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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

`40 muA`

B

`20 muA`

C

`30 muA`

D

`15 muA`

Text Solution

Verified by Experts

The correct Answer is:
B
Using, `I_C=(V_(CE))/R_C=2/(2xx10^3)=10^(-3)`=1 mA
`therefore beta=I_C/I_B,I_B=I_C/beta=10^(-3)/50A=20 muA`
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