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(B) The circuit consists of a network of capacitors. We simplify it step by step from the right side towards the left.$1$. The two capacitors on the f

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$1.6C$

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(B) The circuit consists of a network of capacitors. We simplify it step by step from the right side towards the left.$1$. The two capacitors on the far right are in series. Their equivalent capacitance $C_1$ is given by:$1/C_1 = 1/C + 1/C = 2/C \implies C_1 = C/2$.$2$. This $C_1$ is in parallel with the vertical capacitor $C$ in the middle. Their equivalent capacitance $C_2$ is:$C_2 = C + C/2 = 3C/2$.$3$. This $C_2$ is in series with the top horizontal capacitor $C$. Their equivalent capacitance $C_3$ is:$1/C_3 = 1/C + 1/(3C/2) = 1/C + 2/(3C) = 5/(3C) \implies C_3 = 3C/5$.$4$. Finally,this $C_3$ is in parallel with the leftmost vertical capacitor $C$. The total equivalent capacitance $C_{eq}$ is:$C_{eq} = C + 3C/5 = 8C/5 = 1.6C$.

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

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