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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)=9mA`

B

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

C

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

D

`i_(E)=1mA, i_(B)=1mA`

Text Solution

Verified by Experts

The correct Answer is:
D
`i_(C)=(90)/(100)xxi_(e )implies i_(e)=10xx(100)/(90)=11mA`
`i_(B)=i_(e)-i_(c )=1mA`
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