CHAPTER 21 TRANSFORMERS
Exercise 118, Page 342
1. A transformer has 600 primary turns connected to a 1.5 kV supply. Determine the number of
secondary turns for a 240 V output voltage, assuming no losses.
For a transformer,
N1 V1

N 2 V2
V 
 240 
from which, secondary turns, N 2  N1  2   (600) 
 = 96 turns
 1500 
 V1 
2. An ideal transformer with a turns ratio 2:9 is fed from a 220 V supply. Determine its output
voltage.
N1 2

N2 9
and V1  220 V
N1 V1

N 2 V2
N 
9
from which, output voltage, V2  V1  2    220    = 990 V
2
 N1 
3. A transformer has 800 primary turns and 2000 secondary turns. If the primary voltage is 160 V,
determine the secondary voltage assuming an ideal transformer.
N1
800

N 2 2000
N1 V1

N 2 V2
and V1  160 V
N 
 2000 
from which, output voltage, V2  V1  2   160  
 = 400 V
N
800


 1
4. An ideal transformer with a turns ratio 3:8 has an output voltage of 640 V. Determine its input
voltage.
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266
N1 3

and V2  640 V
N2 8
N1 V1

N 2 V2
N 
 3
from which, input voltage, V1  V2  1    640    = 240 V
8
 N2 
5. An ideal transformer has a turns ratio of 12:1 and is supplied at 192 V. Calculate the secondary
voltage.
N1 12

N2 1
and V1  192 V
N1 V1

N 2 V2
N 
1
from which, output voltage, V2  V1  2   192    = 16 V
 12 
 N1 
6. A transformer primary winding connected across a 415 V supply has 750 turns. Determine how
many turns must be wound on the secondary side if an output of 1.66 kV is required.
N1 V1

N 2 V2
V 
 1660 
from which, secondary turns, N2  N1  2    750  
 = 3000 turns
 415 
 V1 
7. An ideal transformer has a turns ratio of 15:1 and is supplied at 180 V when the primary current
is 4 A. Calculate the secondary voltage and current.
N1 12

, V1  220 V and I1  4 A
N2 1
N1 V1

N 2 V2
N 
1
from which, output voltage, V2  V1  2   180    = 12 V
 15 
 N1 
N1 I 2

N 2 I1
N 
 15 
from which, secondary current, I2  I1  1    4    = 60 A
1
 N2 
8. A step-down transformer having a turns ratio of 20:1 has a primary voltage of 4 kV and a load of
10 kW. Neglecting losses, calculate the value of the secondary current.
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267
N1 20

N2
1
and V1  4000 V
N1 V1

N 2 V2
N 
 1 
from which, output voltage, V2  V1  2    4000    = 200 V
 20 
 N1 
Secondary power = V2 I 2 = 10000
i.e.
200 I 2 = 10000
secondary current, I 2 
from which,
10000
= 50 A
200
9. A transformer has a primary to secondary turns ratio of 1:15. Calculate the primary voltage
necessary to supply a 240 V load. If the load current is 3 A determine the primary current.
Neglect any losses.
N1 V1 I 2


N 2 V2 I1
If
V
1
 1
15 240
If
1 3

15 I1
i.e.
V
1
3
 1 
15 240 I1
1
then primary voltage, V1  240   = 16 V
 15 
 15 
then primary current, I1  3   = 45 A
1
10. A 10 kVA, single-phase transformer has a turns ratio of 12:1 and is supplied from a 2.4 kV
supply. Neglecting losses, determine (a) the full load secondary current, (b) the minimum value
of load resistance which can be connected across the secondary winding without the kVA rating
being exceeded, and (c) the primary current.
10000 = V1 I1  V2 I2 ,
(a)
N1 V1

N 2 V2
N1 12

and V1  2400 V
N2 1
N 
1
from which, output voltage, V2  V1  2    2400    = 200 V
 12 
 N1 
10000 VA = V2 I 2 = 200 I 2
from which, secondary current, I 2 
10000
= 50 A
200
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268
(b) Load resistance, R L 
(c)
V2 200

=4
I2
50
N 
1
from which, primary current, I1  I2  2    50    = 4.17 A
 12 
 N1 
N1 I 2

N 2 I1
11. A 20  resistance is connected across the secondary winding of a single-phase power
transformer whose secondary voltage is 150 V. Calculate the primary voltage and the turns ratio
if the supply current is 5 A, neglecting losses.
Secondary current, I 2 
N1 V1

