# Single Phase Full Wave Controlled Rectifier With R Load MCQ [Free PDF] – Electrical Exams

1. Calculate the value of the Input power factor for a 1-Φ Full wave bridge rectifier if the firing angle value is 39°.

A. .69
B. .59
C. .78
D. .15

The value of the input power factor for the 1-Φ Full wave bridge rectifier is

.9cos(67°)=.69.

The input power factor is a product of the distortion factor and displacement factor.

2. Calculate the value of the fundamental displacement factor for a 1-Φ Full wave bridge rectifier if the firing angle value is 38°.

A. .22
B. .78
C. .33
D. .44

The 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 a 1-Φ Full-wave semi-converter if the firing angle value is 69°.

A. .48
B. .24
C. .82
D. .88

The fundamental displacement factor is the cosine of angle difference between the fundamental voltage and fundamental current. The fundamental displacement factor for a 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

The fundamental component of source current in a 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

The circuit turn-off time for the 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

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

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

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

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