An amplifier is represented by the circuit shown in Fig. Here `r_(i)` is the input resistance of the amplifier and the voltage `V_(i)` is appearing across it. This voltage is amplified by a factor `A_(V)` and appears across the load as voltage `V_(0)`

An external voltage `V_(s)` is applied at the input terminals of the amplifier via series resistance `R_(S)`. What will be the apparent gain `A_(V)(=V_(0)//V_(s))` of the ampilifier in terms of `A_(V), R_(S)` and `r_(i).`

Explanation

In the given circuit, input current `I_(i)` for signal voltage `V_(s)` will be
`I_(i)=V_(s)/(R_(s)+R_(i))`
Input signal voltage drop across `R_(i)` will be
`V_(i)=I_(i)R_(i)=(V_(s).R_(i))/(R_(s)+R_(i))` Now `A_(V)=V_(0)/V_(i)=V_(0)/(V_(s)R_(i)//(R_(s)+R_(i)))=V_(0)/V_(s)((R_(s)+R_(i))/R_(i))`
`V_(0)/V_(s)=A_(V)[R_(i)/(R_(s)+R_(i))]` Now `A_(V)^(')=V_(0)/V_(s)=A_(V)(R_(i)/(R_(s)+R_(i)))`