Two capacitors of capacitance `2muF` and `4muF` respectively are charged to a potential of 12 V. 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 capacitor will be

Options

A. 2 V
B. 3V
C. 4V
D. 6V

Answer

C

Explanation

`Q_(1)=CV_(1)=12xx2=24muC and Q_(2)=VC_(2)=12xx4=48muC`
`therefore V=(Q_(2)-Q_(1))/(C_(1)-C_(2))=(24xx10^(-6))/(6xx10^(-6))=4V`