For the network shown in the figure, the value of the current i is

The circuit given resembles the balanced Wheatstone bridge as $\frac{4}{6}=\frac{2}{3}$

Thus, middle anm containing $4\Omega$ resistance will be ineffective and no current flows through it.
The equivalent circuit is shown as below:

Net resistance of AB and BC,

$R^{\text{'}}=4+2=6\Omega$

Net resistance of AD and DC,

$R^{\text{'}\text{'}}=6+3=9\Omega$

Thus, parallel combination of R' and R" gives

$\mathrm{R}=\frac{{\mathrm{R}}^{\text{'}}\times {\mathrm{R}}^{\text{'}\text{'}}}{{\mathrm{R}}^{\text{'}}+{\mathrm{R}}^{\text{'}\text{'}}}=\frac{6\times 9}{6+9}=\frac{54}{15}=\frac{18}{5}\mathrm{\Omega }$

Hence, current  $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{\mathrm{V}}{18/5}=\frac{5\mathrm{V}}{18}$