LUMS LCAT Physics: Electronics MCQs

LUMS LCAT Physics: Electronics MCQs — sample questions

1. Question 1

   Q1. In a common emitter amplifier, the input resistance is 1 kΩ and load resistance is 3 kΩ. If β = 100, voltage gain is

   • A) 100
   • B) 200
   • C) 300
   • D) 400

   Answer: 300

   Explanation: Voltage gain = β * (R_load / R_input) = 100 * (3 / 1) = 300, using the formula for voltage gain in a common emitter amplifier.

2. Question 2

   Q2. The energy band gap is maximum in

   • A) metals
   • B) superconductors
   • C) semiconductors
   • D) insulators

   Answer: insulators

   Explanation: Insulators have the largest energy band gap, typically > 3 eV, making it difficult for electrons to jump to the conduction band.

3. Question 3

   Q3. The depletion layer in a p-n junction diode consists of

   • A) electrons
   • B) holes
   • C) immobile ions
   • D) both electrons and holes

   Answer: immobile ions

   Explanation: The depletion layer is formed by the immobile ions, which are the result of the diffusion of electrons and holes across the junction.

4. Question 4

   Q4. In a p-type semiconductor, the acceptor level is 0.05 eV above the valence band. The maximum wavelength of light that can create a hole is

   • A) 2480 nm
   • B) 248 nm
   • C) 1240 nm
   • D) 620 nm

   Answer: 2480 nm

   Explanation: Using E = hc / λ, we get λ = hc / E = (6.626 * 10^-34 * 3 * 10^8) / (0.05 * 1.6 * 10^-19) = 2480 nm.

5. Question 5

   Q5. The current gain of a transistor in common base configuration is 0.98. The current gain in common emitter configuration is

   • A) 49
   • B) 50
   • C) 98
   • D) 100

   Answer: 49

   Explanation: Using the formula β = α / (1 - α), we get β = 0.98 / (1 - 0.98) = 49.

6. Question 6

   Q6. The input resistance of a silicon transistor is 1 kΩ. If the base current changes by 10 μA, the collector current changes by 1 mA. The current gain is

   • A) 50
   • B) 100
   • C) 150
   • D) 200

   Answer: 100

   Explanation: Using the formula β = ΔI_c / ΔI_b, we get β = 1 mA / 10 μA = 100.

7. Question 7

   Q7. A transistor is operated in common emitter configuration at V_CC = 2 V such that a change in the base current from 100 μA to 300 μA produces a change in the collector current from 10 mA to 20 mA. The current gain is

   • A) 25
   • B) 50
   • C) 75
   • D) 100

   Answer: 50

   Explanation: Using the formula β = ΔI_c / ΔI_b, we get β = (20 - 10) mA / (300 - 100) μA = 50.

8. Question 8

   Q8. The resistance of a reverse-biased p-n junction diode is

   • A) very low
   • B) low
   • C) high
   • D) very high

   Answer: very high

   Explanation: A reverse-biased p-n junction diode has a very high resistance due to the depletion layer.

9. Question 9

   Q9. The breakdown voltage of a p-n junction diode is 10 V. The maximum reverse current is

   • A) 1 mA
   • B) 10 mA
   • C) 100 mA
   • D) dependent on the diode

   Answer: dependent on the diode

   Explanation: The maximum reverse current depends on the diode's characteristics, not just the breakdown voltage.

10. Question 10

    Q10. The current in a p-n junction diode is due to

    • A) only majority carriers
    • B) only minority carriers
    • C) both majority and minority carriers
    • D) neither majority nor minority carriers

    Answer: both majority and minority carriers

    Explanation: The current in a p-n junction diode is due to both majority and minority carriers, depending on the biasing.

11. Question 11

    Q11. The dynamic resistance of a p-n junction diode is given by

    • A) ΔV / ΔI
    • B) ΔI / ΔV
    • C) V / I
    • D) I / V

    Answer: ΔV / ΔI

    Explanation: The dynamic resistance is given by the ratio of the change in voltage to the change in current.

12. Question 12

    Q12. The ripple factor of a full-wave rectifier is

    • A) 0.48
    • B) 0.812
    • C) 1.21
    • D) 1.414

    Answer: 0.48

    Explanation: The ripple factor for a full-wave rectifier is given by Γ = √((I_rms / I_dc)² - 1) = 0.48.

13. Question 13

    Q13. The efficiency of a full-wave rectifier is

    • A) 40.6%
    • B) 81.2%
    • C) 100%
    • D) 120%

    Answer: 81.2%

    Explanation: The efficiency of a full-wave rectifier is given by η = (0.812) * 100% = 81.2%.

14. Question 14

    Q14. The output frequency of a full-wave rectifier is

    • A) same as the input frequency
    • B) twice the input frequency
    • C) half the input frequency
    • D) four times the input frequency

    Answer: twice the input frequency

    Explanation: The output frequency of a full-wave rectifier is twice the input frequency.

15. Question 15

    Q15. In a bridge rectifier, the number of diodes used is

    • A) 2
    • B) 4
    • C) 6
    • D) 8

    Answer: 4

    Explanation: A bridge rectifier uses 4 diodes to rectify the input AC voltage.

16. Question 16

    Q16. The zener diode is used as

    • A) an amplifier
    • B) a rectifier
    • C) a voltage regulator
    • D) a oscillator

    Answer: a voltage regulator

    Explanation: A zener diode is used as a voltage regulator due to its ability to maintain a constant voltage.

17. Question 17

    Q17. The zener voltage of a zener diode is 6 V. The current through the diode is 10 mA. If the power rating of the diode is 300 mW, the value of the series resistance is

    • A) 200 Ω
    • B) 400 Ω
    • C) 600 Ω
    • D) 800 Ω

    Answer: 400 Ω

    Explanation: Using P = V * I, we get I = P / V = 300 mW / 6 V = 50 mA. The current through the series resistance is 50 mA - 10 mA = 40 mA. Using V = IR, we get R = V / I = (V_in - 6) / 0.04. Assuming V_in = 10 V (not given), R = (10 - 6) / 0.04 = 100 / 0.04 = 400 Ω (approx.).

18. Question 18

    Q18. In a transistor, the base is

    • A) heavily doped
    • B) lightly doped
    • C) moderately doped
    • D) undoped

    Answer: lightly doped

    Explanation: The base of a transistor is lightly doped to minimize the recombination of charge carriers.

19. Question 19

    Q19. The collector current in a transistor is

    • A) always less than the emitter current
    • B) always greater than the emitter current
    • C) always equal to the emitter current
    • D) sometimes less than, sometimes greater than the emitter current

    Answer: always less than the emitter current

    Explanation: The collector current is always less than the emitter current due to the base current.

20. Question 20

    Q20. The common emitter amplifier has

    • A) low input resistance
    • B) high input resistance
    • C) low output resistance
    • D) high voltage gain

    Answer: high voltage gain

    Explanation: The common emitter amplifier has a high voltage gain due to the ratio of the collector resistance to the base resistance.