In `NPN` transistor, `10^(10)` electrons enters in emitter region in `10^(-6)`sc. If `2%` electrons are lost in base region then collector current and current amplification factor `(beta)` respectively are

Options

A. `1.57 mA,49`
B. `1.92 mA,70`
C. `2 mA,25`
D. `2.25 mA,100`

Answer

A

Explanation

`I_(e)=10^(10)xx1.6xx10^(-19)xx1/(10^(-6))=1.6mA ( :' I=Q/t)`
Since `2%` electrons are absorbed by base, hence `98%` electrons reaches the collector i.e., `alpha=0.98`
`implies I_(c)=alphaI_(e)=0.98xx1.6=1.568 mA ~~1.57 mA`