Answer :
The decrease in energy when a 20F capacitor charged to 5V is connected in parallel with an uncharged 30F capacitor is 150J (D).
The question inquires about the change in energy when a fully charged 20F capacitor is connected in parallel with an uncharged 30F capacitor. Initially, the charged capacitor has an energy of 0.5 x 20F x (5V)^2 = 250J.
When the capacitors are connected in parallel, the total capacitance becomes 20F + 30F = 50F, and the charge remains the same, distributing over the larger total capacitance.
The final voltage V across both capacitors is given by the initial charge Q (which is 20F x 5V = 100C) over the total capacitance C, which is V = Q/C = 100C/50F = 2V.
The final energy of the system is 0.5 x 50F x (2V)^2 = 100J.
Therefore, the decrease in energy is 250J - 100J = 150J.
