Single phase Half wave controlled rectifier with Inductive load MCQ [Free PDF] – Objective Question Answer for Single phase Half wave controlled rectifier with Inductive load Quiz

1. Calculate the value of the conduction angle for R-L load if the value of β and α are 19° and 29°.

A. 10°
B. 70°
C. 30°
D. 80°

Answer: A

The conduction angle for the R-L load is

β-α=29°-19°=10°.

R-L load is a current stiff type of load. The current in the circuit only flows from α to β.   

 

2. Calculate the V0 avg for Single-phase Half Wave rectifier for R-L load using the data: Vm=24 V, ∝=30°, β=60°.

A. 1.39 V
B. 8.45 V
C. 4.55 V
D. 1.48 V

Answer: A

In Half-wave controlled rectifier, the average value of the voltage is

Vm(cos(∝)-cos(β))÷2π
=24(cos(30°)- cos(60°))÷6.28=1.39 V.

The thyristor will conduct from ∝ to β.   

 

3. Calculate the I0 avg for the Single-phase Half Wave rectifier for R-L load using the data: Vm=56 V, ∝=15°, β=30°, R=2 Ω.

A. .56 A
B. .44 A
C. .26 A
D. .89 A

Answer: A

In Half-wave controlled rectifier, the average value of the current is

Vm(cos(∝)-cos(β))÷2πR

=56(cos(15°)- cos(30°))÷12.56=.44 A.

The thyristor will conduct from ∝ to β.   

 

4. Calculate the I0 r.m.s for Single-phase Half Wave rectifier for R-L load (Highly inductive) using the data: Vm=110 V, ∝=16°, β=31°, R=1 Ω.

A. 4.56 A
B. 1.82 A
C. 4.81 A
D. 9.15 A

Answer: B

In Half-wave controlled rectifier, the r.m.s value of the current is

Vm(cos(∝) – cos(β))÷2πR
=110(cos(16°) – cos(31°))÷6.28=1.82 A.

The thyristor will conduct from ∝ to β.   

 

5. Calculate the value of the fundamental displacement factor for 1-φ Full-wave semi-converter if the firing angle value is 0o.

A. 1
B. .8
C. .4
D. .2

Answer: A

The 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(0o)=1.   

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6. For highly inductive load current remains continuous.

A. True
B. False

Answer: A

For highly inductive load current remains continuous. The inductor opposes the change in the current. When the current is about to extinct the next conducting cycle starts which currently remains constant.   

 

7. Calculate the value of the Input power factor using the data: Fundamental displacement factor=.96, Distortion factor=.97.

A. .93
B. .84
C. .48
D. .89

Answer: A

The value of the input power factor is a product of the (Fundamental displacement factor)×(Distortion factor)

=g×F.D.F=.97×.96=.93.   

 

8. Displacement factor depends upon the shape of the waveform.

A. True
B. False

Answer: A

The displacement factor depends upon the shape of the waveform. The displacement factor is the cosine of the angle between the fundamental voltage and fundamental current. It depends upon the type of load and converter type.   

 

9. In single-phase RLE load, calculate the voltage across the thyristor when current decays to zero using the data: (Vs)r.m.s=340 V, f=50 Hz, R=2 Ω, E=150 V, β=160°.

A. 45.89 V
B. 74.45 V
C. 54.85 V
D. 84.48 V

Answer: B

In a single-phase RLE load the voltage across the thyristor when the current decays to zero

VT=Vmsin(β)-E
=340×√2sin(160°)-90=74.45 V.   

 

10. In single-phase RLE load, calculate the angle at which conduction starts using the data: (Vs)=14sin(Ωt), f=50 Hz, E=10 V.

A. 45.58°
B. 46.26°
C. 47.26°
D. 49.56°

Answer: A

The angle at which conduction starts when Vmsin(θ)=E. The conduction will remain from ϴ to π-θ.

Θ=sin-(E÷Vm)
=sin-(.71)=45.58°.   

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