In an n-p-n transistor, the collector current is $28 \mathrm{~mA}$. If $80 \%$ of electrons reach the collector, its base current in $\mathrm{mA}$ is

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Aicharya · Accepted Answer

It is given that $80 \% \mathrm{I}_{\mathrm{e}}=\mathrm{I}_{\mathrm{c}}=28 \mathrm{~mA}$ $\begin{aligned} & 	herefore \quad \frac{80}{100} \mathrm{I}_{\mathrm{e}}=28 \mathrm{~mA} \quad \Rightarrow \quad \mathrm{I}_{\mathrm{e}}=35 \mathrm{~mA} \ & 	ext { But, } \mathrm{I}_e=\mathrm{I}_{\mathrm{c}}+\mathrm{I}_{\mathrm{B}} \ & 	herefore \quad \mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{e}}-\mathrm{I}_{\mathrm{c}} \quad \Rightarrow \quad \mathrm{I}_{\mathrm{B}}=35-28=7 \mathrm{~mA} \ & \end{aligned}$ ^

In an n-p-n transistor, the collector current is $28 \mathrm{~mA}$. If $80 \%$ of electrons reach the…

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