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 $4 Ω$ resistance will be ineffective and no current flows through it.
The equivalent circuit is shown as below:

Net resistance of AB and BC,

$R^{'} = 4 + 2 = 6 Ω$

Net resistance of AD and DC,

$R^{''} = 6 + 3 = 9 Ω$

Thus, parallel combination of R' and R" gives

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

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

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