Two capacitors of capacitances `3 mu F` and `6 mu F` are connected in series and connected to `120 V`. The potential difference across `3 mu F` is `V_(0)` and the charge here is `q_(0)`. We have
(A) `q_(0) = 40 mu C`
(B) `V_(0) = 60 V`
( C) `V_(0) = 80 V`
(D) `q_(0) = 240 mu C`.

Options

A. A, C are correct
B. A, B are correct
C. B,D are correct
D. C, D are correct

Answer

D

Explanation

`Q = ((C_(1)C_(2))/(C_(1) + C_(2))) V`