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: Understanding the Concept:
Each capacitor of capacitance \(C\) is charged to \(V\) in parallel, so each carries charge \(q=CV\) and energy \(\dfrac12CV^2\).

Step 2: Energy before:
Total energy \(=n\cdot\dfrac12CV^2=\dfrac12nCV^2\).

Step 3: Reconnect in series:
Each capacitor still holds charge \(q=CV\) and the voltages add (with plates connected positive to negative): \(V_{total}=nV\). The series capacitance is \(\dfrac Cn\).

Step 4: Energy after:
\(U=\dfrac12\cdot\dfrac Cn\cdot(nV)^2=\dfrac12nCV^2\). The energy is unchanged. Option C.

Step 5: Why the other options are wrong.
A and B keep the potential at \(V\), which ignores that the voltages add. D claims the energy rises \(n\) times, but the energy is just the sum of the \(n\) energies already stored.

Final Answer:
The potential becomes nV and the energy stays the same. \[ \boxed{\text{(C) }\text{potential difference becomes }nV\text{ and energy remains the same}} \]