MHTCET 19 Apr 2026 Shift 1 Paper

Concept:When oppositely charged capacitors are connected in parallel, the net charge is the difference of their charges, and energy is stored in the equivalent capacitance.Explanation:Charge on the first capacitor is:$$Q_1 = C V$$Charge on the second capacitor is:$$Q_2 = (3C)(3V) = 9CV$$After disconnecting the batteries, total charge is conserved.Since the positive plate of one capacitor is joined to the negative plate of the other, the net charge is the difference:$$Q_{\text{net}} = 9CV - CV = 8CV$$In parallel, the equivalent capacitance is:$$C_{\text{eq}} = C + 3C = 4C$$The common final potential is:$$V_f = \frac{Q_{\text{net}}}{C_{\text{eq}}} = \frac{8CV}{4C} = 2V$$Final energy stored in the configuration is:$$U_f = \frac{1}{2}C_{\text{eq}}V_f^2 = \frac{1}{2}(4C)(2V)^2 = 8CV^2$$Therefore, the final energy of the configuration is $8.0\, CV^2$.Answer:Option C: $8.0\, C V^2$