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(B) Looking at the circuit, we can identify three parallel branches connected between points $A$ and $B$.$1$. The left branch contains two capacitors

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$3$

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(B) Looking at the circuit, we can identify three parallel branches connected between points $A$ and $B$.$1$. The left branch contains two capacitors of $2 C$ each in series. Their equivalent capacitance is $C_{left} = \frac{2 C 	imes 2 C}{2 C + 2 C} = C$.$2$. The middle branch contains a single capacitor of capacitance $C$. So, $C_{middle} = C$.$3$. The right branch contains a capacitor of $2 C$ in series with the parallel combination of two capacitors of $C$ each. The parallel combination gives $C_{parallel} = C + C = 2 C$. This $2 C$ is in series with the $2 C$ capacitor, so $C_{right} = \frac{2 C 	imes 2 C}{2 C + 2 C} = C$.Since all three branches are in parallel, the total equivalent capacitance is $C_{eq} = C_{left} + C_{middle} + C_{right} = C + C + C = 3 C$.

Find the equivalent capacitance between point $A$ and $B$. (in $C$)

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