In the given figure,the equivalent capacitanc | vedclass

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(A) The circuit consists of two main parts connected in series.$1$. The first part (left side) consists of three capacitors of capacitance $C$ connect

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$1.5$

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(A) The circuit consists of two main parts connected in series.$1$. The first part (left side) consists of three capacitors of capacitance $C$ connected in parallel between the input node and the intermediate node. The equivalent capacitance is $C_1 = C + C + C = 3C$.$2$. The second part (right side) consists of three capacitors of capacitance $C$ connected in parallel between the intermediate node and point $B$. The equivalent capacitance is $C_2 = C + C + C = 3C$.$3$. Now,these two equivalent capacitors $C_1$ and $C_2$ are in series.$4$. The total equivalent capacitance $C_{AB}$ is given by $\frac{1}{C_{AB}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{3C} + \frac{1}{3C} = \frac{2}{3C}$.$5$. Therefore,$C_{AB} = \frac{3C}{2} = 1.5C$.

In the given figure,the equivalent capacitance between points $A$ and $B$ is (in $C$)

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