This set of Power Systems Multiple Choice Questions & Answers (MCQs) focuses on “Per Unit (PU) System”.

1. A power system network is connected as shown in the figure.

Sd1=15+j5 pu

Sd2=25+j15 pu

Zcable = j0.05pu

|V1|=|V2|=1 pu.

The torue angle for the system will be __________

a) 14.4

b) 22.1

c) 16.2

d) 18.2

View Answer

Explanation: As the resistance is zero, losses will be zero.

PG1=PD1+PD2=40 pu

For the equal sharing of load at the station

PG1=PG2=20pu

Real power flow from bus 1 to 2

2. A single phase distributor of 1 km long has resistance and reactance per conductor of 0.1Ω and 0.15 Ωrespectively. If the far end voltage Vb=200V and current is at 100A at 0.8 lag. At the midpoint a current of 100A is tapped at a pf of 0.6 pf with ref to voltage Vm at mid point. The voltage magnitude at M is ________

a) 218V

b) 200V

c) 232V

d) 220V

View Answer

Explanation: Drop in MB= (100ʟ-36.67)(0.1+j0.15)= 18.027ʟ19.44

V

_{M}= 200+18.027 = 218.027 V.

3. A single phase motor is connected to 400V, 50Hz supply. The motor draws a current of 31.7A at a power factor 0.7 lag. The capacitance required in parallel with motor to raise the power factor of 0.9 lag (in micro farads) is __________

a) 94.62

b) 282.81

c) 108.24

d) 46.87

View Answer

Explanation: Active power drawn by the motor=VIcosФ = 400*31.7*0.7 = 8876 W

Reactive power = VIsinФ=400*31.7*sin(45.57.29) = 9055.3 VAR

New power factor=cosθ2 = 0.9

θ

_{2}=cos

^{-1}(0.9)

Q2=8876*tan(25.84) = 4298.855 VAR

Change in reactive power=9055.3-4298.855 = 4756.4 VAR

Qc = V

^{2}/Xc = V

^{2}*2πfC

C=4756.4/(4002*2π*50) = 94.62μF.

4. A single phase motor is connected to 400V, 50Hz supply. The motor draws a current of 31.7A at a power factor 0.7 lag. The additional reactive power (in VAR) to be supplied by the capacitor bank will be ___________

a) 4756

b) 4873

c) 4299

d) 9055.3

View Answer

Explanation: Active power drawn by the motor=VIcosФ = 400*31.7*0.7 = 8876 W

Reactive power = VIsinФ=400*31.7*sin(45.57.29) = 9055.3 VAR

New power factor=cosθ2 = 0.9

θ2=cos-1(0.9)

Q2=8876*tan(25.84) = 4298.855 VAR

Change in reactive power=9055.3-4298.855 = 4756.4 VAR.

5. A 275 kV TL has following line constants A=0.85ʟ5o, B=200ʟ75o. The active power received if the voltage to be maintained is 275kV will be __________

a) 117.63

b) 220

c) 120

d) 115.25

View Answer

6. A 275 kV TL has following line constants A=0.85ʟ5o, B=200ʟ75o. The active power angle such that the voltage to be maintained at the other end will be 275 kV ____________

a) 22

b) 16

c) 18

d) 24

View Answer

7. A power system has a maximum load of 15 MW. Annual load factor is 50%. The reserve capacity of plant is _____ if the plant capacity factor is 40%.

a) 3.75 MW

b) 4.75 MW

c) 18.75 MW

d) 5.75 MW

View Answer

Explanation: LF = (Average Demand)/(Maximum Demand)=0.5

Plant capacity factor =(Average Load)/(Plant Capacity)= 0.5/0.4

Plant capacity= (0.5/0.4)*15 = 18.75 MW

Reserve Capacity = 18.75-15 = 3.75 MW.

8. A 100 MVA synchronous generator operates on full load at a frequency of 50 Hz. The load is suddenly reduced to 50 MW. Due to time lag in governor system, the steam valve begins to close after 0.4s. The change in the frequency is ________(H=5 kW-s/KVA).

a) 1

b) 0.5

c) -1.5

d) 0.8

View Answer

9. A 50 Hz four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8 MJ/MVA. If the mechanical input is suddenly raised to 80 MW for an electrical load of 50MW, then the rotor acceleration is ____________

a) 337.5

b) 3.375

c) 457.5

d) 4.57

View Answer

Explanation: Energy stored = 100*8 =800MJ

Accelerating power = Md

^{2}δ/dt

^{2}

M=GH/180f = 800/(180*50) = 4/45 MJs/elect. Deg

10. A single phase TL has copper conductor of 0.775 cm2 cross section through which 200 kW at UPF at 330 V is to be maintained. If the efficiency of transmission line is 90%, then the minimum length of TL is ___________(in km and take specific resistance to be 1.785 μΩ/cm).

a) 13.6 km

b) 14 km

c) 136 km

d) 16.4 km

View Answer

Explanation: Pr=200 kW, efficiency=0.9

Ps= 200/0.9=222.22 kW

Losses=22.22 kW

Current, I=200000/3300 = 60.60 A

Line losses=2I

^{2}R (for a 2 wire line)

R=22.22/(2*60.602)=3.02 Ω

R=ρl/a

Length, l = (3.025*0.775)/(1.785*10-6) = 13.6 km.

11. A three phase transformer has a nameplate rating of 30 MVA, 230Y/69Y kV with a leakage -reactance of 10% and the transformer connection is wye-wye. Choosing a base of 30MVA and 230 kV on high voltage side, the reactance of transformer in per units is __________

a) 0.1

b) 0.3

c) 0.03

d) 1.5

View Answer

Explanation: The pu value of a transformer does not change.

12. A three phase transformer has a nameplate rating of 30 MVA, 230Y/69Y kV with a leakage -reactance of 10% and the transformer connection is wye-wye. Choosing a base of 30MVA and 230 kV on high voltage side, the high voltage side impedance ____________

a) 1763.3 Ω

b) 158.7 Ω

c) 15.87 Ω

d) 176.3 Ω

View Answer

Explanation: On the high voltage side, Zb=kVb2/MVAb(3-ph) = 2302/30 = 1763.33 Ω.

13. A three phase transformer has a nameplate rating of 30 MVA, 230Y/69Y kV with a leakage -reactance of 10% and the transformer connection is wye-wye. Choosing a base of 30MVA and 230 kV on high voltage side, the low voltage side impedance is ___________

a) 158.7 Ω

b) 176.3 Ω

c) 1763.3 Ω

d) 15.87 Ω

View Answer

Explanation: On the low voltage side, Zb=kV

_{b}

^{2}/MVA

_{b(3-ph)}=692/30 = 158.7 Ω.

14. A three phase transformer has a nameplate rating of 30 MVA, 230Y/69Y kV with a leakage -reactance of 10% and the transformer connection is wye-wye. Choosing a base of 30MVA and 230 kV on high voltage side, the transformer reactance referred to the high voltage side will be _________(in ohms).

a) 176.33 Ω

b) 17.67 Ω

c) 158.7 Ω

d) 15.87 Ω

View Answer

Explanation: Zb=kVb

^{2}/MVA

_{b(3-ph)}= 2302/30 = 1763.33 Ω

X Ω = Xpu*X

_{b(HV)}= 0.1*1763.33 = 176.33 Ω.

**Sanfoundry Global Education & Learning Series – Power Systems.**

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