A `10 muF` capacitor is charged to a potential difference of `50 V` and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes `20` volt. The capacitance of second capacitor is

Options

A. `30muF`
B. `20muF`
C. `15muF`
D. `10muF`

Answer

C

Explanation

`C_(1)=(q_(1))/(V)" "therefore q_(1)=10xx10^(-6)xx50=500xx10^(-6)C`
When they are joined in parallel, their resultant capacity
`=C_(1)+C_(2)` and common potential `V=(q_(1)+q_(2))/(C_(1)+C_(2))`
`therefore C_(1)+C_(2)=(q_(1)+q_(2))/(V)`
`therefore C_(1)+C_(2)=(500xx10^(-6))/(20)=25xx10^(-6)muF`
`therefore C_(2)=(25-10)xx10^(-6)=15muF`