Q132 / 150 · MHT CET 2025 (21 Apr Shift 2) · Physics

In common emitter mode of a transistor, the current gain is 8 . The input impedance is $25 \mathrm{k} \Omega$ and load resistance is $75 \mathrm{k} \Omega$. The power gain is

  • A
    $112$
  • Correct answer
    B
    $192$
  • C
    $224$
  • D
    $316$

Solution

Power gain in a common emitter transistor relates current gain, voltage gain, input impedance, and load resistance. Given current gain β = 8, input impedance Rin = 25 kΩ, and load resistance RL = 75 kΩ, the voltage gain is determined by AV = β × RL/Rin.

$A_V = 8 \times \frac{75}{25} = 8 \times 3 = 24$

The power gain AP is the product of voltage and current gains: $A_P = A_V \times \beta = 24 \times 8 = 192$.

Alternatively, using the direct formula: $A_P = \beta^2 \times \frac{R_L}{R_{in}} = 64 \times 3 = 192$.


Power gain: $\boxed{192}$

Hence, the correct answer is option: 'B'.

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