Power Systems Questions and Answers – Inductance of Composite Conductor Lines…

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%
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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%
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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%
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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
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Answer: a
Explanation: VR = (11.68-10)*100/10 = 16.8 %
Half the VR = 16.8/2 % = 8.4%

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
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Answer: a
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|*( RcosФ_r + XsinФ_r ) …(1)
I = 5000/(cosФ_r*10) …(2)
Solving above eqaution
Ф_r = 18.04°
Cos Ф_r = 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

Answer: a
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|*(RcosФ_r + XsinФ_r) …(1)
I = 5000/(cosФ_r*10) …(2)
Solving above equation
Ф_r = 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

Answer: a
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|*(RcosФ_r + XsinФ_r) …(1)
I = 5000/(cosФ_r*10) …(2)
Solving above equation
Ф_r = 18.04°
Cos Ф_r = 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

Answer: a
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|*(RcosФ_r + XsinФ_r) …(1)
I = 5000/(cosФ_r*10) …(2)
Solving above equation
Ф_r = 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

Answer: a
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|*(RcosФ_r + XsinФ_r) …(1)
I = 5000/(cosФ_r*10) …(2)
Solving above equation
Ф_r = 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

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.