The switch is closed and then reopened after 1.0 s. What is the current through the inductor 0.5 s after the switch is reopened?
Immediately on switch closure, the circuit behaves like the one shown at the left.
This may be determined by determining the Thevenin generator for the circuit above.
Alternatively it may be argued that the 6 ohm resistor has no effect on the energizing time
of the inductor or its final current state,
since the inductor and its resistor are connected directly across the battery.
The final current will be (6.0 V) / (3.0 ohm) = 2.0 A.
The time constant of the circuit is (2.0 H) / (3.0 ohm) = 0.67 s.
The current through the inductor increases as i(t) = (2.0 A) [1 - exp(-t/0.67 s)] after switch closure (t = 0 s). When the switch reopens at 1 s, the current has reached a value of (2.0 A) [1 - exp(-3/2)] = (2.0 A)(0.777) = 1.55 A.
When the switch is reopened, the circuit behaves as the circuit at the left.
The time constant is now (2.0 H) / (9.0 ohm) = 0.22 s.
The inductor deenergizes from its starting current according to i(t) = (1.55 A) exp[-(t-1 s)/0.22 s].
At 0.5 s after the switch reopens; that is, at t = 1.5 s, the current through the inductor is (1.55 A) exp(-0.5/0.22) = (1.55 A)(0.103) = 0.16 A.