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(D) The given circuit is a balanced Wheatstone bridge. In a Wheatstone bridge,if the ratio of capacitors in opposite arms is equal,the central capacit

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$6$

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(D) The given circuit is a balanced Wheatstone bridge. In a Wheatstone bridge,if the ratio of capacitors in opposite arms is equal,the central capacitor $(C_5 = 20 \ \mu F)$ carries no charge and can be removed. Here,$\frac{C_1}{C_3} = \frac{6 \ \mu F}{6 \ \mu F} = 1$ and $\frac{C_2}{C_4} = \frac{6 \ \mu F}{6 \ \mu F} = 1$. Since the ratios are equal,the bridge is balanced. Removing $C_5$,the circuit simplifies to two parallel branches: Branch $1$: $C_1$ and $C_2$ are in series. $C_{s1} = \frac{6 	imes 6}{6 + 6} = 3 \ \mu F$. Branch $2$: $C_3$ and $C_4$ are in series. $C_{s2} = \frac{6 	imes 6}{6 + 6} = 3 \ \mu F$. Total effective capacitance $C_{eq} = C_{s1} + C_{s2} = 3 + 3 = 6 \ \mu F$.

The effective capacitance between $X$ and $Y$ is ... $\mu F$.

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