N 2 V2
V2 150

= 7.5 A , I1  5A and
R 2 20
V2 = 150 V
N 
I 
 7.5 
from which, primary voltage, V1  V2  1   V2  2   150  
 = 225 V
 5 
 N2 
 I1 
Turns ratio,
3
N1 I 2 7.5
 
= 1.5 or
2
N 2 I1
5
or 3:2
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269
Exercise 119, Page 344
1. A 500 V/100 V, single-phase transformer takes a full load primary current of 4 A. Neglecting
losses, determine (a) the full load secondary current, and (b) the rating of the transformer.
(a)
V 
V1 I 2
 500 

from which, full load secondary current, I2  I1  1   (4) 
 = 20 A
V2 I1
 100 
 V2 
(b) Transformer rating = V1 I1  500  4 = 2000 VA = 2 kVA
or transformer rating = V2 I2  100  20 = 2000 VA = 2 kVA
2. A 3300 V/440 V, single-phase transformer takes a no-load current of 0.8 A and the iron loss is
500 W. Draw the no-load phasor diagram and determine the values of the magnetizing and core
loss components of the no-load current.
V1  3300 V , V2  440 V and
IO  0.8A
Core or iron loss = 500 = V1 IO cos O
from which,
cos O 
i.e.
500
= 0.1894
 3300 0.8
500 =  3300  0.8 cos O
and
O  cos 1 0.1894  79.08
The no-load phasor diagram is shown below.
Magnetizing component, IM  IO sin O  0.8sin 79.08 = 0.786 A
Core loss component, IC  IO cos O  0.8(0.1894) = 0.152 A
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3. A transformer takes a current of 1 A when its primary is connected to a 300 V, 50 Hz supply, the
secondary being on open-circuit. If the power absorbed is 120 watts, calculate (a) the iron loss
current, (b) the power factor on no-load, and (c) the magnetizing current.
IO  1A and V1  300 V
(a) Power absorbed = total core loss = 120 = V1 IO cos O
i.e.
and
120 = (300) IO cos O
iron loss current, IC  IO cos O =
(b) Power factor on no-load, cos O 
120
= 0.40 A
300
IC 0.4

= 0.40
IO
1
(c) By Pythagoras, IO 2  IC 2  I M 2 from which,
magnetizing current, IM  IO 2  IC2  12  0.402 = 0.917 A
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Exercise 120, Page 346
1. A 60 kVA, 1600 V/100 V, 50 Hz, single-phase transformer has 50 secondary windings.
Calculate (a) the primary and secondary current, (b) the number of primary turns and (c) the
maximum value of the flux.
V1  1600 V , V2  100 V , f = 50 Hz, N 2  50 turns
(a) Transformer rating = V1 I1  V2 I2  60000 VA
hence,
and
(b)
V1 N1

V2 N 2
primary current, I1 
60000 60000

= 37.5 A
V1
1600
secondary current, I 2 
60000 60000

= 600 A
V2
100
V 
 1600 
from which, primary turns, N1   1  N 2  
  50  = 800 turns
 100 
 V2 
(c) E2  4.44f M N2 from which,
maximum flux,  M 
E2
100
= 9.0 mWb

4.44f N 2 4.44  50  50 
2. A single-phase, 50 Hz transformer has 40 primary turns and 520 secondary turns. The crosssectional area of the core is 270 cm2. When the primary winding is connected to a 300 volt supply,
determine (a) the maximum value of flux density in the core, and (b) the voltage induced in the
secondary winding
(a) From equation (4), e.m.f. E1 = 4.44 f m N1 volts
i.e.
300 = 4.44 (50) m (40)
from which, maximum flux density, m =
300
Wb = 0.033784 Wb
(4.44)(50)(40)
However, m = Bm  A, where Bm = maximum flux density in the core
and A = cross-sectional area of the core
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(see chapter 7)
Bm  270  10
-4
Hence,
= 0.033784
from which, maximum flux density, Bm =
(b)
0.033784
= 1.25 T
270 10 4
N 
V1
N
= 1 from which, V2 = V1  2 
V2
N2
 N1 
 520 
i.e. voltage induced in the secondary winding, V2 = (300) 
 = 3900 V or 3.90 kV
 40 
3. A single-phase 800 V/100 V, 50 Hz transformer has a maximum core flux density of 1.294 T
and an effective cross-sectional area of 60 cm 2 . Calculate the number of turns on the primary and
secondary windings.
Since B 

