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`.

Answer

The current in the resistance R is `I=(E)/(R+r)`
Power consumed in R is `P=PR=((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)rArrR=r`
`(##ALN_PHY_C05_S01_062_A01##)`