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(D) Given: Current gain $\beta = 50$,input resistance $r_i = 1 \ k\Omega = 10^3 \ \Omega$,and peak $AC$ input voltage $v_S = 0.01 \ V$.The base curren

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$500 \ \mu A$

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(D) Given: Current gain $\beta = 50$,input resistance $r_i = 1 \ k\Omega = 10^3 \ \Omega$,and peak $AC$ input voltage $v_S = 0.01 \ V$.The base current $I_B$ is given by the relation $I_B = \frac{v_S}{r_i}$.Substituting the values: $I_B = \frac{0.01 \ V}{10^3 \ \Omega} = 10^{-5} \ A$.In a common-emitter transistor amplifier,the current gain $\beta$ is defined as $\beta = \frac{I_C}{I_B}$.Therefore,the collector current $I_C = \beta 	imes I_B$.Substituting the values: $I_C = 50 	imes 10^{-5} \ A = 500 	imes 10^{-6} \ A = 500 \ \mu A$.

In a common-emitter transistor amplifier,$\beta = 50$ and the input resistance is $1 \ k\Omega$. If the peak value of the $AC$ input voltage is $0.01 \ V$,then the peak value of the $AC$ collector current will be .......

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