DET204 Electrical Machine & Application 1
Tutorial 1 .Answer Transformer Problem
1. The primary winding of transformer is rated at 115 V and the secondary winding is rated at 300
V. The primary winding has 500 turns. If transformer has a full load secondary output of 300 VA
at 300 V. Calculate;
a) The full‐load secondary current
b) The full‐load primary current (neglect all losses)
c) The relationship between the primary current and the secondary current
d) The number of turns on the secondary winding.
Solution;
a) Is = 300VA / 300V = 1A
b) Ip = Is (Vs/Vp) = 1 A ( 300V/115V) = 2.609A
c) Is/Ip = Np/Ns
d) Ns = Np (Vs/Vp) = 500 turns ( 300V/115V) = 1,304 turns
2. The emf per turn for a single phase 2200/220‐V, 50‐Hz transformer is approximately 12 V.
calculate the number of primary and secondary turns.
Solution;
Np = 2200V /12V = 183turns; Ns = 220V/12V = 18 turns
3. Draw a circuit diagram to show how the core loss, in watts, is measured for a 5‐kVA transformer
rated at 2400 to 240 V. The low‐voltage side is used as the primary winding.
Solution;
240V
2400V
4. The core loss of the single phase 2200/220‐V, 50‐Hz transformer is measured using the low‐
voltage side as the primary. The following data are obtained:
Voltmeter reading = 240 V
Exciting current = 1.0 A
Wattmeter reading = 24 W
Calculate:
a) The core loss, in watts.
b) The power factor and phase angle.
c) The in‐phase component of the exciting current required for the core losses.
d) The lagging quadrature (magnetizing) component of the current.
Solution;
a) The core loss = 24 Watts
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b) The power factor = 24W / ( 240V x 1A) = 0.1 lagging
The phase angle = cos‐1(0.1) = 84.16o
c) The in‐phase component of the exciting current = 0.1 x 1 A = 1 A
d) The magnetizing current = 1 A x sin θ = 1A x sin (84.16) = 0.0995A
5. A 5‐kVA, 240/120‐V, 50‐Hz transformer has a core loss of 32W. The transformer has an effective
resistance of 0.05 Ω in the high voltage winding and 0.0125 Ω in the low voltage winding.
Calculate;
a) the efficiency of the transformer at the rated load and unity power factor.
b) the efficiency of the transformer at 50% of the rated load and 0.175 lagging power
factor
Solution;
a) The core loss = 32W
I primary = 5kVA/240V = 20.833 A; The copper losses at high‐voltage winding = (20.833)2
(0.05) = 21.7W
I secondary = 5kVA/120V =41.667 A; The copper losses at high‐voltage winding =
(41.667)2 (0.0125) = 21.7W
The total losses = 32 W + 21.7 W +21.7 W = 75.4W
Po = 5kVA x 1 = 5kW
Efficiency = {Output power / (Output power +total losses)} x 100% = {5kW/ (5000+75.4)}
x100%=98.5%
b) 50% rated load is ;
I primary = (5kVA/240) x 50% = 10.42 A; The copper losses at high‐voltage winding =
(10.42)2 (0.05) =5.429 W
I secondary = (5kVA/120V) x 50% = 20.83A; The copper losses at high‐voltage winding =
(20.83)2 (0.0125) = 5.424W
The total losses = 32 W + 5.429W + 5.424 W = 42.853 W
Po = 5kVA x 0.175 = 875kW
Efficiency = {Output power / (Output power +total losses)} x 100% = {875/ (875+42.853)}
x100%=95.3%
6. A 25‐kVA, 2400/240‐V, 50‐Hz, step‐down transformer has a core loss of 120W. The effective
resistance of the primary winding is 1.9 Ω. The effective resistance of the secondary winding is
0.02 Ω. Calculate;
a) The full load current rating of the high voltage and the low voltage windings. (Neglect
any losses and assume that the input and the output are the same).
b) The total copper losses of the transformer at the rated load.
c) The efficiency of transformer at the rated load and a unity power factor.
