a capacitor of capacitance 4µF is charged to 80V and another capacitor of capacitance 6µF is charged to 30V.when they are connected together,the energy lost by the 4µF capacitor is?
find the energy lost by the capacitor
The formula for Energy lost after connecting two capacitors with rating C1,V1 and c2,v2 respectively
When the two capacitors are connected, charge will be redistributed between the two capacitors so that the potential across both of them is the same.
HERE,Given C1=4uf, v1=80v and C2=6uf , v2=30v..putting the values we can find the energy lost.
Find charge on each capacitor using q = CV; Then find total charge which is q1 + q2= C1 V1 + C2 V2.
This charge will be redistributed between the two capacitors so that the potential across both is same
V1 = V2
q1/C1 = q2/C2 so q1/q2 = C1/C2 = 4/6
Here q1 = 2/5(q)
Similarly q2 = 3/5(q)
So the charge on each capacitor can be found, C of each capacitor is known, If V is found it will be same for both capacitors. Then energy can be found using 1/2CV^2 and loss in energy can also be found.