A n-p-n transisitor 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 it is `0.8 V`. If the current amplification factor is `0.96` and the input resistance of the circuits 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

`R_(L)=800Omega`, `V_(L)=0.8V`
`alpha=0.96`, `r_(i)=192Omega`
`beta=(alpha)/(1-alpha)=(96)/(0.04)=24`
Note : Asked about CE but answer given for CB. Voltage gain for CE,
`A_(V)=beta(R_(L))/(r_(i))=24xx(800)/(192)~~100`
Power gain for CE ,
`A_(P)=betaxxA_(v)=24xx100=2400`
`A_(V)` for CB,
`A_(V)=alpha(R_(L))/(R_(i))=0.96xx(800)/(192)=4`
Power gain for CB
`A_(P)=A_(V)alpha=4xx0.96=3.84`.