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For a p-n-p transistor in CB configurati...

For a `p-n-p` transistor in `CB` configuration, the emitter current `I_(E)` is `1mA` and `alpha = 0.95`. The base current and collector current are

A

`0.95,A,0.05mA`

B

`0.05 mA, 0.95 mA`

C

`9.5 ,A,0.5 mA`

D

`0.5mA,9.5mA`

Text Solution

Verified by Experts

The correct Answer is:
B
`alpha =(I_(c))/(I_(e)),I_(e)=I_(b)+I_(c)`.
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