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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
D
`i_(C)=90/100xxi_(E) =(Deltai_(c))/(Deltai_(b)) implies Deltai_(c)=betaxxDeltai_(b)=80xx250 muA`.
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