A capacitor of capacitance `10 muF` is charged to 100 V. It is now connected to uncharged capacitor in parallel so that the common potential becomes 40 V. The capacitance of the second capacitor is

Options

A. `10 mu F`
B. `5 mu F `
C. `15 mu F `
D. `25 mu F`

Answer

C

Explanation

Common potential `V = (C_1V_1 + C_2V_2)/(C_1 + C_2)`, hence `40 = ((10 muF) xx 100V + C_2 xx0)/((10 mu F + C_2)) rArr C_2 = 15 muF`