Determine the equivalent capacitance between the points a and b for...

The given circuit can be simplified by identifying the parallel and series combinations of the capacitors.

The circuit consists of three parallel branches connected across the terminals $a$ and $b$:

1. The upper branch has capacitors $C_1$ and $C_2$ connected in series. The equivalent capacitance of this branch is $C_{upper} = \dfrac{C_1 C_2}{C_1 + C_2}$.

2. The lower branch also has capacitors $C_1$ and $C_2$ connected in series. The equivalent capacitance of this branch is $C_{lower} = \dfrac{C_1 C_2}{C_1 + C_2}$.

3. The middle branch has a single capacitor $C_3$ connected directly between $a$ and $b$.

Since these three branches are connected in parallel, the total equivalent capacitance $C_{ab}$ is the sum of their individual capacitances:

$C_{ab} = C_{upper} + C_3 + C_{lower}$

$C_{ab} = \dfrac{C_1 C_2}{C_1 + C_2} + C_3 + \dfrac{C_1 C_2}{C_1 + C_2}$

$C_{ab} = C_3 + \dfrac{2 C_1 C_2}{C_1 + C_2}$

Answer: $C_3 + \dfrac{2 C_1 C_2}{C_1 + C_2}$