MHTCET 22 Apr 2025 Shift 2 Paper

Concept:When charged capacitors are reconnected in series without any battery, total charge and energy remain conserved, while the voltages across individual capacitors add up.Explanation:Initially, all $n$ capacitors are in parallel, so each is charged to potential $V$.Charge on each capacitor is $Q = CV$.Total initial energy is $U_i = n \times \frac{1}{2}CV^2 = \frac{1}{2}nCV^2$.After disconnecting and joining them in series, no charge is lost.Each capacitor still holds charge $Q$, so the potential difference across each capacitor remains $V$.In a series combination, the total potential difference is the sum of individual voltages:$$V_{\text{total}} = V + V + \dots + V = nV$$Since the charge and capacitance of each capacitor are unchanged, the energy stored in each remains $\frac{1}{2}CV^2$.Total final energy is $U_f = n \times \frac{1}{2}CV^2 = \frac{1}{2}nCV^2$, which is equal to $U_i$.Therefore, the potential difference becomes $nV$ and the total energy remains the same.Answer:Option C: potential difference becomes $nV$ and energy remains the same.