Two condensers of capacity `3muF and 6muF` respectively are connected in series. The combination is connected across a potential of 6V. The ratio of the energies stored by the condensers will be

Options

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

Answer

B

Explanation

`V_(1)=Q/C_(1)=(VC_(2))/(C_(1)+C_(2))=(6xx6)/(3+6)=4V`
`E_(1)=1/2C_(1)V_(1)^(2)=(3xx10^(-6)xx16)/2=24xx10^(-6)J`
`V_(2)=Q/C_(2)=(VC_(1))/(C_(1)+C_(2))=(6xx3)/(3+6)=2V`
`E_(2)=1/2C_(2)V_(2)^(2)=(6xx10^(-6)xx4)/2=12xx10^(-6)J`
`therefore" "E_(1)/E_(2)=(24xx10^(-6))/(12xx10^(-6))=2/1`