In the circuit shown below, four capacitors are connected between...

From the circuit diagram, nodes X, Y, and Z are connected by ideal wires, meaning they are at the same potential.

Capacitor $C_3$ is connected between nodes X and Y. Since $V_X = V_Y$, the potential difference across $C_3$ is zero, so it is short-circuited and can be removed.

Capacitors $C_1$ and $C_2$ are connected in parallel between terminal A and the common node Z. Their equivalent capacitance is:

$C_{12} = C_1 + C_2 = 12 + 12 = 24\ \mu\text{F}$

This combination is in series with $C_4$ between node Z and terminal B. The total equivalent capacitance $C_{AB}$ is:

$C_{AB} = \dfrac{C_{12} C_4}{C_{12} + C_4}$

$C_{AB} = \dfrac{24 \times 12}{24 + 12} = \dfrac{288}{36} = 8\ \mu\text{F}$

Answer: $8\ \mu\text{F}$