Power Systems Questions and Answers – Symmetrical Component Transformation – 1

This set of Power Systems Multiple Choice Questions & Answers (MCQs) focuses on “Symmetrical Component Transformation – 1”.

1. The receiving end active power for a short transmission line is (where the angles have their usual meanings)
a) \(\frac{|V_s||V_r|}{|Z|} \cos(\theta – \delta) – \frac{|V_r|^2}{|Z|} \cos(\theta)\)
b) \(\frac{|V_s||V_r|}{|Z|} \sin(\theta – \delta) – \frac{|V_r|^2}{|Z|} \cos(\theta)\)
c) \(\frac{|V_s||V_r|}{|Z|} \sin(\theta – \delta) – \frac{|V_r|^2}{|Z|} \sin(\theta)\)
d) \(\frac{|V_s||V_r|}{|Z|} \cos(\theta + \delta) – \frac{|V_r|^2}{|Z|} \cos(\theta)\)
View Answer

Answer: a
Explanation: Receiving end active power for a short transmission line is \(\frac{|V_s||V_r|}{|Z|} \cos(\theta – \delta) – \frac{|V_r|^2}{|Z|} \cos(\theta)\).

2. The receiving end reactive power for a short transmission line is (where the angles have their usual meanings)
a) \(\frac{|V_s||V_r|}{|Z|} \cos(\theta – \delta) – \frac{|V_r|^2}{|Z|} \cos(\theta)\)
b) \(\frac{|V_s||V_r|}{|Z|} \sin(\theta – \delta) – \frac{|V_r|^2}{|Z|} \cos(\theta)\)
c) \(\frac{|V_s||V_r|}{|Z|} \sin(\theta – \delta) – \frac{|V_s|^2}{|Z|} \sin(\theta)\)
d) \(\frac{|V_s||V_r|}{|Z|} \cos(\theta + \delta) – \frac{|V_r|^2}{|Z|} \cos(\theta)\)
View Answer

Answer: c
Explanation: Receiving end reactive power for a short transmission line is
\(\frac{|V_s||V_r|}{|Z|} \sin(\theta – \delta) – \frac{|V_s|^2}{|Z|} \sin(\theta)\)

3. The sending end active power for a 20 km transmission line with Vs as the sending end voltage and Vr as receiving end voltage, can be given by most appropriately
a) \(\frac{|V_s|^2}{|Z|} \cos(\theta) – \frac{|V_s||V_r|}{|Z|} \cos(\theta + \delta)\)
b) \(\frac{|V_r|^2}{|Z|} \cos(\theta) – \frac{|V_s||V_r|}{|Z|} \cos(\theta + \delta)\)
c) \(\frac{|V_s|^2}{|Z|} \cos(\theta) – \frac{|V_s||V_r|}{|Z|} \cos(\theta – \delta)\)
d) \(\frac{|V_s|^2}{|Z|} \sin(\theta) – \frac{|V_s||V_r|}{|Z|} \sin(\theta + \delta)\)
View Answer

Answer: a
Explanation: The sending end active power is \(\frac{|V_s|^2}{|Z|} \cos(\theta) – \frac{|V_s||V_r|}{|Z|} \cos(\theta + \delta)\).
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4. The sending end reactive power for a 20 km transmission line with Vs as the sending end voltage and Vr as receiving end voltage, can be given by most appropriately
a) \(\frac{|V_s|^2}{|Z|} \cos(\theta) – \frac{|V_s||V_r|}{|Z|} \cos(\theta + \delta)\)
b) \(\frac{|V_r|^2}{|Z|} \cos(\theta) – \frac{|V_s||V_r|}{|Z|} \cos(\theta + \delta)\)
c) \(\frac{|V_s|^2}{|Z|} \cos(\theta) – \frac{|V_s||V_r|}{|Z|} \cos(\theta – \delta)\)
d) \(\frac{|V_s|^2}{|Z|} \sin(\theta) – \frac{|V_s||V_r|}{|Z|} \sin(\theta + \delta)\)
View Answer

Answer: d
Explanation: The sending end reactive power is \(\frac{|V_s|^2}{|Z|} \sin(\theta) – \frac{|V_s||V_r|}{|Z|} \sin(\theta + \delta)\).

