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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

0.88

B

0.98

C

0.78

D

0.87

Text Solution

Verified by Experts

The correct Answer is:
B
`I_(c)=98% ` of `I_(e)`
`therefore prop=(I_(c))/(I_(e))=98% =0.98`
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