1. 9EE402.11 1
Department of Technical Education
Andhra Pradesh
Name of the staff member : A. Siva Sankar
Designation : Lecturer
Branch : Electrical and Electronics Engg
Name of the institution : Govt.Polytechnic, Proddatur
Semester : IV
Subject Name : AC Machines I
Subject Code & Title : EE402
Major Topic : Single Phase Transformers
Sub_Topic : Problems On Transformer On Load
Duration : 50 minutes
Teaching Aids : PPT, Animations
Revised by : D.Linga Reddy, HEES
J.N.Govt.polytechnic, Hyd
2. 9EE402.11 2
Recap
• Transformer works on mutual induction
• Ideal transformer on no-load
• Ideal transformer on-load
• Vector diagram
In the previous class, you have learnt about
3. 9EE402.11 3
Objectives
On completion of this period, you would be able
to understand
• TO determine the various currents of a transformer on
load at different power factors
4. 9EE402.11 4
Example1
A 2200/220V, 50Hz single-phase transformer has no-
load primary current of 0.6A and a power factor of
0.275lag when its H.V. side is energized at rated
voltage. If the secondary supplies a load current of
60A at 0.8 power factor lagging, calculate the primary
current and its power factor.
5. 9EE402.11 5
Given Data
V1 =2200V
E2 =220V
I0=0.6A
cosØ0=0.275 lag
I2=60A
cosØ2=0.8 lag
Øo=cos-1 (0.275)= 750
Ø2= cos-1(0.8)=36.860
6. 9EE402.11 6
Required data
• I1 = ?
• Cos Ø1 = ?
Solution
• The primary current component required to
neutralize the effect of secondary current I2 = 60A is
given by
/
0
0 0.6 74
I
0
2 ' 6 36.86
I
2 2
220
60 6
2200
I KI x A
7. 9EE402.11 7
• Primary power factor=cos(400)=0.765 lag
• Primary current = 6.487 A
Solution:
1 0 2
0 0
0
'
0.6 74 6 36.86
4.965 4.175
6.487 40
I I I
j
1 0 2
0 0
0
'
0.6 74 6 36.86
4.965 4.175
6.487 40
I I I
j
1 0 2
0 0
0
'
0.6 74 6 36.86
4.965 4.175
6.487 40
I I I
j
1 0 2
0 0
0
'
0.6 74 6 36.86
4.965 4.175
6.487 40
I I I
j
8. 9EE402.11 8
Example 2
A 400/200V, 1-phase transformer is supplying a
load of 50A at a power factor of 0.866 lagging.
The no-load current of 2A at 0.208 power factor
lagging. Calculate the primary current and
primary power factor.
12. 9EE402.11 12
Primary power factor is the cosine of the angle
between V1 and I1
cos Φ1=cos(-33.33)=0.836lag-------answer
Primary current , I 1=26.38A
answer
-
-
-
-
-
-
-
-
A
33.22
-
26.38
j12.5
-
21.65
j1.956
-
0.416
)
(-30
sin
j
)
cos(-30
25
)
2sin(-78
j
)
2cos(-78
30
-
25
78
-
2
I
I
I
A
78
-
2
I
A
30
-
25
I
0
0
0
0
0
0
0
1
2
0
1
0
0
0
1
2
answer
-
-
-
-
-
-
-
-
A
33.22
-
26.38
j12.5
-
21.65
j1.956
-
0.416
)
(-30
sin
j
)
cos(-30
25
)
2sin(-78
j
)
2cos(-78
30
-
25
78
-
2
I
I
I
A
78
-
2
I
A
30
-
25
I
0
0
0
0
0
0
0
1
2
0
1
0
0
0
1
2
answer
-
-
-
-
-
-
-
-
A
33.22
-
26.38
j12.5
-
21.65
j1.956
-
0.416
)
(-30
sin
j
)
cos(-30
25
)
2sin(-78
j
)
2cos(-78
30
-
25
78
-
2
I
I
I
A
78
-
2
I
A
30
-
25
I
0
0
0
0
0
0
0
1
2
0
1
0
0
0
1
2
answer
-
-
-
-
-
-
-
-
A
33.22
-
26.38
j12.5
-
21.65
j1.956
-
0.416
)
(-30
sin
j
)
cos(-30
25
)
2sin(-78
j
)
2cos(-78
30
-
25
78
-
2
I
I
I
A
78
-
2
I
A
30
-
25
I
0
0
0
0
0
0
0
1
2
0
1
0
0
0
1
2
answer
-
-
-
-
-
-
-
-
A
33.22
-
26.38
j12.5
-
21.65
j1.956
-
0.416
)
(-30
sin
j
)
cos(-30
25
)
2sin(-78
j
)
2cos(-78
30
-
25
78
-
2
I
I
I
A
78
-
2
I
A
30
-
25
I
0
0
0
0
0
0
0
1
2
0
1
0
0
0
1
2
answer
-
-
-
-
-
-
-
-
A
33.22
-
26.38
j12.5
-
21.65
j1.956
-
0.416
)
(-30
sin
j
)
cos(-30
25
)
2sin(-78
j
)
2cos(-78
30
-
25
78
-
2
I
I
I
A
78
-
2
I
A
30
-
25
I
0
0
0
0
0
0
0
1
2
0
1
0
0
0
1
2
answer
-
-
-
-
-
-
-
-
A
33.22
-
26.38
j12.5
-
21.65
j1.956
-
0.416
)
(-30
sin
j
)
cos(-30
25
)
2sin(-78
j
)
2cos(-78
30
-
25
78
-
2
I
I
I
A
78
-
2
I
A
30
-
25
I
0
0
0
0
0
0
0
1
2
0
1
0
0
0
1
2
13. 9EE402.11 13
Example 3
A single phase transformer 1000/250Vn draws primary
current of 25A at 0.707 power factor lagging and
secondary current of 80A at 0.8 power factor lagging.
