# 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

Explanation: Real power = |Vs||Vr|(sinδ)/X

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

Explanation: Real power = |Vs||Vr|(sinδ)/X

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

Explanation: Increasing the voltage level is not always economical and reducing the voltages at stations is easier.

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

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

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°

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

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.

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

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

Explanation: The positive VARs are to be fed to the line at the load end.

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

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.

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