A battery of `epsilon` and internal resistance `r` is used in a circuit with a variable external resistance `R`. Find the value of `R` for which the power consumed in `R` is maximum`.

Explanation

The current in the resistance `R` is
`I = epsilon/(r + R)`.
The power consumed in `R` is
`P = i^2 R = (epsilon^2 R)/(r + R)^2` .
It is maximum when `dP/dR = 0`. We have
dP/dR = epsilon^2 [(r+R)^2 - 2R(r+R)/(r+R)^4]`
It is zero when `(r+R)^2 = 2R(r+R)`
or, `R=r`