# Power Electronics Question and Answers – 1P-Diode Rectifiers HW-1

This set of Power Electronics Multiple Choice Questions & Answers (MCQs) focuses on “1-Phase-Diode Rectifiers HW-1”.

1. In the process of diode based rectification, the alternating input voltage is converted into
a) an uncontrolled alternating output voltage
b) an uncontrolled direct output voltage
c) a controlled alternating output voltage
d) a controlled direct output voltage

Explanation: Rectification is AC to DC. In DIODE biased rectification, control is not possible.

2. In a half-wave rectifier, the
a) current & voltage both are bi-directional
b) current & voltage both are uni-directional
c) current is always uni-directional but the voltage can be bi-directional or uni-directional
d) current can be bi-directional or uni-directional but the voltage is always uni-directional

Explanation: Current is always in one direction only, but voltage can be bi-directional in case of an L load.

3. For a certain diode based rectifier, the output voltage (average value) is given by the equation
1/2π [ ∫Vm sin ωt d(ωt) ] Where the integral runs from 0 to π
The rectifier configuration must be that of a
a) single phase full wave with R load
b) single phase full wave with RL load
c) single phase half wave with R load
d) single phase half wave with RL load

Explanation: Integration is 0 to π from base period of 1/2π so it is a half wave R load.

4. For a single phase half wave rectifier, with R load, the diode is reversed biased from ωt =
a) 0 to π, 2π to 2π/3
b) π to 2π, 2π/3 to 3π
c) π to 2π, 2π to 2π/3
d) 0 to π, π to 2π

Explanation: Diode will be reversed biased in the negative half cycles.

5. For the circuit shown below,

The secondary transformer voltage Vs is given by the expression
Vs = Vm sin ωt
Find the PIV of the diode.
a) √2
b) Vs
c) Vm
d) √2 Vm

Explanation: PIV = √2 Vs = Vm.

6. For the circuit shown below,

The peak value of the load current occurs at ωt = ?
a) 0
b) π
c) 2π
d) Data is insufficient

Explanation: Due to the L nature, load current is maximum when the diode will be com-mutated i.e at π.

7. Find the rms value of the output voltage for the circuit shown below.

Voltage across the secondary is given by Vm sinωt.
a) Vm
b) 2Vm
c) Vm/2
d) Vm2/2

Explanation: The above is a HW diode rectifier, the RMS o/p voltage equation is given by
Vor = √ [ (1/2π) ∫π Vm2sin2ωt. d(ωt) ] Solving above equation we get, Vor = Vm/2.

8. In a 1-Phase HW diode rectifier with R load, the average value of load current is given by
Take Input (Vs) = Vm sinωt
a) Vm/R
b) Vm/2R
c) Vm/πR
d) Zero

Explanation: Vo = √ [(1/2π) ∫π Vm sinωt. d(ωt)] Vo = Vm/π
I = Vo/R = Vm/πR.

9. In the circuit shown below,

The switch (shown in green) is closed at ωt = 0. The load current or capacitor current has the maximum value at ωt =
a) 0
b) π
c) 2π
d) none of the mentioned

Explanation: The instant switch is closed the load current will be zero due to the nature of the capacitor.

10. Find the average value of output current for a 1-phase HW diode rectifier with R load, having RMS output current = 100A.
a) 200R A
b) 100/R√2 A
c) 200/R√2 A
d) 200/Rπ A

Explanation: I(rms) = Vm/2R
Therfore, Vm = 200R
I(avg) = Vm/πR = 200R/πR.

Sanfoundry Global Education & Learning Series – Power Electronics.

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