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In the common emitter configuration of n...

In the common emitter configuration of n-p-n transistor `10^(10)` electrons enter the emitter in 1`mus` and 2% of the electrons are lost of the base. The current gain of the amplifier is

A

2

B

98

C

1

D

49

Text Solution

Verified by Experts

The correct Answer is:
D
`I_(b)=2% ` of ` I_(e), I_(c)=90% ` of `I_(e), beta=?`
`beta=(I_(c))/(I_(b))=(98% " of " I_(e))/(2% " of " I_(b))=49`
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