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= (E)/(R+r) `
Power consumed in R isP ` =I^(2)R =((E)/(R+r))^(2) R `
Power is maximum when ` (dp)/(dR) =0 `
` (dP)/( dR) =E^(2) [((r+R)^(2) -2R(r+R) )/((r+R)^(4))]=0`
` rArr (r+R)^(2) =2R (r+R) " "rArr R =r`