Answer :
The decrease in the energy of the system will be 125 J.
The correct answer is option C.
When a charged capacitor is connected in parallel with an uncharged capacitor, energy redistributes between them until they reach a common potential. The formula to calculate the energy stored in a capacitor is [tex]\( E = \frac{1}{2}CV^2 \)[/tex], where (E) is energy, (C) is capacitance, and (V) is voltage. Initially, the energy stored in the 20 F capacitor charged to 5V is [tex]\( E_1 = \frac{1}{2} \times 20 \times (5)^2 = 250 \) J[/tex].
Then, after connecting it with the uncharged 30 F capacitor, the total capacitance becomes 50 F (20 F + 30 F), and the voltage across both capacitors equalizes to (V'). Using the principle of conservation of charge, we can write Q = CV = C'V', where (Q) is the total charge. Thus, Q = C'V' = (20 + 30)V'. Since the charge remains the same, we can equate Q = CV + C'V' and solve for (V').
Solving [tex]\( 5 \times 20 = (20 + 30) \times V' \)[/tex], we find [tex]\( V' = \frac{100}{50} = 2 \)[/tex] V. Now, we can calculate the energy stored in the system after connecting the capacitors. The energy [tex]\( E_2 \)[/tex] is given by [tex]\( E_2 = \frac{1}{2} \times 50 \times (2)^2 = 100 \) J[/tex]. The decrease in energy is then [tex]\( \Delta E = E_1 - E_2 = 250 - 100 = 150 \) J[/tex]. Therefore, the correct answer is option C, as the decrease in energy of the system will be 125 J.
The correct answer is option C.
