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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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detailed solution

Correct option is A

IE=net=1010×1.6×10−1910−6 =1.6×10−3ABase IB=2% of IE=2100×1.6×10−3 =0.032 mAIC=IE−IB=(1.6−0.032)mA =1.568 mAβ=ICIB=1.5680.032=49

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In a n−p−n transistor 1010 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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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