In case of a bipolar transistor, beta = 45. T | vedclass

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(C) Given: Current gain $\beta = 45$, Collector resistance $R_C = 1 \ k\Omega = 1000 \ \Omega$, Potential drop across collector resistance $V_C = 5 \

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$111$

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(C) Given: Current gain $\beta = 45$, Collector resistance $R_C = 1 \ k\Omega = 1000 \ \Omega$, Potential drop across collector resistance $V_C = 5 \ V$. We know that the collector current $I_C$ is given by $I_C = \frac{V_C}{R_C}$. Substituting the values: $I_C = \frac{5 \ V}{1000 \ \Omega} = 5 	imes 10^{-3} \ A = 5 \ mA$. The relationship between collector current $I_C$ and base current $I_B$ is given by $\beta = \frac{I_C}{I_B}$. Therefore, $I_B = \frac{I_C}{\beta}$. Substituting the values: $I_B = \frac{5 	imes 10^{-3} \ A}{45} = \frac{1}{9} 	imes 10^{-3} \ A \approx 0.111 	imes 10^{-3} \ A$. Converting to microamperes: $I_B \approx 111 \ \mu A$.

In case of a bipolar transistor, $\beta = 45$. The potential drop across the collector resistance of $1 \ k\Omega$ is $5 \ V$. The base current is approximately: (in $\mu A$)

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