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In an n-p-n transistor, the collector current is 10 mA, if 90% of the electrons emitted reach the collector, the emitter current IE and base current IB are given by

a
IE=−1mA, IB=9mA
b
IE=9mA, IB=−1mA
c
IE=1mA, IB=11mA
d
IE=11mA, IB=1mA

detailed solution

Correct option is D

As, collector current, IC=90100×IEwhere, lE = emitter current.⇒    10=0.9×IE⇒    IE=11mAAlso, IE=IB+ICBase current, IB=IE−IC                       =11−10=1mA

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