A
then  M  BM  A  1.294   60 104  = 7.764 mWb
E1  4.44f  M N1
from which, primary turns, N1 
E1
800

4.44 f  M 4.44  50   7.764 10 3 
= 464 turns
E2  4.44f M N2 from which, secondary turns, N 2 
E2
100

4.44 f  M 4.44  50   7.764 103 
= 58 turns
4. A 3.3 kV/110 V, 50 Hz, single-phase transformer is to have an approximate e.m.f. per turn of 22 V
and operate with a maximum flux of 1.25 T. Calculate (a) the number of primary and secondary
turns, and (b) the cross-sectional area of the core
E1
E
= 2 = 22
N1
N2
E
3300
Hence primary turns, N1 = 1 =
= 150
22
22
(a) E.m.f. per turn =
and secondary turns, N2 =
E 2 110
=
=5
22
22
© John Bird Published by Taylor and Francis
273
(b) E.m.f. E1 = 4.44 f m N1
from which, m =
3300
E1
=
= 0.0991 Wb
4.44f N 1
(4.44)(50)(150)
Now flux, m = Bm  A, where A is the cross-sectional area of the core,
hence area, A =
0.0991
m
=
= 0.07928 m 2 or 792.8 cm 2
1.25
Bm
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274
Exercise 121, Page 347
1. A single-phase transformer has 2400 turns on the primary and 600 turns on the secondary. Its noload current is 4 A at a power factor of 0.25 lagging. Assuming the volt drop in the windings is
negligible, calculate the primary current and power factor when the secondary current is 80 A at a
power factor of 0.8 lagging.
Let I1 ' be the component of the primary current which provides the restoring m.m.f.
I1 ' N1  I2 N 2
Then
i.e.
I1 '(2400)  (80)(600) from which,
I1 ' 
80  600 
2400
= 20 A
If the power factor of the secondary is 0.8, then cos 2  0.8
from which,
2  cos 1 0.8  36.87
If the power factor at no load is 0.25, then cos O  0.25
from which,
O  cos 1 0.25  75.52
In the phasor diagram shown below, I 2 = 80 A at an angle 2  36.87 to V2 and I1 '  20 A and is
shown anti-phase to I 2
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275
The no load current, IO = 4 A is shown at an angle O  75.52 to V1
Current I1 is the phasor sum of I1 ' and IO and by calculation:
Total horizontal component, I1 cos 1  IO cos O  I1 'cos 2
= (4)(0.25) + (20)(0.8) = 1 + 16 = 17 A
Total vertical component, I1 sin 1  IO sin O  I1 'sin 2
= (4)(sin 75.52) + (20)(sin 36.87) = 15.87 A
Hence, magnitude of I1 = 172  15.872 = 23.26 A
and
 15.87 
1  15.87 
tan 1  
 and 1  tan 
  43.03
 17 
 17 
Hence, power factor = cos O = cos 43.03 = 0.73
© John Bird Published by Taylor and Francis
276
Exercise 122, Page 350
1. A transformer has 1200 primary turns and 200 secondary turns. The primary and secondary
resistances are 0.2  and 0.02  respectively and the corresponding leakage reactances are 1.2 
and 0.05  respectively. Calculate (a) the equivalent resistance, reactance and impedance
referred to the primary winding, and (b) the phase angle of the impedance.
2
V 
 1200 
(a) Equivalent resistance, R e  R1  R 2  1   0.2  0.02 
 = 0.92 
 200 
 V2 
2
2
V 
 1200 
Equivalent reactance, X e  X1  X 2  1   1.2  0.05 
 = 3.0 
 200 
 V2 
2
Equivalent impedance, Ze  R e 2  Xe 2  0.922  3.02 = 3.138  or 3.14 
(b) cos e 
Re
0.92

Ze 3.138
 0.92 
and phase angle of impedance, e  cos 1 
 = 72.95
 3.138 
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Exercise 123, Page 350
1. A 6 kVA, 100 V/500 V, single-phase transformer has a secondary terminal voltage of 487.5 V
when loaded. Determine the regulation of the transformer.
Regulation =
=
no load sec ondary voltage  ter min al voltage on l oad
 100%
no load sec ondary voltage
500  487.5
12.5
100% 
100% = 2.5%
500
500
2. A transformer has an open circuit voltage of 110 volts. A tap-changing device operates when the
regulation falls below 3%. Calculate the load voltage at which the tap-changer operates.
Regulation =
Hence,
from which,
and
3=
no load sec ondary voltage  ter min al voltage on l oad
 100%
no load sec ondary voltage
110  V2
100%
110
3(110)
 110  V2
100
V2  110 
3(110)
= 106.7 V = voltage at which the tap-changer operates.
100
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Exercise 124, Page 352
1. A single-phase transformer has a voltage ratio of 6:1 and the h.v. winding is supplied at 540 V.
The secondary winding provides a full load current of 30 A at a power factor of 0.8 lagging.
Neglecting losses, find (a) the rating of the transformer, (b) the power supplied to the load, (c) the
primary current.
V1 6

