The transfer ratio $β$ of a transistor is 50. The input resistance of the transistor when used in common – emitter configuration is 1kΩ. The peak value of the collector AC current for an AC input voltage of 0.01 V peak is

500 μA

Given that,

$V_I = 0.01$ volt

$R_I = 1 kΩ = 10^3 Ω$

$∴I_b=\frac{V_I}{R_I}=\frac{0.01}{1×10^3}=0.01×10^3Ω=0.01 mA$

Further, $I_C = βI_b = 50 × 0.01 mA$

= 0.5 mA = 500 μA