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Homework Help: RL Circuit

  1. Apr 20, 2008 #1
    Circuit is shown in attachement. At t = 0s, switch S is closed. Find a function for the current through the inductor starting from t = 0s.

    Using Kirchoff's method, I find that the maximum current through the inductor is .500 A as t approaches infinite and that the total current is 1.50 A. I took in the assumption that from t = 0s and onward, the current through the path with the inductor is proportionally 1/3 of the total current. Then I set up an equation:

    E - 3IR1 - IR2 - L*dI/dt = 0, in which E is the emf of the battery, I is the current through the inductor, R1 is the resistor on the top left (4.00 Ohm), R2 is the resistor on the top right (8.00 Ohm), L is the inductance, and dI/dt is a derivative of the current. R1 carries the total current, so its current would be 3 times the current in R2. Using this equation, I set up a differential equation, receiving:
    I = .500*(1 - e^(-20t)), in which the maximum current corresponds to the maximum current I got from Kirchoff's method.
    However, the answer I'm supposed to get is I = .500*(1 - e^(-10t))...I didn't see anything wrong in my method. Any help?

    Attached Files:

    Last edited: Apr 20, 2008
  2. jcsd
  3. Apr 20, 2008 #2
    Any help...
  4. Apr 20, 2008 #3
    That's not true. The current through the inductor directly after closing the switch is 0, but the total current is not 0.
  5. Apr 20, 2008 #4
    So then how would I model a situation like this? I keep bumping into answers that differ from the one I'm supposed to get.
  6. Apr 20, 2008 #5
    you use Kirchhofs laws like in a resistor network, and use L*(dI/dt) for the potential difference across the inductor
  7. Apr 20, 2008 #6
    ...Isn't that sort of what I did?
  8. Apr 20, 2008 #7


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    Homework Helper

    Hi Gear300,

    Since the current through the inductor is not one third the current through the battery at all times , you'll need to write down your three equations from Kirchoff's rule again, keeping i1, i2, and i3 as unknowns. (Your differential equation had the current through the inductor explicitly as one third the current through the battery.)

    Then use two of the equations to eliminate all currents except the current through the inductor. At that point you'll have a differential equation you can solve.

    It looks like you've done most of this process; just with the wrong currents.
  9. Apr 21, 2008 #8
    I see. Thanks, so steady-state conditions dont hold at each point in time...alright.
  10. Apr 22, 2008 #9
    Wait a minute...I sort of ran into a problem here. Each time I come up with a differential equation, I end up with 2 variable currents. I can eliminate the one I dont need by replacing it with values from equations I get through the Kirchoff's method, but I end up dealing with a 0; the integration still isn't the answer I need...heh, looks like I got into another loop...any help?
  11. Apr 22, 2008 #10
    Actually...nevermind that last statement...I just realized that replacing values from Kirchoff's method wouldn't hold valid.
  12. Apr 22, 2008 #11
    Nevermind...I have at last realized my mistakes...problem solved.
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