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(C) Given that $90 \%$ of the electrons emitted reach the collector,the current gain $\alpha$ (common base) is defined as the ratio of collector curre

Vedclass · Accepted Answer

$\alpha=0.9, \beta=9$

Vedclass · Answer

(C) Given that $90 \%$ of the electrons emitted reach the collector,the current gain $\alpha$ (common base) is defined as the ratio of collector current to emitter current. Since $90 \%$ of the emitter current $(I_E)$ reaches the collector $(I_C)$,we have $I_C = 0.9 \ I_E$. Thus,$\alpha = \frac{I_C}{I_E} = 0.9$. The current gain $\beta$ (common emitter) is related to $\alpha$ by the formula $\beta = \frac{\alpha}{1 - \alpha}$. Substituting the value of $\alpha$: $\beta = \frac{0.9}{1 - 0.9} = \frac{0.9}{0.1} = 9$. Therefore,$\alpha = 0.9$ and $\beta = 9$.

In an $n-p-n$ transistor amplifier,the collector current is $9 \ mA$. If $90 \%$ of the electrons emitted reach the collector,then

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