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In CE NPN transistor 10^(10) electrons e...

In CE NPN transistor `10^(10)` electrons enter the emitter in `10^(-6)` s when it is connected to battery. About 5% electrons recombine with holes in the base. The current gain of the transistor is ______
`(e = 1.6 xx 10^(-19) C)`

A

0.98

B

19

C

49

D

0.95

Text Solution

Verified by Experts

The correct Answer is:
B
Current gain common emitter `B=(i_c)/(i_B)`
`B=(95% "of " i_E)/(5% " of " i_E)`
`B=95/5=19`
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