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A transistor connected in common emitter...

A transistor connected in common emitter mode, the voltage drop across the collector is 2 V and `beta` is 50, the base current if `R_C` is `2 kOmega` is

A

`10 muA`

B

`15muA`

C

`20muA`

D

`25 muA`

Text Solution

Verified by Experts

The correct Answer is:
D
`I_c= V_c/R_l = 2.5 /(2xx 10^3)`
`I_(c)=1.25A`
`beta=(I_(c))/(I_(b))`
`therefore I_(b)=(I_(c))/(beta)=(1.25)/(50)=25muA`
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