and V1  540 V
V2 1
hence,
V2 
540
= 90 V and I 2 = 30 A
6
(a) Rating of transformer = V2 I2  90  30 = 2700 VA or 2.7 kVA
(b) Power supplied to load = V I cos  = (2700)(0.8) since power factor = cos  = 0.8
= 2.16 kW
(c)
V1 I 2

V2 I1
V 
1
from which, primary current, I1  I2  2    30    = 5 A
6
 V1 
2. A single-phase transformer is rated at 40 kVA. The transformer has full-load copper losses of
800 W and iron losses of 500 W. Determine the transformer efficiency at full load and 0.8 power
factor.
Efficiency =
output power input power  losses
losses

 1
input power
input power
input power
Full-load output power = V I cos  = (40)(0.8) = 32 kW
Total losses = 800 + 500 = 1.3 kW
Input power = output power + losses = 32 + 1.3 = 33.3 kW
Hence, efficiency,   1 
1.3
= 0.961 or 96.10%
33.3
3. Determine the efficiency of the transformer in problem 2 at half full-load and 0.8 power factor.
© John Bird Published by Taylor and Francis
279
Half full load power output =
1
 40  0.8  = 16 kW
2
Copper loss (or I2 R loss) is proportional to current squared
2
1
Hence, copper loss at half full load =    800  = 200 W
2
Iron loss == 500 W (constant)
Total loss = 200 + 500 = 700 W or 0.7 kW
Input power at half full load = output power at half full load + losses = 16 + 0.7 = 16.7 kW
Hence, efficiency,   1 
losses
0.7
 1
= 0.9581 or 95.81%
input power
16.7
4. A 100 kVA, 2000 V/400 V, 50 Hz, single-phase transformer has an iron loss of 600 W and a fullload copper loss of 1600 W. Calculate its efficiency for a load of 60 kW at 0.8 power factor.
Efficiency =
output power input power  losses
losses

 1
input power
input power
input power
Full-load output power = V I cos  = (100)(0.8) = 80 kW
Load power = 60 kW
Hence the transformer is at
60 3
 full load
80 4
2
3
Hence, copper loss at 3/4 load =   1600  = 900 W
4
Total losses = 900 + 600 = 1.5 kW
Input power = output power + losses = 60 + 1.5 = 61.5 kW
Hence, efficiency,   1 
1.5
= 0.9756 or 97.56%
61.5
© John Bird Published by Taylor and Francis
280
5. Determine the efficiency of a 15 kVA transformer for the following conditions:
(i) full-load, unity power factor (ii) 0.8 full-load, unity power factor (iii) half full-load, 0.8
power factor. Assume that iron losses are 200 W and the full-load copper loss is 300 W
(i) Full load power output = V I cos  = (15)(1) = 15 kW
Losses = 200 + 300 = 500 W or 0.5 kW
Input power at full load = output power + losses = 15 + 0.5 = 15.5 kW
Hence, efficiency,   1 
losses
0.5
 1
= 0.9677 or 96.77%
input power
15.5
(ii) At 0.8 full load, unity power factor, output power = 0.8  15 = 12 kW
Losses =  0.8   300   200  192  200 = 392 W or 0.392 kW
2
Input power at 0.8 full load = output power at 0.8 full load + losses
= 12 + 0.392 = 12.392 kW
Hence, efficiency,   1 
0.392
= 0.9684 or 96.84%
12.392
(iii) At 0.5 full load and 0.8 power factor, output power = 0.5  15  0.8= 6 kW
Losses =  0.5   300   200  75  200 = 275 W or 0.275 kW
2
Input power at 0.5 full load = output power at 0.5 full load + losses
= 6 + 0.275 = 6.275 kW
Hence, efficiency,   1 
0.275
= 0.9562 or 95.62%
6.275
6. A 300 kVA transformer has a primary winding resistance of 0.4  and a secondary winding
resistance of 0.0015 . The iron loss is 2 kW and the primary and secondary voltages are 4 kV and
200 V respectively. If the power factor of the load is 0.78, determine the efficiency of the
transformer (a) on full load, and (b) on half load.
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281
(a) Rating = 300 kVA = V1 I1 = V2 I2
Hence primary current, I1 =
400 10 3
300 10 3
=
= 75 A
V1
4000
and secondary current, I2 =
300 10 3
300 10 3
=
= 1500 A
V2
200
Total copper loss = I12 R1 + I22 R2, (where R1 = 0.4  and R2 = 0.0015 )
= (75)2(0.4) + (1500)2(0.0015)
= 2250 + 3375 = 5625 watts
On full load, total loss = copper loss + iron loss
= 5625 + 2000
= 7625 W = 7.625 kW
Total output power on full load = V2 I2 cos 2
= (300  103)(0.78) = 234 kW
Input power = output power + losses = 234 kW + 7.625 kW = 241.625 kW

