When the switch is closed for a long time, the inductor acts as short for the DC current
Applying KCL at the node
Current IL will be
When the switch is opened
Applying KVL
Characteristic equation
Solution
At t=0, i(0) = - 0.1875A
Output voltage across the inductor will be
The time constant of the circuit is
τ = 1/5.33 = 0.1876
At five time constant , that is t = 5τ = 5(0.1876) = 0.938s