The equivalent capacitance between $\mathrm{A}$ and $\mathrm{B}$ in the given figure is

Question

Aicharya · Accepted Answer

For symmetry potential difference across each capacitor of capacity is zero.

$\begin{aligned} & \frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{1}+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots \\ & =1+\frac{1}{3}\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots\right) \\ & =1+\frac{1}{3}\left[\frac{1}{1-\frac{1}{3}}\right]=1+\frac{1}{3} \times \frac{3}{2}=\frac{3}{2} \\ & \mathrm{C}_{\mathrm{eq}}=\frac{2}{3} \mu \mathrm{F} \text { then } \mathrm{C}_{\mathrm{AB}}=2 \times \frac{2}{3}=\frac{4}{3} \mu \mathrm{F}\end{aligned}$

The equivalent capacitance between $\mathrm{A}$ and $\mathrm{B}$ in the given figure is

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