This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Halfwave Rectifier”.

1. DC average current of a half wave rectifier output is ___________

(Where I_{m} is the maximum peak current of input)

a) 2I_{m}/ᴨ

b) I_{m}/ᴨ

c) I_{m}/2ᴨ

d) 1.414I_{m}/ᴨ

View Answer

Explanation: Average DC current of half wave rectifier is I

_{m}/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 I

_{m}. This is equal to I

_{m}/ᴨ.

Thus this is calculated by Area of graph of current/Perion.

I

_{DC}=1/2π∫idx between 0 to 2π.

This is equal to I

_{m}/ᴨ.

2. DC power output of half wave rectifier is equal to ____________

(I_{m} is the peak current and R_{L} is the load resistance)

a) (2I_{m}^{2}/ ᴨ^{2})R_{L}

b) (I_{m}^{2}/2 ᴨ^{2})R_{L}

c) (I_{m}^{2}/ ᴨ^{2})R_{L}

d) (4I_{m}^{2}/ ᴨ^{2})R_{L}

View Answer

Explanation: Average DC power of half wave rectifier output is (I

_{m}

^{2}/ ᴨ

^{2})R

_{L}. Since power is VDC * IDC,

= I

_{m}/ᴨ

_{x}V

_{m}/ᴨ = V

_{m}I

_{m}/ ᴨ

^{2}

We know V

_{m}= I

_{m}R

_{L}. Therefore, power = (I

_{m}

^{2}/ ᴨ

^{2})R

_{L}.

3. Ripple factor of half wave rectifier is _________

a) 1.414

b) 1.21

c) 1.3

d) 0.48

View Answer

Explanation: Ripple factor of a rectifier is the measure of the effectiveness of a power supply filter

in reducing the ripple voltage. It is calculated by taking ratio of RMS AC component of output voltage to DC component of output voltage.

r = √I

_{rms}

^{2}– I

_{DC}

^{2}/IDC

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

Explanation: The ripple frequency of output and input is the 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

Explanation: Transformer utilization factor is the ratio of DC output power to the AC rating of the secondary winding. AC rating is the product of RMS voltage across winding and RMS current through the winding, expressed in volt-amp. This factor indicates the effectiveness of transformer usage by a 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) 3.6V

b) 4.3V

c) 5V

d) 5.7V

View Answer

Explanation: PIV is the maximum reverse 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 V

_{m}. Therefore, PIV is 5V.

7. Efficiency of half wave rectifier is __________

a) 50%

b) 81.2%

c) 40.6%

d) 45.3%

View Answer

Explanation: Efficiency of a rectifier is a measure of the ability of a rectifier to convert input power into DC power. Mathematically it is equal to the ratio of DC output power to the total input power and efficiency = 40.6xR

_{L}/R

_{F}+R

_{L}%. Its maximum value is 40.6 %.

8. In a half wave rectifier, the input sine wave is 200sin100 ᴨt Volts. The average output voltage is __________

a) 57.456V

b) 60.548V

c) 75.235V

d) 63.661V

View Answer

Explanation: The equation of sine wave is in the form E

_{m}sin wt.

Therefore, E

_{m}=200

Hence output voltage is E

_{m}/ᴨ. That is 200/ᴨ = 63.6619V.

9. In a half wave rectifier, the input sine wave is 200sin200 ᴨt Volts. If load resistance is of 1k then the average DC power output of half wave rectifier is __________

a) 3.25W

b) 4.05W

c) 5.02W

d) 6.25W

View Answer

Explanation: The equation of sine wave is in the form E

_{m}sin wt.

On comparing E

_{m}= 200

Power = E

_{m}

^{2}/ ᴨ

^{2}R

_{L}= 200/ ᴨ

^{2}x 1000.

10. In a half wave rectifier, the input sine wave is 250sin100 ᴨt Volts. The output ripple frequency of rectifier will be __________

a) 100Hz

b) 200Hz

c) 50Hz

d) 25Hz

View Answer

Explanation: The equation of sine wave is in the form E

_{m}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.

**Sanfoundry Global Education & Learning Series – Analog Circuits.**

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