Power Electronics Question and Answers – 1P-Diode Rectifier FW-1 and Answers

Power Electronics Multiple Choice Questions on “1-Phase-Diode Rectifiers FW-1”.

1. A single-phase full wave mid-point type diode rectifier requires __________ number of diodes whereas bridge type requires _________
A. 1,2
B. 2,4
C. 4,8
D. 3,2
Answer: B
Clarification: A bridge type requires 4 diodes which are connected in a bridge, and the mid-point has 2 diodes.

2. A single-phase full wave rectifier is a
A. single pulse rectifier
B. multiple pulse rectifier
C. two pulse rectifier
D. three pulse rectifier
Answer: C
Clarification: It is a two-pulse rectifier as it generates 2 pulses per cycle.

3. The below shown circuit is that of a

A. full wave B-2 type connection
B. full wave M-2 type connection
C. half wave B-2 type connection
D. half wave M-2 type connection
Answer: B
Clarification: Full wave M-2 type employs a transformer and two diodes.

4. In a 1-phase full wave bridge rectifier with M-2 type of connection has secondary side voltage Vs = Vm sin ωt, with R load & ideal diodes.
The expression for the average value of the output voltage can be given by
A. 2Vm/π
B. Vm/π
C. Vm/√2
D. 2Vm/√2
Answer: A
Clarification: The voltage waveform is a pulsating voltage with peak value Vm & symmetrical about π.
Vo = (1/π) ∫^π Vm sinωt d(ωt)

5. The below shown circuit is that of a

A. full wave B-2 type connection
B. full wave M-2 type connection
C. half wave B-2 type connection
D. half wave B-2 type connection
Answer: A
Clarification: Full wave B-2 type uses 4 diodes in a bridge connection.

6. In a 1-phase full wave bridge rectifier with M-2 type of connection has secondary side voltage Vs = Vm sin ωt,
with R load & ideal diodes.
The expression for the rms value of the output voltage can be given by
A. Vm/π
B. Vm/√2
C. Vm
D. Vm²
Answer: B
Clarification: The voltage waveform is a pulsating voltage with peak value Vm & symmetrical about π.
Vo = (1/π) ∫^π Vm² sin²ωt d(ωt) = Vm/√2.

7. For the circuit shown below, find the power delivered to the R load

Where,
Vs = 230V
Vs is the secondary side single winding rms voltage.
R = 1KΩ
A. 46 W
B. 52.9 W
C. 67.2 W
D. 69 W
Answer: B
Clarification: Power delivered to the load is V(rms).I(rms) = V(rms)²/R
Where, for the given circuit, V(rms) = Vs.

8. The PIV experienced by the diodes in the mid-point type configuration is
A. Vm
B. 2Vm
C. 4Vm
D. Vm/2
Answer: B
Clarification: In the m-2 type connection, each diode experiences a reverse voltage of 2Vm.

9. For the circuit shown below, find the value of the average output current.

Where,
Vs = 230V
R = 1KΩ
Vs is the secondary side single winding rms voltage.
A. 100mA
B. 107mA
C. 200mA
D. 207mA
Answer: D
Clarification: I = Vo/R
Vo = 2Vm/π.

10. In the circuit, let Im be the peak value of the sinusoidal source current. The average value of the diode current for the below given configuration is

A. Im
B. Im/2
C. Im/π
D. Im/√2
Answer: B
Clarification: The peak current through the diodes is same as that passing from the load. Each diode pair conducts for 180 degrees. Hence, 1/2 of a cycle. Average current = Im/2.