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

• A

  19 mA
• Correct answer
  B

  38 mA
• C

  9.5 mA
• D

  48 mA

Solution

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, $\begin{aligned} & I_E=I_C+I_B \Rightarrow \frac{I_C}{0.95}=I_C+2 \times 10^{-3} \\ \Rightarrow & \frac{I_C}{0.95}-I_C=2 \times 10^{-3} \\ \Rightarrow & I_C\left(\frac{1}{0.95}-1\right)=2 \times 10^{-3} \Rightarrow I_C \times \frac{1}{19}=2 \times 10^{-3} \\ \Rightarrow & I_C=38 \times 10^{-3} \mathrm{~A}=38 \mathrm{~mA} \end{aligned}$

Hence, the correct answer is option: 'B'.