250+ TOP MCQs on Traction Drives – Power Factor and Harmonics and Answers

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This set of Electric Drives online quiz on “Traction Drives – Power Factor and Harmonics”.

1. Full form of SVC is ___________
A. Static Var Compensator
B. Static Variable Counter
C. State Value Counter
D. Static Value Compensator
Answer: A
Clarification: The full form of SVC is Static Var Compensator. It is a FACTS controller used for reactive power compensation. It maintains the voltage profile.

2. Full form of STATCOM is ___________
A. Static Synchronous Counter
B. Static Synchronous Compensator
C. State Synchronized Compensator
D. State System Counter
Answer: B
Clarification: The full form of STATCOM is a Static synchronous compensator. It is a voltage source converter. It maintains the voltage profile of the system. It is a FACTS controller.

3. The circuit which operates signals is called as ___________
A. Track circuit
B. Fellow circuit
C. Ground circuit
D. Unit circuit
Answer: A
Clarification: The circuit which operates signals is called track circuit. Its supply consists of low AC or DC voltage supply. Running rail is used as the return line for the track circuit.

4. SVC is used to maintain the power factor above __________
A. .2
B. .5
C. .8
D. .6
Answer: C
Clarification: SVC stands for static var compensator. It is the FACTS controller used for reducing the harmonics from the system. It is used for reactive power compensation.

5. Which is the most severe type of fault?
A. LL
B. LLG
C. LLL
D. LLLG
Answer: D
Clarification: LLLG fault is the most severe type of fault. The order of severity is maximum for LLLG fault. LG fault is more severe at the generator terminals.

6. Skin effect increases at higher frequencies.
A. True
B. False
Answer: A
Clarification: Skin effect increases with the increase in frequency. The value of skin depth decreases with the increase in frequency.

7. If induction motor air gap power is 11 KW and mechanically developed power is 2 KW, then rotor ohmic loss will be _________ KW.
A. 8
B. 10
C. 9
D. 11
Answer: C
Clarification: Rotor ohmic losses are due to the resistance of armature windings. Net input power to the rotor is equal to the sum of rotor ohmic losses and mechanically developed power. Rotor ohmic losses = Air gap power – Mechanical developed power = 11-2 = 9 KW.

8. If induction motor rotor power is 26 KW and gross developed power is 21 KW, then rotor ohmic loss will be _________ KW.
A. 5
B. 6
C. 7
D. 8
Answer: A
Clarification: Rotor ohmic losses are due to the resistance of armature windings. Net input power to the rotor is equal to the sum of rotor ohmic losses and mechanically developed power. Rotor ohmic losses=Air gap power-Mechanical developed power=26-21=5 KW.

9. Calculate the active power in 14 μH inductors if the THD value is 58.1%.
A. 5 W
B. 9 W
C. 4 W
D. 0 W
Answer: D
Clarification: The inductor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the inductor so the angle between V & I is 90°. P = VIcos90° = 0 W.

10. Calculate the power factor using the data: P=15 W, S=25 VA.
A. 0.5
B. 0.9
C. 0.4
D. 0.6
Answer: D
Clarification: The power factor is the ratio of active power and apparent power. The value of the power factor is P÷S=15÷25=.6.

11. Calculate the active power in 56 μF capacitor if the VRF value is 5.1%.
A. 6 W
B. 0 W
C. 9 W
D. 2 W
Answer: B
Clarification: The capacitor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the capacitor so the angle between V & I is 90°. P = VIcos90° = 0 W.

12. Calculate the apparent power using the data: P=3 W, Q=4 VAR.
A. 5 VA
B. 6 VA
C. 9 VA
D. 2 VA
Answer: A
Clarification: The apparent power is the product of the voltage and current conjugate phasor. The value of apparent power is (9+16).5 = 5 VA.

13. Calculate the value of impedance using the data: R=2 Ω, XL=4 Ω.
A. 5.5 Ω
B. 5.4 Ω
C. 2.5 Ω
D. 4.4 Ω
Answer: D
Clarification: The value of impedance is {(R)2+(XL)2).5 = (4+16).5 = 4.4 Ω. It is total impedance offered by an AC circuit. It is expressed in Ω.

14. Which type of grounding is used to suppress the capacitive effect?
A. Solid grounding
B. Reactance grounding
C. Resistance grounding
D. Peterson coil
Answer: D
Clarification: Peterson coil or resonance grounding is used to suppress the capacitive effect. The frequency in resonance grounding is ω2 = 1÷(3LC..

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