Question

Initially, 'n' identical capacitors are joined in parallel, are charged to potential 'V'. Now they are separated and joined in series. Then

• A. potential difference and total energy of the combination 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:

Charge each capacitor in parallel, then reconnect in series: voltages add, energy stays the same.

Answer 1

The Correct Option is C

Step 1: Add the energies directly:
In series, each capacitor still holds $\dfrac12CV^2$, because charge and capacitance of each are unchanged.

Step 2: Total:
The total energy is the sum $n\times\dfrac12CV^2$, which equals the earlier total.

Step 3: Voltage:
Each capacitor still has potential $V$ across it, so in series they add up to $nV$. Option C.

Final Answer:
Series gives nV across the pack, and the stored energy is unchanged. \[ \boxed{\text{(C) }\text{potential difference becomes }nV\text{ and energy remains the same}} \]