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(D) The given circuit can be redrawn as a balanced Wheatstone bridge. In a balanced Wheatstone bridge,the potential difference across the central capa

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$4$

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(D) The given circuit can be redrawn as a balanced Wheatstone bridge. In a balanced Wheatstone bridge,the potential difference across the central capacitor is zero,so it can be removed from the circuit. After removing the central capacitor,the two upper capacitors ($4 \mu F$ and $4 \mu F$) are in series,and the two lower capacitors ($4 \mu F$ and $4 \mu F$) are in series. The equivalent capacitance of the upper branch is $C_1 = \frac{4 	imes 4}{4 + 4} = 2 \mu F$. The equivalent capacitance of the lower branch is $C_2 = \frac{4 	imes 4}{4 + 4} = 2 \mu F$. These two branches are in parallel,so the total equivalent capacitance is $C_{eq} = C_1 + C_2 = 2 \mu F + 2 \mu F = 4 \mu F$.

The equivalent capacitance between points $X$ and $Y$ in the given figure is . . . . . . . (in $\mu F$)

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