Question

A capacitor of capacity 2 $ \mu F $ is charged up to a potential 14 V and then connected in parallel to an uncharged capacitor of capacity 5 $ \mu F $. The final potential difference across each capacitor will be

• A. 6 V
• B. 4 V
• C. 8 V
• D. 14 V

Hint:

When capacitors are connected in parallel, they share the same potential difference, and the total charge is distributed between them according to their capacitances.

Answer 1

The Correct Option is B

To determine the final potential difference across each capacitor when they are connected in parallel, we follow these steps:

1. Calculate the initial charge on the charged capacitor using the formula: \( Q = C \times V \).

   Here, the capacitance \( C_1 \) is 2 µF and the initial potential difference \( V_1 \) is 14 V.

   \( Q_1 = 2 \, \mu\text{F} \times 14 \, \text{V} = 28 \, \mu\text{C} \).

2. After the capacitors are connected in parallel, they share the same potential difference, but the total charge is conserved.

   Total charge \( Q_{\text{total}} = Q_1 = 28 \, \mu\text{C} \).

3. Calculate the equivalent capacitance when they are connected in parallel: \( C_{\text{eq}} = C_1 + C_2 \).

   Here, \( C_2 \) is 5 µF, so:

   \( C_{\text{eq}} = 2 \, \mu\text{F} + 5 \, \mu\text{F} = 7 \, \mu\text{F} \).

4. Find the final potential difference using the formula: \( V_{\text{final}} = \frac{Q_{\text{total}}}{C_{\text{eq}}} \).

   \( V_{\text{final}} = \frac{28 \, \mu\text{C}}{7 \, \mu\text{F}} = 4 \, \text{V} \).

Therefore, the final potential difference across each capacitor is 4 V.

Answer 2

When two capacitors are connected in parallel, the potential difference across them will be the same. Initially, the 2 \( \mu F \) capacitor is charged to 14 V. The charge on this capacitor is: \[ Q = C \times V = 2 \, \mu F \times 14 \, V = 28 \, \mu C \] When it is connected in parallel to the uncharged 5 \( \mu F \) capacitor, the total charge is shared between the two capacitors. The total capacitance of the parallel combination is: \[ C_{\text{total}} = 2 \, \mu F + 5 \, \mu F = 7 \, \mu F \] The final potential difference across each capacitor is: \[ V_{\text{final}} = \frac{Q}{C_{\text{total}}} = \frac{28 \, \mu C}{7 \, \mu F} = 4 \, V \]
Thus, the final potential difference across each capacitor is 4 V.