Two condensers `C_(1) and C_(2)` in a circuit are joined as shown in figure. The potential of point A is `V_(1)` and that of B is `V_(2)`. The potential of point D will be

Options

A. `1/2(V_(1)+V_(2))`
B. `(C_(1)V_(2)+C_(2)V_(1))/(C_(1)+C_(2))`
C. `(C_(1)V_(1)+C_(2)V_(2))/(C_(1)+C_(2))`
D. `(C_(2)V_(1)-C_(1)V_(2))/(C_(1)+C_(2))`

Answer

C

Explanation

`V_(1)-V_(2)` will divided between `C_(1)` and `C_(2)` in series. Potential difference aross `C_(1)` is
`V_(1)-V_(D)=(C_(1)(V_(1)-V_(2)))/(C_(1)+C_(2))`
or `V_(D)=(C_(1)V_(1)+C_(2)V_(q2))/(C_(1)+C_(2))`.