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Inductor Transient Question

  1. Oct 28, 2012 #1
    1. The problem statement
    See attached question - Inductor transient question.jpg
    So I have this inductor transient question and can't seem to get the right answer. Basically I don't know how to determine the current in steady state after switch operation.


    2. Relevant equations
    tau = L/C
    iL = a+be-t/tau
    V=IR
    3. The attempt at a solution
    tau = 4/8 = 0.5sec
    iL(0-) = 6A
    6A = a + b
    then I'm not sure what happens after switch operation.

    Any help appreciated thanks!
     
  2. jcsd
  3. Oct 28, 2012 #2

    gneill

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    Staff: Mentor

    How did you arrive at 6A for the initial inductor current? Show your work.
     
  4. Oct 28, 2012 #3
    Your initial current is not right.

    One way to solve after the switch is opened is to model the inductor as an inductor at rest with a parallel current source carrying the inductor's initial current.

    Another way seems to be what you are doing. You know the response will be first order and follow an exponential. The trick is to get the exponential equation right and to do that you have to look at your initial and final conditions. At time 0, iL = iL(0-). At time infinity, iL = ?. I know you know this one :)

    Then see if you can fit an exponential response to those two conditions.
     
  5. Oct 29, 2012 #4
    Does io(0-) = 4.8 Amps sound better?
    6 was a typo. I went 12x(4/10). Total current x resistance of interest/total resistance

    Yeah definately forgot to put that one in iL(∞) = 0


    So that would give me an answer of io(t) = 4.8e-2t ??
    Is that right?
    I'm not sure how the inductor would act at io(0+)
    I only learned that for capacitors. Would this effect it?
    And is my new initial current now correct, or did I not take the inductor into account?
     
  6. Oct 29, 2012 #5

    gneill

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    Staff: Mentor

    No, that's not right. The total resistance as seen by the voltage source (before the switch opens) is not 10Ω. Also, when determining the current through a particular resistor that is in parallel with another, the ratio you want is to place the resistance that's NOT of interest over the total resistance. If you think about it, you expect the lower resistance of a pair to carry more of the current, so it should have the higher ratio value...
    Inductors will not let their current change instantaneously, much like capacitors won't let their voltage change instantaneously. So the current through the inductor at t = 0+ must be the same as it was at t = 0-.
     
  7. Oct 29, 2012 #6
    You've got the right form of the equation but still the wrong initial current.

    We're assuming the circuit is at steady state just before the switch is thrown. This means all the currents and voltages in the circuit are constant.

    Take a look at the equation governing the V/I relationship of an inductor: V=L di/dt. If all currents are constant, what does that say about the voltage across the inductor? Does that make it easier to find the steady current flowing through the inductor?


    ===== a little extra

    I just want to add (and then correct a mistake I made above!), when you know the response is first order, you always know the voltage or current will follow either Ae-t/τ or A(1-e-t/τ). You just need to know what the voltage/current is at t=0 and what it is at t->∞ to know which of these the result is. That, added to knowledge of what τ is -- L/R in RL circuits and RC in RC circuits, is a very useful shorthand to get circuit behaviour for first order circuits. (The R in the time constant equations is the R seen by the inductor or capacitor).

    If you don't have a trick at hand for higher order circuits, you have to solve the usual way. Incorporating initial conditions for inductors and capacitors comes directly from the Laplace transform. Inductors: v=L di/dt -> V(s) = sL I(s) - Li(0), Capacitors: i = C dv/dt -> I(s) = sC V(s) - Cv(0). Notice how in the laplace domain, an inductor looks like a resistor of value sL in series with a battery of value -Li(0) (The voltage V(s) appears across the two in series). Also a capacitor looks like an admittance of sC-- equivalently an impedance of 1/sC -- with a current source of -Cv(0) in parallel.

    I don't know if you've studied that yet, but above I said you could replace the inductor with an inductor in parallel with a current source to model its initial condition. Actually

    I should have said the inductor could be modelled as an unenergized inductor in series with a battery of -Li(0) volts

    (the sign acts to generate the initial current direction). If you put that into the circuit, you could solve the usual way for circuit behaviour including initial conditions.
     
    Last edited: Oct 29, 2012
  8. Oct 29, 2012 #7
    I didn't think it was. I just don't know how to sub the inductor into it.


    Ahh so the initial voltage through the resistor is zero!
    Do I then treat that branch as if it is no longer connected?

    Sorry you are probably putting the answer pretty much in front of me and I'm not seeing it.


    So obviously we are going to end up with a negative due to the size of the inductor.

    Ok so we then have 24V = sL x 12A - 4i(o)
    Now sL = 1??
    therefore i(o) = -3
    therefore -3e-2t?
    Im not sure how i got sL = 1 though?

    By the way thanks both of you for the help so far.
     
  9. Oct 29, 2012 #8

    gneill

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    Staff: Mentor

    The DC steady-state resistance of an ideal inductor is zero. That means that "after a long time", when all transients have died away, an inductor looks like a short circuit: zero ohms.

    To find the steady-state conditions of a given circuit, replace inductors with wires (shorts) and remove all capacitors (opens). Then calculate the currents and potentials as required.
     
  10. Oct 29, 2012 #9
    Thanks that makes sense. I remembered that for capacitors, just forgot for inductors.
    So that would then give me 4Ω||4Ω (parallel) then in series with the 2Ω, which would then give me a total 4Ω resistor. And since current runs through each resistor in series equally that gives me 24/4 = 6A. Wait now we are back to the start.

    I'm sorry I'm really not seeing the answer!
     
  11. Oct 29, 2012 #10
    One of those 2 ohm resistors is 4Ω||4Ω. So that 6A passing through the 2 ohm equivalent resistor is splitting between those two branches.
     
  12. Oct 29, 2012 #11
    Ahh of course, I then thought that went in series with the 2 ohm, but if I use source transformation I get two parallel 2ohm resistance, thus giving me two separate branches.
    Thanks that gets me the right answer I think.
    i(0) = 3A (initially)
    Thanks heaps for your help. :)
     
  13. Oct 29, 2012 #12
    V= IR, the voltage is across the resistor, the current is through the resistor
    V = L di/dt, the voltage is across the inductor, the current is through the inductor

    If di/dt = 0, then V = voltage across the inductor is zero which means it looks like a wire.

    No, it has nothing to do with the size of the inductor. The sign is just falling out of the math. The important thing is the initial voltage has to drive the initial current in the right direction and that is how you determine what the sign is.

    V = L di/dt
    V(s) = sL I(s) - Li(0)

    The voltage is across the inductor and that looks like a resistance of sL in series with a voltage of Li(0). So it is valid to replace an inductor with non-zero initial current with an unenergized inductor in series with a voltage source.

    This is what would be done if the circuit was too complicated to know that form the answer is. You would replace your state elements (inductors and capacitors) with equivalent circuits that contain initial conditions and then solve with the usual KCL, KVL.

    But the path you've chosen is the easier one with first order circuits. You already know the answer so you are fitting initial and final conditions to it to nail down the constants.
     
  14. Oct 29, 2012 #13
    yes that's it, 3A :)
     
  15. Oct 29, 2012 #14
    Ok awesome, I think I understand now. Thanks heaps :) I'll go try another transient with an inductor and see how i go.
     
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