Calculate the no-load current of transformer and its
phase with respect to the voltage from phasor diagram.
17. 9EE402.11 17
Therefore, the magnitude of the no-load
current is 5.92A and its phase is -73.520
0
0
0
1
2
1
0
73.52
-
5.92
j5.68
-
1.68
j12
16
-
j17.68
-
17.68
36.8
-
20
-
45
-
25
I
-
I
I
Solution
0
0
0
1
2
1
0
73.52
-
5.92
j5.68
-
1.68
j12
16
-
j17.68
-
17.68
36.8
-
20
-
45
-
25
I
-
I
I
0
0
0
1
2
1
0
73.52
-
5.92
j5.68
-
1.68
j12
16
-
j17.68
-
17.68
36.8
-
20
-
45
-
25
I
-
I
I
0
0
0
1
2
1
0
73.52
-
5.92
j5.68
-
1.68
j12
16
-
j17.68
-
17.68
36.8
-
20
-
45
-
25
I
-
I
I
0
0
0
1
2
1
0
73.52
-
5.92
j5.68
-
1.68
j12
16
-
j17.68
-
17.68
36.8
-
20
-
45
-
25
I
-
I
I
18. 9EE402.11 18
Example 4
The secondary of a single phase transformer is supplying a
current of 300A at a power factor of 0.8 lagging. The no-load
current of the transformer is 5 A at a power factor of 0.2
lagging. Calculate the primary current and its power factor.
Assume the ratio of transformation to be 3 and neglect
voltage drops due to the impedance of the windings.
22. 9EE402.11 22
•
=5L-78.460 +100L-36.860-
= 1-j4.89+80-j60
=81-j64.89
=103.78L-38.730
• Therefore, the magnitude of the Primary current is
103.78A and its power factor is cos(38.730)=0.78lag
1
2
0
1 I
I
I
23. 9EE402.11 23
Summary
We have discussed about
• The concept of various currents in the windings of a
transformer
• The procedures to solve problems on currents and
power factors of a transformer assuming the windings
have no leakage impedances
24. 9EE402.11 24
1. What is done to balance the m.m.f setup due to the
secondary current
A) The primary voltage is increased
B) The core flux increases immediately
C) The current in the primary increases
D) All of these
Quiz
25. 9EE402.11 25
2. Which of the following applies to an ideal transformer
A) Coefficient of coupling is unity
B) Permeability of the core is infinite
C) Copper and core losses are zero
D) All of these
Quiz
26. 9EE402.11 26
3. The no-load power factor angle is close to 900,
because
A) I0<Iw
B) Iw> Iµ
C) Iw<Iµ
D) I0<Iµ
Quiz
27. 9EE402.11 27
Frequently asked questions
1. A 2200/200V single phase transformer takes 1A at the
HT side on no-load at a power factor of 0.385 lagging.
Calculate the iron losses. If the load of 50A at a power
factor of 0.8 lagging is taken from the secondary of the
transformer. Calculate the actual primary current and
its power factor.
[ 847W, 5.444A, 0.74 lag ]
28. 9EE402.11 28
2. A 400/200V, single phase transformer is
supplying a load of 25 A at a power factor of
0.866 lagging. On no-load the current and power
factor are 2A and 0.208 respectively. Calculate
the current taken from the supply.
[ 13.9A, -36.10]
3. A 1600/200V, 50Hz single-phase transformer
supplies a load of 20 kW at a power factor of 0.8
lagging and takes a magnetising current of 2A at
a power factor of 0.2 lagging. Calculate the
primary current and its power factor
[17.15A, 0.753 lag]
Frequently asked questions
29. 9EE402.11 29
4. A 100kVA, 2400/240 V, 50 Hz single-phase
transformer has an no-load primary current of
0.64A and core loss of 700 W. If a load of 40A at
0.8 power factor lagging on its L.V side, then
calculate the primary current and its power factor
[4.58A, 0.762 lagging]
Frequently asked questions
