Question

In common emitter amplifier, input resistance is \( 1000 \, \Omega \), peak value of input signal voltage is 5 mV and \( \beta = 60 \). The peak value of output current is

• A. \( 0.5 \times 10^{-4} \, \text{A} \)
• B. \( 3 \times 10^{-4} \, \text{A} \)
• C. \( 2 \times 10^{-5} \, \text{A} \)
• D. \( 1 \times 10^{-5} \, \text{A} \)

Hint:

In common emitter amplifiers, use the relationship \( I_{\text{out}} = \beta \times \frac{V_{\text{in}}}{R_{\text{in}}} \) to calculate the output current.

Answer 1

The Correct Option is B

Step 1: Formula for output current.
The output current is related to the input voltage and the current gain \( \beta \) by the equation: \[ I_{\text{out}} = \beta \times \frac{V_{\text{in}}}{R_{\text{in}}} \] Substituting the values: \[ I_{\text{out}} = 60 \times \frac{5 \times 10^{-3}}{1000} = 3 \times 10^{-4} \, \text{A} \]
Step 2: Conclusion.
Thus, the correct answer is (B) \( 3 \times 10^{-4} \, \text{A} \).