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

Options

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

Answer

A

Explanation

`alpha=0.98`
`beta=(alpha)/(1-alpha)=(0.98)/(1-0.98)=49`
`A_(p)=beta^(2)(R_(L))/(R_("in"))`
`6.0625xx10^(6)=(49)^(2)xx(500xx10^(3))/(R_(in))`
`R_("in")=198Omega`