A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.
Advertisement Remove all ads
Solution
Let 'q' be the charge on the charged capacitor. Energy stored in it is
`U=q^2/(2C)`
When another similar uncharged capacitor is connected, the net capacitance of the system is C' = 2C
The charge on the system is constant. So, the energy stored in the system now is
`U'=q^2/(2(C"))`
`=>U'=q^2/(2(2C))`
`=>U'=q^2/(4C)`
Thus, the required ratio is
`(U')/U=(q^2/(4C))/(q^2/(2C))`
`=>(U')/U=1/2`
Concept: Capacitors and Capacitance
Is there an error in this question or solution?
Advertisement Remove all ads
APPEARS IN
Advertisement Remove all ads
Advertisement Remove all ads