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Power Systems MCQ Set 1

1. For an ac three-phase four wire system, having a voltage of 415V, with load of 120 kW and resistance of 0.05Ω at power factor of 0.75. The incurred copper losses of the system is ____
a) 4.955 kW
b) 47 kW
c) 49 kW
d) 7.9 kW

View Answer

Answer: a [Reason:] Copper losses for a three-phase four wire system will be, power-systems-assessment-questions-answers-q1

2. The ratio of area of cross section of the wire of a 3-phase 3-wire and that of dc 2-wire mid point earthed system will be ___
power-systems-questions-answers-distribution-system-3-q2

View Answer

Answer: a [Reason:] The ratio of area of cross section of the conductor used in 3-ph 3-wire to dc 2-wire will be power-systems-questions-answers-distribution-system-3-q3-exp as the power factor will come into picture.

3. Distributors fed at both ends has an advantage of ____
a) continuous supply
b) fault isolation
c) being economical
d) all of the mentioned

View Answer

Answer: d [Reason:] All the points mentioned are the benefits of the dual fed distribution system.

4. Why is a ring main distribution system preferred to a radial system?
1. Voltage drop in feed is more
2. Supply is reliable
3. Power factor is higher
a) 2
b) 1,2
c) 1,2,3
d) 2,3

View Answer

Answer: a [Reason:] Ring main system has less voltage drop and power factor at the consumer level is not high.

5. For a given power delivered, if the working voltage of a distribution line is decreased to ‘x’ times, the cross sectional area ‘a’ of distributor line would reduce to _____
power-systems-questions-answers-distribution-system-3-q5

View Answer

Answer: d [Reason:] The cross-sectional area has to be increased not reduced after reducing the voltage levels.

6. For a given power delivered, if the working voltage of a distribution line is increased to ‘x’ times, the cross sectional area ‘a’ of distributor line would reduce to _______
power-systems-questions-answers-distribution-system-3-q5

View Answer

Answer: a [Reason:] Assuming the power at both the times remain constant. power-systems-questions-answers-distribution-system-3-q6

7. The distribution feeding system of ring main system has ________
a) one feeder
b) 2 feeders
c) 4 feeders
d) cannot be said

View Answer

Answer: a [Reason:] Ring main uses only one feeder.

8. A substation records a dip in the voltage received by 15%. To overcome this issue, the booster employed at the substation is _____
a) a series wound generator coupled to dc shunt motor
b) a series wound motor
c) a shunt motor operating at lagging power factor
d) a shunt motor coupled to differential motor

View Answer

Answer: a [Reason:] A dc series motor coupled with the shunt motor has the ability to boost the voltage at its input supply.

9. A substation records a dip in the voltage received by 15%. To overcome this issue, the booster employed at the substation which should have a _______
a) low voltage and high current generator
b) high voltage and low current generator
c) medium voltage but high current generator
d) any of the mentioned

View Answer

Answer: a [Reason:] The booster need to have the low voltage and high current profile so that it can easily boost the voltage levels at its output.

10. Single core cables for an ac systems are not provided with armouring.
a) True
b) False

View Answer

Answer: a [Reason:] Because of the armouring of the cables leads to eddy current in the power cables adding more to the losses.

Power Systems MCQ Set 2

1. A 10 MVA, 11 kV alternators has positive, negative and zero sequence reactance of 25%, 40%, 10% respectively. The resistance that must be placed in the neutral such that zero sequence of fault current for a single phase to ground fault will not exceed the rated line current in pu is ______
a) 0.22
b) 0.11
c) 0.16
d) 0.97

View Answer

Answer: a [Reason:] power-systems-interview-questions-answers-q1

2. A 10 MVA, 11 kV alternator has positive, negative and zero sequence reactance of 25%, 40%, 10% respectively. The resistance that must be placed in the neutral such that fault current for a single phase to ground fault will not exceed the rated line current in pu is _______
a) 0.97
b) 0.22
c) 0.87
d) 0.45

