Electrical Technology MEC4105
Tutorial 1: Single Phase Transformer - Solution
1. A 200 kVA, 6600V/400V, 50 Hz single-phase transformer has 80 turns on the secondary. Calculate:
(a) the approximate values of the primary and secondary currents;
(b) the approximate number of primary turns;
(c) the maximum value of the flux.
200 103
Ans. (a) Primary current 
30.3A
6600
200 103
Secondary current 
500A
400
6600
(b) Primary tu rns 80 
1320 turns
400
6600
(c) E 4.44fNmax  max 
0.0225 Wb
4.44 50 1320
2. The primary winding of a single-phase transformer is connected to a 230V 50 Hz supply. The
secondary winding has 1500 turns. If the maximum value of the core flux is 0.00207Wb, determine:
(a) the number of turns on the primary winding;
(b) the secondary induced voltage;
(c) the net cross-sectional core area if the flux density has a maximum value of 0.465 telsa.
230
Ans. (a) Primary turns 
500 turns
4.44 50 0.00207
1500
(b) Secondary induced voltage 230 
690 Volts
500
Flux
0.00207
(c) Cross sectional Area 

4452mm2
flux density
0.465
3. A 3200/400 single-phase transformer has winding resistance and reactance of 3and 13
respectively in the primary and 0.02and 0.065in the secondary. Express these in terms of
(a) primary alone ,
(b) secondary alone.
Ans. (a) Turn ratio = 8, to transfer impedance from LV side to HV side, the impedance is
multiplied by 82 = 64, R2’= 0.02 ｘ 64 = 1.28Ω, X2’= 0.065 ｘ 64 = 4.16Ω
R1eq + j X1eq = (R1 + R2’) + j(X1 + X2’) = (3+1.28) + j(13+4.16) = 4.28+j 17.16Ω
(b) Turn ratio = 8, to transfer impedance from HV side to LV side, the impedance is
divided by 8 2 = 64, R1’= 3 ÷64 = 0.0469Ω, X1’= 13 ÷64 = 0.203Ω
R2eq + j X2eq = (R1’+ R2) + j(X1’+ X2)
= (0.0469+0.02) + j(0.203+0.065) = 0.067+j 0.268Ω
4. The primary winding of a transformer has 500 turns and is supplied at a voltage of 2000 V r.m.s. at a
frequency of 50 Hz. Estimate the maximum value of the flux through the core.
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2000
Ans. E 4.44fN max  max 
0.018 Wb
4.44 50 500
5. A single-phase transformer has 1000 turns on the primary and 200 turns on the secondary. The noload current is 3A at a power factor 0.2 lagging. Calculate the primary current and power factor
when the secondary current is 280 A at a power factor of 0.8 lagging. Assume the voltage drop in the
windings to be negligible.
200
Ans. Secondary current referred to the primary side = 280 
56 A
1000
Power factor = 0.8 lagging → Φ = － 36.9o, i.e. I 2 ' 5636.9
I1 I o I 2 ' 378 .556 36.9 
0.5981 j2.9398 44.7823 j33.6235
45. 3804 j36.5633 58.28 38 .9 The power factor of I1 = Cos –38.9o = 0.78 lagging
6. A single-phase transformer has a primary voltage of 2000V, a secondary voltage of 440 V and a full
load-output of 20 kVA. The secondary winding has 130 turns. Calculate the number of primary turns
and the primary and secondary full-load currents, neglecting losses.
Ans.
2000
Primary turns 130 
591 turns
440
20 10 3
20 103
Primary current 
10A
Secondary current 
45.5A
2000
440
7. The voltage on the secondary of a single-phase transformer is 200 V when supplying a load of 8 kW
at a power factor of 0.8 lagging. The secondary resistance is 0.04and the secondary leakage
reactance is 0.8, calculate the induced e.m.f. in the secondary winding.
8000
Ans. V2 = 200, I 2 
50A , Cos Φ = 0.8, Sin Φ = 0.6
200 0.8
For secondary circuit only, voltage regulation = I2R2Cos Φ + I2X2 SinΦ
= 50 x ( 0.04 x 0.8 + 0.8 x 0.6 ) = 25.6 Volts
Therefore, the induced e.m.f. E2 = 200 + 25.6 = 225.6 Volts.
8. If the transformer of Question 7 has 500 primary turns and 50 secondary turns,
(a) find the induced e.m.f. in the primary winding.
(b) if the primary resistance is 4and the primary leakage reactance is 70, estimate the primary
terminal voltage. The magnetizing current can be ignored.
Ans. (a) Turns ratio = 10, induced e.m.f. E1 = 10 x E2 = 2256 Volts
(b) For Primary circuit only, voltage regulation = I1R1eqCos Φ
+ I1X1eq SinΦ
I1 = 50 / 10 = 5A, Volt drop in primary winding = 5 ( 4 x 0.8 +70 x 0.6 ) = 226 Volts
Therefore primary terminal voltage = 2256 + 226 = 2482 Volts
9. A farm, whose electrical load can be represented by a resistance of 1in series with an inductive
reactance of 1, is supplied from an 11000V single-phase line through a transformer of turns ratio
50:1. The resistance and leakage reactance of the transformer are 125and 250respectively when
referred to its primary and its magnetizing current may be neglected. Determine :
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(a) the magnitude of the current taken from the secondary terminals of the transformer ;
(b) the potential difference between the secondary terminals,
(c) the magnitude and power-factor of the primary current. ans. (a) 145 A ; (b) 205 V, (c) 2.9 A, 0.7
2
1 
Ans. (a) Primary resistance referred to the secondary side = 125 
  0.05
50 
2
1 
Primary leakage reactance referred to the Secondary side = j2 50 
  j0 .1
50 
Total impedance (including load) referred to the secondary side
= (1+0.05) + j (1 +0.1) = 1.05 + j 1.1 Ω 1.520746.3

