Two capacitors `C_(1)` and `C_(2)` in a circuite are joined as shown. If `V_(A)=V_(1),V_(B)=V_(2)`, the potemntial of point `P` is

Options

A. `(1)/(2)(V_(1)-V_(2))`
B. `(C_(2)V_(1)+C_(1)V_(2))/(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'=((C_(1))/(C_(1)+C_(2)))(V_(1)-V_(2))`
`V_(p)-V_(B)=(C_(1)V_(1)-C_(1)V_(2))/(C_(1)+C_(2))`
`V_(p)-V_(2)=(C_(1)V_(1)-C_(1)V_(2))/(C_(1)+C_(2))`
`V_(p)=(C_(1)V_(1)-C_(1)V_(2))/(C_(1)+C_(2))+V_(2)`
`=(C_(1)V_(1)+C_(2)V_(2))/(C_(1)+C_(2))`