# Power Electronics Questions and Answers – Miscellaneous

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This set of Power Electronics Multiple Choice Questions & Answers (MCQs) focuses on “Miscellaneous”.

1. Solid State Relays (SSRs) have a
a) coil and contact arrangement
b) optocoupler
c) scr
d) none of the mentioned

Explanation: Coil and contact arrangement is used in mechanical relays, SSRs have a optocoupler which connects the control circuit to the power circuit via light sensitive devices.

2. The converter circuit which employs turn on and turn off when the voltage and/or current through the device is zero at the instant of switching is ____________
a) a conventional converter
b) a resonant converter
c) a zero switching circuit
d) none of the mentioned

Explanation: Resonant converters are used to turn on and turn off when the voltage and/or current through the device is zero at the instant of switching.

3. Induction heating is a ___________ type of heating
a) zero frequency
b) high frequency
c) power frequency
d) none of the mentioned

Explanation: As eddy current is proportional to the square of the supply frequency, induction heating is a high frequency heating.

4. The factors governing the induction heating are
a) resistivity
b) relative permeability
c) magnetic field intensity
d) all of the mentioned

Explanation: Induction heating depends on all of the above given factors.

5. The reverse recovery time of a diode is trr = 3 μs and the rate of fall of the diode current (di/dt) = 30 A/μs. Determine the storage charge.
a) 145 μs
b) 135 μs
c) 0
d) none of the mentioned

Explanation: Storage charge = (1/2) x (di/dt) x (trr)2 = 135 μs.

6. For a SCR, conduction angle is 120° when average on-state current is 20 A. When the conduction angle is halved the earlier value, the on-state average current will be?
a) 5 A
b) 40 A
c) 10 A
d) 20 A

Explanation: When the conduction angle is halved, the device will conduct twice then it was conducting earlier. Hence, I = 2x 20 = 40 A.

7. A single-phase full bridge diode rectifier delivers power to a constant load current of 10 A. The average and rms values of the source currents will be respectively.
a) 5 A, 10 A
b) 10 A, 10 A
c) 5 A, 5 A
d) 10 A, 5A

Explanation: As the load current is continuous, Iavg = Irms = 10 A.

8. TRIAC is used in
a) chopper
b) speed control of induction machine
c) speed control of universal motor
d) none of the mentioned

Explanation: TRIAC is used in speed control of universal motor.

9. The ratio Vrms/ Vdc is known as
a) Form factor
b) Ripple factor
c) Utilization factor
d) None of the mentioned

Explanation: Vrms/ Vdc = FF.

10. Determine the loss in the snubber circuit, if C = 0.545 μF and supply is 200 V, 10 kHz.
a) 233 W
b) 133 W
c) 333 W
d) 233 W

Explanation: Snubber loss Ps = (1/2) x C x V2 x f = 133.1 W.

Sanfoundry Global Education & Learning Series – Power Electronics.

To practice all areas of Power Electronics, here is complete set of 1000+ Multiple Choice Questions and Answers.