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1. DC average current of a center taped full wave rectifier
(Where Im is the maxImum peak current of input)
a) 2Im/ᴨ
b) Im/ᴨ
c) Im/2ᴨ
d) 1.414Im/ᴨ

View Answer

Answer: a [Reason:] Average DC current is the average current in the output of rectifier. For a center taped full wave rectifier, if a sinusoidal input is given output will contain only one half cycle repeatedly. So average current will be twice as half wave rectifier.

2. DC power output of center taped full wave rectifier is equal to
(Im is the peak current and RL is the load resistance)
a) (2Im2/ ᴨ2)RL
b) (Im2/2 ᴨ2)RL
c) (Im2/ ᴨ2)RL
d) (4Im2/ ᴨ2)RL

View Answer

Answer: d [Reason:] DC output power is the power output of the rectifier. It is equal to VDCIDC. We know VDC for a center taped rectifier is 2Vm/ᴨ and IDC for a center tap rectifier is 2Im/ᴨ. We also know VDC = IDC/RL. Hence output power is (4Im2/ ᴨ2)RL.

3. Ripple factor of center taped full wave rectifier is
a) 1.414
b) 1.21
c) 1.3
d) 0.48

View Answer

Answer: d [Reason:] Ripple factor of a rectifier is the measure of effectiveness of a power supply filter in reducing the ripple voltage. It is calculated by taking ratio of ripple voltage to DC output voltage. For a center taped full wave rectifier, it is 0.482.

4. If input frequency is 50Hz then ripple frequency of center taped full wave rectifier will be equal to
a) 100Hz
b) 50Hz
c) 25Hz
d) 500Hz

View Answer

Answer: a [Reason:] Since in the output of center taped rectifier one half cycle is repeated hence frequency will twice as that of input frequency. That is 100Hz.

5. Transformer utilization factor of a center taped full wave rectifier is equal to
a) 0.623
b) 0.678
c) 0.693
d) 0.625

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Answer: c [Reason:] Transformer utilization 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 center taped full wave rectifier it is equal to 0.693.

6. If peak voltage on a center taped full wave rectifier circuit is 5V and diode cut-in voltage is 0.7, then peak inverse voltage on diode will be
a) 4.3V
b) 9.3V
c) 5.7V
d) 10.7V

View Answer

Answer: b [Reason:] PIV is the maxImum reveres bias voltage that can be appeared across a diode in the circuit. If PIV rating of the diode is less than this value breakdown of diode may occur. For a center tap full wave rectifier PIV of diode is 2Vm – VD. Therefore, PIV is 10-0.7V.

7. Efficiency of center taped full wave rectifier is
a) 50%
b) 81.2%
c) 40.6%
d) 45.3%

View Answer

Answer: b [Reason:] Efficiency of a rectifier is the effectiveness of rectifier to convert AC to DC.It is obtained by taking ratio of DC power output to maxImum AC power delivered to load. It is usually expressed in percentage. For center taped full wave rectifier, it is 81.2%.

8. In an center taped full wave rectifier, the input sine wave is 20sin500 ᴨt. The average output voltage is
a) 12.73V
b) 6.93V
c) 11.62V
d) 3.23V

View Answer

Answer: a [Reason:] The equation of sine wave is in the form Em sin wt. Therefore, Em=20 Hence output voltage is 2Em/ ᴨ. That is 40/ ᴨ.

9. In an center taped full wave rectifier, the input sine wave is 200sin50 ᴨt. If load resistance is of 1k then average DC power output of half wave rectifier is
a) 12.56W
b) 16.20W
c) 4.02W
d) 8.04W

View Answer

Answer: b [Reason:] The equation of sine wave is in the form Em sin wt. On comparing Em = 200 Power = 4Em2/ ᴨ2RL = 800/ ᴨ2x 1000 = 16.20W.

10. In an center taped full wave rectifier, the input sine wave is 250sin100 ᴨt. The output ripple frequency of rectifier will be
a) 50Hz
b) 200Hz
c) 100Hz
d) 25Hz

View Answer

Answer: c [Reason:] The equation of sine wave is in the form Em sin wt. Therefore, w = 100ᴨ that is, frequency f = w/2ᴨ = 50Hz Since output of center taped full wave rectifier have double the frequency of input, output frequency is 100Hz.