losses 
Efficiency,  = 1 
  100%
 input power 
7.625 

= 1 
  100% = 96.84%
 241.625 
(b) Since the copper loss varies as the square of the current, then total
2
1
copper loss on half load = 5625    = 1406.25 W
2
Hence total loss on half load = 1406.25 + 2000
= 3406.25 W or 3.40625 kW
Output power on half full load =
1
(234) = 117 kW
2
Input power on half full load = output power + losses
= 117 kW + 3.40625 kW = 120.40625 kW
Hence efficiency at half full load,
© John Bird Published by Taylor and Francis
282

losses 
 = 1 
  100%
 input power 
3.40625 

= 1 
  100% = 97.17%
 120.40625 
7. A 250 kVA transformer has a full load copper loss of 3 kW and an iron loss of 2 kW. Calculate
(a) the output kVA at which the efficiency of the transformer is a maximum, and (b) the
maximum efficiency, assuming the power factor of the load is 0.80.
(a) Let x be the fraction of full load kVA at which the efficiency is a maximum.
The corresponding total copper loss =  3kW  x 2
At maximum efficiency, copper loss = iron loss
Hence,
3x 2  2
from which,
x2 
2
3
and
x=
2
= 0.8165
3
Thus, the output kVA at maximum efficiency = 0.8165  250 = 204.1 kVA
(b) Total loss at maximum efficiency = 2  2 = 4 kW
Output power = 204.1  0.8 = 163.3 kW
Input power = output power + losses = 163.3 + 4 = 167.3 kW
Hence, maximum efficiency,   1 
4
= 0.9761 or 97.61%
167.3
© John Bird Published by Taylor and Francis
283
Exercise 125, Page 355
1. A transformer having a turns ratio of 8:1 supplies a load of resistance 50 . Determine the
equivalent input resistance of the transformer.
2
N 
8
The equivalent input resistance, R1 =  1  R L =   (50) = 3200 Ω = 3.2 k
1
 N2 
2
2. What ratio of transformer turns is required to make a load of resistance 30  appear to have a
resistance of 270 
2
N 
R1   1  R L
 N2 
2
i.e.
N 
270 =  1   30 
 N2 
2
from which,
 N1 
270
9

 
30
 N2 
N1
 9 3
N2
and
i.e. the turns ratio required is 3:1
3. Determine the optimum value of load resistance for maximum power transfer if the load is
connected to an amplifier of output resistance 147  through a transformer with a turns ratio of
7:2
The equivalent input resistance R1 of the transformer needs to be 147  for maximum power transfer.
2
N 
R1 =  1  R L from which, RL = R1
 N2 
2
 N2 
2

 = 147   = 12 
7
 N1 
2
4. A single-phase, 240 V/2880 V ideal transformer is supplied from a 240 V source through a cable
of resistance 3 . If the load across the secondary winding is 720  determine (a) the primary
current flowing and (b) the power dissipated in the load resistance.
The circuit is as shown below.
© John Bird Published by Taylor and Francis
284
N1 V1
240
1



N 2 V2 2880 12
(a)
2
N 
1
Equivalent input resistance, R1   1  R L     720  = 5 
 12 
 N2 
2
Total input resistance, R IN  R  R1 = 3 + 5 = 8 
Hence, primary current, I1 
N1 I 2

N 2 I1
(b)
from which,
V1
240

= 30 A
R IN
8
N 
1
I2  I1  1    30    = 2.5 A
 12 
 N2 
Power dissipated in load, P = I 2 2 R L   2.5   720  = 4500 W or 4.5 kW
2
5. A load of resistance 768  is to be matched to an amplifier which has an effective output
resistance of 12 . Determine the turns ratio of the coupling transformer.
2
N 
R1   1  R L
 N2 
and
2
N 
hence 12 =  1   768 
 N2 
and
12  N1 


