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(D) The given circuit is a Wheatstone bridge. We check the ratio of the capacitors in the arms: $\frac{C_1}{C_2} = \frac{3}{4}$ and $\frac{C_3}{C_4} =

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$\frac{14}{3}\,\mu F$

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(D) The given circuit is a Wheatstone bridge. We check the ratio of the capacitors in the arms: $\frac{C_1}{C_2} = \frac{3}{4}$ and $\frac{C_3}{C_4} = \frac{6}{8} = \frac{3}{4}$.Since the ratios are equal,the bridge is balanced.In a balanced Wheatstone bridge,the potential difference across the central capacitor $(2\,\mu F)$ is zero,so it can be removed from the circuit.Now,the $3\,\mu F$ and $6\,\mu F$ capacitors are in series,and the $4\,\mu F$ and $8\,\mu F$ capacitors are in series.The equivalent capacitance of the upper branch is $C_{up} = \frac{3 	imes 6}{3 + 6} = \frac{18}{9} = 2\,\mu F$.The equivalent capacitance of the lower branch is $C_{low} = \frac{4 	imes 8}{4 + 8} = \frac{32}{12} = \frac{8}{3}\,\mu F$.These two branches are in parallel,so the total effective capacitance is $C_{AB} = C_{up} + C_{low} = 2 + \frac{8}{3} = \frac{6 + 8}{3} = \frac{14}{3}\,\mu F$.

The effective capacitance between $A$ and $B$ in the shown figure is (all capacitances are in $\mu F$):

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