In an n-p-n transistor,200 electrons enter th | vedclass

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(B) The total charge $q$ entering the emitter is given by $q = n 	imes e = 200 	imes 1.6 	imes 10^{-19} \ C = 3.2 	imes 10^{-17} \ C$.The emitter

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$3.2 	imes 10^{-9} \ A$ and $99$

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(B) The total charge $q$ entering the emitter is given by $q = n 	imes e = 200 	imes 1.6 	imes 10^{-19} \ C = 3.2 	imes 10^{-17} \ C$.The emitter current $I_e$ is calculated as $I_e = \frac{q}{t} = \frac{3.2 	imes 10^{-17} \ C}{10^{-8} \ s} = 3.2 	imes 10^{-9} \ A$.Given that $1 \%$ of electrons are lost in the base,the base current $I_b = 0.01 	imes I_e$.The collector current $I_c$ is the remaining current,so $I_c = I_e - I_b = 0.99 	imes I_e$.The current amplification factor $\beta$ is defined as $\beta = \frac{I_c}{I_b} = \frac{0.99 	imes I_e}{0.01 	imes I_e} = 99$.

In an $n-p-n$ transistor,$200$ electrons enter the emitter in $10^{-8} \ s$. If $1 \%$ of electrons are lost in the base,then the current that enters the emitter and the current amplification factor are respectively $\left[e=1.6 \times 10^{-19} \ C\right]$

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