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. 0.88
B. 0.78
C. 0.98
D. 0.68

Answer

C

Explanation

`"Current"=("Charge")/("time")`
The emitter current, `I_(E )` is given by
`I_(E)=(N e)/(t)=(10^(10)xx(1.xx10^(-19)))/(10^(-16))=1.6xx10^(-3)A=1.6mA`
The base current, `I_(B)` is givne by
`I_(B)=(2)/(100)xx1.6mA=0.032mA`
In a transistor, `I_(E)=I_(B)+I_(C )`
`therefore" "I_(C )=I_(E)-I_(B)=1.6mA-0.032mA=1.568mA`
Current transfer ratio `=(I_(C ))/(I_(E ))=(1.567)/(1.6)=0.98`