This set of Power Systems Multiple Choice Questions & Answers (MCQs) focuses on “Inductance of Composite Conductor Lines – 3”.
1. A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. The voltage regulation will be _______
a) 16.8 %
b) 18.8%
c) 21.75%
d) 12.8%
View Answer
Explanation: VR = (11.68-10)*100/10 = 16.8 %.
2. A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. Identify the transmission line and the voltage regulation.
a) STL, 11.68%
b) MTL, 11.68
c) STL, 21.5%
d) MTL, 14.2%
View Answer
Explanation: It is a short transmission line.
Current, I = 5000/(10*0.8)=625 A
Vs = |Vr|+|I|*(RcosФr + XsinФr)
= 10000+625(0.39*0.8+3.96*0.6)
= 11.68kV
VR = (11.68-10)*100/10 = 16.8 %.
3. A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. The half the voltage regulation will be _____________
a) 8.4 %
b) 16.8 %
c) 14.2%
d) 10.5%
View Answer
Explanation: VR = (11.68-10)*100/10 = 16.8 %
Half the VR = 16.8/2 % = 8.4%.
4. A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. The new sending end voltage at the half the voltage regulation is _____________
a) 10.84 kV
b) 11.84 kV
c) 8.84 kV
d) 16.2 kV
View Answer
5. A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. If the voltage regulation is reduced to 50%, then the power factor at this operation mode will be _________________
a) 0.95
b) 0.92
c) 0.74
d) 0.90
View Answer
Explanation: VR = (11.68-10)*100/10
VR = 16.8 %
Half the VR = 16.8/2 %
Half the VR = 8.4%

(10.84-10)*1000 = |I|*( RcosФr + XsinФr ) …(1)
I = 5000/(cosФr*10) …(2)
Solving above eqaution
Фr = 18.04°
Cos Фr = 0.9508, lagging.
6. A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. If the voltage regulation is reduced to 50%, then the power factor angle at this operation mode will be _____________
a) 18.04°
b) 8.04°
c) 21.06°
d) 12°
View Answer
Explanation: VR = (11.68-10)*100/10
VR = 16.8 %
Half the VR = 16.8/2 %
Half the VR = 8.4%

(10.84-10)*1000 = |I|*(RcosФr + XsinФr) …(1)
I = 5000/(cosФr*10) …(2)
Solving above equation
Фr = 18.04°.
7. A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. If the voltage regulation is reduced to 50%, then the power factor at this operation mode will be ____________
a) 0.95, lagging
b) 0.92, leading
c) 0.95, lagging
d) 0.90, leading
View Answer
Explanation: VR = (11.68-10)*100/10
VR = 16.8 %
Half the VR = 16.8/2 %
Half the VR = 8.4%

(10.84-10)*1000 = |I|*(RcosФr + XsinФr) …(1)
I = 5000/(cosФr*10) …(2)
Solving above equation
Фr = 18.04°
Cos Фr = 0.9508, lagging.
8. A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. If the voltage regulation is reduced to 50%, then the power factor angle at this operation mode will be ________
a) 18.04°, lagging
b) 18.04°, leading
c) 21.06°, leading
d) 21.06°, lagging
View Answer
Explanation: VR = (11.68-10)*100/10
VR = 16.8 %
Half the VR = 16.8/2 %
Half the VR = 8.4%

(10.84-10)*1000 = |I|*(RcosФr + XsinФr) …(1)
I = 5000/(cosФr*10) …(2)
Solving above equation
Фr = 18.04°, lagging.
9. A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. If the voltage regulation is reduced to 50%, then the receiving end current at this operation mode will be _____________
a) 526 A
b) 549 A
c) 521 A
d) 580 A
View Answer
Explanation: VR = (11.68-10)*100/10
VR = 16.8 %
Half the VR = 16.8/2 %
Half the VR = 8.4%

(10.84-10)*1000 = |I|*(RcosФr + XsinФr) …(1)
I = 5000/(cosФr*10) …(2)
Solving above equation
Фr = 18.04°, lagging
I = 526 A.
10. Suppose the transmission line is loaded with its surge impedance, the receiving-end voltage is greater than sending end voltage.
a) True
b) False
View Answer
Explanation: It will be equal.
Sanfoundry Global Education & Learning Series – Power Systems.
To practice all areas of Power Systems, here is complete set of 1000+ Multiple Choice Questions and Answers.