In a npn transistor `10^(10)` electrons enter the emitter in `10^(–6)` s. 4% of the electrons are lost in the base.
The current transfer ratio will be

Options

A. 0.98
B. 0.97
C. 0.96
D. 0.94

Answer

C

Explanation

`rArr cot 60^(@) = e cot 30^(@) rArr e = (cos 60^(@))/( cot 30^(@))`
`rArr e = ((1)/(sqrt(3)))/(sqrt(3)) rArr e =(1)/(3)`
`n_(C) = (96)/(100) xx 10^(10) =0.96 xx 10^(10)`
Emitter current `I _(E) = (n_(E) xx e)/( t)`
Collector current `I_(C) = (n_(c) xx e)/(t)`
Current transfer ratio .
`alpha = (I_(C))/(I_(E)) =(n_(C))/(n_(E)) = (0.96 xx 10^(10))/(10^(10)) = 0.96`