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In an NPN transistor the collector curre...

In an `NPN` transistor the collector current is `10 mA`. If `90%` of electrons reach collector, the emitter current `(i_(E))` and base current `(i_(B))` are given by

A

`I_(E) = 1mA , I_(B) = 11 mA`

B

`I_(E) = 11 mA l I_(B) = 1mA`

C

`I_(E) = -1mA , I_(B) = 9mA`

D

`I_(E) = 9mA , I_(B) = -1 mA`

Text Solution

Verified by Experts

The correct Answer is:
B
`I_(C)=90%I_(E)=(90)/(100)I_(E)rArr I_(E)=(100)/(90)xxI_(C)=11mA`
`I_(B)=I_(E)-I_(C)=11mA-10mA=1mA`.
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