# Equivalent Resistance

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.

• Hint: transform the Norton circuit composed of I1 and R1 to its Thevenin form. – Alfred Centauri Oct 6 '13 at 2:52
• 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 – codedude Oct 6 '13 at 2:54
• 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. – Alfred Centauri Oct 6 '13 at 2:55
• Indeed! 0||15 = 0. – Alfred Centauri Oct 6 '13 at 3:02
• There won't be any current through L1, all the current will flow through the short(least resistance path) which is parallel to R1. – AKR Oct 6 '13 at 9:23

## 1 Answer

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 :) I1 0 0 10m
R1 0 0 15
R2 0 N001 47
L1 N001 0 10m
.tran 0 10m 0 1u
.backanno
.end

• You have neatly transcribed the comment into an answer. – KalleMP Aug 1 '16 at 8:44