1. **Problem statement:** Determine the total capacitance and the maximum potential difference for three capacitors each of 30 µF arranged as two in parallel and one in series with this parallel group between points A and B. 2. **Formula and rules:** - For capacitors in parallel: $$C_{parallel} = C_1 + C_2 + \cdots$$ - For capacitors in series: $$\frac{1}{C_{series}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots$$ - The maximum voltage rating for capacitors in series is divided among them, but for parallel capacitors, the voltage rating remains the same. 3. **Step-by-step solution:** (i) Calculate the equivalent capacitance of the two capacitors in parallel: $$C_{parallel} = 30 + 30 = 60\ \mu F$$ (ii) Now, this 60 µF capacitor is in series with the third 30 µF capacitor. Calculate total capacitance: $$\frac{1}{C_{total}} = \frac{1}{60} + \frac{1}{30} = \frac{1}{60} + \frac{2}{60} = \frac{3}{60}$$ $$C_{total} = \frac{60}{3} = 20\ \mu F$$ (iii) Maximum potential difference: - Each capacitor is rated 6 V max. - The two capacitors in parallel share the same voltage, so max voltage across parallel group is 6 V. - The single capacitor in series also has max 6 V. - Total max voltage across A and B is sum of voltages across series capacitors: $$V_{max} = 6 + 6 = 12\ V$$ **Final answers:** - Total capacitance = 20 µF - Maximum potential difference = 12 V