# MCQs on Power Systems

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01․ The voltage and current instantaneous values are given as 5 sin (ωt + 30°) pu and 2 sin (ωt - 15°) pu respectively. Find the per unit active power?
5 pu
3.535 pu
5.353 pu
7 pu

The voltage and current instantaneous values are given as V = Vm sin (ωt + θ1) pu I = Im sin (ωt + θ2) pu Apparent power in pu (Spu) = (Vm *Im )/2 Active power in pu (Ppu) = (Vm *Im )/2*cos (θ12) Reactive power (Qpu) = (Vm *Im )/2*sin (θ12) From given data, Active power in pu (Ppu) = 5*2/2 * cos (30+15) = 3.535 pu

02․ The angular moment of 10 MVA generator is 0.2. Find the new value of angular moment for 20 MVA base?
0.2 pu
0.1 pu
0.25 pu
0.4 pu

Angular moment (M) in pu = M/Sb Sb = Base MVA Mpun = Mpuo* (MVA)bo/(MVA)bn Here, "n" represents the new values "o" represents the old values Mpun = 0.2*10/20 = 0.1 pu

03․ The per unit impedance Z(pu) is given by
ZΩ* (kVA)b/(KV)²b
ZΩ* (MVA)b/((KV)²b *100)
ZΩ* (MVA)b*100/(KV)²b
ZΩ* (MVA)b/(KV)²b

Zpu = ZΩ* (MVA)b/(KV)²b Where, Zpu = Per unit impedance ZΩ = Impedance in Ω (MVA)b = Base MVA (KV)b = Base voltage

04․ The per unit impedance of a circuit element is 0.30. If the base kV and base MVA are halved, then the new value of the per unit impedance of the circuit element will be
0.30
0.60
0.0030
0.0060

Zpu = ZΩ* (MVA)b/(kV)²b Where, Zpu = Per unit impedance ZΩ = Impedance in Ω (MVA)b = Base MVA (kV)b = Base voltage Zpun = Zpuo * (MVA)bn/(MVA)bo * (kV)²bo/(kV)²bn Here, "n" represents the new values "o" represents the old values Zpun = 0.30* 1/2 *4 = 0.60 pu

05․ The per unit value of a 2 ohm resistor at 100 MVA and 10 kV base voltage is
4 pu
2 pu
0.5 pu
0.2 pu

Zpu = ZΩ* (MVA)b/(KV)²b Where, Zpu = Per unit impedance ZΩ = Impedance in Ω (MVA)b = Base MVA (KV)b = Base voltage Zpu = 2 * 100/10² = 2 pu

06․ For a 500 Hz frequency excitation, a 100 km long power line will be modelled as
shunt capacitor
long transmission line
medium transmission line
any of the above

Classification of transmission lines: Short transmission line = Length(L) < 80 km and product of length(L) and frequency(f) L.f< 4000 Medium transmission line = 80 km ≤ L ≤ 200 Km and 4000 ≤ L.f ≤ 10000 Long transmission line = L > 200 km and L.f >10000

07․ The charging current of a 400 kV line is more than that of a 220 kV line of the same length?
True
False
either 1 or 2
none of the above

Charging current Ic= V/Xc Where, V = Voltage Xc = Charging reactance = 1/(2πfl) l = length of the transmission line Charging current Ic= V*2πfl Ic ∝ V Therefore, the given statement is correct statement.

08․ If in a line, resistance and reactance are found to be equal and regulation is zero, then load will have
unity power factor
zero power factor
0.707 leading power factor
0.707 lagging power factor

Voltage regulation= (Vs - Vr)/Vr = Ir(R cosφ ± X sinφ)/Vr For lagging load: Voltage regulation = Ir(R cosφ + X sinφ)/Vr For leading load: Voltage regulation = Ir(R cosφ - X sinφ)/Vr For unity load power factor: Voltage regulation = Ir(R cosφ)/Vr Therefore, zero voltage regulation is possible only for leading power factor. From given data Cosφ = X/Z = 1/√2 = 0.707 leading

09․ The voltage regulation of a line at full load 0.8 power factor is 11. A long transmission line under no load conditions, for a good voltage profile needs
shunt resistance at receiving end
shunt reactor at receiving end
shunt capacitor at receiving end
all of the above

Under no load conditions or light load conditions of a transmission line, receiving end voltage is greater than sending ending voltage will occur in case of medium and transmission lines. This is called ferranti effect. This is due to capacitance effect is dominating compare to inductance of transmission line under no load only. To over come this problem, shunt reactors are connected at receiving end.

10․ The load current in short circuit calculations are neglected because
short circuit currents are much larger than load currents
short circuit currents are greatly out of phase with load currents
both 1 and 2
nether 1 nor 2

The load current in short circuit calculations are neglected because 1. Short circuit currents are much larger than load currents 2. Short circuit currents are greatly out of phase with load currents Therefore, both statements are correct statements.

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