Q132 / 180 · AIIMS 2019 (26 May) · Physics

In an $n-p-n$ transistor (in common emitter mode) $10^{10}$ electrons enter the emitter in $10^{-}$ ${ }^6$ s. Only $2 \%$ of the electrons are lost in the base. Calculate the current amplification factor. (Charge on electron is $1.6 \times 10^{-19} \mathrm{C}$ ).

  • A
    $20$
  • B
    $72$
  • C
    $22.5$
  • Correct answer
    D
    $49$

Solution

$i_e=\frac{q}{t}=\frac{10^{10} \times 1.6 \times 10^{-19}}{10^{-6}}=1.6 \mathrm{~mA}$ Base current, $i_b=2 \%$ of $i_e=\left(\frac{2}{100}\right) \times 1.6 \mathrm{~mA}=0.032 \mathrm{~mA}$ $\therefore \quad i_c=i_e-i_b=1.568 \mathrm{~mA}$ Therefore current amplification factor, $\beta=\frac{i_c}{i_b}=\frac{1.568}{0.032}=49$

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

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