if two capacitors of capacities `C_(1)=4muF and C_(2)=1muF` are connected in series, the ratio of the potential drops across the capacitors `C_(1) and C_(2)` is

Options

A. `1:4`
B. `4:1`
C. `1:2`
D. `2:1`

Answer

A

Explanation

`V=(Q)/(C)` since they are in series. Q is same for both.
`therefore (V_(1))/(V_(2))=(C_(2))/(C_(1))=(1)/(4)`