250+ TOP MCQs on Semiconductor Electronics – Application of Junction Diode as a Rectifier | Class12 Physics

Physics MCQs for IIT JEE Exam 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
Answer: a
Clarification: 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
Answer: c
Clarification: 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
Answer: b
Clarification: For a half-wave rectifier,
I_rms=(frac {I_m}{2}); I_dc=(frac {I_m}{pi })
r = (sqrt {frac {(frac {I_m}{2} )^2}{(frac {I_m}{pi } )^2} – 1})
r=1.21

4. The ripple frequency of a full-wave rectifier is twice to that of a half-wave rectifier.
a) True
b) False
Answer: a
Clarification: 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})
Answer: d
Clarification: 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
Answer: a
Clarification: For a full-wave rectifier,
I_rms=(frac {I_m}{sqrt {2}}); I_dc=(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 }{2sqrt {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 R_L = 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
Answer: c
Clarification: The required equation is as follows:
I_peak=I_rms × √2=(frac {V_{rms} times sqrt {2}}{R_L+2r_p})
I_peak=(frac {360 times sqrt {2}}{5000 + 200})
I_peak=(frac {360 times 1.414}{5200})
I_peak=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
Answer: b
Clarification: Given: I₀ = 0.095 A
The required equation is ➔ I_DC = (frac {(2 times I_O)}{pi })
I_DC = (frac {(2 times 0.095)}{3.14})
I_DC = 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
Answer: d
Clarification: Given: I₀ = 0.092 A
The required equation is ➔ I_rms = (frac {I_O}{sqrt {2}})
I_rms = (frac {0.092}{1.414})
I_rms = 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
Answer: a
Clarification: Given: E_rms = 300 V
The required equation is ➔ P.I.V. = 2 × E₀ or P.I.V. = 2√2 × E_rms
P.I.V. = 2√2 × 300 V
P.I.V. = 848.52 V ≈ 849 V.

Physics MCQs for IIT JEE Exam,