Q159 / 160 · AP EAMCET 2024 (21 May Shift 2) · Physics
In $n-p-n$ transistor circuit, the collector current is 10 mA . If $95 \%$ of the electrons emitted reach the collector, then the base current is nearly
- A5.3 mA
- B53 mA
- C35 mA
- Correct answerD0.53 mA
Solution
In n-p-n transistor,
$\begin{aligned}
& \mathrm{I}_{\mathrm{C}}=10 \mathrm{~mA}, \mathrm{I}_{\mathrm{C}}=95 \% \text { of } \mathrm{I}_{\mathrm{E}}=\frac{95}{100} \mathrm{I}_{\mathrm{E}} \\
& \therefore \mathrm{I}_{\mathrm{E}}=\frac{100 \mathrm{I}_{\mathrm{C}}}{95}=\frac{100 \times 10}{95}=10.53 \mathrm{~mA}
\end{aligned}$
Now, $\mathrm{I}_{\mathrm{E}}^{\prime}=\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}$
$\therefore \quad \mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{E}}-\mathrm{I}_{\mathrm{C}}=10.53-10=0.53 \mathrm{~mA}$
Now, $\mathrm{I}_{\mathrm{E}}^{\prime}=\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}$
$\therefore \quad \mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{E}}-\mathrm{I}_{\mathrm{C}}=10.53-10=0.53 \mathrm{~mA}$
Hence, the correct answer is option: 'D'.