In the circuit shown, all the side resistances are of value $2\mathrm{\Omega }$ and all inner resistances are of value $4\mathrm{\Omega }.$. Find the current via branch ${\mathrm{R}}^{\mathrm{\prime }}=4\mathrm{\Omega }.$

Here BD is the line of symmetry. One part is the mirror image of the other part about BD. Hence ${\mathrm{V}}_{\mathrm{B}}={\mathrm{V}}_O={\mathrm{V}}_{\mathrm{D}}$. The resistances between B, O and O, D will become ineffective and the connections at B, D can be removed.

$\therefore {\mathrm{V}}_{\mathrm{AC}}=\frac{20}{3}\mathrm{V}$

current in any one of the four branches across A an C is $\mathrm{I}=\frac{{\mathrm{V}}_{\mathrm{AC}}}{\mathrm{R}}=\frac{20/3}{8}=\frac{20}{24}=0.8334\mathrm{A}$