With all the impedance referred to the secondary, the transformer becomes an ideal transformer,
1 
therefore the e.m.f. at secondary = 11000 
 220 Volts
50 
220
Secondary currents =
144.7 A
1 .520746.3
Impedance of load = 1 + j 1 Ω,
power factor of the load = Cos 45o = Sin 45o = 0.7071 lagging
Volt drop across transformer impedance = I2Req2Cos Φ
+ I2Xeq2 SinΦ
,
= 144.7 (0.05 x 0.7071 + 0.1 x 0.7071) = 15.35 Volts
V2 = E2 –Volt drop = 220 –15.35 = 204.65 Volts
1
Neglecting No-load component, I1 I 2 ' 144.7  2.9A
50
Power factor of the primary current is the same as the secondary current, i.e. 0.7 lagging
10. A 4000/400V , 10 kVA transformer has primary and secondary winding resistance of 13 and
0.15respectively. The leakage reactance referred to the primary is 45, the magnetizing reactance
referred to the primary is 6000, and the resistance corresponding to the core loss is 12000.
Determine
(a) total resistance referred to the primary and the values of all impedances referred to the secondary.
(b) the input current when the secondary terminals are open-circuited.
(c) the input current when the secondary load current is 25 A at a power factor of 0.8 lagging.
2
4000 
Ans. (a) Secondary resistance referred to primary 0.15 

 15
400 
Total resistance referred to the primary = R1eq = 13 + 15 = 28 Ω
2
400 
R 2eq 28 
 0.28
4000 
2
400 
RC ' 12000 
 120 
4000 
2
400 
X 2eq 45 
 0. 45
4000 
2
400 
X M ' 6000
 j 60 
4000 
(b) During open circuit, only the no load current flow, IO 2 = IC 2+ IM2
V 4000 
V
4000 
IC  1 
0.3333A IM  1 
j0.6666A
RC 12000 
X M 6000 
0. 6666 
I O  0.33332 0.6666 2 tan 1 
A