View Answer

Answer: a [Reason:] power-systems-questions-answers-inductance-single-phase-two-wire-line-2-q2

3. A 10 MVA, 11 kV alternator has positive, negative and zero sequence reactance of 25%, 40%, 10% respectively. The resistance that must be placed in the neutral such that fault current for a single phase to ground fault will not exceed the rated line current in ohms is ____
a) 11.7
b) 12.8
c) 2.66
d) 5.2

View Answer

Answer: a [Reason:] power-systems-questions-answers-inductance-single-phase-two-wire-line-2-q3

4. (I) Unsymmetrical spacing configurations cause the line interference.
(II) Unsymmetrical spacing causes the voltage induction in the communication lines.
a) (I) and (II) are valid
b) Only (I) is valid
c) Only (II) is valid
d) Both are invalid

View Answer

Answer: a [Reason:] Both the statements are correct as the unsymmetrical spacing causes the unbalanced voltage generation.

5. (I) Unsymmetrical spacing configurations cause the line interference.
(II) Unsymmetrical spacing causes the voltage induction in the communication lines.
The above problems can be eliminated by ________
a) transposition
b) parallel lines
c) three phase
d) all of the mentioned

View Answer

Answer: a [Reason:] Transposition meaning exchange of conductors will cause the net voltage induced in the other lines to be zero as the net inductance remains constant.

6.Transposition is the mean to balance ______
a) Inductance
b) Voltage
c) Current
d) All of the mentioned

View Answer

Answer: a [Reason:] Transposition primarily balances the inductances in the line thus causing a constant inductance throughout the line.

7. Over the length of one transposition cycle of a power line, the total flux linkages of a nearby telephone line are zero for the unbalanced three phase currents.
a) True
b) False

View Answer

Answer: b [Reason:] For an unbalanced supply, there will be non zero flux zero flux linkages in the telephone lines.

8. Over the length of one transposition cycle of a power line, the total flux linkages of a nearby telephone line are zero for the balanced three phase currents.
a) The above statement is valid
b) The above statement is invalid
c) It can’t be concluded
d) The above statement is valid for all conditions.

View Answer

Answer: a [Reason:] Its valid only with balanced systems.

Power Systems MCQ Set 3

1. If we account for the effect of the presence of the capacitance of the transmission line, then the capacitance _____
a) increases
b) decreases
c) remains same
d) increases hyperbolically

View Answer

Answer: a [Reason:] Earth effects increases the capacitance of the system.

2. While estimating the line parameters, the term self GMD is used for the calculation of ________
a) inductance
b) capacitance
c) conductance
d) inductance, capacitance and conductance

View Answer

Answer: a [Reason:] Self GMD is used in the calculation of the inductance.

3. The ground effect causes line capacitance to _______
a) Increase by 0.2%
b) Increase by 2%
c) Decrease by 0.2%
d) Decrease by 2%

View Answer

Answer: a [Reason:] The effect of the ground is not very significant on the capacitance.

4. If the spacing between the phase conductors is increased, the line capacitance will ___
a) decrease
b) increase
c) remain same
d) depends on the conductor

View Answer

Answer: a [Reason:] power-systems-interview-questions-answers-experienced-q4

5. If the length of the transmission line is increased, the line charging current in the TL will _______
a) increase
b) decrease
c) remain same
d) not be affected

View Answer

Answer: a [Reason:] If the length of the TL increases, then the line admittance will decrease, so the line charging current will decrease.

6. On applying the alternating sine wave as the input to the line, the line capacitance draws _______
a) leading sinusoidal current
b) lagging sinusoidal current
c) current in phase with voltage
d) depends on the receiving end load

View Answer

Answer: a [Reason:] The line current will be leading sinusoidal current.

7. The capacitance becomes significant for the ____
a) long transmission line
b) medium transmission line
c) short transmission line
d) cannot be determined

View Answer

Answer: a [Reason:] Capacitance parameter is highly important for long transmission line due to the charging current flowing in the LTL.

