The input resistance of a common emitter transistor amplifier, if the output resistance is `500 kOmega`, the current gain `alpha=0.98` and the power gain is `6.0625 xx 10^(6)` is

Options

A. `198Omega`
B. `300Omega`
C. `100Omega`
D. `400Omega`

Answer

A

Explanation

`A_(P)=beta^(2)(R_(L))/(r_(i))`, `beta=(alpha)/(1-alpha)=(0.98)/(1-0.98)=49`
`:.A_(P)=(49)^(2)xx(500xx10^(3))/(R_(1))impliesR_(1)=198Omega`.