0.74563.4
0.3333 
(c) When the secondary current is 25A at p.f. 0.8 lagging,
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I 2 2536.9A I 2 ' 2. 536.9 A2 j1.5A IO 0.3333j0.6666 A
,
,
I1 IO I 2 ' 
2 0.3333
j 
1.5 0.66663.1843
A
11. The primary and secondary windings of a 40kVA 6600/250V single-phase transformer have
resistances of 10and 0.02respectively. The leakage reactance of the transformer referred to the
primary is 35. Neglect the no-load current, calculate:
ans. (a) 257 V, (b) 2.2 %, 3.7 %
(a) the primary voltage required to circulate full load current when the secondary is short-circuited,
(b) the full load regulation at (i) unity (ii) 0.8 lagging power factor.
2
Ans.
6600 
(a) R2 ' 0.02
 13.94,
250 
Z1eq R 1eq X1eq 
10 13.94 
j35 23.94 j35Ω 42.455.6
Full load primary current = 40000/6600 = 6.06A
Primary voltage required to circulate full load current = 6.06 x 42.4 = 257 Volts
(b) At full load unity power factor, V.R. = 6.06 x (23.94 x 1 + 35 x 0) = 145Volts
Percentage voltage regulation = 145/6600 x 100% = 2.2%
At full load 0.8 p.f. lagging, V.R. = 6.06 x (23.94 x 0.8 + 35 x 0.6 ) = 243.3 Volts
Percentage voltage regulation = 243.3/6600 x 100% = 3.7%
12. A 40kVA 6000V/240V power transformer has the following parameters :
Resistance of primary winding = 10 ; Resistance of secondary winding = 0.017
Reactance referred to primary side = 35. Calculate:
(a) the primary voltage and power required to circulate full-load current when the secondary is shortcircuited ,
(b) the percentage voltage regulation and secondary terminal voltage for load of 33kVA at 0.65
leading power factor.
Ans. (a) The problem ask for primary voltage therefore we have to transfer all the impedance
to the primary:
2
6000 
Secondary resistance referred to the primary = 0.017 

10. 625
240 
Total equivalent impedance referred to the primary = (10 + 10.625) + j 35 Ω
= 20.625 + j35Ω = 40.625∠ 59.5o
40000
Primary full load current 
6.6667A
6000
Primary voltage required to circulate full load current = 6.6667 x 40.625 = 270.83 V
Power required = I2R1eq = 6.66672 x 20.625 = 916.66W
33000
(b) With a load of 33kVA, primary current 
5.5A
6000
p.f. = 0.65 → Cos Φ = 0.65, Sin Φ = 0.76
Voltage regulation = I1Req1Cos Φ + I1Xeq1 SinΦ ,
5.5 
20.625 0.65 35 0.76 
% Voltage regulation 
100% 1.209%
6000
E 1 = V1 –V.R. = 6000 –(–72.54) = 6072.54 Volt
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N 2 
240 
E 2 E1 
N 
6072 6000 242.9 Volts


 1
13. A 50 kVA transformer, which steps down from 6600V to 220V, has a primary resistance of 10and
a secondary resistance of 0.01, with the leakage reactance neglected, calculate:
(a) the total resistance referred to the secondary;
(b) the total resistance referred to the primary;
(c) the full-load copper loss.
2
220 
Ans. (a) Primary resistance referred to the secondary 1 0 

 0 .0111
6600 
Total equivalent resistance referred to the secondary = 0.01 + 0.01111 = 0.02111Ω
2
6600 
(b) Total equivalent resistance referred to the primary 0.021111 

 19 
220 
50000 
(c) Full load primary current = 

7 .58A
6600 
Full load copper loss = I2R1eq = 7.582 x 19 = 1091.7 W
14. The following results were obtained on a 50 kVA single-phase transformer:
Open circuit test -
Short circuit test -
Primary voltage, 3300 V
Primary voltage, 124 V
Secondary voltage, 400 V
Primary current, 15.3 A,
Primary power, 430 W
At Secondary current, full load value, Primary power, 525 W
Calculate: (a) the efficiencies at full load and at half load for 0.7 power factor;
(b) the voltage regulations for power factor 0.7 lag, 0.7 leading,
(c) the secondary terminal voltages corresponding to (a) & (b).
Ans. (a) The transformer on open circuit test is at full voltage and on short circuit test is at full
current, therefore full load iron loss = 430W, full load copper loss = 525 W

Output power

Full load efficiency 
Output power Iron Loss Copper Loss 
100 %


50000 0.7



100% 97.34%
50000 0.7 430 525 
At half load, the iron loss remains at 430W but the copper loss is only 1/4 of the copper loss at full
load, since copper loss is proportional to the square of the current.


