I'm given this circuit which is in steady state: RL circuit

At some point the switch closes. What happens to the voltage across the inductor and the current through it?

Well, the way I see it is that if the circuit reached steady state, the inductor is a short so adding a resistor in parallel with it does nothing as no current will run though the resistor which means no voltage will build up at the junction between the inductor and two resistor. The voltage across the inductor is 0V and the current through it is 20/10k = 2mA no matter when I check after reaching steady state.

  • \$\begingroup\$ Sounds reasonable \$\endgroup\$ – Andy aka Feb 7 '14 at 13:11
  • \$\begingroup\$ Ummmmm, you are correct in your steady-state analysis. But, I believe at the moment the switch is closed a new dynamic is established: The SS 2 mA current in the inductor has established a magnetic field which will force the inductors current to remain at 2 mA. \$\endgroup\$ – FiddyOhm Feb 13 '14 at 23:04

Let's try it out:


simulate this circuit – Schematic created using CircuitLab

If you edit that schematic and run a time-domain simulation with time from 0 to 1 second with 0.001 second intervals, you'll notice the current through R1 is indeed 2mA. If you take out the inductor, the two resistors act as a voltage divider and the voltage between them is 12V.

So yes, your analysis is correct.

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