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
`= ("Change in output power")/("Change in input power")`
`= (DeltaV_(c) xx Deltai_(c))/(DeltaV_(i) xx Deltai_(b)) = ((DeltaV_(c))/(DeltaV_(i))) xx ((Deltai_(c))/(Deltai_(b)))`
`= A_(V) xx beta_(AC)`
where `A_(V)` is voltage gain and `beta_(AC)` is `AC` current gain. Also,
`A_(V) = beta_(AC) xx` Resistance gain `(=(R_(0))/(R_(i)))`
Given, `A_(V) = 50, R_(0) = 200 Omega, R_(i) = 100 Omega`
Hence, `50 = beta_(AC) xx (200)/(100)`
`:. beta_(AC) = 25`
Now, `AC` power gain `= A_(V) xx beta_(AC)`
`= 50 xx 25 = 1250`