# Electronic Devices and Circuits Questions and Answers – Full-wave Rectifier

This set of Electronic Devices and Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Full-wave Rectifier”.

1. Efficiency of a centre tapped full wave rectifier is _________
a) 50%
b) 46%
c) 70%
d) 81.2%

Explanation: 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 load. 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) 56.2V
b) 78.5V
c) 95.4V
d) 34.4V

Explanation: The ripple voltage (Vϒ)RMS = ϒ * VDC.
Here ϒ is ripple factor of a full wave rectifier and it is 0.482.

VDC = (2/π) * VRMS* √2 = 0.636 * 220 * √2 = 198V
Hence, (Vϒ)RMS = 0.482 * 198 = 95.4V.

3. A full wave rectifier delivers 50W to a load of 200Ω. If the ripple factor is 2%, calculate the AC ripple across the load.
a) 2V
b) 5V
c) 4V
d) 1V

Explanation: We know that, PDC=VDC2/RL. So, VDC=(PDC*RL)1/2=100001/2=100V.
Here, ϒ=0.02
ϒ=VAC/VDC=VAC/100.So, VAC=0.02*100=2V.

4. A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. 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

Explanation: The AC power input PIN=IRMS2(RF+Rr).
IRMS=Im/√2=Vm/(Rf+RL)√2=30/(1500+10)*1.414=13.5mA
So, PIN=(13.5*10-3)2*(1500+10)=275.2mW.
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5. In a centre tapped full wave rectifier, RL=1KΩ and for diode Rf=10Ω. The primary voltage is 800sinωt with transformer turns ratio=2. The ripple factor will be _________

a) 54%
b) 48%
c) 26%
d) 81%

Explanation: The ripple factor ϒ= [(IRMS/IAVG)2 – 1]1/2. IRMS =Im /√2=Vm/(Rf+RL)√2=200/1.01=198.
(Secondary line to line voltage is 800/2=400. Due to centre tap Vm=400/2=200)
IRMS=198/√2=140mA, IAVG=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

Explanation: In the output of the centre tapped rectifier, one of the half cycle is repeated. 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

Explanation: 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)
b)
c)
d)

Explanation: 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

Explanation: 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=2Vm-Vd = 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

Explanation: The equation of sine wave is in the form Vmsinωt. 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, fout = 100Hz.

Sanfoundry Global Education & Learning Series – Electronic Devices and Circuits.

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