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

Power Electronics Multiple Choice Questions 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
Answer: B
Clarification: 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
Answer: C
Clarification: 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
Answer: C
Clarification: 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
Answer: B
Clarification: Diode will be reversed biased in the negative half cycles.

5. For the circuit shown below,
power-electronics-mcqs-1p-diode-rectifiers-hw-1-q5
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
Answer: C
Clarification: PIV = √2 Vs = Vm.

6. For the circuit shown below,
power-electronics-mcqs-1p-diode-rectifiers-hw-1-q6
The peak value of the load current occurs at ωt = ?
A. 0
B. π
C. 2π
D. Data is insufficient
Answer: B
Clarification: 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.
power-electronics-mcqs-1p-diode-rectifiers-hw-1-q7
Voltage across the secondary is given by Vm sinωt.
A. Vm
B. 2Vm
C. Vm/2
D. Vm2/2
Answer: C
Clarification: 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
Answer: C
Clarification: Vo = √ [(1/2π) ∫π Vm sinωt. d(ωt)]
Vo = Vm/π
I = Vo/R = Vm/πR.

9. In the circuit shown below,
power-electronics-mcqs-1p-diode-rectifiers-hw-1-q9
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
Answer: A
Clarification: 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
Answer: D
Clarification: I(rms) = Vm/2R
Therfore, Vm = 200R
I(avg) = Vm/πR = 200R/πR.

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