A condenser having a capacity of `6 muF` is charged to `100 V` and is then joined to an uncharged condenser of `14 muF` and then removed. The ratio of the charges on `6 muF` and `14 muF` and the potential of `6 muF` will be

Options

A. `(6)/(14)` and `50` volts
B. `(14)/(6)` and `30` volts
C. `(6)/(14)` and `30` volts
D. `(14)/(6)` and `0` volts

Answer

C

Explanation

Let `q_(1), q_(2)` be the charges on two condensers
`:. V = (q_(1))/(6) = (q_(2))/(14) rArr (q_(1))/(q_(2)) = (6)/(14) = (3)/(7)`
Also `q_(1) + q_(2) = 600 rArr q_(1) + (14)/(6)q_(1) = 600`
`rArr q_(1) = (600)/(20) xx 6`
`:. V = (q_(1))/(6) = (600)/(20) = 30` volt