5. The simplified ABCD representation of a 40 km transmission line is best given as
a) \(\begin{matrix}
1 & z \\
0 & 1
\end{matrix}\)
b) \(\begin{matrix}
1 & 0 \\
z & 1
\end{matrix}\)
c) \(\begin{matrix}
z & z \\
0 & 1
\end{matrix}\)
d) \(\begin{matrix}
1 & z \\
0 & z
\end{matrix}\)
View Answer

Answer: a
Explanation: From the given length, it is a short TL. Hence,
From the given length, it is a short TL
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6. For a 35 km transmission line having a lumped impedance of the line as 20 ohms, is required to be shown in the ABCD form, it is given as
a) \(\begin{matrix}
1 & 20 \\
0 & 1
\end{matrix}\)
b) \(\begin{matrix}
1 & 0 \\
2 & 1
\end{matrix}\)
c) \(\begin{matrix}
1 & 0 \\
20 & 1
\end{matrix}\)
d) \(\begin{matrix}
20 & 0 \\
1 & 1
\end{matrix}\)
View Answer

Answer: a
Explanation: From the given length, it is a short TL. Hence,
From the given length, it is a short TL in given figure

7. If it is tried to represent the active and reactive power on a circle, then the radius would be
a) \(\frac{|V_s||V_r|}{|B|}\)
b) \(\frac{|V_s||V_r|}{|Z|} \sin \alpha\)
c) \(\frac{|V_s|^2}{|B|}\)
d) None of the mentioned
View Answer

Answer: a
Explanation:
The radius would be A in given figure

8. The maximum power delivered to the load for short transmission line is at
a) β=α
b) β>α
c) β=δ
d) β>δ
View Answer

Answer: c
Explanation: Maximum power occurs for β=δ.

9. The maximum real active power delivered to the load is defined most accurately by
a) \(\frac{|V_s||V_r|}{|B|} – \frac{|V_r|^2 |D|}{|B|} \cos(\beta – \alpha)\)
b) \(\frac{|V_s||V_r|}{|B|} – \frac{|V_s|^2 |D|}{|B|} \cos(\beta – \alpha)\)
c) \(\frac{|V_s||V_r|}{|B|} – \frac{|V_r|^2 |D|}{|B|} \cos(\beta + \alpha)\)
d) \(\frac{|V_s||V_r|}{|D|} – \frac{|V_r|^2 |B|}{|D|} \cos(\beta – \alpha)\)
View Answer

Answer: a
Explanation: The maximum real active power delivered to the load is \(\frac{|V_s||V_r|}{|B|} – \frac{|V_r|^2 |D|}{|B|} \cos(\beta – \alpha)\).
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10. For a given power system, its zero and maximum regulation will occur at the impedance angle of
a) 45
b) 90
c) 0
d) 60
View Answer

Answer: a
Explanation: At θ=45°, ZVR and maximum VR coincide.

11. The charging currents due to shunt admittance can be neglected for ______ transmission line?
a) short
b) long
c) medium
d) all of the mentioned
View Answer

Answer: a
Explanation: The shun admittance for lines more than 100 km become very prominent and can not be neglected.

12. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify the missing term marked as ’?’.
Vs = ?*Vr+B*Ir
Is = C*Vr+D*Ir
1+YZ equations are given by the below set of equations based on the line diagram
a) 1+YZ
b) Z
c) Y
d) 1
View Answer

Answer: a
Explanation: Using KVL to the line diagram,
Vs = (1+YZ)*Vr+Z*Ir
Is = Y*Vr+Ir.

13. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify the missing term marked as ’?’.
Vs = A*Vr+?*Ir
Is = C*Vr+D*Ir
1+YZ equations are given by the below set of equations based on the line diagram
a) 1+YZ
b) Z
c) Y
d) 1
View Answer

Answer: b
Explanation: Using KVL to the line diagram,
Vs = (1+YZ)*Vr+Z*Ir
Is = Y*Vr+Ir.

14. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify the missing term marked as ’?’.
Vs = A*Vr+B*Ir
Is = ?*Vr+D*Ir
1+YZ equations are given by the below set of equations based on the line diagram
a) 1+YZ
b) Z
c) Y
d) 1
View Answer

Answer: c
Explanation: Using KVL to the line diagram,
Vs = (1+YZ)*Vr+Z*Ir
Is = Y*Vr+Ir.

15. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify the missing term marked as ’?’.
Vs = A*Vr+B*Ir
Is = C*Vr+?*Ir
1+YZ equations are given by the below set of equations based on the line diagram
a) 1+YZ
b) Z
c) Y
d) 1
View Answer

Answer: d
Explanation: Using KVL to the line diagram,
Vs = (1+YZ)*Vr+Z*Ir
Is = Y*Vr+Ir.

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.

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Manish Bhojasia - Founder & CTO at Sanfoundry
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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