A capacitor of capacitance `10 m F` is charged up a potential difference of `2V` and then the cell is removed. Now it is connected to a cell of `emf 4V` and is charged fully. In both cases the polarities of the two cells are in the same directions. Total heat produced in the second charging process is :

Options

A. 10 mJ
B. 20 mJ
C. 40 mJ
D. 80 mJ

Answer

B

Explanation

Energy stored in capacitor when it is charged upto `2V = (1)/(2) 10 2^(2) = 20 mu J = u_(1)` (suppose)
Energy stored in capacitor when it is charged upto `4V =(1)/(2) 10 4^(2) = 80 mu J = u_(2)` (suppose)
Increase in charge `= 40 + 20 = 20 mu C`
Energy drawn from cell `= 20 xx 4 = 80 mu J = u` (suppose)
Heat produced `d =u_(1) + u-u_(2)`
`=20 +80 - 80 = 20 muJ`.