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In a $n - p - n$ transistor $10^{10}$ electrons enter the emitter is $10^{- 6} \sec$ $, 2 %$ of the electrons are lost in the base then the current gain is common emitter configuration is

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$49$

$98$

$0.49$

$0.98$

answer is A.

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Detailed Solution

$I_{E} = \frac{n e}{t} = \frac{10^{10} \times 1.6 \times 10^{- 19}}{10^{- 6}}$

$= 1.6 \times 10^{- 3} A$

Base $I_{B} = 2 % o f I_{E}$

$= \frac{2}{100} \times 1.6 \times 10^{- 3} = 0.032 m A$

$I_{C} = I_{E} - I_{B} = (1.6 - 0.032) m A$ $= 1.568 m A$

$β = \frac{I_{C}}{I_{B}} = \frac{1.568}{0.032} = 49$

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