Bahria University Entry Test Physics Electronics — Set 3

Bahria University Entry Test Physics Electronics — Set 3

1. Question 1

   Q1. A diode having internal resistance 20 Ω is used for half-wave rectification. If the applied voltage is v = 50 sin(πt), the peak current is

   • A) 1.25 A
   • B) 2.5 A
   • C) 1 A
   • D) 2 A

   Answer: 2.5 A

   Explanation: Peak current = Vpeak / (Rf + R), where Rf = 20 Ω and Vpeak = 50 V, so Ipeak = 50 / 20 = 2.5 A.

2. Question 2

   Q2. The frequency of output signal of a LC oscillator is

   • A) 1 / (2π√LC)
   • B) 1 / (2πLC)
   • C) 1 / √(LC)
   • D) √(LC) / (2π)

   Answer: 1 / (2π√LC)

   Explanation: The resonant frequency of LC circuit is given by f = 1 / (2π√LC).

3. Question 3

   Q3. The resistance of a germanium diode at 300 K changes from 100 Ω to 50 Ω when its temperature increases by 10 K. The bandgap energy is

   • A) 0.25 eV
   • B) 0.5 eV
   • C) 0.67 eV
   • D) 1 eV

   Answer: 0.67 eV

   Explanation: Using the formula for temperature dependence of resistance, Eg = k * ΔT / (ΔR / R), Eg = 0.67 eV.

4. Question 4

   Q4. A transistor is operated in CE configuration at Vc = 2 V such that a change in base current from 100 μA to 200 μA produces a change in collector current from 9 mA to 16.5 mA. The current gain is

   • A) 75
   • B) 85
   • C) 90
   • D) 100

   Answer: 75

   Explanation: Current gain (β) = ΔIc / ΔIb = (16.5 - 9) mA / (200 - 100) μA = 7.5 mA / 100 μA = 75.

5. Question 5

   Q5. The minimum number of NAND gates required to make a NOT gate is

   • A) 1
   • B) 2
   • C) 3
   • D) 4

   Answer: 1

   Explanation: One NAND gate with both inputs tied together acts as a NOT gate.

6. Question 6

   Q6. A Zener diode is used as a

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

   Answer: voltage regulator

   Explanation: Zener diode is used for voltage regulation due to its ability to maintain a constant voltage.

7. Question 7

   Q7. The forbidden energy gap in a silicon crystal is

   • A) 1.1 eV
   • B) 0.7 eV
   • C) 1.5 eV
   • D) 2.5 eV

   Answer: 1.1 eV

   Explanation: The bandgap energy for silicon is approximately 1.1 eV.

8. Question 8

   Q8. The input signal given to a CE amplifier having a voltage gain of 150 is Vi = 2 cos(15t + π/3). The corresponding output signal will be

   • A) 300 cos(15t + π/3)
   • B) 300 cos(15t + 4π/3)
   • C) 75 cos(15t + 2π/3)
   • D) 2 cos(15t + π/3)

   Answer: 300 cos(15t + 4π/3)

   Explanation: Output voltage = voltage gain * input voltage, with a 180° phase shift, so Vo = 300 cos(15t + 4π/3).

9. Question 9

   Q9. The current gain of a transistor in CB configuration is 0.98. The change in collector current for a change in emitter current of 5 mA is

   • A) 4.9 mA
   • B) 3.9 mA
   • C) 2.9 mA
   • D) 1.9 mA

   Answer: 4.9 mA

   Explanation: α = ΔIc / ΔIe, so ΔIc = α * ΔIe = 0.98 * 5 mA = 4.9 mA.

10. Question 10

    Q10. A transistor has a current gain of 30. If the collector resistance is 6 kΩ, input resistance is 1 kΩ, the voltage gain is

    • A) 120
    • B) 150
    • C) 180
    • D) 200

    Answer: 180

    Explanation: Voltage gain = β * (Rc / Ri) = 30 * (6 kΩ / 1 kΩ) = 180.

11. Question 11

    Q11. The Boolean expression for the output of an XOR gate is

    • A) A + B
    • B) A * B
    • C) A ⊕ B
    • D) A = B

    Answer: A ⊕ B

    Explanation: XOR gate output is given by A ⊕ B, which is 1 when A and B are different.

12. Question 12

    Q12. The output of a logic gate is 0 when both inputs are 1. The gate is

    • A) AND
    • B) NAND
    • C) NOR
    • D) OR

    Answer: NAND

    Explanation: NAND gate produces output 0 when both inputs are 1.

13. Question 13

    Q13. The frequency of oscillation of a Wien bridge oscillator is

    • A) 1 / (2πRC)
    • B) 1 / (2π√(LC))
    • C) 1 / (RC)
    • D) 1 / (√(LC))

    Answer: 1 / (2πRC)

    Explanation: Wien bridge oscillator frequency is given by f = 1 / (2πRC).

14. Question 14

    Q14. A silicon diode requires a minimum current of 1 mA to be above the knee voltage (0.7 V) of its I-V characteristic curve. When it is connected in series with a resistance R across a voltage source V, the voltage across the diode is 0.7 V when the current through the circuit is 1 mA. The value of R is

    • A) 1 kΩ
    • B) 2 kΩ
    • C) 3 kΩ
    • D) 4 kΩ

    Answer: 3 kΩ

    Explanation: Using Ohm's law, R = (V - 0.7) / I = (V - 0.7) / 1 mA = 3 kΩ for V = 4 V.

15. Question 15

    Q15. In a CE amplifier, the output voltage is given by V_o = -βR_L/R_i * V_i. What is β?

    • A) Current gain
    • B) Voltage gain
    • C) Power gain
    • D) Input resistance

    Answer: Current gain

    Explanation: β is the ratio of collector current to base current, representing current gain in a CE amplifier.

16. Question 16

    Q16. A Zener diode is used for voltage regulation. What is the condition for regulation?

    • A) V_in > V_z
    • B) V_in < V_z
    • C) I_z = 0
    • D) R_L = 0

    Answer: V_in > V_z

    Explanation: Zener regulation occurs when input voltage exceeds Zener voltage, allowing current to flow through the Zener diode.

17. Question 17

    Q17. The gain of an op-amp is given by A = V_o / (V+ - V-). What is the typical value of A?

    • A) 10
    • B) 100
    • C) 10^5
    • D) 10^8

    Answer: 10^5

    Explanation: Op-amp gain is typically very high, around 10^5, due to its high input impedance and low output impedance.

18. Question 18

    Q18. The cutoff frequency of an RC low-pass filter is given by f_c = 1 / (2πRC). What happens to f_c if R is doubled?

    • A) f_c is halved
    • B) f_c is doubled
    • C) f_c remains the same
    • D) f_c becomes zero

    Answer: f_c is halved

    Explanation: f_c is inversely proportional to R, so doubling R halves f_c.

19. Question 19

    Q19. In a series RLC circuit, resonance occurs when ωL = 1 / ωC. What is ω?

    • A) Resonant frequency
    • B) Angular frequency
    • C) Damping factor
    • D) Quality factor

    Answer: Angular frequency

    Explanation: ω is the angular frequency, related to frequency f by ω = 2πf.

20. Question 20

    Q20. The CMRR of an op-amp is given by CMRR = 20log(A_d / A_c). What does A_d represent?

    • A) Common-mode gain
    • B) Differential gain
    • C) Input resistance
    • D) Output resistance

    Answer: Differential gain

    Explanation: A_d is the differential gain, representing the gain for differential input signals.