8. The method of the images is applied to find the ______
a) calculation of capacitance
b) calculation of inductance
c) calculation of resistance
d) calculation of effect of the earth on capacitance

View Answer

Answer: d [Reason:] Method of images given by Lord Kelvin is used to calculate the effect of the earth on capacitance.

Power Systems MCQ Set 4

1. A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. The voltage regulation will be _______
a) 16.8 %
b) 18.8%
c) 21.75%
d) 12.8%

View Answer

Answer: a [Reason:] VR = (11.68-10)*100/10 = 16.8 %.

2. A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. Identify the transmission line and the voltage regulation.
a) STL, 11.68%
b) MTL, 11.68
c) STL, 21.5%
d) MTL, 14.2%

View Answer

Answer: a [Reason:] It is a short transmission line. Current, I = 5000/(10*0.8)=625 A Vs = |Vr|+|I|*(RcosФr + XsinФr) = 10000+625(0.39*0.8+3.96*0.6) = 11.68kV VR = (11.68-10)*100/10 = 16.8 %.

3. A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. The half the voltage regulation will be _______
a) 8.4 %
b) 16.8 %
c) 14.2%
d) 10.5%

View Answer

Answer: a [Reason:] VR = (11.68-10)*100/10 = 16.8 % Half the VR = 16.8/2 % = 8.4%.

4. A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. The new sending end voltage at the half the voltage regulation is _______
a) 10.84 kV
b) 11.84 kV
c) 8.84 kV
d) 16.2 kV

View Answer

Answer: a [Reason:] VR = (11.68-10)*100/10 = 16.8 % Half the VR = 16.8/2 % = 8.4% power-systems-interview-questions-answers-freshers-q4

5. A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. If the voltage regulation is reduced to 50%, then the power factor at this operation mode will be _____
a) 0.95
b) 0.92
c) 0.74
d) 0.90

View Answer

Answer: a [Reason:] VR = (11.68-10)*100/10 VR = 16.8 % Half the VR = 16.8/2 % Half the VR = 8.4% power-systems-interview-questions-answers-freshers-q4 (10.84-10)*1000 = |I|*( RcosФr + XsinФr ) …(1) I = 5000/(cosФr*10) …(2) Solving above eqaution Фr = 18.04° Cos Фr = 0.9508, lagging.

6. A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. If the voltage regulation is reduced to 50%, then the power factor angle at this operation mode will be _______
a) 18.04°
b) 8.04°
c) 21.06°
d) 12°

View Answer

Answer: a [Reason:] VR = (11.68-10)*100/10 VR = 16.8 % Half the VR = 16.8/2 % Half the VR = 8.4% power-systems-interview-questions-answers-freshers-q4 (10.84-10)*1000 = |I|*(RcosФr + XsinФr) …(1) I = 5000/(cosФr*10) …(2) Solving above equation Фr = 18.04°.

7. A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. If the voltage regulation is reduced to 50%, then the power factor at this operation mode will be ______
a) 0.95, lagging
b) 0.92, leading
c) 0.95, lagging
d) 0.90, leading

View Answer

Answer: a [Reason:] VR = (11.68-10)*100/10 VR = 16.8 % Half the VR = 16.8/2 % Half the VR = 8.4% power-systems-interview-questions-answers-freshers-q4 (10.84-10)*1000 = |I|*(RcosФr + XsinФr) …(1) I = 5000/(cosФr*10) …(2) Solving above equation Фr = 18.04° Cos Фr = 0.9508, lagging.

8. A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. If the voltage regulation is reduced to 50%, then the power factor angle at this operation mode will be ________
a) 18.04°, lagging
b) 18.04°, leading
c) 21.06°, leading
d) 21.06°, lagging

View Answer

Answer: a [Reason:] VR = (11.68-10)*100/10 VR = 16.8 % Half the VR = 16.8/2 % Half the VR = 8.4% power-systems-interview-questions-answers-freshers-q4 (10.84-10)*1000 = |I|*(RcosФr + XsinФr) …(1) I = 5000/(cosФr*10) …(2) Solving above equation Фr = 18.04°, lagging.

