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

Answer

D

Explanation

`V_(i)=I_(B)R_(B)+V_(BE)`
`20=I_(B)xx(500xx10^(3))+0`
`I_(B)=20/(500xx10^(3))=40 muA`
`V_(CC)=I_(C)R_(C)+V_(CE)`
`20=I_(C)xx(4xx10^(3))+0`
`I_(C)=50xx10^(-3)=5mA`
`beta=(I_(C))/(I_(B))=(5xx10^(-3))/(40xx10^(-6))=125`