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In a transistor connected in common emit...

In a transistor connected in common emitter mode, `R_C=4 kOmega, R_1=1 kOmega, I_C=1 mA and I_B=20 muA`. The voltage gain is

A

50

B

200

C

`1/50`

D

`1/200`

Text Solution

Verified by Experts

The correct Answer is:
B
`A_(v)=(beta_(ac)R R_(c))/(R_(1))=(I_(e))/(I_(B)).(R_(C))/(R_(1))=(10^(-3)xx4xx10^(3))/(20xx10^(-6)xx10^(3))=200`
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