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

Your objective is to find out potential difference across the capacitor first. Applying Kirchoff.s junction law at A and B
`At A: 2 + 1 - I_(1)= 0, I_(1)= 3A`
`At B: I_(2) + 1 - 2 = 0, I_(2) =1 A`
Now, applying KVL in the mesh ADCBA (mesh means smallest loop or closed path)`V_(AB) = 20 V`
So, energy stored in the capacitor
`U=(1)/(2)CV^(2)=(1)/(2)xx(4xx10^(-6))xx(20)^(2)=8xx10^(-4)J`