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In the following common emitter circuit ...

In the following common emitter circuit if `beta=100, V_(CE)=7 V, V_(BE)`=Negligible, `R_(C)=2 kOmega` then `I_(B)=?`

A

`0.01 mA`

B

`0.04 mA`

C

`0.02 mA`

D

`0.03 mA`

Text Solution

Verified by Experts

The correct Answer is:
B
`V=V_(CE)+I_(C)R_(L)`
`implies 15=7+I_(C)xx2xx10^(3) implies i_(C)=4mA`
`:' beta=(i_(C))/(i_(B)) implies i_(B)=4/100=0.04 mA`
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