250+ TOP MCQs on Solid-State Devices – Transistors and Answers

Electric Drives Multiple Choice Questions on “Solid-State Devices – Transistors”.

1. The base of a transistor is lightly doped.
A. True
B. False
Answer: A
Clarification: The base of a transistor is lightly doped because it carries the electrons from the emitter and passed it to the collector region. There should be less recombination in the base region.

2. Who is larger in size?
A. Base
B. Collector
C. Emitter
D. P-N Junction
Answer: A
Clarification: Collector is larger in size because it receives the electrons from the base region. If it is not bigger in sizer there will be crowding of electrons which causes high power dissipation.

3. Calculate the collector current using the data: I_E=14 mA, I_B=7 mA.
A. 7 mA
B. 8 mA
C. 9 mA
D. 10 mA
Answer: A
Clarification: The collector current in the transistor is I_E-I_B=14-7=7 mA. The KCL equation I_E=I_B+I_c is always in every configuration of the transistor.

4. Calculate the current gain using the data: I_c=25 mA, I_B=10 mA.
A. 3.5
B. 2.5
C. 1.4
D. 7.8
Answer: B
Clarification: The current gain is represented as β in electronic circuits. The current gain is the ratio of the collector current and base current. β=I_c÷I_B=25÷10=2.5.

5. Calculate the value of ∝ using the data: I_c=18 mA, I_E=17 mA.
A. 1.05
B. 1.47
C. 1.89
D. 2.55
Answer: A
Clarification: The value of ∝ is I_c÷I_E. The current gain is the ratio of the collector current and base current. ∝=I_c÷I_E=18÷17=1.05.

6. In a transistor, I_c=47 mA and I_E=53 mA. The value of β is _____________
A. 8.48
B. 7.83
C. 9.26
D. 3.15
Answer: B
Clarification: The value of β is I_c÷I_B. The value of I_B=I_E-I_c=53-47=6 mA. The value of β is I_c÷I_B=47÷6=7.83.

7. Choose the correct relationship between β and ∝.
A. β=∝/(1-∝)
B. β=1/(1+∝)
C. β=1/(1-∝)
D. β=1/(∝)
Answer: A
Clarification: The relationship between β and ∝ is β=∝/(1-∝). The current gain is the ratio of the collector current and base current. The value of β is I_c÷I_B.

8. A Transistor is a current operated device.
A. True
B. False
Answer: A
Clarification: Transistor is a current operated device. The value of collector current is α times of the emitter current. It is a current controlled device.

9. Calculate the value of β for α=.8.
A. 2
B. 8
C. 3
D. 4
Answer: D
Clarification: The relationship between β and ∝ is β=∝/(1-∝)=.8÷.2=4. The current gain is the ratio of the collector current and base current. The value of β is I_c÷I_B.

10. Which one of the following is correct?
A. I_E=I_B+I_c
B. I_E=I_B-I_c
C. I_E=I_c
D. I_E=I_B
Answer: A
Clarification: The KCL equation that holds correct is I_E=I_B+I_c. At any point emitter current is the sum of the base current and collector current.

11. In a transistor, I_B=2 mA and I_c=3 mA. The value of β is _____________
A. 1.5
B. 5.8
C. 9.6
D. 3.5
Answer: A
Clarification: The value of β is I_c÷I_B. The value of I_B and I_c are 2 mA, 3 mA. The value of β is I_c÷I_B=3÷2=1.5.

12. In a transistor, β=75 and I_c=53 mA. The value of I_B is _____________
A. .70 mA
B. .25 mA
C. .26 mA
D. .15 mA
Answer: A
Clarification: The value of β is I_c÷I_B. The value of I_B=I_E-I_c. The value of I_B is I_c÷β=53÷75=.70 mA. The relationship between β and ∝ is β=∝/(1-∝).

13. Calculate the value of V_peak if crest factor=5 and V_r.m.s=4 V.
A. 20 V
B. 30 V
C. 40 V
D. 50 V
Answer: A
Clarification: The value of the crest factor is V_peak÷V_r.m.s. The value of V_peak is (crest factor) × V_r.m.s=5×4=20 V. It signifies the ratio of V_peak and V_r.m.s.

14. Calculate the value of α for β=.2.
A. .85
B. .15
C. .23
D. .16
Answer: D
Clarification: The relationship between β and ∝ is ∝=β/(1+β)=.2÷1.2=.16. The current gain is the ratio of the collector current and base current. The value of α is I_c÷I_E.

15. Calculate the work done in stretching the wire of volume=12 m³ and the values of stress and strain are 28 N/m² and 7.
A. 1176 J
B. 1544 J
C. 2542 J
D. 8788 J
Answer: A
Clarification: The work done in stretching a wire is .5×Stress×Strain×Volume. The value of the work done is .5×12×28×7=1176 J.

16. Calculate the compensator rating required for cos(Φ)=.35.
A. .936 P.U
B. .154 P.U
C. .481 P.U
D. .289 P.U
Answer: A
Clarification: The compensator rating can be calculated using the relation Q_P.U=√1-cos²(∅)=√1-.1225=.936. This per unit value VAR compensator is required to improve the power factor of the system.