First, the initial values of iR(0+) and iL(0+)

Before the switch is in a steady state, the inductance is regarded as a short circuit.

And the current of the inductor cannot change suddenly. therefore

iL(0+)= iL(0-)= 10/(2+3)= 2(A)

When t=0+, iL(0+) is regarded as the current source (as shown in the figure).

Obtained by superposition method

iR(0+)= 10/(6+3)-2×3/(6+3)= 4/9A

Secondly, the inductance is considered as short circuit in the steady state after switching, as shown in the figure.

3Ω+(2Ω//6Ω)=4.5Ω

10V/4.5ω= 20/9 (ampere)

iR(∞)= 20/9×2/(2+6)= 5/9(A)

iL(∞)=(20/9)×6/(2+6)=5/3 (A)

Third, the time constant τ=L/R

r = 2+(6//3)= 4ω

τ=L/R= 1/4 (s)

Four,