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

Options

A. `I_(B)=40muA`, `I_(C)=10mA`, `beta=250`
B. `I_(B)=25muA`, `I_(C)=5mA`, `beta=250`
C. `I_(B)=20muA`, `I_(C)=5mA`, `beta=250`
D. `I_(B)=40muA`, `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))=40muA`
`V_("CC")=I_(C)R_(C)+V_(CE)`
`20=I_(C)xx(4xx10^(3))+0`
`I_(C)=5xx10^(-3)=5mA`
`beta=(I_(C))/(I_(B))=(5xx10^(-3))/(40xx10^(-6))=125`.