A battery has an emf E and internal resistance r. A variable resistance R is connected across the terminals of the battery. Find the value of R such that (a) the current in the circuit id maximum (b) the potential difference across the terminals is maximum.

Explanation

Current in the circuit, `I = E/(R + r)`
Power deliverd to the resistance R is,
`P = I^(2) xx R = (E^(2)R)/(R +r)^(2)`
Power delivered to the load will be maximum when `R/(R +r)^(2)` is maximum, i.e.,
`(d)/(dR) [R (R + r)^(-2)]=0`
or `[(R +r)^(-2) + R(-2) (R + r)^(-3)]=0`
or `(R + r)^(-3) [r +R]=0`
As `(R + r)^(-3)!= 0` for finite value of R,
So, r-R =0 or R=r
When R = r, the power transferred to the load is
`= E^(2)r/(r+r)^(2) = E_(2)/4r = max`.