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

Answer

B

Explanation

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