Solving Current in RL Circuit: Find i(t=6s)

[pillanoid]

Homework Statement

The voltage v(t) shown in the circuit attached ("Circuit 8.3") is 10 V for the time 0<t<2 (s), -10 V for the time 2<t<5 (s), and 0 all other times. If iL(0) = 0, what is i(t=6s)?

Homework Equations

i(t) = i(t=infinity) + (i(t_o)-i(t=infinity))*exp(-(t-t_o)/$$\tau$$) {1}
$$\tau$$ = L/R_equivalent {2}

The Attempt at a Solution

My understanding is that the current is found using {1}. However, after the voltage source is turned off, it becomes a short circuit, meaning that all of the current goes through the left side of the circuit, and that the inductor does not see any equivalent resistance. This would mean that the time constant $$\tau$$ is infinity.

I don't think that this makes sense, and if it does, I don't know what it means with respect to the problem. Am I approaching the problem correctly, or is there some other equivalent resistance in the circuit that I am not considering here?

 

[KataKoniK]

You are correct in your understanding that the current can be found using equation {1}. However, in this case, the equivalent resistance cannot be ignored as it is a crucial component in calculating the time constant \tau. The equivalent resistance can be found by combining the resistors in the circuit using series and parallel combination rules. Once the equivalent resistance is determined, it can be used in equation {2} to calculate the time constant. From there, you can use equation {1} to find the current at t=6s.

I hope this helps clarify the approach to the problem. If you have any further questions, please don't hesitate to ask. Best of luck with your calculations!