9. A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. If the voltage regulation is reduced to 50%, then the receiving end current at this operation mode will be _______
a) 526 A
b) 549 A
c) 521 A
d) 580 A

View Answer

Answer: a [Reason:] VR = (11.68-10)*100/10 VR = 16.8 % Half the VR = 16.8/2 % Half the VR = 8.4% power-systems-interview-questions-answers-freshers-q4 (10.84-10)*1000 = |I|*(RcosФr + XsinФr) …(1) I = 5000/(cosФr*10) …(2) Solving above equation Фr = 18.04°, lagging I = 526 A.

10. Suppose the transmission line is loaded with its surge impedance, the receiving-end voltage is greater than sending end voltage.
a) True
b) False

View Answer

Answer: b [Reason:] It will be equal.

Power Systems MCQ Set 5

1. What is the value of charging current in short transmission lines?
a) Less than medium transmission lines
b) Equal to medium transmission lines
c) More than medium transmission lines
d) More than long transmission lines

View Answer

Answer: c [Reason:] Line to Earth capacitance of short transmission line is less than that in medium and long transmission line and is negligable. Because of this the charging current of short transmission line is less than that in medium and long transmission line and is negligable.

2. What is the value of Inductive reactance in short transmission lines?
a) More at reciving end
b) More at sending end
c) Uniformly distributed over entire length
d) More in middle of sending end and reciving end

View Answer

Answer: c [Reason:] Actually the inductive reactance of line is uniformly distributed over its entire length. To make calculations simple this reactance is assumed to be lumped and connected in series with the line.

3. Regulation of short transmission lines depends:
a) Only on line resistance
b) Only on line inductance
c) Only on line capacitance
d) On line inductance and line resistance

View Answer

Answer: d [Reason:] The expression of voltage regulation of short transmission line shows that, its voltage regulation depends on line resistance and line inductive reactance. Capacitance of short transmission line is negligible so it does not cause its effecting it.

4. What is the percentage voltage regulation of short transmission line if its sending end and reciving end voltages are 160 KV and 132 KV respectively?
a) 30 %
b) 21.21 %
c) 12.12 %
d) 38.22 %

View Answer

Answer: b [Reason:] % Voltage Regulation = (Sending end voltage -Reciving end voltage) ÷ Reciving end voltage = [( 160 – 132 ) × 100] ÷ 132 = 21.21 %.

5. A single phase transmission line is transmitting 1,100 KW power at 11 KV and at unity power fector. If it has a total resistance of 5 Ω, what is the efficiency of the transmission line?
a) 80 %
b) 89.65 %
c) 97.24 %
d) 99.54 %

View Answer

Answer : c [Reason:] Efficiency (η) = Power delivered ÷ (Power Delivered + line losses) Line current (I) = P/V = (1100 × 1000/11000) = 100 Amp. Line losses = I2 R = (100)2 5 = 50 KW η = [11000/(11000+50)] × 100 = 99.54 %.

6. Voltage regulation of a transmission line should be ________
a) Minimum
b) Maximum
c) Greater than 50 %
d) Less than 50 %

View Answer

Answer : a [Reason:] More value of voltage regulation means more voltage fluctuations and this is undesirable. So the value of voltage regulation should be as low as possible.

7. Which of the following is like equivalent circuit of short transmission line?
a) Series RLC circuit
b) Parallel RLC circuit
c) Series RL circuit
d) Parallel RL circuit

View Answer

Answer: c [Reason:] Due to smaller distance and lower line voltage, the shunt capacitance effects are extremely small and also the shunt conductance effects are very low. So both of them are neglected and equivalent circuit contains only series inductance(L) and resistance(R).

8. In single phase transmission lines resistance and inductance are considered only up to neutral.
a) True
b) False

View Answer

Answer: b [Reason:] In case of single phase transmission lines the total loop resistance and inductance is to be taken into account. Resistance and inductance only up to neutral are taken into account for 3 phase transmission lines.

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