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