MCQ-6
This set of Class 12 Physics Chapter 14 Multiple Choice Questions & Answers (MCQs) 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
Answer: a
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
Answer: c
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
Answer: b
Explanation: For a half-wave rectifier,
Irms=Im2; Idc=Imπ
r = (Im2)2(Imπ)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
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=acinputpowerfromtransformersecondarydcpowerdeliveredtoload
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=dcpowerdeliveredtoloadacinputpowerfromtransformersecondary
View Answer
Answer: d
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=dcpowerdeliveredtoloadacinputpowerfromtransformersecondary
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
Explanation: For a full-wave rectifier,
Irms=Im2√; Idc=2Imπ
Form factor=IrmsIdc
Form factor=Im2√2Imπ
Form factor=π22√=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
Answer: c
Explanation: The required equation is as follows:
Ipeak=Irms × √2=Vrms×2√RL+2rp
Ipeak=360×2√5000+200
Ipeak=360×1.4145200
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
Answer: b
Explanation: Given: I0 = 0.095 A
The required equation is ➔ IDC = (2×IO)π
IDC = (2×0.095)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
Answer: d
Explanation: Given: I0 = 0.092 A
The required equation is ➔ Irms = IO2√
Irms = 0.0921.414
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
Answer: a
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.