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(D) Given: Current amplification factor $\beta = 50$,Input resistance $R_i = 1 	ext{ k}\Omega = 10^3 \Omega$,Input voltage $V_i = 0.01 	ext{ V}$.Usi

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$500 \mu	ext{A}$

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(D) Given: Current amplification factor $\beta = 50$,Input resistance $R_i = 1 	ext{ k}\Omega = 10^3 \Omega$,Input voltage $V_i = 0.01 	ext{ V}$.Using Ohm's law for the input circuit,the base current $I_B$ is given by:$I_B = \frac{V_i}{R_i} = \frac{0.01 	ext{ V}}{10^3 \Omega} = 10^{-5} 	ext{ A}$.The relationship between collector current $I_C$ and base current $I_B$ is $I_C = \beta 	imes I_B$.Substituting the values:$I_C = 50 	imes 10^{-5} 	ext{ A} = 500 	imes 10^{-6} 	ext{ A} = 500 \mu	ext{A}$.

The current amplification factor of a transistor is $50$. The input resistance when used in common emitter mode is $1 \text{ k}\Omega$. The peak value of the collector current for an a.c. input voltage of $0.01 \text{ V}$ peak is:

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