In the circuit shown in the figure, the input voltage `V_(i)` is `20V,V_(BE)=0` and `V_(CE)=0`. The values of `I_(B),I_(C)` and `beta` are given by:

Options

A. `I_(B) = 40 muA, I_(C ) = 10 mA, beta = 250`
B. `I_(B) = 25 muA, I_( C) = 5 mA, beta = 200`
C. `I_(B) = 20 muA, I_( C)= 5mA, beta = 250`
D. `I_(B) = 40 muA, I_( C) = 5 mA, beta = 125`

Answer

D

Explanation

Given `V_(BE)=0`
`V_(CE)=0`
`therefore" Collector current,"`
`I_(C )=((20-0))/(4xx10^(3))`
or `I_(C)=5xx10^(-3)A=5mA`

Input voltage,
`V_(i)=V_(BE)+I_(B)R_(B)`
`or V_(i)=0+I_(B)R_(B)`
or `20+I_(B)xx500xx10^(3)`
`therefore" "I_(B)=(20)/(500xx10^(3))=40muA`
`therefore" Current gain, "beta=(I_(C ))/(I_(B))=(5xx10^(-3))/(40xx10^(-6))=125`