Solution: When t=0-, the inductance is equivalent to a short circuit. UL=0, controlled current source 0.25UL=0, which is equivalent to an open circuit.

At this time, it is equivalent to three 2 Ω resistors connected in parallel across the 6A current source, so iL(0-)=6/3=2(A).

Switch theorem: iL(0+)=iL(0-)=2A.

When t=∞, the inductance is equivalent to a short circuit. UL(∞)=0, and the controlled current source 0.25UL is zero.

So: iL(∞)=6/2=3(A).

Turn on the current source, apply a voltage u from the inductance disconnection, and let the inflow current be I.

Obviously: U=UL.

KVL:U = 2I+2×(I+0.25 U)= 2I+2I+0.5U .

0.5U=4I, REQ = U/I = 8 (ω), and the time constant of the circuit is: τ=L/Req=0.5/8=0.0625(s).

il(t)=il(∞)+[il(0+)-il(∞)]e^(-t/τ)=3+(2-3)e^(-t/0.0625)=3-e^(- 16t)？ (A) Fifty percent.

ul(t)=ldil(t)/dt=0.5×[3-e^(- 16t)]'=0.5×( 16e^(- 16t))=8e^(- 16t)(v)。