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(D) The circuit can be simplified by identifying nodes with the same potential due to symmetry. Let the intermediate nodes be $P$ and $Q$. By analyzin

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$3 C$

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(D) The circuit can be simplified by identifying nodes with the same potential due to symmetry. Let the intermediate nodes be $P$ and $Q$. By analyzing the symmetry,the circuit can be redrawn as two parallel branches connected in series. Each branch consists of capacitors in parallel. The equivalent capacitance of the first part is $C_1 = 3 C + 2 C + C = 6 C$. The equivalent capacitance of the second part is $C_2 = 3 C + 2 C + C = 6 C$. Since these two parts are in series,the total equivalent capacitance $C_{AB}$ is given by: $\frac{1}{C_{AB}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{6 C} + \frac{1}{6 C} = \frac{2}{6 C} = \frac{1}{3 C}$. Therefore,$C_{AB} = 3 C$.

The network of six capacitors is as shown in the figure. The equivalent capacitance between $A$ and $B$ is

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