Determine the equivalent capacitance between the points A and B for...

The given circuit (a) is a Wheatstone bridge.

Let the capacitors be $C_1 = 1\ \mu\text{F}$, $C_2 = 2\ \mu\text{F}$, $C_3 = 3\ \mu\text{F}$, $C_4 = 6\ \mu\text{F}$, and the central capacitor be $C_5 = 5\ \mu\text{F}$.

Checking the balance condition of 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, no charge will flow through the central capacitor $C_5$, and it can be removed from the circuit.

The circuit now simplifies to two parallel branches. The top branch has $C_1$ and $C_3$ in series, and the bottom branch has $C_2$ and $C_4$ in series.

Equivalent capacitance of the top branch:

$C_{top} = \dfrac{C_1 C_3}{C_1 + C_3} = \dfrac{1 \times 3}{1 + 3} = \dfrac{3}{4} = 0.75\ \mu\text{F}$

Equivalent capacitance of the bottom branch:

$C_{bottom} = \dfrac{C_2 C_4}{C_2 + C_4} = \dfrac{2 \times 6}{2 + 6} = \dfrac{12}{8} = 1.5\ \mu\text{F}$

The total equivalent capacitance between A and B is the sum of the capacitances of these parallel branches:

$C_{AB} = C_{top} + C_{bottom} = 0.75 + 1.5 = 2.25\ \mu\text{F}$