In the network shown the current flowing through the resistance 2R is

${\text{R}}_{\text{1}}={\text{R}}_{\text{2}}={\text{R}}_{\text{3}}={\text{R}}_{\text{4}}=\frac{\text{R}}{\text{4}}$

$\frac{\text{1}}{{\text{R}}_{\text{P}}}=\frac{1}{{\text{R}}_{\text{1}}}+\frac{1}{{\text{R}}_{\text{2}}}+\frac{1}{{\text{R}}_{\text{3}}}+\frac{1}{{\text{R}}_{\text{4}}}$

$\frac{\text{1}}{{\text{R}}_{\text{P}}}=\frac{1}{\frac{\text{R}}{\text{4}}}+\frac{1}{\frac{\text{R}}{\text{4}}}+\frac{1}{\frac{\text{R}}{\text{4}}}+\frac{1}{\frac{\text{R}}{\text{4}}}$

$\frac{\text{1}}{{\text{R}}_{\text{P}}}=\frac{4}{\text{R}}+\frac{4}{\text{R}}+\frac{4}{\text{R}}+\frac{4}{\text{R}}$

$\frac{\text{1}}{{\text{R}}_{\text{P}}}=\frac{16}{\text{R}}$

${\text{R}}_{\text{P}}=\frac{\text{R}}{16}$

${\text{R}}^{\text{1}}=\text{R}+\frac{4\text{R}\times 2\text{R}}{6\text{R}}$                         

$=\text{R}+\frac{\text{8R}}{\text{6}}=\text{R}+\frac{\text{4R}}{\text{3}}$

$=\text{R}\left[1+\frac{4}{3}\right]$

${\text{R}}^{\text{1}}=\frac{\text{7R}}{3}$

$\text{V}={\text{IR}}^{\text{1}}$

$\text{I}=\frac{\text{V}}{\text{R}}=\frac{\text{E}}{\frac{\text{7R}}{\text{3}}}$

$\text{I}=\frac{\text{3E}}{\text{7R}}$

$I_t=I_1+I_2$

$\frac{\text{3E}}{\text{7R}}=\frac{\text{V}}{\text{2R}}+\frac{\text{V}}{\text{4R}}$

$\frac{\text{3E}}{\text{7R}}=\frac{\text{V}}{\text{R}}\left[\frac{3}{4}\right]\text{v}=\frac{4\text{E}}{7}$                    

$\text{v}=\text{IR}\text{V}={\text{I}}_{\text{2}}\left(2\text{R}\right)$

$=\frac{\text{4E}}{\text{7}}=i_2\times 2\text{R}$

${\text{i}}_2=\frac{\text{2E}}{\text{7R}}$