A battery of emf $\epsilon$and internal resistance r is hooked up to a variable "load" resistance,R . If you want to deliver the maximum possible power to the load, what resistance R should you choose? (You can't change e and R , of course.)

The emf of the battery is$\epsilon$ .

Internal resistance of the battery is$r$ .

Load resistance is R .

Consider the circuit diagram shown below.

The current in the above circuit is given by,

$\mathit{i}\mathbf{=}\frac{\mathbf{E}}{\mathbf{R}\mathbf{+}\mathbf{r}}$

The power deliver to the load is given by,

$\mathit{P}\mathbf{=}{\mathit{i}}^{\mathbf{2}}\mathit{R}$

Substitute $\frac{E}{R+r}$for into above equation.

$p={\left(\frac{E}{R+r}\right)}^{2}R$ …… (1)

The maximum power can be delivered to the load under the condition$\frac{dP}{dR}=0$.

Now differentiate the equation (1) with respect to R,

$$\begin{gathered} \frac{dP}{dR}=\frac{{\left(R+r\right)}^2{E}^2+E^{2}R\times 2\left(R\times r\right)}{{\left(R\times r\right)}^4} \\ \frac{dP}{dR}=\frac{{\left(R+r\right)}^2{E}^2+2E^2\left(R\times r\right)}{{\left(R\times r\right)}^4} \end{gathered}$$

Substitute the above equation equal to zero.

$$\begin{gathered} \frac{dP}{dR}=0 \\ \frac{{\left(R+r\right)}^2{E}^2-2E^2{R}^4\left(R+r\right)}{\left(R+r\right)}=0 \\ \left(R+r\right)E^2\left[R+r-2R\right]=0 \\ \left(R+r\right)E^2\left[r-R\right]=0 \end{gathered}$$

Solve further as,

$$\begin{gathered} r=R=0 \\ R=r \end{gathered}$$

Hence the maximum power is delivered to the load when the internal resistance of the battery is equal to the load resistance.