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(D) Given:$\beta = 50$Input resistance $R_i = 1 	ext{ k}\Omega = 10^3 \Omega$Peak input voltage $V_i = 0.01 	ext{ V}$Step $1$: Calculate the peak va

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$500$

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(D) Given:$\beta = 50$Input resistance $R_i = 1 	ext{ k}\Omega = 10^3 \Omega$Peak input voltage $V_i = 0.01 	ext{ V}$Step $1$: Calculate the peak value of base current $(\Delta I_B)$:$\Delta I_B = \frac{V_i}{R_i} = \frac{0.01 	ext{ V}}{10^3 \Omega} = 10^{-5} 	ext{ A}$Step $2$: Calculate the peak value of collector current $(\Delta I_C)$:Using the relation $\beta = \frac{\Delta I_C}{\Delta I_B}$,we get:$\Delta I_C = \beta 	imes \Delta I_B$$\Delta I_C = 50 	imes 10^{-5} 	ext{ A}$$\Delta I_C = 5 	imes 10^{-4} 	ext{ A} = 500 	imes 10^{-6} 	ext{ A} = 500 \mu	ext{A}$

In a common emitter configuration,a transistor has $\beta=50$ and input resistance $1 \text{ k}\Omega$. If the peak value of a.c. input is $0.01 \text{ V}$,then the peak value of collector current is: (in $\mu\text{A}$)

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