I've been tackling some problems related to natural and step responses for RL & RC circuits, but I think I've hit a major roadblock for this RL circuit. I'm trying to find the time constant and current i3(t). While I'm certain that this type of circuit is first-order, I'm honestly stumped as to how the inductor, voltage source, and the current source change (or stay the same) for t<0, t=0, and t>0 conditions. enter image description here

I'm certain that the equivalent resistance, Req, will be required for the time constant, so I attempted to do superposition (removing sources) at t=0 in order to find Req, knowing that it is relative to the way it is viewed from the inductor. Upon redrawing the circuit diagram (for t=0) and using thevenin theorem to solve for Req, I ended up with Req = 1.59 ohms. I'm not 100% certain whether this is correct, but I found that R3 is || to the series combination of (R4 + R5), followed by adding R2 to that result, then making that parallel to R1, leading to 1.59 ohms. enter image description here However, what I'm puzzled about is how to solve for the current i3(t) across the R3 resistor. My way of thinking tells me that i3(t) includes the steady states for both i3(t=0) and i3(t= infinity), in which the first one is found by having the voltage source intact (and all of the resistors too), but the inductor does short circuit at t<0 (right?). i3(t=infinity) should be able to be found for by short circuiting the inductor, replacing the voltage source with a wire, and having the current source remain intact. I have a good feeling that this is one way to find i3(t), as the general equation for it is:

i3(t) = i3(t=infinity) + [i3(t<0) - i3(t=infinity)] e^(-t/tau)*

Are there perhaps potential gaps I have in my approach of solving for i3(t<0) and i3(t)?

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    \$\begingroup\$ R2 and R3 are not in parallel from the perspective of the inductor. You need to find the equivalent resistance that the inductor sees after the switch is closed. \$\endgroup\$ – Elliot Alderson Dec 2 '18 at 16:49
  • \$\begingroup\$ So, if R2 and R3 aren't in parallel, then R2 (upon redrawing the circuit) must be parallel to (R1, R4, and R5), since they are all in series relative to R2 then? In that sense, R3 must be in series instead with [R2 || (R1 + R4 + R5)] right? When I took that approach, I ended up with a equivalent resistance of 6 ohms, which led to a small time constant of (L/Req) = 0.000167 ms. Isn't that perhaps quite small? \$\endgroup\$ – Retrovert Dec 3 '18 at 4:16
  • \$\begingroup\$ @ElliotAlderson Also, considering that the equivalent resistance has been found to be higher than R3 series resistor, does the inductor no longer act as a short circuit, for t = 0+? Part of me thinks that with the current source being introduced into the circuit, both voltage/current sources change the dynamics of the inductor, or am I wrong in thinking this way? \$\endgroup\$ – Retrovert Dec 3 '18 at 9:19
  • \$\begingroup\$ No R1 is not in series with R4 and R5 for purposes of finding Req...one end of the inductor is connected between R1 and R4. The inductor does not act as a short circuit for t > 0, but the inductor current is continuous from t=0- to t=0+. I have no idea what you mean by "change the dynamics" of the inductor. \$\endgroup\$ – Elliot Alderson Dec 3 '18 at 13:15
  • \$\begingroup\$ @ElliotAlderson Wait, R1 isn't in series with R4/R5? How so? From what I see on the diagram, one end of the inductor is connected between R1 & R2, while the other end is between R3 & the voltage source (which is replaced by a wire). Am I misinterpreting that? As for the inductor current being continuous for both t<0 and t>0, does this suggest that iL(0-) = iL(0+) then? Oh, apologies for that remark, I'm just uncertain as to whether both the voltage and current sources affect the inductor current. So if I understand it right, is finding iL(t) no longer needed to solve for "i3" current? \$\endgroup\$ – Retrovert Dec 3 '18 at 13:58

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