The equivalent capacitance between A and B is | vedclass

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(D) In the given circuit,let us analyze the connections. The four capacitors,each of capacitance $C$,are connected such that the first capacitor is in

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$\frac{4C}{3}$

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(D) In the given circuit,let us analyze the connections. The four capacitors,each of capacitance $C$,are connected such that the first capacitor is in parallel with the series combination of the other three capacitors.Let the nodes be defined. The top wire connects the left plate of the first capacitor to the right plate of the fourth capacitor.This simplifies the circuit into two branches connected in parallel between points $A$ and $B$:$1$. The first branch contains a single capacitor of capacitance $C$.$2$. The second branch contains three capacitors of capacitance $C$ each,connected in series.The equivalent capacitance of the three capacitors in series is given by $\frac{1}{C_{eq1}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C}$,which implies $C_{eq1} = \frac{C}{3}$.Now,these two branches are in parallel,so the total equivalent capacitance $C_{eq}$ is the sum of the capacitances of the two branches:$C_{eq} = C + C_{eq1} = C + \frac{C}{3} = \frac{4C}{3}$.

The equivalent capacitance between $A$ and $B$ is

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