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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

`0.95mA, 0.05mA`

B

`0.05mA, 0.95mA`

C

`9.5 mA, 0.5mA`

D

`0.5 mA, 9.5mA`

Text Solution

Verified by Experts

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