A part of a circuit in steady state along with the currents flowing in the branches,the values of resistances,etc, is shown in figure.Calculate the energy stored in the capacitor.

Explanation

To get the energy stored in the capacitor,we shall calculate the potential difference between the points P and Q. In steady state, there is no current in the capacitor branch. Applying Kirchhoff's junctions law at P, the current in the `(5 Omega-1 Omega)`branch will be 3 A and hence
`V_(P)-V_(S)=(6Omega)xx3A=18V.`
Applying the same theorem at Q, the current in the `(2Omega)`resistor will be 1A towards Q so that
`V_(S)-V_(Q)=(2Omega)xx1A=2 V.`
`Thus, V_(P)-V_(Q)=(V_(P)-V_(S))+(V_(S)-V_(Q))=20 V.`
The energy stored in the capacitor,brgt`=(1)/(2)CV^(2)=(1)/(2)xx4(mu)Fxx400V^(2)`
`800(mu)J.`