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When two charged capacitors are connected, charge redistributes to equalize potential. This is a frequently tested JEE topic with a key result: energy is always lost as heat.

Same polarity connection (positive to positive): Total charge is conserved: $Q_{total}$ = C1V1 + C2V2. Common potential: $V_f$ = $\frac{C1V1 + C2V2}{(C1 + C2)}$. Final charges: Q1 = C1*$V_f$, Q2 = C2*$V_f$.

Opposite polarity connection (positive to negative): Net charge: $Q_{net}$ = C1V1 - C2V2 (or |C1V1 - C2V2|). Common potential: $V_f$ = $\frac{Q_net}{C1 + C2}$. The polarity of $V_f$ is determined by which capacitor had more charge.

Energy loss (same polarity): $U_{lost}$ = C1C2(V1 - V2)^2 / (2(C1 + C2)). For opposite polarity: replace (V1-V2) with (V1+V2). This is always positive (except when V1 = V2, giving zero loss).

Maximum fractional loss occurs when C1 = C2 and the second capacitor is initially uncharged: exactly 50% of the initial energy is lost. The loss is independent of the connecting wire's resistance — even with zero resistance, the same energy is dissipated (as electromagnetic radiation in the ideal case).

This is analogous to a perfectly inelastic collision: momentum (charge) is conserved, but kinetic energy (electrical energy) is not. The fraction lost depends on the "mass ratio" (capacitance ratio) and "velocity difference" (voltage difference).