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Q.

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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a

0.99 and 99

b

0.8 and 49

c

0.7 and 50

d

0.9 and 90

answer is D.

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Detailed Solution

IE=108×1.6×1019108A=1.6mA.IB=1100IE     =0.01IE        and  IC=99%ofIE=0.99IE.       β=IcIB=0.99IE0.01IE=99.

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