Two capacitances of capacity `C_(1)`and `C_(2)` are connected in series and potential difference `V` is applied across it. Then the potential difference across `C_(1)` will be

Options

A. `V/2`
B. `((C_(2)^(2))/(C_(1)^(2)+C_(2)^(2)))V`
C. `((C_(1))/(C_(1)+C_(2)))V`
D. `((C_(2))/(C_(1)+C_(2)))V`

Answer

D

Explanation

In series, the charge on both capacitors is equal Let the steady state charge on each capacitor be Q.
Then, `Q=C_(eq)V=((C_(1)C_(2))/(C_(1)+C_(2)))V`
So, potential difference across the capacitor of capacitance `C_(1), V_(1)=(Q)/(C_(1))=((C_(2))/(C_(1)+C_(2)))V`.