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In a common emitter transistor amplifier...

In a common emitter transistor amplifier, the output resistance is `500Omega` and the current gain `beta=49`. If the power gain of the amplifier is `5xx10^(6)`, the input resistance is

A

`325Omega`

B

`165Omega`

C

`198Omega`

D

`240Omega`

Text Solution

Verified by Experts

The correct Answer is:
D
Given, `R_(o)=500kOmega, beta=49 and P=5xx10^(6)`
We have, `P=beta^(2)(R_(o))/(R_(i))`
`5xx10^(6)=((49)^(2)xx500 xx 10^(3))/(R_(i))`
`R_(i)=(49xx49xx500)/(5xx10^(6))=240Omega`
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