Question

The voltage gain of a common emitter amplifier is 60. If the output and input resistances are 4 k\(\Omega\) and 2 k\(\Omega\) respectively, the power gain of the amplifier is:

• A. \(60\)
• B. \(2\)
• C. \(30\)
• D. \(1800\)

Hint:

Verify calculations and provided options when discrepancies occur, especially in examinations or quizzes.

Answer 1

The Correct Option is D

The voltage gain (\(A_v\)) of the common emitter amplifier is given as \(A_v = 60\). The input resistance is given as \(R_i = 2 \, \text{k}\Omega = 2000 \, \Omega\). 
The output resistance is given as \(R_o = 4 \, \text{k}\Omega = 4000 \, \Omega\). The power gain (\(A_p\)) of an amplifier is the ratio of the output power to the input power: \[ A_p = \frac{P_o}{P_i} \] We can express power in terms of voltage and resistance, \(P = \frac{V^2}{R}\). Thus, \[ P_o = \frac{V_o^2}{R_o} \quad \text{and} \quad P_i = \frac{V_i^2}{R_i} \] Substituting these into the power gain equation: \[ A_p = \frac{V_o^2 / R_o}{V_i^2 / R_i} = \left(\frac{V_o}{V_i}\right)^2 \times \frac{R_i}{R_o} \] We know that voltage gain \(A_v = \frac{V_o}{V_i}\). Therefore, \[ A_p = (A_v)^2 \times \frac{R_i}{R_o} \] Substituting the given values: \[ A_p = (60)^2 \times \frac{2 \, \text{k}\Omega}{4 \, \text{k}\Omega} \] \[ A_p = (60)^2 \times \frac{2}{4} \] \[ A_p = 3600 \times \frac{1}{2} \] \[ A_p = 1800 \] The power gain of the amplifier is 1800. 
Correct Answer: (4) 1800