If a battery of emf `E` and internal resistance `r` is connected across a load of resistance `R`. Shot that the rate at which energy is dissipated in `R` is maximum when `R = r` and this maximur power is `P = E^2//4r`.

Answer

A

Explanation

`i=E/(R+r)impliesP="power across" R=i^2R`
`P=(E)/(R+r)^2R`…………i
For power to be maximum
`(dP)/(dR)=0`
By putting `(dP)/(dR)=0` we get, `R=r`
Further, by putting `R=r` in Eqn i
We get `P_(max)=E^2/(4r)`