A battery of emf 2V and internal resistance 1 $\mathrm{\Omega }$ is connected across terminals A and B of the circuit shown in figure.

In the simplified circuit, the circuit is a balanced Wheatstone bridge and a branch of 20/29 $\mathrm{\Omega }$ and R is parallel with this balanced bridge for maximum Power.

$\mathrm{r}={\mathrm{R}}_{\text{external }}$

$\begin{array}{l}1=\frac{1}{(3+6)}+\frac{1}{(20/29+\mathrm{R})}+\frac{1}{(4+8)}\\ \mathrm{R}=\frac{16}{29}\mathrm{\Omega }\end{array}$

Maximum power developed in the external circuit is

${\mathrm{P}}_{max}={\mathrm{i}}^2\mathrm{r}={\left(\frac{2}{1+1}\right)}^2\times 1=1\mathrm{W}$

Current through the upper branch

$\begin{array}{l}{\mathrm{i}}_1=\mathrm{i}\left[\frac{\left(\frac{20}{29}+\mathrm{R}\right)(4+8)}{9+\left(\frac{20}{29}+\mathrm{R}\right)+12}\right]\\ {\mathrm{i}}_2=\mathrm{i}\left[\frac{\left(\frac{20}{29}+\mathrm{R}\right)(3+6)}{9+\left(\frac{20}{29}+\mathrm{R}\right)+12}\right]\end{array}$

Therefore, ${\mathrm{i}}_1/{\mathrm{i}}_2$ is independent of R.