A `npn` transistor is connected in common emitter configuration in a given amplifier. A load resistance of `800 Omega`is connected in the collector circuit and the voltage drop across `0.96` and the input resistance of the circuit is `192 Omega`, the voltage gain and the power gain of the amplifier will respectively be :

Options

A. `4,3.84`
B. `3.69,3.84`
C. `4,4`
D. `4,3.69`

Answer

A

Explanation

`beta=0.96, R_(L)=800 Omega, R_(i n)=192 Omega`
`A_(V)=beta(R_(L))/(R_(i n))=(0.96)(800/192)=4`
`A_(p)=beta^(2) (R_(L))/(R_(i n))=3.84`