L 2 henry and R 1/3 ohm Since the switch was in the A position for a long time before it was switched to the B position at t 0, there is no current in the inductor at t 0+ At t 0+, 1 amp V/1Ω + (1/L)(∫ Vdt) The voltage across the inductor will be V(t) 1(e^ -(t/τ) for 0+ ≤ t 2 seconds where τ L/1Ω L since the resistor is 1Ω The current through the inductor will be i (t) (1 - e^-(t/τ) amps from 0+ ≤ t 2 seconds where τ L/1Ω L since the resistor is 1Ω At t 2- seconds i (2-) (1 - e^-(2/L) amps (1 - e^-(1)) since t/τ 2/L 1 at t 2-, L must be 2 Henry At t 2 seconds the switch is thrown back to A. The inductor current will be i( t2+) (1 - e^-1)(e^-(t/τ) where τ L/R 2/R i(6) (1 - e^-1)(e^-(6/τ) (1 - e^-1)(e^-(1) so 6/τ 1 6/(2/R) 3R R 1/3 ohm
