For the network shown in the figure,the value | vedclass

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(B) The given circuit is a Wheatstone bridge. Let the nodes be $A, B, C, D$. The bridge is balanced if the ratio of resistances in opposite arms is eq

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$\frac{5V}{18}$

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(B) The given circuit is a Wheatstone bridge. Let the nodes be $A, B, C, D$. The bridge is balanced if the ratio of resistances in opposite arms is equal. Here,the ratio of resistances is $\frac{4}{2} = 2$ and $\frac{6}{3} = 2$. Since the ratios are equal,the bridge is balanced.In a balanced Wheatstone bridge,no current flows through the central $4\,\Omega$ resistor. Thus,it can be removed from the circuit.Now,the circuit consists of two parallel branches: one with $(4\,\Omega + 2\,\Omega) = 6\,\Omega$ and the other with $(6\,\Omega + 3\,\Omega) = 9\,\Omega$.The equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{9} = \frac{3+2}{18} = \frac{5}{18}$.Therefore,$R_{eq} = \frac{18}{5}\,\Omega$.The total current $i$ from the battery is $i = \frac{V}{R_{eq}} = \frac{V}{18/5} = \frac{5V}{18}$.

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

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