In an n-p-n transistor, 108 electrons enter the emitter in 10-8 sec. If 1% electrons are lost in the base, the fraction of current that enters the collector and current amplification factor are respectively
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answer is 4.
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Detailed Solution
IE=108×1.6×10−1910−8 A=1.6mA.IB=1100IE =0.01IE and IC=99% of IE=0.99 IE.∴ β =IcIB=0.99 IE0.01 IE=99.