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In an $n-p-n$ transistor, $95 \%$ of emitted electrons reach the collector. If the base current is $2 \mathrm{~mA}$, then collector current is
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Verified Answer
The correct answer is:
38 mA
Given, base current in $n-p-n$ transistor,
$I_B=2 \mathrm{~mA}=2 \times 10^{-3} \mathrm{~A}$
Since, $95 \%$ of emitted electrons reach the collector, hence
Collector current,
But we know that,
$\begin{aligned}
& I_E=I_C+I_B \Rightarrow \frac{I_C}{0.95}=I_C+2 \times 10^{-3} \\
\Rightarrow & \frac{I_C}{0.95}-I_C=2 \times 10^{-3} \\
\Rightarrow & I_C\left(\frac{1}{0.95}-1\right)=2 \times 10^{-3} \Rightarrow I_C \times \frac{1}{19}=2 \times 10^{-3} \\
\Rightarrow & I_C=38 \times 10^{-3} \mathrm{~A}=38 \mathrm{~mA}
\end{aligned}$
$I_B=2 \mathrm{~mA}=2 \times 10^{-3} \mathrm{~A}$
Since, $95 \%$ of emitted electrons reach the collector, hence
Collector current,
But we know that,
$\begin{aligned}
& I_E=I_C+I_B \Rightarrow \frac{I_C}{0.95}=I_C+2 \times 10^{-3} \\
\Rightarrow & \frac{I_C}{0.95}-I_C=2 \times 10^{-3} \\
\Rightarrow & I_C\left(\frac{1}{0.95}-1\right)=2 \times 10^{-3} \Rightarrow I_C \times \frac{1}{19}=2 \times 10^{-3} \\
\Rightarrow & I_C=38 \times 10^{-3} \mathrm{~A}=38 \mathrm{~mA}
\end{aligned}$
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