# Electric Drives Questions and Answers – Solid-State Switching Circuits – Single Phase, Full-Wave, AC/DC Conversion for Resistive Loads

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This set of Electric Drives Questions and Answers for Aptitude test focuses 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

Explanation: 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

Explanation: 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

Explanation: 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.
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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

Explanation: 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

Explanation: 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.
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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

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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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