Power Electronics Questions and Answers – AC Voltage Controllers-3

This set of Power Electronics Multiple Choice Questions & Answers (MCQs) focuses on “AC Voltage Controllers-3”.

1. The below given circuit has Vs = 230V and R = 20 Ω. Find the value of the average output voltage at the R load for a firing angle of 45°.

a) 224 V
b) -15.17 V
c) 15.17 V
d) –224 V

Explanation: Vo = [(√2 x 230) x (cos45 – 1)]/2π = -15.17 V.
Negative value is due to the fact that the average value in the positive half cycle is less than that in the negative half cycle.

2. The below given circuit has Vs = 230V and R = 20 Ω. Find the value of the average output load current at the R load for a firing angle of 45°.

a) – 0.7585
b) 0.7585
c) -0.6396
d) -0.5

Explanation: Vo = [(√2 x 230) x (cos45 – 1)]/2π = -15.17 V.
Io = (Vo)/R = -0.7585
Negative value is due to the fact that the average value in the positive half cycle is less than that in the negative half cycle.

3. A single phase voltage controller has input of 230 V and a load of 15 Ω resistive. For 6 cycles on and 4 cycles off, determine the rms output voltage.
a) 189 V
b) 260 V
c) 156 V
d) 178 V

Explanation: Vrms = Vo x √k
k = (6/6+4) = 6/10 = 0.6
Vrms = √0.6 x 230 = 178.157 V.

4. A single phase voltage controller has input of 230 V and a load of 15 Ω resistive. For 6 cycles on and 4 cycles off, determine the input pf.
a) 0.6
b) 0.7746
c) 0.855
d) 0.236

Explanation: k = (6/6+4) = 6/10 = 0.6
input pf = √0.6 = 0.7746.

5. A single phase voltage controller has input of 230 V and a load of 15 Ω resistive. For 6 cycles on and 4 cycles off, determine the power delivered to the load.
a) 2.1 W
b) 2.1 kW
c) 516 W
d) 5.16 kW

Explanation: Vrms = Vo x √k
k = (6/6+4) = 6/10 = 0.6
Vrms = √0.6 x 230 = 178.157 V
P = (Vrms)2/R = 2116 W = 2.116 kW.
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6. A single phase voltage controller has input of 230 V and a load of 15 Ω resistive. For 6 cycles on and 4 cycles off, determine the average value of SCR current.
a) 21.68 A
b) 200 mA
c) 4.14 A
d) 2.07 A

Explanation: Peak current Im = (230 x √2)/15 = 21.681 A
k = (6/6+4) = 6/10 = 0.6
Avg current = (k Im)/π = 4.14 A.

7. Pulse gating is suitable for

Explanation: In RL loads with pulse gating, the incoming SCR may be fired during the interval when it is reversed biased by the outgoing SCR, thus it won’t get turn on even after the outgoing SCR is not reveres basing the incoming SCR is the pulse width is over before forward biasing the SCR.

8. In continues gating
a) overlap angle is very high
b) SCR is heated up
c) size of the pulse transformer is small
d) commutation cannot be achieved effectively

Explanation: As the gating is applied for a longer duration, the device is heated up.

9. High frequency gating uses a
a) train of pulses
b) continuous gating block
c) carrier signal
d) none of the above

Explanation: In high frequency gating a train of pulses are used to overcome the thermal problems due to continuous gating.

10. A single-phase voltage controller, using one SCR in anti parallel with a diode, feeds a load R and Vs = 230 V. For a firing angel of 90° for the SCR, the PMMC voltage connected across R would read
a) 0
b) 51.8 V
c) –51.8 V
d) –36.82 V

Explanation: As firing angle is 90, there is ideally be no conduction in the positive half. Hence, the average value will be zero.
Vo = (√2 Vs)/2π x (cos90 – 1) = – 51.8 V.

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