Q147 / 200 · NEET 2022 (Phase 2) · Physics

The collector current in a common base amplifier using $n$ - $p$-n transistor is $24 \mathrm{~mA}$. If $80 \%$ of the electrons released by the emitter is accepted by the collector, then the base current is numerically:

  • A
    $6 \mathrm{~mA}$ and leaving the base
  • B
    $3 \mathrm{~mA}$ and leaving the base
  • Correct answer
    C
    $6 \mathrm{~mA}$ and entering the base
  • D
    $3 \mathrm{~mA}$ and entering the base

Solution

As per question, we have, $\mathrm{I}_{\mathrm{c}}=24 \mathrm{~mA}$, and $\alpha=0.8$ Now, $\quad \mathrm{I}_{\mathrm{E}}=\frac{\mathrm{I}_{\mathrm{c}}}{\alpha}=\frac{24}{0.8}=30 \mathrm{~mA}$ As, $\quad \mathrm{I}_{\mathrm{E}}=\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}$ Hence, $\quad \mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{E}}-\mathrm{I}_{\mathrm{C}}$ $=6 \mathrm{~mA}$ (where, the symbols have their usual meanings)

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

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