0.5 50000 0.7
Half load efficiency 
100% 96.9%
2
0.5 50000 0.7 430 525 
0.5

(b) From short circuit test, I12R1eq = 525W → R 1eq  525 2 2.2427Ω

15.3
124
8.1046Ω

15.3
Z1eq 
=>
X1eq  Z1 eq R1eq  8.10462 2.2427 2 7.7881Ω
2
2
Voltage regulation at full load 0.7 p.f. lagging = I1Req1Cos Φ
+ I1 Xeq1 SinΦ
Power factor = 0.7 → Cos Φ = 0.7, Sin Φ = 0.7141,
50000
15.1515A
3300
Full load primary current 
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108.05
V.R. = 15.1515 x (2.2427 x 0.7 + 7.7881 x 0.7141) = 108.05 V ; % V.R. 
100% 3.3%
3300
V.R. at 0.7 p.f. leading = 15.1515 x (2.2427 x 0.7 –7.7881 x 0.7141)= - 60.48 Volts
60.48
% V.R. 
100% 1.83%
3300
(c) At full load 0.7 p.f. lagging, E1 = V1 –Volt drop = 3300 –108.05 = 3192 Volts
Now all the impedance has been transferred to the primary, therefore the transformer
N 2 
400
becomes an ideal transformer, V2 E 2 E1 
N 
3192 3300 386.9Volts
1
Similarly, at full load 0.7 power factor leading,
E 1 = V1 –Volt drop = 3300 –(-60.48) = 3360.48 Volts
N 2 
400
V2 E1 
N 
3360.48 3300 407.3 Volts
 1
15. A 250kVA, 4160 / 480V, 60Hz single-phase transformer has the following parameters :
R1 = 0.09, R 2 = 0.0012, Rm = 31.6k, X1 = 1.7, X2 = 0.0226, Xm = 3.24k
The transformer is step-down.
(a)
(b)
(c)
(d)
(e)
Calculate the values of the transformer parameters referred to the primary side.
Hence sketch the equivalent circuit with all values referred to the primary.
Calculate the primary voltage for rated load at 0.76 lagging power factor.
Calculate the transformer efficiency for (c) if core loss obtained from the no-load test is 547W.
Find the voltage regulation of this transformer operating at 0.76 leading power factor.
Ans. (a) R1 = 0.09, R2 = 0.0012, Rm = 31.6k, X1 = 1.7, X2 = 0.0226, Xm = 3.24k
2
N 1 
4160 
R2 ' R2 
 0 .09
N 
 0 .0012
480 
 2
2
2
=>
R1eq = R1 +R2’= 0.09 + 0.09 = 0.18Ω
N1 
4160 
X 2 ' X 2 
 j1 .7 =>
N 
 0.0226 
480


 2
2
X1eq = X1 +X 2’= 1.7 + 1.7 = j 3.4 Ω
250000
(c) Full load secondary current 
520.8A ,
480
p.f. = 0.76 → Cos Φ = 0.76, Sin Φ = 0.65
2
N 2 
480 
R 2 eq R1eq 
N 
0.18 4160  0.0024 ;


 1
2
2
X 2 eq
N 2 
480 
X 1eq 
N 
3.4 4160  j0.0453


 1
2
At 0.76 p.f. leading, V.R. = 520.8 x (0.0024 x 0.76 - 0.0453 x 0.65) = -14.39 V
E 2 = V2 + Volt drop across Z2eq = 480 + (–14.39) = 465.61 Volts
Since all the impedance is transferred to the secondary, the transformer becomes an ideal transformer:


N1
4160
V1 = E1 E 2 
4035 Volts
 
465.61 
N 2 
480
(d) Given core loss = 547 W, full load copper loss = 520.82 x 0.0024 = 651 W
Total transformer loss at full load = iron loss + copper loss = 1198 W
output power
250000 0.76
Efficiency 

100% 99.37%
output power losses 250000 0.76 1198
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14.35
(e) Voltage regulation at p.f. = 0.76 leading = - 14.35 Volts ; % V.R. 
100% 3%
480
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