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

Applying Kirchhoff's first law at junctin M, we get the
current `I_1 3A.` Applying kirchhoff's firest law at junction P,
we get the current `I_2 =1A.`

Moving the loop form MNOP, we get
`V_m -5i 1 I 1 - 2 I 2 V -p or V_m v p 6i +2l 20V`
energy stored in the capacitor is
`(1)/I(2)CV^2 (1)/(2) xx4xx10^(-6) xx20xx20=8xx10^(-4)J`