# Electric Drives Questions and Answers – Solid-State Switching Circuits – Single Phase, Half-Wave, AC/DC Conversion for Resistive Loads

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This set of Electric Drives Multiple Choice Questions & Answers (MCQs) focuses on “State Switching Circuits – Single Phase, Half-Wave, AC/DC Conversion for Resistive Loads”.

1. In Half-wave controlled rectifier calculate the average value of the voltage if the supply is 13sin(25t) and firing angle is 13°.
a) 4.08 V
b) 4.15 V
c) 3.46 V
d) 5.48 V

Explanation: In Half-wave controlled rectifier, the average value of the voltage is Vm(1+cos(∝))÷2π=13(1+cos(13°))÷6.28=4.08 V. The thyristor will conduct from ∝ to π.

2. Calculate the extinction angle in purely inductive load if the firing angle is 13°.
a) 328°
b) 347°
c) 349°
d) 315°

Explanation: The extinction angle in the purely inductive load is 2π-∝=360°-13°=347°. The extinction angle is the angle at which current in the circuit becomes zero. The average value of the voltage in a purely inductive load is zero.

3. Calculate the conduction angle in purely inductive load if the firing angle is 165°.
a) 78°
b) 55°
c) 30°
d) 19°

Explanation: The conduction angle in the purely inductive load is β-α=2(π-∝)=2(180°-165°)=30°. The conduction angle is the angle for which the current exists in the circuit. The average value of the voltage in a purely inductive load is zero.

a) True
b) False

Explanation: R-L-C underdamped loads are generally leading power factor loads. They do not require forced commutation. Anti-Parallel diodes help in the commutation process.

5. In Half-wave uncontrolled rectifier calculate the average value of the voltage if the supply is 3sin(5t).
a) .95 V
b) .92 V
c) .93 V
d) .94 V

Explanation: In Half-wave uncontrolled rectifier, the average value of the voltage is Vm÷π=3÷π=.95 V. The diode will conduct only for the positive half cycle. The conduction period of the diode is π.

6. In Half-wave uncontrolled rectifier calculate the r.m.s value of the voltage if the supply is 89sin(41t).
a) 91.5 V
b) 44.5 V
c) 25.1 V
d) 15.1 V

Explanation: In Half-wave uncontrolled rectifier, the r.m.s value of the voltage is Vm÷2=89÷2=44.5 V. The diode will conduct only for the positive half cycle. The conduction period of the diode is π.

7. In Half-wave uncontrolled rectifier calculate the power dissipation across the 8 Ω resistor if the supply is 29sin(22t).
a) 26.2 W
b) 24.2 W
c) 26.1 W
d) 29.1 W

Explanation: In Half-wave uncontrolled rectifier, the r.m.s value of the voltage is Vm÷2=29÷2=14.5 V. The diode will conduct only for the positive half cycle. Power dissipation across the resistor is V2r.m.s÷R=14.52÷8=26.2 W.

8. The conduction period of diode in Half-wave uncontrolled rectifier for resistive load is ______________
a) π
b) 2π
c) 3π
d) 4π

Explanation: The conduction period of the diode in Half-wave uncontrolled rectifier for the resistive load is π. For the negative A.C supply diode will be reverse biased.

9. In Half-wave uncontrolled rectifier calculate the average value of the current for 3 Ω resistive load if the supply is 34sin(11t).
a) 3.6 A
b) 2.6 A
c) 2.5 A
d) 3.1 A

Explanation: In Half-wave uncontrolled rectifier, the average value of the current is Vm÷πR=34÷3π=3.6 A. The diode will conduct only for the positive half cycle. The conduction period of the diode is π.

10. In Half-wave controlled rectifier calculate the average value of the current for 2.5 Ω resistive load if the supply is sin(5.2t) and firing angle is 26°.
a) 0.8 V
b) 0.15 V
c) 0.12 V
d) 0.21 V

Explanation: In Half-wave controlled rectifier, the average value of the current is Vm(1+cos(∝))÷2πR=(1+cos(26°))÷6.28×2.5=.12 V. The thyristor will conduct from ∝ to π.

11. Calculate the circuit turn-off time for Half-wave controlled rectifier for a ω=5 rad/sec for resistive load.
a) .62 sec
b) .42 sec
c) .58 sec
d) .64 sec

Explanation: The value of the circuit turn-off for Half-wave controlled rectifier for a ω=5 rad/sec for the resistive load is a π÷Ω=π÷5=.62 sec.

12. Calculate the string efficiency if the de-rating factor is .429.
a) 48.1 %
b) 57.1 %
c) 47.8 %
d) 46.5 %

Explanation: The string efficiency is calculated for series and parallel connection of SCRs. The value of string efficiency is 1-(De-rating factor)=1-.429=57.1 %.

13. Calculate the output frequency for the six-pulse converter if the supply frequency is 10 Hz.
a) 40 Hz
b) 30 Hz
c) 60 Hz
d) 80 Hz

Explanation: The output of a six-pulse converter consists of six pulses in one cycle. The output frequency of the six pulse converter is 6×supply frequency=6×10=60 Hz.

14. Calculate the pulse number if the supply frequency is 2π and the output frequency is π÷6.
a) 4
b) 12
c) 16
d) 8

Explanation: The pulse number can be calculated using the ratio of input frequency to the output frequency. The value of pulse number (P) is 2π÷π÷6=12. It is a twelve-pulse converter.

15. Volt-sec balance method is based on the principle of the energy of conservation.
a) True
b) False 