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

Options

A. `325Omega`
B. `165Omega`
C. `240Omega`
D. `225Omega`

Answer

C

Explanation

Here, Current gain, `beta=49`
Output resistance, `R_(o)=500kOmega=500xx10^(3)Omega`
Power gain `=5xx10^(6)`, Input resistance, `R_(i)=?`
Power gain `=beta^(2)xx(R_(o))/(R_(i))`
`5xx10^(6)=(49)^(2)xx(5xx10^(5))/(R_(i))`
`R_(i)=((49)^(2)xx5xx10^(5))/(5xx10^(6))~~240Omega`