Solution;
a) The full load current rating of the high voltage = 25kVA/2400V = 10.417 A ; The copper
losses at high‐voltage winding =(10.417)2 (1.9)= 206.176W
The full load current rating of the low voltage windings = 25kVA/240V =104.167 A ; The
copper losses at high‐voltage winding = (104.167)2 (0.02)=217.015 W
b) The total copper losses of the transformer at the rated load= 206.176W + 104.167W =
310.343W
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c) The total losses = 120W + 310.343W = 430.343W
The output power = 25kVA x 1 = 25kW
The efficiency of transformer at the rated load and a unity power factor = {Output
power / (Output power +total losses)} x 100% = {25kW/ (25000+430.343)} x100%=98.3%
7. A 50‐kVA, 4600/230‐V, 50‐Hz, single‐phase, step‐down transformer is designed so that at the
condition of full load and unity power factor, the core loss equal the copper losses. The copper
losses are equally divided between the two windings. At full load the efficiency is 96.5%.
Calculate;
a) The core loss
b) The total copper losses
c) The rated current of the primary winding and secondary windings. (Neglect the losses
and assume that the input and the output are the same).
d) The effective resistance of the primary and the secondary windings.
Solution;
a) The efficiency = Output power / (Output power +total losses)} x 100% = 96.5%
The output power = 50kVA x 1 = 50kW
Hence, 96.5% ={Output power / (Output power +total losses)} x 100% = {50kW/
(50000+total losses)} x100%=98.5%
And, (50,000 +total losses) = 50,000 / 0.965 = 51,813.5 W
The total losses = 51,813.5 ‐ 50,000
If, the core loss = the total copper losses
And, the total losses = the core loss + the total copper loss = 2 x the core loss
So, the core loss = (1/2) x 1,813.5 W = 906.75W
b) The total copper losses = 2 the primary copper loss = 2 the secondary copper loss = the
primary copper loss + the secondary copper loss
The total copper loss = 906.75W
The primary copper loss = (1/2) x 906.75W = 453.375W
The secondary copper loss = (1/2) x 906.75W = 453.375W
c) The full load current rating of the high voltage = 50kVA/4600V = 10.47 A
The full load current rating of the low voltage windings = 50kVA/230V =217.4 A
d) The primary copper loss = 437.5W = (10.47)2 (Rh)
So, the effective resistance of the primary windings = 437.5 / (10.47)2 = 3.99Ω
The secondary copper loss = 437.5W= (217.4)2 (Rl) =217W
So, the effective resistance of the secondary windings = 437.5 / (217.4)2 = 0.00926Ω
8. The following data were obtained during the short circuit test for the 20‐kVA, 4800/240‐V, 50‐
Hz:
Voltmeter reading = 160 V
Exciting current = 4.2 V
Wattmeter reading = 280 W
Another test shows the core loss to be 120W.
Calculate:
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a) The efficiency of the transformer at the rated load and unity power factor.
b) The efficiency of the transformer at 50% of the rated load and 0.8 lagging power factor.
Solution;
a) at the rated load and unity power factor
The core loss = 120W
Rop = Psc/Isc2 = 280/(4.2)2 = 15.873 Ω
Ros = Rop x (Ns/Np)2 = 15.873 x ( 240/4800)2 = 0.0397Ω
The primary rated current = 20kVA / 4800V = 4.167A
The primary copper losses = (4.167A)2 x 15.873 Ω =275.617 W
The secondary rated current = 20kVA / 240V = 83.3A
The secondary copper losses = (83.3A)2 x 0.0397 Ω =275.474 W
The total losses = 120W + 275.617W + 275.474W = 671.091W
The output transformer is =20kVA x 1 = 20 kW
The efficiency of the transformer = {20,000/20000+671.091)} x 100% = 96.75%
b) at 50% of the rated load and 0.8 lagging power factor
The primary rated current = (20kVA / 4800V ) x 50% = 2.08A
The primary copper losses = (2.08A)2 x 15.873 Ω =68.67 W
The secondary rated current = (20kVA / 240V) x 50% = 41.65A
The secondary copper losses = (41.65A)2 x 0.0397 Ω =68.87 W
The total losses = 120W + 68.67W + 68.87W = 257.54W
The output transformer is =20kVA x 0.8 = 16 kW
The efficiency of the transformer = {16000 /(16000+257.54)} x 100% = 98.4%
9. A 30‐kVA, 2400/240‐V, 50‐Hz transformer is used as step‐down transformer. The core loss is 150
W, the rated primary current is 12.5 A, the resistance of the primary winding is 1.5 Ω, the rated
secondary current is 125 A, and the resistance of the secondary winding is 0.015 Ω. At the rated
kVA capacity and a 0.75 lagging power factor, determine
a) The total copper losses
b) The efficiency of the transformer.
