# Physics Questions and Answers – Semiconductor Electronics – Application of Junction Diode as a Rectifier

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This set of Physics MCQs for IIT JEE Exam focuses on “Semiconductor Electronics – Application of Junction Diode as a Rectifier”.

1. What is a rectifier used for?
a) Convert ac voltage to dc voltage
b) Convert dc voltage to ac voltage
c) Measure resistance
d) Measure current

Explanation: A rectifier is based on the fact that a forward bias p-n junction conducts and a reverse bias p-n junction does not conduct electricity. The rectifier is used to convert alternating current voltage (ac) to direct current voltage (dc).

2. How many main types of rectifiers are there?
a) 1
b) 5
c) 2
d) 4

Explanation: Rectifier is a device that does the process of rectification. This means that rectifiers straighten the direction of the current flowing through it. There are mainly 2 types of rectifiers, namely, full-wave rectifiers and half-wave rectifiers.

3. What is the ripple factor for a half-wave rectifier?
a) 2.0
b) 1.21
c) 0.482
d) 0.877

Explanation: For a half-wave rectifier,
Irms=$$\frac {I_m}{2}$$; Idc=$$\frac {I_m}{\pi }$$
r = $$\sqrt {\frac {(\frac {I_m}{2} )^2}{(\frac {I_m}{\pi } )^2} – 1}$$
r=1.21
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4. The ripple frequency of a full-wave rectifier is twice to that of a half-wave rectifier.
a) True
b) False

Explanation: Yes, this is a true statement. The ripple frequency is doubled in a full-wave rectifier because we have to rectify both the positive and negative sides of the waveform. For example, if the input frequency is 50 Hz, then the ripple frequency of a full-wave rectifier is 100 Hz.

5. Identify the expression for rectification efficiency from the following.
a) η=$$\frac {ac \, input \, power \, from \, transformer \, secondary}{dc \, power \, delivered \, to \, load}$$
b) η=dc power delivered to load × ac input power from transformer secondary
c) η=dc power delivered to load + ac input power from transformer secondary
d) η=$$\frac {dc \, power \, delivered \, to \, load}{ac \, input \, power \, from \, transformer \, secondary}$$

Explanation: The rectification efficiency tells us what percentage of the total input ac power can be converted into useful dc output power. The expression for rectification efficiency is given as:
η=$$\frac {dc \, power \, delivered \, to \, load}{ac \, input \, power \, from \, transformer \, secondary}$$

6. What is the form factor for a full-wave rectifier?
a) 1.11
b) 1.57
c) 2.62
d) 0.453

Explanation: For a full-wave rectifier,
Irms=$$\frac {I_m}{\sqrt {2}}$$; Idc=$$\frac {2I_m}{\pi}$$
Form factor=$$\frac {I_{rms}}{I_{dc}}$$
Form factor=$$\frac {\frac {I_m}{\sqrt {2}}} {\frac {2I_m}{\pi}}$$
Form factor=$$\frac {\pi }{2\sqrt {2}}$$=1.11

7. An alternating voltage of 360 V, 50 Hz is applied to a full-wave rectifier. The internal resistance of each diode is 100 W. If RL = 5 kW, then what is the peak value of output current?
a) 0.9 A
b) 0.07 A
c) 0.097 A
d) 1.097 A

Explanation: The required equation is as follows:
Ipeak=Irms × √2=$$\frac {V_{rms} \times \sqrt {2}}{R_L+2r_p}$$
Ipeak=$$\frac {360 \times \sqrt {2}}{5000 + 200}$$
Ipeak=$$\frac {360 \times 1.414}{5200}$$
Ipeak=0.097 A

8. Find the value of output direct current if the peak value of output current is given as 0.095 A.
a) 0.6
b) 0.060
c) 0.05
d) 6.06

Explanation: Given: I0 = 0.095 A
The required equation is ➔ IDC = $$\frac {(2 \times I_O)}{\pi }$$
IDC = $$\frac {(2 \times 0.095)}{3.14}$$
IDC = 0.060 A.

9. What is the rms value of output current if the peak value of output current is given as 0.092 A?
a) 0.65 A
b) 6.5 A
c) 0.45 A
d) 0.065 A

Explanation: Given: I0 = 0.092 A
The required equation is ➔ Irms = $$\frac {I_O}{\sqrt {2}}$$
Irms = $$\frac {0.092}{1.414}$$
Irms = 0.065 A

10. Calculate the value of peak reverse voltage (P.I.V.) if the full-wave rectifier has an alternating voltage of 300 V.
a) 849 V
b) 800 V
c) 750 V
d) 870 V

Explanation: Given: Erms = 300 V
The required equation is ➔ P.I.V. = 2 × E0 or P.I.V. = 2√2 × Erms
P.I.V. = 2√2 × 300 V
P.I.V. = 848.52 V ≈ 849 V.

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