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1. DC average current of a half wave rectifier output is
(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: b [Reason:] Average DC current of half wave rectifier is Im/ᴨ . Since output of half wave rectifier contains only one half of the input. The average value is the half of the area of one half cycle of sine wave with peak Im. This is equal to Im/ᴨ.

2. DC power output of half 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: c [Reason:] Average DC power of half wave rectifier output is (Im2/ ᴨ2)RL. Since power is VDC * IDC, = Im/ᴨ x Vm/ᴨ = VmIm/ ᴨ2 We know Vm = Im RL. Therefore, power = (Im2/ ᴨ2)RL.

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

View Answer

Answer: b [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 half wave rectifier, it is 1.21.

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

View Answer

Answer: b [Reason:] The ripple frequency of output and input is same since one half cycle of input is passed and other half cycle is blocked. So effectively frequency is the same.

5. Transformer utilization factor of a half wave rectifier is equal to
a) 0.267
b) 0.287
c) 0.297
d) 0.256

View Answer

Answer: b [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 half wave rectifier it is low and equal to 0.287.

6. If peak voltage on a half wave rectifier circuit is 5V and diode cut-in voltage is 0.7, then peak inverse voltage on diode will be
a) 5.7V
b) 3.6V
c) 4.3V
d) 5V

View Answer

Answer: c [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 half wave rectifier PIV of diode is Vm – VD. Therefore, PIV is 5-0.7V.

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

View Answer

Answer: c [Reason:] Efficiency of a rectifier is the effectiveness of rectifier to convert AC to DC. For half wave rectifier it is only 40.6%.

8. In an half wave rectifier, the input sine wave is 200sin100 ᴨt. The average output voltage is
a) 57.456V
b) 60.548V
c) 75.235V
d) 63.694V

View Answer

Answer: d [Reason:] The equation of sine wave is in the form Em sin wt. Therefore, Em=200 Hence output voltage is Em/ ᴨ. That is 200/ ᴨ.

9. In an half wave rectifier, the input sine wave is 200sin200 ᴨt. If load resistance is of 1k then average DC power output of half wave rectifier is
a) 3.25W
b) 4.05W
c) 5.02W
d) 6.25W

View Answer

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

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

View Answer

Answer: c [Reason:] The equation of sine wave is in the form Em sin wt. On comparing equation w = 100 ᴨ rad/s We know w=2 ᴨ x frequency. Therefore, frequency = 50 Hz. Ripple frequency and input frequency are the same.