250+ TOP MCQs on Full-wave Rectifier and Answers

Electronic Devices and Circuits Multiple Choice Questions on “Full-wave Rectifier”.

1. Efficiency of a centre tapped full wave rectifier is _________
A. 50%
B. 46%
C. 70%
D. 81.2%
Answer: D
Clarification: Efficiency of a rectifier is the effectiveness to convert AC to DC. It’s obtained by taking ratio of DC power output to maximum AC power delivered to loaC. It’s usually expressed in percentage. For centre tapped full wave rectifier, it’s 81.2%.

2. A full wave rectifier supplies a load of 1KΩ. The AC voltage applied to diodes is 220V (rms). If diode resistance is neglected, what is the ripple voltage?
A. 0.562V
B. 0.785V
C. 0.954V
D. 0.344V
Answer: C
Clarification: The ripple voltage is (V_ϒ)_RMS=ϒV_DC /100.
V_DC=0.636*V_RMS* √2=0.636*220* √2=198V and ripple factor ϒ for full wave rectifier is 0.482.
Hence, (V_ϒ)_RMS=0.482*198 /100=0.954V.

3. A full wave rectifier delivers 50W to a load of 200Ω. If the ripple factor is 2%, calculate the AC ripple across the loaC.
A. 2V
B. 5V
C. 4V
D. 1V
Answer: A
Clarification: We know that, P_DC=V_DC²/R_L. So, V_DC=(P_DC*RL)^1/2=10000^1/2=100V.
Here, ϒ=0.02
ϒ=V_AC/V_DC=V_AC/100.So, V_AC=0.02*100=2V.

4. A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have R_f=10Ω, R_r=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input.
A. 368.98mW
B. 275.2mW
C. 145.76mW
D. 456.78mW
Answer: B
Clarification: The AC power input P_IN=I_RMS²(R_F+R_r).
I_RMS=I_m/√2=V_m/(R_f+R_L)√2=30/(1500+10)*1.414=13.5mA
So, P_IN=(13.5*10-3)2*(1500+10)=275.2mW.

5. In a centre tapped full wave rectifier, R_L=1KΩ and for diode Rf=10Ω. The primary voltage is 800sinωt with transformer turns ratio=2. The ripple factor will be _________
2018/02/electronic-devices-circuits-questions-answers-full-wave-rectifier-q5″>
A. 54%
B. 48%
C. 26%
D. 81%
Answer: B
Clarification: The ripple factor ϒ= [(I_RMS/I_AVG)² – 1]^1/2. I_RMS =I_m /√2=V_m/(R_f+R_L)√2=200/1.01=198.
(Secondary line to line voltage is 800/2=400. Due to centre tap V_m=400/2=200)
I_RMS=198/√2=140mA, I_AVG=2*198/π=126mA. ϒ=[(140/126)²-1]^1/2=0.48. So, ϒ=48%.

6. If input frequency is 50Hz for a full wave rectifier, the ripple frequency of it would be _________
A. 100Hz
B. 50Hz
C. 25Hz
D. 500Hz
Answer: A
Clarification: In the output of the centre tapped rectifier, one of the half cycle is repeateC. The frequency will be twice as that of input frequency. So, it’s 100Hz.

7. Transformer utilization factor of a centre tapped full wave rectifier is_________
A. 0.623
B. 0.678
C. 0.693
D. 0.625
Answer: C
Clarification: Transformer utilisation factor is the ratio of AC power delivered to load to the DC power rating. This factor indicates effectiveness of transformer usage by rectifier. For a half wave rectifier, it’s low and equal to 0.693.

8. In the circuits given below, the correct full wave rectifier is _________
A. 2018/02/electronic-devices-circuits-questions-answers-full-wave-rectifier-q8″>
B. 2018/02/electronic-devices-circuits-questions-answers-full-wave-rectifier-q8a”>
C. 2018/02/electronic-devices-circuits-questions-answers-full-wave-rectifier-q8b”>
D. 2018/02/electronic-devices-circuits-questions-answers-full-wave-rectifier-q8c”>
Answer: C
Clarification: When the input is applied, a full wave rectifier should have a current flow. The flow should be in the same direction for both positive and negative half cycles. Only the third circuit satisfies the above condition.

9. If the peak voltage on a centre tapped full wave rectifier circuit is 5V and diode cut in voltage is 0.7. The peak inverse voltage on diode is_________
A. 4.3V
B. 9.3V
C. 5.7V
D. 10.7V
Answer: B
Clarification: PIV is the maximum reverse bias voltage that can be appeared across a diode in the given circuit, if PIV rating is less than this value of breakdown of diode will occur. For a rectifier, PIV=2V_m-V_d = 10-0.7 = 9.3V.

10. In a centre tapped full wave rectifier, the input sine wave is 250sin100t. The output ripple frequency will be _________
A. 50Hz
B. 100Hz
C. 25Hz
D. 200Hz
Answer: B
Clarification: The equation of sine wave is in the form V_msinωt. So, by comparing we get ω=100. Frequency, f =ω/2=50Hz. The output of centre tapped full wave rectifier has double the frequency of inpu. Hence, f_out = 100Hz.