Find the capacitance between points A and B of a circuit...

The given circuit forms a Wheatstone bridge with the following capacitors:

$C_1 = 1 \mu\text{F}$ (between A and C)
$C_2 = 2 \mu\text{F}$ (between A and D)
$C_3 = 3 \mu\text{F}$ (between C and B)
$C_4 = 6 \mu\text{F}$ (between D and B)
$C_5 = 5 \mu\text{F}$ (between C and D)

Checking the balance condition for the Wheatstone bridge:
$\dfrac{C_1}{C_2} = \dfrac{1}{2}$
$\dfrac{C_3}{C_4} = \dfrac{3}{6} = \dfrac{1}{2}$

Since $\dfrac{C_1}{C_2} = \dfrac{C_3}{C_4}$, the bridge is balanced. Therefore, the potential at node C is equal to the potential at node D ($V_C = V_D$). No charge will flow through the $5 \mu\text{F}$ capacitor, and it can be removed from the circuit.

The circuit now simplifies to two parallel branches:
Upper branch (series combination of $C_1$ and $C_3$):
$C_{upper} = \dfrac{C_1 C_3}{C_1 + C_3} = \dfrac{1 \times 3}{1 + 3} = \dfrac{3}{4} = 0.75 \mu\text{F}$

Lower branch (series combination of $C_2$ and $C_4$):
$C_{lower} = \dfrac{C_2 C_4}{C_2 + C_4} = \dfrac{2 \times 6}{2 + 6} = \dfrac{12}{8} = 1.5 \mu\text{F}$

The equivalent capacitance between points A and B is the sum of the capacitances of these parallel branches:
$C_{AB} = C_{upper} + C_{lower} = 0.75 + 1.5 = 2.25 \mu\text{F}$