Question

Initially \( n \) identical capacitors are joined in parallel and are charged to potential V. Now they are separated and joined in series. Then}

• A. potential difference and total energy remain the same.
• B. potential difference remains the same and energy increases n times.
• C. potential difference becomes nV and energy remains the same.
• D. potential difference is nV and energy increases n times.

Hint:

In series, voltages add ($V_s = nV_p$); Energy remains constant if no charge is lost during reconnection.

Answer 1

The Correct Option is C

Step 1: Parallel Connection
Each capacitor has charge $q = CV$ and potential $V$. Total energy $U = n \left( \frac{1}{2}CV^2 \right)$.
Step 2: Series Connection
The potential differences add up: $V_{total} = V + V + ....... + V = nV$.
Step 3: Energy Conservation
Since the capacitors are isolated after charging, the total work/energy stored in the electric fields remains conserved. $U_{series} = \sum \frac{1}{2}CV^2 = n(\frac{1}{2}CV^2)$.
Final Answer: (C)