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(D) The given circuit is a balanced Wheatstone bridge. In a balanced Wheatstone bridge,the central capacitor (between $B$ and $D$) does not contribute

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$6$

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(D) The given circuit is a balanced Wheatstone bridge. In a balanced Wheatstone bridge,the central capacitor (between $B$ and $D$) does not contribute to the charge flow,so it can be removed. Now,the two upper capacitors of $6\ \mu F$ each are in series,giving an equivalent capacitance of $C_1 = (6 	imes 6) / (6 + 6) = 3\ \mu F$. Similarly,the two lower capacitors of $6\ \mu F$ each are in series,giving an equivalent capacitance of $C_2 = (6 	imes 6) / (6 + 6) = 3\ \mu F$. These two combinations ($C_1$ and $C_2$) are in parallel with each other. Therefore,the total effective capacitance is $C_{eq} = C_1 + C_2 = 3\ \mu F + 3\ \mu F = 6\ \mu F$.

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

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