In an n-p-n transistor 200 electrons enter the emitter in $10^{-8}$ second. If $1 \%$ electrons are lost in…

In an n-p-n transistor 200 electrons enter the emitter in $10^{-8}$ second. If $1 \%$ electrons are lost in the base, then the current that enters the emitter and the current amplification factor are respectively $\left[\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\right]$
  1. $2 \times 10^{-10} \mathrm{~A}$ and 49
  2. $3.2 \times 10^{-9} \mathrm{~A}$ and 99
  3. $1.6 \times 10^{-19} \mathrm{~A}$ and 90
  4. $1.7 \times 10^{-11} \mathrm{~A}$ and 70

Solution

$\begin{aligned} & \mathrm{q}=200 \times 1.6 \times 10^{-19} \mathrm{C}, \mathrm{t}=10^{-8} \mathrm{~s} \\ & \therefore \text { Emitter current } \mathrm{I}_{\mathrm{e}}=\frac{\mathrm{q}}{\mathrm{t}}=\frac{3.2 \times 10^{-17}}{10^{-8}}=3.2 \times 10^{-9} \mathrm{~A} \\ & \mathrm{I}_{\mathrm{b}}=\frac{1}{100} \cdot \mathrm{I}_{\mathrm{e}} \\ & \mathrm{I}_{\mathrm{c}}=\frac{99}{100} \cdot \mathrm{I}_{\mathrm{e}} \end{aligned}$ $\therefore$ Current amplification factor $\beta=\frac{I_c}{I_b}=99$

Asked in: MHT CET 2021 (22 Sep Shift 1)

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