In an npn transistor,10^{10} electrons enter | vedclass

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(D) The emitter current $I_{E}$ is given by the rate of flow of charge: $I_{E} = \frac{n \cdot e}{t}$.Given $n = 10^{10}$ electrons,$e = 1.6 	imes 10

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$0.99$ and $99$

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(D) The emitter current $I_{E}$ is given by the rate of flow of charge: $I_{E} = \frac{n \cdot e}{t}$.Given $n = 10^{10}$ electrons,$e = 1.6 	imes 10^{-19} \ C$,and $t = 10^{-8} \ s$.$I_{E} = \frac{10^{10} 	imes 1.6 	imes 10^{-19}}{10^{-8}} = 1.6 	imes 10^{-1} \ A = 0.16 \ A$.Since $1\%$ of electrons are lost in the base,the base current $I_{B} = 0.01 \ I_{E}$.The collector current $I_{C}$ is the remaining current: $I_{C} = I_{E} - I_{B} = I_{E} - 0.01 \ I_{E} = 0.99 \ I_{E}$.The fraction of current entering the collector is $\frac{I_{C}}{I_{E}} = 0.99$.The current amplification factor $\beta$ is defined as $\beta = \frac{I_{C}}{I_{B}} = \frac{0.99 \ I_{E}}{0.01 \ I_{E}} = 99$.

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

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