768  N 2 
2
N1
12 1


N2
768 8
Hence, the turns ratio of the coupling transformer is 1:8
6. An a.c. source of 20 V and internal resistance 20 k is matched to a load by a 16:1 single-phase
transformer. Determine (a) the value of the load resistance and (b) the power dissipated in the
load.
© John Bird Published by Taylor and Francis
285
The circuit is shown below.
(a) For maximum power transfer, R 1 needs to be 20 k
2
2
N 
N 
1
R 1   1  R L from which, load resistance, R L  R1  2   (20000)   = 78.13 
 16 
 N2 
 N1 
2
(b) Total input resistance when source is connected to the matching transformer is R IN  R1 ,
i.e. 20 k + 20 k = 40 k
Primary current, I1 
N1 I 2

N 2 I1
V
20

= 0.5 mA
40000 40000
from which,
N 
 16 
I2  I1  1    0.5 103    = 8 mA
1
 N2 
Power dissipated in load, P = I2 2 R L  8 103   78.13 = 5 mW
2
© John Bird Published by Taylor and Francis
286
Exercise 126, Page 357
1. A single-phase auto transformer has a voltage ratio of 480 V:300V and supplies a load of
30 kVA at 300 V. Assuming an ideal transformer, calculate the current in each section of the
winding.
Rating = 30 kVA = V1 I1  V2 I2
Hence, primary current, I1 
and
30 103
= 62.5 A
480
secondary current, I 2 
30 103
= 100 A
300
Hence, current in common part of winding = I2  I1 = 100 – 62.5 = 37.5 A
2. Calculate the saving in the volume of copper used in an auto transformer compared with a
double-wound transformer for (a) a 300 V:240 V transformer, and (b) a 400 V:100 V
transformer.
(a) For a 300 V:240 V transformer, x =
V2 240

= 0.80
V1 300
From equation (20.12), volume of copper in auto transformer
= (1 – 0.80)(volume of copper in a double-wound transformer)
= (0.20)(volume of copper in a double-wound transformer)
Hence, saving is 80%
(b) For a 400 V:1000 V transformer, x =
V2 100

= 0.25
V1 400
From equation (20.12), volume of copper in auto transformer
= (1 – 0.25)(volume of copper in a double-wound transformer)
= (0.75)(volume of copper in a double-wound transformer)
Hence, saving is 25%
© John Bird Published by Taylor and Francis
287
Exercise 127, Page 358
1. A three-phase transformer has 600 primary turns and 150 secondary turns. If the supply voltage
is 1.5 kV determine the secondary line voltage on no-load when the windings are connected
(a) delta-star, (b) star-delta.
(a) For a delta connection, VL  VP
hence, primary phase voltage, VP1 = 1.5 kV = 1500 V
N 
 150 
Secondary phase voltage, VP2 = VP1  2   (1500) 
 = 375 V
 600 
 N1 
For a star connection, VL  3 VP
3  375  = 649.5 V
hence, secondary line voltage =
(b) For a star connection, VL  3 VP or VP 
Primary phase voltage, VP1 
VL1
3

VL
3
1500
= 866.0 V
3
For a delta connection, VL  VP
N1 V1

N 2 V2
N 
 150 
from which, secondary phase voltage, VP2  VP1  2   (866.0) 

 600 
 N1 
= 216.5 V = secondary line voltage
© John Bird Published by Taylor and Francis
288
Exercise 128, Page 353
1. A current transformer has two turns on the primary winding and a secondary winding of 260
turns. The secondary winding is connected to an ammeter with a resistance of 0.2 , the
resistance of the secondary winding is 0.3 . If the current in the primary winding is 650 A,
determine (a) the reading on the ammeter, (b) the potential difference across the ammeter, and
(c) the total load in VA on the secondary.
N 
 2 
(a) Reading on ammeter, I2  I1  1   (650) 
 =5A
 260 
 N2 
(b) P.d. across ammeter = I2 R 2  (5)(0.2) = 1 V
(c) Total resistance of secondary circuit = 0.2 + 0.3 = 0.5 
Induce e.m.f. in secondary = (5)(0.5) = 2.5 V
Total load on secondary = (2.5)(5) = 7.5 VA
© John Bird Published by Taylor and Francis
289