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In a transistor 10^(8) electrons enter a...

In a transistor `10^(8)` electrons enter at the emitter in `10^(-4)` s, out of which `2%` electron go to the base. The current transfer ratio in common base configuration is

A

98

B

2

C

0.98

D

0.2 mA

Text Solution

Verified by Experts

The correct Answer is:
C
`I_(b)=2% ` of `I_(e), I_(c)=98% ` of `I_(e), alpha=?`
`alpha=(I_(c))/(I_(e))=(98% " of " I_(e))/(I_(e))=98% =0.98`
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