Answer :
The decrease in energy of the system will be 125 J when a 20F charged capacitor is connected in parallel with a 30F uncharged capacitor.
The decrease in energy of the system will be 125 J. When a 20F capacitor charged to 5V is connected in parallel with an uncharged 30F capacitor, the total energy stored in the system initially is 250 J. After connecting them in parallel, the total capacitance becomes 50F and the voltage remains 5V.
Using the formula for energy stored in a capacitor, the final energy stored in the system is 125 J. The decrease in energy is given by the initial energy minus the final energy, which equals 125 J.
