Electronics MCQs set 2 for Bahria University Entry Test Physics — 20 solved questions.
Q1. In a p-n junction, the depletion layer is due to
Answer: immobile ions
Explanation: Depletion layer is formed due to immobile ions left behind after majority carriers diffuse across the junction.
Q2. The voltage gain of an amplifier with negative feedback is
Answer: inversely proportional to feedback fraction
Explanation: Voltage gain with feedback = A / (1 + βA), for large A, gain is inversely proportional to β.
Q3. A diode having a forward resistance of 50 Ω is used for half-wave rectification. The input voltage is 220 V (rms). The load resistance is 1 kΩ. The output dc current is
Answer: 0.099 A
Explanation: Idc = Im / π = Vm / (π * (Rf + Rl)) = 220√2 / (π * (50 + 1000)) = 0.099 A.
Q4. The bandwidth of an amplifier is
Answer: inversely proportional to gain
Explanation: Bandwidth = f2 - f1, gain-bandwidth product is constant, hence bandwidth is inversely proportional to gain.
Q5. A transistor is operated in common emitter configuration. The collector supply voltage is 10 V and the voltage drop across a resistor of 1 kΩ in the collector circuit is 0.7 V. The collector current is
Answer: 0.7 mA
Explanation: Ic = Vc / Rc = 0.7 V / 1 kΩ = 0.7 mA.
Q6. In a CE amplifier, the output voltage is
Answer: out of phase with input voltage by 180°
Explanation: CE amplifier inverts the input signal, hence output is 180° out of phase.
Q7. The emitter current in a transistor is
Answer: equal to the sum of base and collector currents
Explanation: Ie = Ib + Ic, emitter current is the sum of base and collector currents.
Q8. A Zener diode is used for
Answer: stabilization
Explanation: Zener diode is used for voltage stabilization due to its constant voltage drop in reverse bias.
Q9. The barrier potential in a p-n junction is
Answer: 0.7 V
Explanation: Barrier potential for Si is approximately 0.7 V.
Q10. The common-mode rejection ratio (CMRR) of a differential amplifier is
Answer: the ratio of differential gain to common-mode gain
Explanation: CMRR = Ad / Ac, where Ad is differential gain and Ac is common-mode gain.
Q11. The input impedance of a FET is
Answer: high
Explanation: FET has high input impedance due to its insulated gate.
Q12. The output of a logic gate is 1 when both inputs are 0. The gate is
Answer: NAND
Explanation: NAND gate produces 1 output when both inputs are 0.
Q13. A silicon diode is forward-biased with a voltage of 0.7 V. The current is 10 mA. The dynamic resistance is
Answer: 2.5 Ω
Explanation: Dynamic resistance = ηVt / I = 25 mV / 10 mA = 2.5 Ω.
Q14. The gain-bandwidth product of an amplifier is
Answer: constant
Explanation: Gain-bandwidth product is a constant for an amplifier.
Q15. The collector current in a transistor is 10 mA. If 90% of the electrons emitted reach the collector, the emitter current is
Answer: 11.11 mA
Explanation: Ie = Ic / 0.9 = 10 mA / 0.9 = 11.11 mA.
Q16. The voltage gain of a CE amplifier with a load resistance of 1 kΩ is 100. If the load resistance is increased to 2 kΩ, the voltage gain becomes
Answer: 200
Explanation: Voltage gain is proportional to load resistance, hence gain doubles when Rl is doubled.
Q17. The current gain of a transistor is 50. If the base current is 50 μA, the collector current is
Answer: 2.5 mA
Explanation: Ic = β * Ib = 50 * 50 μA = 2.5 mA.
Q18. In a p-n junction, the depletion region is due to
Answer: immobile ions
Explanation: Depletion region is formed due to immobile ions left behind after recombination of majority carriers.
Q19. The current gain of a transistor in CE mode is
Answer: always greater than 1
Explanation: In CE mode, current gain (β) = ΔIc / ΔIb, which is typically greater than 1.
Q20. The output of a logic gate is 1 when all its inputs are 0. The gate is
Answer: NAND
Explanation: NAND gate produces output 1 when all inputs are 0, as per its truth table.