Solid Sate DC Drives MCQ [Free PDF] – Objective Question Answer for Solid Sate DC Drives Quiz

21. In a 3-Φ semi-controlled rectifier calculate the average value of the voltage if the supply is 440 V and the firing angle is 22°.

A. 571.5 V
B. 572.8 V
C. 548.3 V
D. 524.1 V

Answer: B

In 3-Φ Semi-controlled rectifier, the average value of the voltage is

3Vml(1+cos(∝))÷2π

=3×440×√2(1+cos(22°))÷6.28=572.8 V. 

 

22. Calculate the circuit turn-off time for 3-Φ Fully controlled rectifier if the firing angle is 20° and the supply frequency is 60 Hz.

A. 8.8 msec
B. 7.4 msec
C. 10.1 msec
D. 6.5 msec

Answer: C

The circuit turn-off time for 3-Φ Fully controlled rectifier is
(240°-α)÷ω.

The value of circuit turn-off time for ∝ < 60° is

(240°-20°)÷6.28×60=10.1 msec. 

 

23. Calculate the circuit turn-off time for 3-Φ Fully controlled rectifier if the firing angle is 110° and the supply frequency is 50 Hz.

A. 3.8 msec
B. 5.2 msec
C. 9.3 msec
D. 8.7 msec

Answer: A

The circuit turn-off time for 3-Φ Fully controlled rectifier is
(180°-α)÷ω;

The value of circuit turn-off time for ∝ ≥ 60° is

(180°-110°)÷6.28×50=3.8 msec. 

 

24. Calculate peak-peak voltage if Vmax=80 V and Vmin=20 V.

A. 60 V
B. 50 V
C. 70 V
D. 10 V

Answer: A
Peak-Peak voltage is equal to the difference between the maximum

and minimum voltage. It is mathematically represented as

Vp-p=Vmax-Vmin=80-20=60 V. 

 

25. Calculate the value of the Crest factor if Vpeak=12 V and Vr.m.s=24 V.

A. .2
B. .3
C. .4
D. .5

Answer: D

The value of the crest factor is Vpeak÷Vr.m.s=12÷24=.5. It signifies the peak value is .5 times the r.m.s value. 

 

26. Calculate the output voltage of the Buck converter if the supply voltage is 11 V and the duty cycle value is .4.  [/bg_collapse]

A. 4.4 V
B. 2.2 V
C. 4.8 V
D. 6.4 V

Answer: A

The output voltage of the buck converter is
Vo = Vin×(D.=11×.4=4.4 V.

The value of the duty cycle is less than one which makes the Vo < Vin. The buck converter is used to step down the voltage. 

 

27. Calculate the output voltage of the Boost converter if the supply voltage is 8 V and the duty cycle value is .6.

A. 40 V
B. 20 V
C. 48 V
D. 51 V

Answer: B

The output voltage of the boost converter is
Vo = Vin ÷ (1-D. = 8÷.4 = 20 V.

The value of the duty cycle is less than one which makes the Vo > Vin as denominator value decreases and becomes less than one. The boost converter is used to step up the voltage. 

 

28. Calculate the output voltage of the Buck-Boost converter if the supply voltage is 78 V and the duty cycle value is .1.

A. 7.2 V
B. 4.5 V
C. 8.6 V
D. 5.1 V

Answer: C

The output voltage of the buck-boost converter is

Vo = D×Vin ÷ (1-D.=.1(78)÷.9=8.6 V.

It can step up and step down the voltage depending upon the value of the duty cycle. If the value of the duty cycle is less than .5 it will work as a buck converter and for a duty cycle greater than .5 it will work as a boost converter. 

 

29. The principle of the Boost converter can be applied for regenerative braking.

A. True
B. False

Answer: A

The Buck converter is used in motoring mode but a Boost converter can operate only in braking mode because the characteristics are in the second quadrant only. 

 

30. The unit of angular acceleration is Joule.

A. True
B. False

Answer: B

Angular acceleration is defined as a derivate of angular velocity with respect to time. It is generally written as α. The unit of angular velocity in rad/sec and of time is second so the unit of angular acceleration in rad/s2

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