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

«
»

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
View Answer

Answer: a
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.
advertisement

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

Answer: b
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°
View Answer

Answer: c
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.

4. R-L-C underdamped loads are generally lagging power factor loads.
a) True
b) False
View Answer

Answer: b
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
View Answer

Answer: a
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.
advertisement

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
View Answer

Answer: b
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
View Answer

Answer: a
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
View Answer

Answer: a
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
View Answer

Answer: 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.
advertisement

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
View Answer

Answer: c
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
View Answer

Answer: a
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 %
View Answer

Answer: b
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
View Answer

Answer: c
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.
advertisement

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
View Answer

Answer: b
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
View Answer

Answer: a
Explanation: Volt-sec balance method states that the net area over the cycle will be zero. It is based on the principle of the energy of conservation. The amount of charging is equal to discharging.

Sanfoundry Global Education & Learning Series – Electric Drives.

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

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
advertisement
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn