
80
400
I
N
N
I P
S
P
=×==
Ω=





+=





+= 022.0
400
80
3.001.0
22
1
2
12
N
N
RRRe
V
N
N
V
P
S
P 440
400
80
2200 =×=×=Secondary voltage on no-load
440
022.02.227
load-noonvoltagesecondary
secondarytoreferred
resistanceequivalent
currentsecondary
load-full
×
=






×





=unitperdropResistance
1.14%unitper0.0114 ==

33. 2200
975.145.45
ltageprimary vo
primarytoreferred
resistanceleakageequivalent
currentprimary
load-full
×
=






×





=unitperdropractanceLeakage
Ω=





+=





+= 975.1
80
400
035.01.1
22
2
1
21
V
V
XXXe
%08.4unitper0408.0 ==
primarytoreferredresistanceleakageequivalent
A
V
P
currentprimaryloadFull 45.45
2200
10100 3
=
×
==−