In an n-p-n transistor `10^(10)` electrons enter the emitter in `10^(-6)`s. If 2% of the electrons are lost in the base, find the current transfer ratio and the current amplification factor.

Explanation

`I_(e)=(n_(e)xxe)/(t)`
and `I_(c)=(n_(c)xxe)/(t)=((98//100)n_(c)xxe)/(t)=(98)/)100)I_(e)`
`therefore` Current transfer ratio, `alpha=(I_(c))/(I_(e))=(98)/(100)=0.98`
Current amplification factor, `beta=(alpha)/(1-alpha)=(0.98)/(1-0.98)=49`