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

This set of Power Electronics Multiple Choice Questions & Answers (MCQs) focuses 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

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: Full wave B-2 type uses 4 diodes in a bridge connection.
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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

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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.

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