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(B) The circuit consists of three vertical branches connected in parallel. However,looking at the structure,we simplify from right to left.$1$. The tw

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$1.6 C$

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(B) The circuit consists of three vertical branches connected in parallel. However,looking at the structure,we simplify from right to left.$1$. The two capacitors on the far right are in series: $C_{eq1} = \frac{C 	imes C}{C + C} = \frac{C}{2}$.$2$. This $C_{eq1}$ is in parallel with the middle vertical capacitor $C$: $C_{eq2} = C + \frac{C}{2} = \frac{3C}{2}$.$3$. This $C_{eq2}$ is in series with the top horizontal capacitor $C$: $C_{eq3} = \frac{C 	imes (3C/2)}{C + 3C/2} = \frac{3C^2/2}{5C/2} = \frac{3C}{5} = 0.6C$.$4$. Finally,this $C_{eq3}$ is in parallel with the leftmost vertical capacitor $C$: $C_{eq} = C + 0.6C = 1.6C$.

What is the equivalent capacitance of the system of capacitors between $A$ and $B$?

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