I(0)=9*3/(3+6)=3A, and u 1(0)=-9V,

u 1(∞)=0

τ=L/R=4/(6+6//3)=4/8=0.5S

u 1(t)=-9e^(-t/τ)=-9^(-2t)

Primary energy storage of inductance WL =1/2 (L * I 2) =1/2 (4 * 9) =18j.

After stabilization, all the original energy is released and absorbed by the resistor, in which the energy absorbed by the 3-ohm 6-ohm parallel network is 1/4, and the energy absorbed by the 3-ohm resistor is 2/3 of 1/4 (according to calculation, the energy consumed by the 3-ohm resistor is w3 =18 * (1/).

Complete the solution