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.

Options

A. `2xx10^(-10)` A and 49
B. `1.6xx10^(-19)` A and 90
C. `1.7xx10^(-11) ` A and 70
D. `3.2xx10^(-9)` A and 99

Answer

D

Explanation

Let `I_E` be the emitter current .
Then, `I_E=(dq)/(dt) = (200xx1.6xx10^(-19))/10^(-8)=3.2xx10^(-9)A`
Since 1% electrons are lost in the base ,
`therefore` Base current `(I_B)=I_E/100` and the remaining `99/100 I_E` will enter the collector
`therefore I_C=99/100 I_E`
and the current amplification factor `beta=I_C/I_B=(0.99I_E)/(0.01 I_E)`=99