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(A) In the given circuit,the first three capacitors are connected in parallel. Let their equivalent capacitance be $C_1$. Since they are in parallel,$

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$1.5\,C$

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(A) In the given circuit,the first three capacitors are connected in parallel. Let their equivalent capacitance be $C_1$. Since they are in parallel,$C_1 = C + C + C = 3C$.Similarly,the last three capacitors are also connected in parallel. Let their equivalent capacitance be $C_2$. Thus,$C_2 = C + C + C = 3C$.Now,these two groups (each of capacitance $3C$) are connected in series between points $A$ and $B$.The effective capacitance $C_{eq}$ is given by the series combination formula: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$.Substituting the values: $\frac{1}{C_{eq}} = \frac{1}{3C} + \frac{1}{3C} = \frac{2}{3C}$.Therefore,$C_{eq} = \frac{3C}{2} = 1.5C$.

All capacitors used in the diagram are identical and each is of capacitance $C$. Then the effective capacitance between the points $A$ and $B$ is

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