250+ TOP MCQs on Solid-State Switching Circuits – Single Phase, Full-Wave, AC/DC Conversion for Resistive Loads and Answers

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Electric Drives Questions and Answers for Aptitude test on “State Switching Circuits – Single Phase, Full-Wave, AC/DC Conversion for Resistive Loads “.

1. Calculate the value of the Input power factor for 1-Φ Full wave bridge rectifier if the firing angle value is 39°.
A. .69
B. .59
C. .78
D. .15
Answer: A
Clarification: The value of the input power factor for 1-Φ Full wave bridge rectifier is .9cos(67°)=.69. The input power factor is a product of distortion factor and displacement factor.

2. Calculate the value of the fundamental displacement factor for 1-Φ Full wave bridge rectifier if the firing angle value is 38°.
A. .22
B. .78
C. .33
D. .44
Answer: B
Clarification: Fundamental displacement factor is the cosine of angle difference between the fundamental voltage and fundamental current. D.F=cos(∝)=cos(38°)=0.78.

3. Calculate the value of the fundamental displacement factor for 1-Φ Full wave semi-converter if the firing angle value is 69°.
A. .48
B. .24
C. .82
D. .88
Answer: C
Clarification: Fundamental displacement factor is the cosine of angle difference between the fundamental voltage and fundamental current. The fundamental displacement factor for 1-Φ Full wave semi-converter is cos(∝÷2)=cos(34.5°)=.82.

4. Calculate the fundamental component of source current in 1-Φ Full wave bridge rectifier for load(Highly inductive) current=3.14 A.
A. 2.82 A
B. 1.45 A
C. 3.69 A
D. 4.78 A
Answer: A
Clarification: The fundamental component of source current in 1-Φ Full wave bridge rectifier is 2√2Io÷π. It is the r.m.s value of the fundamental component. Is1(r.m.s) = 2√2Io÷π=2√2=2.82 A.

5. Calculate the circuit turn-off time for 1-Φ Full wave bridge rectifier for α=145°. Assume the value of ω=5 rad/sec.
A. 84.9 msec
B. 94.5 msec
C. 101.2 msec
D. 87.2 msec
Answer: D
Clarification: The circuit turn-off time for 1-Φ Full wave bridge rectifier is (π-α)÷ω. The value of circuit turn-off time is (π-145°)÷5=87.2 msec.

6. Calculate the fundamental component of source current in 1-Φ Full wave bridge rectifier for the load(Highly inductive) current=78 A.
A. 78 A
B. 45 A
C. 69 A
D. 13 A
Answer: A
Clarification: The fundamental component of source current in 1-Φ Full wave bridge rectifier is Io. It is the r.m.s value of the source current. Is(r.m.s)=Io=78 A.

7. Calculate the r.m.s value of source current in 1-Φ Full wave semi-converter for the load (Highly inductive) current=51.2 A and α=15°.
A. 10.53 A
B. 14.52 A
C. 44.92 A
D. 49.02 A
Answer: D
Clarification: The r.m.s value of source current in 1-Φ Full wave semi-converter is Io√π-α÷π. It is the r.m.s value of the source current. I(r.m.s) = Io√π-α÷π = 51.2(√.916) = 49.02 A.

8. Calculate the r.m.s value of thyristor current in 1-Φ Full wave semi-converter for the load (Highly inductive) current=2.2 A and α=155°. (Asymmetrical configuration)
A. .58 A
B. .57 A
C. .51 A
D. .52 A
Answer: B
Clarification: The r.m.s value of source current in 1-Φ Full wave semi-converter is Io√π-α÷2π. It is the r.m.s value of the thyristor current. I(r.m.s) = Io√π-α÷2π=2.2(√.069)=.57 A.

9. Calculate the r.m.s value of diode current in 1-Φ Full wave semi-converter for the load (Highly inductive) current=5.1 A and α=115°. (Asymmetrical configuration)
A. 4.21 A
B. 4.61 A
C. 4.71 A
D. 4.52 A
Answer: B
Clarification: The r.m.s value of diode current in 1-Φ Full wave semi-converter is Io√π+α÷2π. It is the r.m.s value of the diode current. I(r.m.s) = Io√π+α÷π=5.1(√.819)=4.61 A.

10. Calculate the average value of thyristor current in 1-Φ Full wave semi-converter for the load (Highly inductive) current=25.65 A and α=18°. (Asymmetrical configuration)
A. 11.54 A
B. 12.15 A
C. 15.48 A
D. 14.52 A
Answer: A
Clarification: The average value of thyristor current in 1-Φ Full wave semi-converter is Io(π-α÷2π). It is the average value of the thyristor current. Iavg = Io(π-α÷2π)=25.65(.45)=11.54 A.

11. Calculate the average value of diode current in 1-Φ Full wave semi-converter for the load (Highly inductive) current=75.2 A and α=41°. (Asymmetrical configuration)
A. 46.16 A
B. 42.15 A
C. 41.78 A
D. 41.18 A
Answer: A
Clarification: The average value of diode current in 1-Φ Full wave semi-converter is Io(π+α÷2π). It is the average value of the diode current. Iavg = Io(π+α÷2π)=75.2(.61)=46.16 A.

12. Calculate the average value of diode current in 1-Φ Full wave semi-converter for the load (Highly inductive) current=5.2 A and α=11°. (F.D configuration)
A. .32 A
B. .31 A
C. .25 A
D. .27 A
Answer: B
Clarification: The average value of diode current in 1-Φ Full wave semi-converter is Io(α÷π). It is the average value of the diode current. Iavg= Io(α÷π)=5.2(.061)=.31 A.

13. Calculate the r.m.s value of diode current in 1-Φ Full wave semi-converter for the load (Highly inductive) current=.2 A and α=74°. (F.D configuration)
A. .154 A
B. .248 A
C. .128 A
D. .587 A
Answer: C
Clarification: The r.m.s value of diode current in 1-Φ Full wave semi-converter is Io√(α÷π). It is the r.m.s value of the diode current. Ir.m.s = Io√(α÷π)=.2√(.41)=.128 A.

14. Diodes in 1-Φ Full wave semi-converter protects the thyristor from short-circuiting.
A. True
B. False
Answer: A
Clarification: Diodes in 1-Φ Full wave semi-converter protects the thyristor from short-circuiting. They provide the gap from (α, π+α) to avoid the conduction of one leg thyristors.

15. The problem of short-circuiting in 1-Φ Full wave semi-converter is very common.
A. True
B. False
Answer: A
Clarification: The problem of short-circuiting in 1-Φ Full wave semi-converter is very common. Diodes protect the thyristor from short-circuiting. They provide the gap from (α, π+α) to avoid the conduction of one leg thyristors.

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