A 20F capacitor is charged to 5 V and isolated. It is then connected in parallel with an uncharged 30 F capacitor. The decrease in the energy of the system will be

Options

A. 25 j
B. 100 J
C. 125 J
D. 150 J

Answer

(d)

Explanation

`Delta U = 1/2(C_(1)C_(2)/C_(1) + C_(2))(V_(1) - V_(2))^(2) = 1/2 (20 xx 30 / 20 +30)(5 - 0)^(2)`
`= 1/2 (600/50)|(25) = 150 J`