Question

Two capacitors of capacities 10 µF and 20 µF are connected in series across a 200 V source. The charged capacitors are disconnected from the source and reconnected in parallel with their positive plates together. The potential difference across each capacitor is:

• A. 200 V
• B. 400 V
• C. \( \frac{400}{9} \, \text{V} \)
• D. 800 V

Hint:

When capacitors are connected in series, the total capacitance is found using \( \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} \), and the total charge is the same. When they are connected in parallel, the potential across each is the same.

Answer 1

The Correct Option is C

When the capacitors are connected in series, the total capacitance is given by: \[ \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} \] Thus, \( C_{\text{total}} = \frac{20}{3} \, \mu\text{F} \). The total charge when connected in series is: \[ Q_{\text{total}} = C_{\text{total}} \times V = \frac{20}{3} \times 200 = \frac{4000}{3} \, \mu\text{C} \] Now, reconnecting the capacitors in parallel, the potential difference across each capacitor is: \[ V = \frac{Q_{\text{total}}}{C_1 + C_2} = \frac{\frac{4000}{3}}{30} = \frac{400}{9} \, \text{V} \] Thus, the potential difference across each capacitor is \( \frac{400}{9} \, \text{V} \).