Switch theorem: iL(0+)=iL(0-)=5A. The equivalent circuit at t=0+ is as follows.
Let the 3 Ω resistance current be I, and the 2 Ω resistance current can be obtained from KCL: I+I-0.5i = I-0.5i. ..
At this time, i=iL=5A, so the resistance current of 2ω is: I-0.5×5=I-2.5(A).
KVL:2×(I-2.5)+3I= 10, I=3(A). That is, I(0+)=3A.
When t=∞, the inductance is equivalent short-circuited, the 3 Ω resistance is short-circuited, and the current I(∞)=0. The equivalent circuit is as follows:
At this time, the 2 Ω resistance current is: i-0.5i=0.5i=0.5iL, and the direction is correct. According to KVL:
2×0.5iL= 10, iL= 10(A, that is, iL(∞)= 10A.
Find the equivalent resistance at both ends of the inductor. The equivalent circuit is as follows:
KCl: u/2+u/3+0.5i = i, so: 5U=3i, r = u/i = 3/5 = 0.6 (ω).
Time constant: τ=L/R=0.3/0.6=0.5(s).
Therefore: il (t) = il (∞)+[il (0+)-il (∞)] e (-t/τ) =10+(5-10) e (-t/0.5) = 65438+ (A) Fifty percent.
I(t)=0+(3-0)e^(-t/0.5)=3e^(-2t)？ (A) Fifty percent.
The energy consumed by t>0, 3Ω resistor is: W=∫(0, ∞)I? Rdt=∫(0,∞)[3e^(-2t)]？ ×3dt=27∫(0,∞)e^(-4t)dt=-27/4∫(0,∞)e^(-4t)d(-4t)=(-27/4)×[e^(-4t)]|(0,∞)=(-27/4)×(0- 1)=27/4=6.75(w)。