A common emitter amplifier has a voltage gain of `50`, an input impedence of `100 Omega` and an output impedence of `200 Omega`. The power gain of the of the amplifier is

Options

A. `500`
B. `1000`
C. `1250`
D. `100`

Answer

C

Explanation

`AC` power gain is the ratio of change in output power to the change in input power.
`AC` power gain
`=(Chang e i n output power)/(Chang e i n i nput power)`
`=(DeltaV_(c)xxDeltai_(C))/(DeltaV_(i)xxDeltai_(b))=((DeltaV_(c))/(DeltaV_(I)))xx((Deltai_(C))/(Deltai_(b)))=A_(V)xxbeta_(AC)`
where `A_(V)` is voltage gain and `(beta)_(AC)` is `AC` current gain.
Also, `A_(V)=beta_(AC) implies ` resistance gain `=((R_(0))/(R_(i)))`
Given, `A_(V)=50,R_(0)=200 Omega,R_(i)=100 Omega`
Hence, `50=B_(AC)=200/100`
`beta_(AC)=25`
Now, `AC` power gain. `=A_(V)xxbeta_(AC)`
`=50xx25=1250`