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 arm containing $4 \Omega$ resistance will be ineffective and no current flows through it. The equivalent circuit is shown as below : Net resistance of $A B$ and $B C$, $R^{\prime}=4+2=6 \Omega$ Net resistance of $A D$ and $D C$, $R^{\prime \prime}=6+3=9 \Omega$ Thus, parallel combination of $R^{\prime}$ and $R^{\prime \prime}$ gives $\begin{gathered}R=\frac{R^{\prime} \times R^{\prime \prime}}{R^{\prime}+R^{\prime \prime}} \\=\frac{6 \times 9}{6+9}=\frac{54}{15}=\frac{18}{5} \Omega\end{gathered}$ Hence, current $i=\frac{V}{R}=\frac{V}{18 / 5}=\frac{5 V}{18}$