I'm working on the following homework problem. (Yes, this is homework. :p ) I'm supposed to find the current through L1 as a function of time.


simulate this circuit – Schematic created using CircuitLab

Using current division I've identified the current at zero time.

$$ I_L (0) = 0.01(15/(47+15)) = 2.42mA$$

However, I'm a little confused on how to find the equivalent resistance (since the Time Constant = Inductance/Equivalent Resistance.) Are they in parallel with each other? I guess the reason I'm confused is that my textbook said to find the equivalent resistance from the view of the inductor.

Note: The empty wire parallel to the inductor had a switch that was closed at t=0.

  • \$\begingroup\$ Hint: transform the Norton circuit composed of I1 and R1 to its Thevenin form. \$\endgroup\$ Oct 6, 2013 at 2:52
  • \$\begingroup\$ I had that done in my notes but that doesn't get rid of that middle empty cross connection. That's what's screwing with my mind right now \$\endgroup\$
    – codedude
    Oct 6, 2013 at 2:54
  • \$\begingroup\$ Is the wire placed or removed at t = 0? It the wire is placed at t=0, the resistance seen by the inductor is obvious. \$\endgroup\$ Oct 6, 2013 at 2:55
  • 1
    \$\begingroup\$ Indeed! 0||15 = 0. \$\endgroup\$ Oct 6, 2013 at 3:02
  • 3
    \$\begingroup\$ There won't be any current through L1, all the current will flow through the short(least resistance path) which is parallel to R1. \$\endgroup\$
    – AKR
    Oct 6, 2013 at 9:23

1 Answer 1


Looks like a trick question. Circuit is shorted, so all the current will flow through the wire. Current through \$L_1\$ is 0. Ran the SPICE sim just to be sure. Results and netlist below :)

SPICE sim results

I1 0 0 10m
R1 0 0 15
R2 0 N001 47
L1 N001 0 10m
.tran 0 10m 0 1u
  • \$\begingroup\$ You have neatly transcribed the comment into an answer. \$\endgroup\$
    – KalleMP
    Aug 1, 2016 at 8:44

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