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In an n-p-n transistor 200 electrons enter the emitter in $10^{-8}$ second. If $1 \%$ electrons are lost in the base, then the current that enters the emitter and the current amplification factor are respectively $\left[\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\right]$
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Verified Answer
The correct answer is:
$3.2 \times 10^{-9} \mathrm{~A}$ and 99
$$
\begin{aligned}
& \mathrm{q}=200 \times 1.6 \times 10^{-19} \mathrm{C}, \mathrm{t}=10^{-8} \mathrm{~s} \\
& \therefore \text { Emitter current } \mathrm{I}_{\mathrm{e}}=\frac{\mathrm{q}}{\mathrm{t}}=\frac{3.2 \times 10^{-17}}{10^{-8}}=3.2 \times 10^{-9} \mathrm{~A} \\
& \mathrm{I}_{\mathrm{b}}=\frac{1}{100} \cdot \mathrm{I}_{\mathrm{e}} \\
& \mathrm{I}_{\mathrm{c}}=\frac{99}{100} \cdot \mathrm{I}_{\mathrm{e}}
\end{aligned}
$$
$\therefore$ Current amplification factor $\beta=\frac{I_c}{I_b}=99$
\begin{aligned}
& \mathrm{q}=200 \times 1.6 \times 10^{-19} \mathrm{C}, \mathrm{t}=10^{-8} \mathrm{~s} \\
& \therefore \text { Emitter current } \mathrm{I}_{\mathrm{e}}=\frac{\mathrm{q}}{\mathrm{t}}=\frac{3.2 \times 10^{-17}}{10^{-8}}=3.2 \times 10^{-9} \mathrm{~A} \\
& \mathrm{I}_{\mathrm{b}}=\frac{1}{100} \cdot \mathrm{I}_{\mathrm{e}} \\
& \mathrm{I}_{\mathrm{c}}=\frac{99}{100} \cdot \mathrm{I}_{\mathrm{e}}
\end{aligned}
$$
$\therefore$ Current amplification factor $\beta=\frac{I_c}{I_b}=99$
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