Solution;
a) The primary copper losses = (12.5A)2 x (1.5 Ω) = 234.375W
The secondary copper losses = (125A) 2 x (0.015 Ω) = 234.375W
The total copper losses = 234.375W + 234.375W = 468.75 W
b) The total loss = 150W + 468.75W = 618.75W
The output transformer = 30kVA x 0.75 = 22.5kW
The efficiency of the transformer= {22500/(22500+618.75)} x 100% = 97.3%
10. A 20‐kVA standard distribution transformer is rated at 2400/4800 V on the high‐voltage side,
and 120/240 V on the low‐voltage side. The transformer has an efficiency of 97% when it’s
operating at the rated load as a step‐down transformer supplying a non‐inductive lighting load.
a) Draw a diagram to show the internal and external transformer connections required to
step‐down the voltage from 2400 V to a 120/240‐V, single phase three‐wire service.
Show the polarity markings for the transformer terminals for additive polarity.
b) What is the full‐load secondary current if the load is balanced?
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c) What is the power input to the primary side at the rated load?
d) If the no‐load voltage of the 240‐V secondary is 4 V more than the full‐load voltage,
what is the percentage voltage regulation?
Solution;
a) Draw a diagram to show the internal and external transformer connections required to
step‐down the voltage from 2400 V to a 120/240‐V, single phase three‐wire service.
Show the polarity markings for the transformer terminals for additive polarity.
b) The full‐load secondary current = 20kVA / 120V = 166.67a if the load is balanced.
c) The power input to the primary side at the rated load = 20kVA
d) If the no‐load voltage of the 240‐V secondary is 4 V more than the full‐load voltage,
what is the percentage voltage regulation?
VS ,nl − VS , fl
VR =
X 100% ; VR = {(244‐240)/240} x 100% = 1.67%
VS , fl
11. List the data commonly found on a transformer nameplate.
Solution;
Manufacturer’s Name
Volt‐Amperage:
HV winding:
Frequency:
Additive polarity:
Serial no.:
Type no.:
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Single‐phase
LV winding:
Percent impedance:
Temperature rise:
Continuous duty
Gallons of insulating oil:
Model no.:
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DET204 Electrical Machine & Application 1
12. A 50‐kVA single phase transformer has a secondary voltage of 240 V. The nameplates indicate
that the transformer has an impedance of 2.35%. What is the short circuit current for this
transformer?
Solution;
Irated = 50kVA/ 240V = 208.3A
Isc = Irated / %Z= 208.3A / 0.0235 = 8,866A
13. A 20kVA, 50Hz, 2000/200V distribution transformer have data as follows :
OC test with HV open circuited; 200V; 4A; 120W
SC test with LV short circuited; 60V; 10A; 300W
a) With full load on the LV side at rated voltage, calculate the excitation voltage and the
voltage regulation on the HV side. The load power factor is;
i. 0.8 lagging
ii. 0.8 leading
b) The transformer supplies full load current at 0.8 lagging power factor with 2000V on the HV
side. Calculate the voltage at the load terminals and the operating efficiency.
Solution;
a)
3Ω
VH
‐4
5.2Ω
‐4
(0.03x10 )
(0.04x10 )
VL’
Referred to HV
20 × 1000VA
= 100 A
200V
20 × 1000VA
VL ' = 2000V ; I L ' =
= 10 A
2000V
VH = VL '+ I L ' (RH cos φ ± X H sin φ )
RH = 3Ω
; X H = 5.2Ω
VL = 200V
i)
; IL =
cos φ = 0.8 lagging, sin φ = 0.6
VH = 2000 + 10(3 × 0.8 + 5.2 × 0.6) = 2055.2V
2055.2 − 2000
Voltage regulation =
× 100% = 2.76%
2000
ii)
cos φ = 0.8 leading, sin φ = 0.6
VH = 2000 + 10(3 × 0.8 − 5.2 × 0.6) = 1992.8V
1992.8 − 2000
× 100% = −0.36%
Voltage regulation =
2000
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b) I L (FL ) = 100 A ; 0.8 lagging pf
VL ' = VH '− I L ' (RH cos φ ± X H sin φ )
VL ' = 2000 − 10(3 × 0.8 + 5.2 × 0.6) = 1944.8V
VL = 194.48V
Output, Po = VL I L cosφ = 194.48 x100 x0.8 = 15558.4 W
Loss, PLoss = Pi + Pc = 120W + {(10)2 x 3} 420 W
⎡
420
⎤
η = ⎢1 −
× 100%⎥ = 97.38%
⎣ 15558.4 + 420
⎦
14. A transformer on no‐load has a core loss of 50W flows a current of 2A (rms) and has an induced
emf of 230V (rms). Neglect winding resistance and leakage flux.
