In the assembly shown in the figure, each capacitor possesses a...

The given circuit consists of three parallel branches between points A and B, where each branch contains three capacitors connected in series.

Let the capacitance of each individual capacitor be $C = 2 \mu\text{F}$.

For a single branch, the equivalent capacitance $C_{\text{branch}}$ of the three capacitors in series is calculated as:

$\dfrac{1}{C_{\text{branch}}} = \dfrac{1}{C} + \dfrac{1}{C} + \dfrac{1}{C} = \dfrac{3}{C}$

$C_{\text{branch}} = \dfrac{C}{3} = \dfrac{2}{3} \mu\text{F}$

The total equivalent capacitance $C_{AB}$ of the three parallel branches is the sum of their individual capacitances:

$C_{AB} = C_{\text{branch}} + C_{\text{branch}} + C_{\text{branch}} = 3 \times \left(\dfrac{2}{3}\right) = 2 \mu\text{F}$

When a battery of $60 \text{ V}$ is connected across A and B, the potential difference across each parallel branch is $60 \text{ V}$.

Since each branch consists of three identical capacitors in series, the total potential difference of $60 \text{ V}$ is divided equally among them.

The potential difference $V_c$ across each individual capacitor is:

$V_c = \dfrac{60}{3} = 20 \text{ V}$