Consider the following network of capacitors. The capacitances are marked in the figure. If emnergy stored in the capacitor of capacitance `6muF` is E, then what will be the total amount of energy stored in the system?

Explanation

We can refer to the charge distribution in the following figure:

In Q is the total charge supplied by the battery then charges on individual capacitors are shown in the figure.
Energy stored in `6muF` is E:
`E=(Q^(2))/(2(6)) rArr Q^(2)=12E " " ...(i)`
Energy stored in `2muF` capacitor, `E_(1)=((Q//3)^(2))/(2(2))=(Q^(2))/(36)`
Using equation (i) we get the following:
`E_(1)=(Q^(2))/(36) =(12E)/(36)=(E )/(3) " " ...(ii)`
Energy stored in `4muF`:
`E_(2)=((2Q//3)^(2))/(2(4))=(Q^(2))/(18)=(12E)/(18)=2E//3" " ...(iii)`
Total stored energy `=E+(E )/(3)+(2E )/(3)=2E`