A `100muF` capacitor is charged to `100 V`. After the charging, battery is disconnected. The capacitor is then connected in parallel to another capacitor. The final voltage is `20 V`. Calculate cap the capacity of second capacitor.

Answer

D

Explanation

Charge `q=CV =10^4muC`
In parallel common potential is given by
`V=("total charge")/("total capacity")`
`20=((10^4muC))/((C+100)muC)`
Solving this equation we get
`C=400muF`