For the network shown in fig, the value of n current is

The circuit is a balanced Wheatstone bridge. Therefore, the resistance 4$\mathrm{\Omega }$ is  and can be eliminated. The resistances 4 $\mathrm{\Omega }$ and 2$\mathrm{\Omega }$ are in series and their equivalent resistance is 6$\mathrm{\Omega }$. Similarly, the equivalent resistance of 6$\mathrm{\Omega }$ and 3$\mathrm{\Omega }$ resistances is 9 $\mathrm{\Omega }$ The equivalent resistance of the circuit is given by
$\begin{array}{l}{\mathrm{R}}_{\mathrm{eq}}=\frac{6\times 9}{6+9}=\frac{18}{5}\\ \therefore \mathrm{i}=\frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{\mathrm{V}}{(18/5)}=\frac{5\mathrm{V}}{18}\end{array}$