MHTCET 10 May 2024 Shift 1 Paper

Concept:When two identical capacitors are connected in parallel, they redistribute charge to a common potential, causing a loss of stored energy.Explanation:Initial energy of the system is$$U_i = \frac{1}{2}CV_1^2 + \frac{1}{2}CV_2^2 = \frac{1}{2}C\left(V_1^2+V_2^2\right)$$After connecting the positive ends together and negative ends together, the total charge is conserved.Total charge $Q = CV_1 + CV_2$ and total capacitance is $2C$.So the common potential is$$V = \frac{CV_1 + CV_2}{2C} = \frac{V_1+V_2}{2}$$Final energy of the combined system is$$U_f = \frac{1}{2}(2C)V^2 = \frac{1}{4}C\left(V_1+V_2\right)^2$$Decrease in energy$$\Delta U = U_i - U_f = \frac{1}{2}C\left(V_1^2+V_2^2\right) - \frac{1}{4}C\left(V_1+V_2\right)^2$$Simplifying,$$\Delta U = \frac{1}{4}C\left(V_1-V_2\right)^2$$Answer:The decrease in energy of the combined system is $\frac{1}{4}C\left(V_1-V_2\right)^2$.Therefore, the correct option is D.