\(N\) identical capacitors are joined in parallel and the combination is charged to a potential \(V\). Now if they are separated and then joined in series then energy of combination will :
remain same and potential difference will also remain same
remain same and potential difference will become \(NV\)
increase \(N\) times and potential difference will become \(NV\)
increase \(N\) time and potential difference will remains same
Solution:
When connected in series, the charges do not change, so the total energy stored remains the same: \(U = N \times \left(\frac{1}{2} C V^2\right)\). However, the individual potential differences add up, so the total potential difference becomes \(NV\).
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