This set of Power Systems Multiple Choice Questions & Answers (MCQs) focuses on “Switchgear and Protection – 4”.
1. An isolated generator connected to a turbine with its continuous maximum power of 20 MW, 50 Hz. Generator connected with two loads of 8 MW, each operate at 50 Hz.. Generator has 5% droop characteristic. If an additional load of 6 MW is added then the change in the frequency will be ____________
a) 0.25
b) 0.75
c) 0.5
d) 1
View Answer
Explanation:
2. For the given system the Ybus matrix is show below.
a) \(j\begin{bmatrix}
1.8 & 0.2 & 0 \\
0.2 & 1.8 & -2 \\
0 & -2 & 1.8
\end{bmatrix}\)
b) \(j\begin{bmatrix}
1.8 & -2 & 0 \\
-2 & 1.8 & 0.2 \\
0 & 0.2 & 1.8
\end{bmatrix}\)
c) \(j\begin{bmatrix}
4.5 & 5 & 0 \\
5 & 7 & 2 \\
0 & 2 & 7
\end{bmatrix}\)
d) None of the mentioned
View Answer
Explanation: Y11=j2+(1/5j) = 1.8j
Y12=-(1/5j) = 0.2j; Y13=0
Hence option ‘a’ is most applicable.
3. If short TL is delivering a lagging power factor load, sending end p.f. would be ________
a) (VrcosФr+IR)/Vs
b) (VrcosФr+IRsinФr)/Vs
c) (VrsinФr+IR)/Vs
d) (VssinФr+IRcosФr)/Vs
View Answer
Explanation: The phasor diagram is given below
4. If all the sequence voltages at the fault point in a power system are equal, then fault is __________
a) LLG fault
b) LG fault
c) Three phase to ground fault
d) Line to Line fault
View Answer
Explanation: It is LLG fault.
5. The below mentioned graph is a recorded data of the voltage fluctuation of the network, then it can be concluded that these faults are __________
a) L-G
b) A three phase motor short
c) Feeder fault
d) Any of the mentioned
View Answer
Explanation: The sudden dip is the LG fault in the graph as the dips are not very large and the system is maintaining its stability.
6. The inertia constant of two groups of machines which do not swing together are 4 and 20 MJ/MVA, the equivalent inertia constant of the system is _________
a) 2.85
b) 14
c) 6
d) 12
View Answer
Explanation: For the incoherent system, the equivalent inertia = H1H2/(H1+H2).
7. A fault current of 1500A passing through the primary side of a 400/5 CT. On the secondary side an inverse time over current relay is connected whose current setting is 75%, the plug setting is __________
a) 5
b) 3.75
c) 6
d) 4
View Answer
Explanation: \(\text{PSM} = \frac{I_f}{C_S \times C_T \text{ ratio} \times \text{rated secondary current}} = \frac{1500}{0.75 \times \left( \frac{400}{5} \right)} = 5\).
8. The three phase line is working at 50 Hz with conductors rearranged as shown. The Cph of TL at 132 kV is ___________
a) 8.75*10-9
b) 8.75*10-12
c) 26.2*10-12
d) 26.2*10-9
View Answer
Explanation: \(C_{ph} = \frac{0.242}{\log \left( \frac{gmd}{gmr} \right)} = \frac{0.242 \times 10^{-12}}{\log \left( \frac{\sqrt[3]{2 \times 2 \times 3}}{0.4 \times 10^{-12}} \right)} = 8.74 \times 10^{-9} \, \text{F/m}\).
9. For a 100 MVA, 11 kV three phase Alternator observes a three phase fault at terminals of it. If the fault current is 2000 A, the pu value of positive sequence reactance ______
a) 2.62
b) 0.46
c) 1.31
d) 0.92
View Answer
Explanation: PU fault current = 2000/5248.63 = 0.381 pu
0.381 = 1/X
X= 1/0.381 =2.62 pu.
10. A lossless TL having SIL of 3000 MW is provided with uniform distributed series capacitance compensation of 25% and shunt compensation of 30% then the SIL of compensated TL will be _______________
a) 3949.9
b) 3289
c) 2784.9
d) 2877.6
View Answer
Explanation: SIL=(V2)/Z
SIL α 1/Z
\(Z_{\text{comp}} = Z_s \sqrt{\frac{1 – k_{se}}{(1 + k_{sh})(1 – k_{lsh})}}\)
Kse = 0.25; Klsh = 0; Kcsh = 0.3,
Zcomp =0.7595
SILcomp = 0.7595*3000 = 3949.96 MVA.
11. A lossless TL having SIL of 3000 MW is provided with uniform distributed series capacitance compensation of 25% and shunt compensation of 30% then the compensated impedance of TL will be(in pu) ____________
a) 0.7595
b) 2.88
c) 0.647
d) 3.27
View Answer
Explanation: SIL=(V2)/Z
SIL α 1/Z
\(Z_{\text{comp}} = Z_s \sqrt{\frac{1 – k_{se}}{(1 + k_{sh})(1 – k_{lsh})}}\)
Kse = 0.25; Klsh = 0; Kcsh = 0.3,
Zcomp =0.7595.
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