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(D) The given circuit is a balanced Wheatstone bridge. Let the capacitors be $C_1 = 6 \mu F$,$C_2 = 6 \mu F$,$C_3 = 6 \mu F$,$C_4 = 6 \mu F$,and $C_5

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$6$

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(D) The given circuit is a balanced Wheatstone bridge. Let the capacitors be $C_1 = 6 \mu F$,$C_2 = 6 \mu F$,$C_3 = 6 \mu F$,$C_4 = 6 \mu F$,and $C_5 = 20 \mu F$. We observe that $\frac{C_1}{C_3} = \frac{6}{6} = 1$ and $\frac{C_2}{C_4} = \frac{6}{6} = 1$. Since $\frac{C_1}{C_3} = \frac{C_2}{C_4}$,the bridge is balanced. Therefore,no charge flows through the capacitor $C_5$ $(20 \mu F)$,and it can be removed from the circuit. Now,$C_1$ and $C_2$ are in series,and $C_3$ and $C_4$ are in series. The equivalent capacitance of the upper branch ($C_1$ and $C_2$) is $C' = \frac{C_1 	imes C_2}{C_1 + C_2} = \frac{6 	imes 6}{6 + 6} = \frac{36}{12} = 3 \mu F$. The equivalent capacitance of the lower branch ($C_3$ and $C_4$) is $C'' = \frac{C_3 	imes C_4}{C_3 + C_4} = \frac{6 	imes 6}{6 + 6} = \frac{36}{12} = 3 \mu F$. Finally,$C'$ and $C''$ are in parallel. The net effective capacitance is $C_{eq} = C' + C'' = 3 \mu F + 3 \mu F = 6 \mu F$.

What is the effective capacitance between points $X$ and $Y$ in the given circuit? (All capacitors are in $\mu F$)

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