In a common emitter transistor amplifier, the output resistance is 500 `Omega` and the current gain `beta` = 49. If the power gain of the amplifier is `5 xx 10^(6)`, the input resistance is

Options

A. 325 `Omega`
B. 165 `Omega`
C. 198 `Omega`
D. 240 `Omega`

Answer

D

Explanation

Given `R_(0) = 500 k Omega, beta = 49 and P = 5 xx 10^(6)`
we have `P = beta^(2)(R_(o))/(R_(i))`
`5 xx 10^(6) = ((49)^(2) xx 500)/(R_(i))`
`R_(i) = ((49)^(2) xx 500)/(5 xx 10^(6))`
`= 240 Omega`