Current Source Inverter MCQ [Free PDF] – Objective Question Answer for Current Source Inverter Quiz

21. In a Half-wave uncontrolled rectifier calculate the r.m.s value of the voltage if the supply is 89sin(41t).   [/bg_collapse]

A. 91.5 V
B. 44.5 V
C. 25.1 V
D. 15.1 V

Answer: B

In a 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.

 

22. In a 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

Answer: A

In a 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.   

 

23. The conduction period of diode in Half-wave uncontrolled rectifier for resistive load is ______________

A. π
B. 2π
C. 3π
D. 4π

Answer: A

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

 

24. In a 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

Answer: A

In a 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.   

 

25. In a 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 the firing angle is 26°.

A. 0.8 V
B. 0.15 V
C. 0.12 V
D. 0.21 V

Answer: C

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.   

 

26. 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

Answer: A

The value of the circuit turn-off for the Half-wave controlled rectifier for a ω=5 rad/sec for the resistive load is a

π÷Ω=π÷5=.62 sec.   

 

27. Calculate the string efficiency if the de-rating factor is .429.

A. 48.1 %
B. 57.1 %
C. 47.8 %
D. 46.5 %

Answer: B

The string efficiency is calculated for the series and parallel connection of SCRs.

The value of string efficiency is 1-(De-rating factor)=1-.429=57.1 %.   

 

28. 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

Answer: C

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.   

 

29. 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

Answer: B

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

 

30. Volt-sec balance method is based on the principle of the energy of conservation.

A. True
B. False

Answer: A

The 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.   

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