Calculate:
a) The no‐load power factor
b) Core loss current
c) Magnetizing current
Solution
a) Power factor, cos θ o =
50
=0.108 lagging
2 × 230
b) Core loss current, Ii = Io cos θ o
= 2 x 0.108 = 0.216 A
c) Magnetizing current, Im = Io sin θ o
= 2 x sin (cos‐10.108) = 1.988A
15. A 25kVA transformer has 500 turns on the primary and 59 turns on the secondary winding. The
primary is connected to 3000V, 50Hz supply.
Calculate:
a) The current ratings for the primary and secondary windings
b) The voltage induced in the secondary winding
c) The maximum flux in the core.
Solution;
a). a =
N2
50
1
=
=
N1 500 10
P = I1 V1
25kVA = I1 (3000V)
25kVA
=8.33A
3000V
I
I
8.33
a = 1 ; I2 = 1 =
= 83.3 A
1
I2
a
10
I1 =
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b) a =
V2
1
; V2 = a × V1 =
× 3000 = 300V
V1
10
c) E1 = 4.44 fφ m N1.
3000 = 4.44 × 50 × 500 × φ m
φ m = 27mWb
16. A single‐phase 50 Hz transformer is required to step‐down from 2200V to 250V. The maximum
flux in the core is 0.0216Wb.
Calculate:
a) The number of turns of the primary and secondary winding.
b) The turn ratio
Solution;
a) E1 = 4.44 fφ m N1.
2200 = 4.44 × 50 × 0.0216 × N1
N1 = 459turns
E2 = 4.44 fφ m N 2.
250 = 4.44 × 50 × 0.0216 × N 2
N 2 = 52turns
b) a =
V2
250
=
V1 2200
17. An ideal transformer is rated at 2400/120V, 9.6kVA and has 50 turns on the secondary side.
Calculate:
a) Turns ratio
b) The number of turns on the secondary side
c) The current ratings for the primary and secondary windings.
Solution;
a) a =
V2
120
1
=
=
V1 2400 20
b) N1 =
N2
= 50 × 20 = 1000 turns
a
c) S = V1.I 2 = 9.6k .VA
So, I1 =
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S 9,600
=
= 4A
V1 2,400
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DET204 Electrical Machine & Application 1
I2 =
S 9600
=
= 80 A
120
V2
18. A single‐phase transformer has 400 primary and 1000 secondary turns. If the primary is
connected to a 520V, 50Hz.
Calculate:
a) turns ratio
b) the voltage included in the secondary winding
Solution;
a) a =
V2 N 2 1000
=
=
= 2.5
V1 N1
400
b) V2 = aV1 = 2.5 × 520 = 1300 Volt
19. A 20kVA, 50Hz, 2000/200V distribution transformer has a leakage impedance of 0.42 + j 0.52Ω in
high voltage (HV) side winding and 0.004 + j 0.05Ω in low voltage (LV) side winding, when seen
from the LV side, the shunt branch admittance yo is 0.002 –j 0.015 mho (at rated voltage and
frequency)
Draws the equivalent circuit referred to:
a) HV side
b) LV side
indicating all impedance on the circuit.
Solution;
0.42 + j 0.52Ω
0.004 + j 0.05Ω
10:1
1
2
0.002 –j 0.015 mho
1’
2’
(HV) side
(LV) side
Transformation ratio; 1 = N1 = 2000 = 10
a N2
200
a) Z 2 ' = (10) (0.004 + j 0.005) = 0.4 + j 0.5Ω
2
yo ' =
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1
(10)2
(0.002 − j 0.015) = (0.002 − j 0.015) × 10 −2 Ω
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DET204 Electrical Machine & Application 1
0.42 + j 0.52Ω
0. 4 + j 0.5Ω
10:1
1
2
‐2
(0.002 –j 0.015) x10
h
1’
2’
(HV) side
b) Z1 ' =
1
(10)2
(LV) side
(0.042 − j 0.052) = (0.0042 − j 0.0052) ohm
0.0042 + j 0.0052Ω
1
0.004 + j 0.005Ω
10:1
2
(0.002 –j 0.015)mho
1’
(HV) side
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2’
(LV) side
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