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In a n-p-n transistor 10^(10) electrons ...

In a `n-p-n` transistor `10^(10)` electrons enter the emitter in `10^(-6)s`. `2%` of the elecrons are lost in the base. The current transfer ratio and the current amplification factor will be

A

`0.98,49`

B

`98,0.49`

C

`0.49,98`

D

`49,0.49`

Text Solution

Verified by Experts

The correct Answer is:
A
`alpha=(I_(C ))/(I_(B))=(0.98)`
`beta=(I_(C ))/(I_(B))=(0.98)/(0.02)=49`
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