# Power Systems Questions and Answers – Comparison of Faults in Three Phase System – 1

This set of Power Systems Multiple Choice Questions & Answers (MCQs) focuses on “Comparison of Faults in Three Phase System – 1”.

1. A three phase transformer has a name plate details of 30 MVA and voltage rating of 230Y kV/69∆ kV with a leakage reactance of 10% and the transformer connection via wye-delta. Taking a base of 230 kV on the high voltage side, turns ratio of the windings is ______
a) 2
b) 1.5
c) 6
d) 4

Explanation: n= VHV/VLV = (230/√3)/69=1.9245.

2. A three phase transformer has a name plate details of 30 MVA and voltage rating of 230Y kV/69∆ kV with a leakage reactance of 10% and the transformer connection via wye-delta. Taking a base of 230 kV on the high voltage side, the transformer reactance on the LV side is
___________
a) 176.33 Ω
b) 1763.3 Ω
c) 47.6 Ω
d) 15.87 Ω

Explanation: Zb(HV) = kVb(HV))2/MVAb
Zb(HV) = 2302/30=1763.33 Ω
XΩ =0.1*1763.33 = 176.33 Ω.

3. A three phase transformer has a name plate details of 30 MVA and voltage rating of 230Y kV/69∆ kV with a leakage reactance of 10% and the transformer connection via wye-delta. Taking a base of 230 kV on the high voltage side, the transformer reactance referred to the low voltage side in ohms is __________
a) 47.61 Ω
b) 15.87 Ω
c) 176.33 Ω
d) 157.8 Ω

Explanation: Zb(HV) = (kVb(HV)2/MVAb = 2302/30=1763.33 Ω
XΩ(HV) =0.1*1763.33 = 176.33 Ω
XΩ(LV) =176.33/(1.92452) = 47.61 Ω.

4. A three phase transformer has a name plate details of 30 MVA and voltage rating of 230Y kV/69∆ kV with a leakage reactance of 10% and the transformer connection via wye-delta. Taking a base of 230 kV on the high voltage side, the transformer reactance referred to the low voltage side in ohms is
a) 0.1
b) 0.2
c) 0.198
d) 0.4

Explanation: XΩ(LV) = XΩ(HV) /n2
= 176.33/(230/69)2 =15.87 Ω
Xpu= 15.87/158.7 = 0.1.

5. A 200 bus power system has 160 PQ bus. For achieving a load flow solution by N-R in polar coordinates, the minimum number of simultaneous equation to be solved is ___________
a) 359
b) 329
c) 360
d) 320

Explanation: Total buses = 200
PQ buses = 160
PV buses = 200-160 = 40
Slack bus = 1
Total number of equation = (40-1)*1 + (160*2) = 359.
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6. Two alternators A and B having 5% speed regulation are working in parallel at a station. Alternator A is rated at 15 MW while B is at 20 MW. When the total load to be shared is 12 MW, then how much of the load will be shared by the alternator B?
a) 6.85 MW
b) 5.14 MW
c) 6 MW
d) 7 MW

Explanation: 1 α15 ;
P2 α 20;

P1 = 0.75 P2 …(1)
P1 +P2 = 12 …(2)
Solving above, P2 = 6.85 MW.

7. Two alternators A and B having 5% speed regulation are working in parallel at a station. Alternator A is rated at 15 MW while B is at 20 MW. When the total load to be shared is 12 MW, then how much of the load will be shared by the alternator B?
a) 6.85 MW
b) 5.14 MW
c) 6 MW
d) 7 MW

Explanation: 1 α15 ;
P2 α 20;

P1 = 0.75 P2 …(1)
P1 +P2 = 12 …(2)
Solving above equations, P1 = 5.14 MW.

8. A 400 V, 50 Hz three phase balanced source ripples to a star connected load whose rating is S(=300+j400) kVA. The rating of the delta connected capacitor bank needed to bring p.f. to 0.9 lagging is _______ KVAR.
a) 254.7
b) 25.4
c) 84.9
d) 284.5

Explanation: Qc = P[tanθ1 – tanθ2] =300[(4/3) – cos-10.9] = 254.72 kVAR.

9. A 400 V, 50 Hz three phase balanced source ripples to a star connected load whose rating is S(=300+j400) kVA. A delta connected capacitor bank needed to bring p.f. to 0.9 lagging. The operating power factor of the system is
a) 0.8
b) 4/3
c) 3/4
d) 0.6

Explanation: cosФ = 3/5 = 0.6.

10. A given system to be analysed was found with the below phasor representation of the system voltages. Which of the symmetrical components will be present in the mentioned system?
a) Positive sequence components
b) Negative sequence components
c) Zero sequence components
d) All of the mentioned

Explanation: In an unbalanced system, all the symmetrical components will be present.

11. The phasor operator which is used to depict the unbalanced phase voltages into three phase quantities, provides a rotation of
a) 120o counter clockwise
b) 120o clockwise
c) 90o counter clockwise
d) 90o clockwise

Explanation: ‘α’ operator used for the conversion has the counter clockwise rotation of the quantity and it has unity magnititude.

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