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(A) To find the effective capacitance between points $A$ and $B$,we analyze the circuit step by step.$1$. The circuit contains five capacitors,each of

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$2 C$

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(A) To find the effective capacitance between points $A$ and $B$,we analyze the circuit step by step.$1$. The circuit contains five capacitors,each of capacitance $C$.$2$. By simplifying the circuit using nodal analysis or symmetry,we observe that the circuit can be reduced to a simpler equivalent circuit.$3$. As shown in the equivalent circuit diagram,the combination simplifies to two capacitors of $2C$ in series,which are in parallel with a capacitor of $C$.$4$. The series combination of two $2C$ capacitors gives an equivalent capacitance of $C_{s} = \frac{2C 	imes 2C}{2C + 2C} = \frac{4C^2}{4C} = C$.$5$. This $C$ is then in parallel with the remaining capacitor $C$,so the total effective capacitance is $C_{eq} = C + C = 2C$.

The effective capacitance between points $A$ and $B$ shown in the circuit is

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