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Driven RL circuit, why at infinity is this the current 2, and yet at 0+, its 0?

  1. Mar 26, 2006 #1
    Hello everyone i'm confused...i was just doing RL and RC circuits without a driven source, they made sense after doing them a few 100 times. Now i'm at Driven RL circuits. My professor said:
    iL(infinity) = 2mA
    iL(0+) = 0.

    From my understanding, i thought at infinity, the switch has been closed for a long period of time. If the switch is closed for a long period of time that means there would be no current flowing through the inductor, because you have to have changing current. So why isn't THe current of the inductor 0 at time infinity?>

    At time 0+ that means the switch has just been closed, meaning your going to have a change in current. So wouldn't that be when the current through the inductor is 2mA?

    Am i interpretting the unit step wrong?
    2u(t) ?
    this formula doesn't ask for time 0-, that means before the switch is closed, now at that point i can see how she would get iL(0-) = 0, becuase the current is totally disconnected!

    Any help on explaining this would be great! :)
    Also by doing it her way, she got the right answer in the back of the book.

    http://suprfile.com/src/1/bx0wa2/lastscan.jpg [Broken]

    THen she found VL, the voltage on the inductor and got the opposite, she wrote:
    VL(0) = 2 mA* 3k = 6v
    vL(infinity) = 0;
    VL = 6e^(-t/5E-6);
    x = 15mH/3k = 5E-6;

    She also wrote this which i think explains why,

    Current in inductor and Voltage on a capacitor can NOT change instantly
    Voltage on inductor and current in a capacitaor can change indstantly

    So is this why she has iL(0+) = 0? it takes time for the 2 mA to go through the inductor?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Mar 26, 2006 #2


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    No, after the switch has been closed for a very long time, an inductor will act as it it was an ideal wire! So in that case, all the current will go through the inductor and none through the resistor (since the current has the choice of flowing through a branch offering no resistance at all, it will go that way). At t=0+, the change of current is huge (theoretically infinite) so the inductor offers an inifinite resistance, so all the current will flow through the resistor.
  4. Mar 26, 2006 #3
    Thanks again for clearing up that! :biggrin:
    What if i replaced that inductor with a capacitor? With sourcless circuits I treated the capaictor as if it was an open circuit if no current has been flowing or if its been flowing for along time.

    nrqed, is this the case only for source driven RL/RC circuits? or does it also apply to sourceless RC/RL?
  5. Mar 26, 2006 #4


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    In general, you have the following (for DC circuits):

    If a circuit has been in a state for a very long time, inductors will act like ideal wires (R=0) and capacitors will act like infinite resistances (no current flows through their branch. They have a potential difference across them and Q= C V).

    If change is made (by closing, opening a switch) then an inductor will act (at t=0+) like an infinite resistance. For capacitors, it depends if they were initially charged or uncharged. If they were initially uncharged, at t=0+ they will act like ideal wires (R=0) (Of course, this is just an analogy..no current flows through them, but for the purpose of calculating solving the circuit they can be treated like ideal wires at t=0+).
    If they were initially charged, they can be treated (still, at t=0+) like batteries with a potential difference V=Q_0/C where Q_0 is the inital charge.

  6. Mar 26, 2006 #5
    Thanks a ton Patrick, excellent summary!
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