Two capacitors of capacitances 3`muF and 6muF` are charged to a potential of 12V each. They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each will be

Options

A. 4V
B. 6V
C. zero
D. 3V

Answer

A

Explanation

Capacitance,
`C_(1)=4muF,C_(2)=6muF`.
Potential, `V_(1)=12V,V_(2)=12V`
`therefore` Charge `q_(1)=C_(1)V_(1)`
`=3xx10^(-6)xx12C`
`q_(2)=C_(2)V_(2)=6xx10^(-6)xx12C`
The arrangement is shown in the figrue.
Here, net charge after connection
`=q_(2)-q_(1)=6xx10^(-6)xx12-3xx10^(-6)xx12`
let V be the common potential difference after connection.
Then `V(C_(1)+C_(2))=q_(2)-q_1`
or `V(3xx10^(-6)+6xx10^(-6))`
`=6xx10^(-6)xx12-3xx10^(-6)xx12`
or, `V=(12xx3xx10^(-6))/(9xx10^(-6))=4V`.