Power Systems Questions and Answers – Line Compensation – 2

This set of Power Systems Questions and Answers for Entrance exams focuses on “Line Compensation – 2”.

1. If a line is considered with negligible power losses, then the real power transmitted will be proportional to voltage drop and the reactive power drop is proportional to cosδ.
a) True
b) False
View Answer

Answer: b
Explanation: Real power = |Vs||Vr|(sinδ)/X
Real power is |Vs||Vr|(sinδ)/X in given figure

2. If a line is considered with negligible power losses, then the real power transmitted will be proportional to ______ and the reactive power drop is proportional to ____
a) sinδ, voltage drop across line
b) cosδ, voltage drop across line
c) δ, voltage drop across line
d) voltage drop across line, sinδ
View Answer

Answer: a
Explanation: Real power = |Vs||Vr|(sinδ)/X
Real power is |Vs||Vr|(sinδ)/X in given figure

3. Choose the most feasible method for raising the power to be delivered at the reactive end.
(i)Reducing the line reactance
(ii)Raising the voltage level
a) (i)
b) (ii)
c) (i), (ii)
d) None of the methods
View Answer

Answer: a
Explanation: Increasing the voltage level is not always economical and reducing the voltages at stations is easier.
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4. The lagging reactive power delivered by a line is proportional to the line _____ and independent of _____
a) voltage,δ
b) δ, voltage
c) voltage, sinδ
d) voltage drop, cos δ
View Answer

Answer: a
Explanation: The lagging reactive power delivered by a line is proportional to the line voltage and independent of δ.

5. Assume a 275 kV transmission line having the following line constants:A = 0.8∠5° ; B = 200∠75°
For obtaining 275 kV voltage at the receiving end obtaining unity power factor, the power transmitted will be _____________
a) 124 MW
b) 117.2 MW
c) 116 MW
d) 110 MW
View Answer

Answer: a
Explanation: Qr = 0
0=(275*275)/200 sin⁡(75-δ)-0.8/200 sin⁡(75-5)
0=378sin⁡(75-δ)-284.26
δ=26.23°
Pr = (275*275)/200 cos⁡(75-δ)-0.8/200 cos⁡(75-5)
= 124 MW.

6. Assume a 275 kV transmission line having the following line constants:A = 0.8∠5° ; B = 200∠75°
For obtaining 275 kV voltage at the receiving end obtaining unity power factor, the power angle required to be set as __________
a) 26.23°
b) 22°
c) 25°
d) 24°
View Answer

Answer: a
Explanation: Qr = 0
0=(275*275)/200 sin⁡(75-δ)-0.8/200 sin⁡(75-5)
0=378sin⁡(75-δ)-284.26
δ=26.23°.

7. Assume a 275 kV transmission line having the following line constants:A = 0.8∠5° ; B = 200∠75°
If a load is connected at receiving end at unity power factor but maintaining the same voltage profile. Will this system now need any type of compensation?
a) Yes
b) No
c) Cannot be said
d) Never
View Answer

Answer: a
Explanation: Yes this system will require a compensation system to be employed as the net reactive power will not be zero due to the load mentioned.
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8. Assume a 275 kV transmission line having the following line constants:A = 0.8∠5° ; B = 200∠75°
If a load is connected at receiving end at unity power factor but maintaining the same voltage profile. Will this system now need any type of compensation? If so, then which one?
a) Yes, capacitive
b) No
c) Yes, inductive
d) Yes, inductive and capacitive
View Answer

Answer: a
Explanation: yes, it will need compensation. As ∆Q≠0 and it will be less than zero.

9. Assume a 275 kV transmission line having the following line constants:A = 0.8∠5° ; B = 200∠75°
If a load is connected at receiving end at unity power factor but maintaining the same voltage profile. Then _________ compensation will be needed at ______
a) capacitive, receiving end
b) inductive, sending end
c) capacitive, sending end
d) inductive, receiving end
View Answer

Answer: a
Explanation: The positive VARs are to be fed to the line at the load end.
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10. Assume a 275 kV transmission line having the following line constants:A = 0.85∠5° ; B = 200∠75°
If a load is connected at receiving end at unity power factor but maintaining the same voltage profile. The receiving end voltage if the compensation equipment is not installed will be __________
a) 245 kV
b) 280 kV
c) 255 kV
d) 272 kV
View Answer

Answer: a
Explanation: Pr = 150 MW, Qr = 0
|Vs| = 275 kV, |Vr| = ?
150 = (275*|Vr|)/200 cos⁡(75-δ)-0.8/200 cos⁡(75-5)
0=(275*|Vr|)/200 sin⁡(75-δ)-0.8/200 sin⁡(75-5)
Solving the quadratic equations above and retaining the higher value for the voltage, |Vr| = 245 kV.

Sanfoundry Global Education & Learning Series – Power Systems.

To practice all areas of Power Systems for Entrance exams, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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