The transfer ratio $\beta$ of a transistor is 50. The input resistance of the transistor when used in common emitter configuration is $1k\Omega$. The peak value of the collector ac current for an ac input voltage of 0.01 V peak is
Answer & explanation
Correct answer: option 3
$I_B = \frac{V_i}{R_i} =\frac{0.01}{10^{-3}}= 10^{-5}A = 0.01mA$
$I_c = \beta I_B = 50 \times 0.01mA = 0.5mA = 500 \mu A$