In the circuit shown here `C_(1)=6muF,C_(2)=3 muF` and battery B = 20 V. The switch `S`, is first closed. It is then opened and afterwards `S_(2)` is closed. What is the charge finally on `C_(2)` ?

Options

A. `120 muC`
B. `80 muC`
C. `40 muC`
D. `20 muC`

Answer

C

Explanation

After closing `S_(1)`, charge on `C_(1)` is `q=6xx20=120 muC`. Now, `S_(1)` is opene. On closing `S_(2)`, Charge q will be distributed between `C_(1)` and `C_(2)` according to their capacitances. So charge on `C_(2)` is
`q_(2) = (C_(2)q)/(C_(1) + C_(2)) = (3 xx 120)/(3+6) = 40 muC`