The current gain beta of a transistor is 50. | vedclass

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(D) Given: Current gain $\beta = 50$, Input resistance $R_i = 1\,k\Omega = 1000\,\Omega$, Input peak voltage $V_i = 0.01\,V$.First, calculate the peak

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$500$

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(D) Given: Current gain $\beta = 50$, Input resistance $R_i = 1\,k\Omega = 1000\,\Omega$, Input peak voltage $V_i = 0.01\,V$.First, calculate the peak input base current $(I_b)$:$I_b = \frac{V_i}{R_i} = \frac{0.01\,V}{1000\,\Omega} = 10^{-5}\,A$.Using the relation for current gain in common emitter configuration:$\beta = \frac{I_c}{I_b}$Therefore, the peak collector current $(I_c)$ is:$I_c = \beta 	imes I_b = 50 	imes 10^{-5}\,A$.Converting to microamperes $(\mu A)$:$I_c = 50 	imes 10^{-5} 	imes 10^6\,\mu A = 500\,\mu A$.

The current gain $\beta$ of a transistor is $50$. The input resistance of the transistor when used in the common emitter mode is $1\,k\Omega$. The peak value of the collector current for an input peak voltage of $0.01\,V$ is ...... $\mu A$.

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