In a NPN transistor,10^8 electrons enter the | vedclass

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(A) The emitter current $I_E$ is proportional to the number of electrons entering the emitter per unit time.Given that $1\%$ of electrons are lost in

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$0.99$ and $99$

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(A) The emitter current $I_E$ is proportional to the number of electrons entering the emitter per unit time.Given that $1\%$ of electrons are lost in the base,the collector current $I_C$ is $99\%$ of the emitter current $I_E$.Thus,the fraction of current entering the collector is $\frac{I_C}{I_E} = \frac{99}{100} = 0.99$.The base current $I_B$ is $1\%$ of the emitter current,so $I_B = 0.01 I_E$.The current amplification factor $\beta$ is defined as $\beta = \frac{I_C}{I_B}$.Substituting the values,$\beta = \frac{0.99 I_E}{0.01 I_E} = 99$.Therefore,the fraction is $0.99$ and $\beta = 99$.

In a $NPN$ transistor,$10^8$ electrons enter the emitter in $10^{-8} \ s$. If $1\%$ of electrons are lost in the base,the fraction of current that enters the collector and the current amplification factor $\beta$ are respectively:

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