In an n-p-n transistor circuit,the collector | vedclass

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(D) In an $n-p-n$ transistor,the relationship between emitter current $(I_E)$,collector current $(I_C)$,and base current $(I_B)$ is given by $I_E = I_

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$0.53$

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(D) In an $n-p-n$ transistor,the relationship between emitter current $(I_E)$,collector current $(I_C)$,and base current $(I_B)$ is given by $I_E = I_B + I_C$.Given that $95\%$ of the emitted electrons reach the collector,we have $I_C = 0.95 	imes I_E$.Given $I_C = 10 	ext{ mA}$,we can find $I_E$ as:$I_E = \frac{I_C}{0.95} = \frac{10}{0.95} \approx 10.53 	ext{ mA}$.Now,calculating the base current:$I_B = I_E - I_C = 10.53 	ext{ mA} - 10 	ext{ mA} = 0.53 	ext{ mA}$.

In an $n-p-n$ transistor circuit,the collector current is $10 \text{ mA}$. If $95\%$ of the electrons emitted reach the collector,then the base current is nearly: (in $mA$)

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