Two capacitors of capacitances `3muF` 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.
(A) the common potential will be `12V`
(B) the loss in energy will be `576muJ`
(c) the common potential will be `4V`
(D) the loss in energy will be `144 muJ`

Answer

4

Explanation

Charge stored in capacitor `3muF=36muC`
Charge stored on capacitor `6muF=72muC`
After the two capacitors are connected let us assume that Q charge flows from the positive plate of capacitor `6muF` to negative plate of capacitor `3muF=36muF`.
At steady state, potential difference across the two capacitors is equal Thereofre, `(-36+Q)/(3)=(72-Q)/(6)`
`implies Q=48, therefore V_("common")=(48-36)/(3)=4V`