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(D) The given circuit can be analyzed by identifying the nodes. The capacitors $C_1, C_2, C_3, C_4$ are all equal to $6\,\mu F$.By observing the circu

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$6$

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(D) The given circuit can be analyzed by identifying the nodes. The capacitors $C_1, C_2, C_3, C_4$ are all equal to $6\,\mu F$.By observing the circuit,we can see that it forms a Wheatstone bridge structure between points $A, B, C,$ and $D$.Specifically,the ratio of capacitances $\frac{C_1}{C_3} = \frac{6}{6} = 1$ and $\frac{C_2}{C_4} = \frac{6}{6} = 1$.Since the ratios are equal,the bridge is balanced.In a balanced Wheatstone bridge,the potential difference across the central capacitor $C_5$ is zero,so no charge flows through it.Thus,$C_5$ can be removed from the circuit.Now,the circuit consists of two parallel branches: one with $C_1$ and $C_2$ in series,and the other with $C_3$ and $C_4$ in series.Equivalent capacitance of the upper branch $(C_{upper})$ = $\frac{C_1 	imes C_2}{C_1 + C_2} = \frac{6 	imes 6}{6 + 6} = \frac{36}{12} = 3\,\mu F$.Equivalent capacitance of the lower branch $(C_{lower})$ = $\frac{C_3 	imes C_4}{C_3 + C_4} = \frac{6 	imes 6}{6 + 6} = \frac{36}{12} = 3\,\mu F$.Since these two branches are in parallel,the total effective capacitance $C_{eq} = C_{upper} + C_{lower} = 3 + 3 = 6\,\mu F$.

What is the effective capacitance between points $X$ and $Y$ in $\mu F$?

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