A `20F` capacitor is charged to `5V` 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. `200 J`
C. `125 J`
D. `150 J`

Answer

D

Explanation

Charge stored on `20 F` capacitor `= 20 xx 5 = 100 C`
When `20F` capacitor is connected to `30F` capacitor,
`V = (q)/(C_(1) + C_(2)) = (100)/(20 + 30) = 2V`
Energy of the system will be `E_(f) = (1)/(2) xx (30 + 20) xx (2)^(2)`
`= 100 J`
Energy before isolation `E_(i) = (1)/(2) xx 20 xx (5)^(2) = 250 J`
Therefore, decrease in energy `= 250 - 100 = 150 J`