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Homework Help: Inductance and RL-circuits.

  1. Mar 24, 2014 #1
    1. The problem statement, all variables and given/known data

    Compute I1, I2 and I3 instantly after the switch is opened in the below attached circuit.


    2. Relevant equations

    3. The attempt at a solution

    I don't know L and I don't know E. All I know is the currents while the switch is closed.

    I can't find tau, which I probably don't even need as it's the instant it's opened.

    How can I use only the currents while the switch is closed, to find the currents the instant the switch is opened?

    Sorry for not having too much work to show here, but I can't seem to find anything useful to solve this problem.

    Any feedback is appreciated. Thanks.

    Attached Files:

  2. jcsd
  3. Mar 24, 2014 #2


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    Staff: Mentor

    Start to work this problem by assuming some value for E, call it 1V. In the end, you may be able to cancel out the voltage or ignore it once you have the currents i(t).

    So if E = 1V, then what voltage(s) do you have across the 3 resistors? What value of resistance does that set for each?

    What voltage is across the inductor at t=0?

    Now open the switch at t=0, and solve for the voltage across the inductor v(t), and that gives you the i(t) for each resistor...

    [EDIT/ADD] -- just treat the bottom of the circuit as a common "ground" node.
  4. Mar 24, 2014 #3

    1V across R1, 1V across R2 and 1V across R3 and the inductor combined. I'm not sure how an inductor affects voltage when in series with a resistor, and I don't know the relative size on the two.

    Unfortunately, look at my above answer :uhh:
  5. Mar 24, 2014 #4


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    Staff: Mentor

    What is the resistance of an ideal inductor? (You can Google it if you haven't learned that already...)
  6. Mar 24, 2014 #5
    It doesn't have any resistance, but I was thinking maybe other factors (inductance) would affect the voltage on the resistor in series with the inductor.

    So, no voltage across the inductor, 1v across all the resistors?
  7. Mar 24, 2014 #6


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    Staff: Mentor

    Correct. Assuming that the switch was closed for a long time before t=0, then everything is steady state until the switch is opened.

    What equation should you use for the voltage across the inductor as a function of time? Can you work out the voltages and currents now?
  8. Mar 24, 2014 #7
    If I had two resistors in parallell (R1 is removed):

    VL = -(1+[itex]\frac{R2}{R3}[/itex]) * E

    Perhaps I should use a thevenin equivalent circuit?

    Edit: wrong formula. VL = -Vie-t/tau'

    Sorry, I got confused.
  9. Mar 24, 2014 #8
    Sorry for the double post, but apparently I can't edit my above post.

    The equation for voltage across the inductor as a function of time: VL = - E e-t/tau

    VL = 1ve-t/tau

    Tau = L/R

    I don't know L nor R. I can decide R if I apply my 1V, but I still won't know L?

    R1 = 200Ω
    R2 = 333.33Ω
    R3 = 100 Ω
  10. Mar 24, 2014 #9


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    Staff: Mentor

    I wouldn't worry about the L. It will just show up in your final answers.

    You have the initial current Ii for each of the 3 legs. You are solving for the I(t) for each of the 3 legs, so I wouldn't combine the first two resistors at this point.

    I'd also be inclined to use the original differential equation for the inductor voltager as a function of the changing inductor current, but it may simplify pretty quickly to the equation you have listed. At the very least, it will let you find tau in terms of L.
  11. Mar 24, 2014 #10
    I'm confused as to where to go from here. I understand what you're saying, but I'm not certain what's right and what's wrong.

    To proceed on my own, I need to know (or be pointed in the direction of) how I can find tau - if I'm using the correct approach.
  12. Mar 24, 2014 #11


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    Staff: Mentor

    Call the bottom node ground, and write the 2 KCL equations for the top left and right nodes. Use the differential equation for the inductor V,I relationship, and solve the KCL equations. If you want to simplify the left side into one resistor for now, you can do that and go back at the end to separate out the two currents.

    Can you show us the 2 KCL equations you get?
  13. Mar 24, 2014 #12
    I'm pretty sure this isn't what you're looking for, but rather the start:

    Left: I = IR1 + IR2R3
    Right: IR2R3 = IR2 + IR3

    Is this the differential equation you're referring to?

    v(t) = L * [itex]\frac{dt(t)}{dt}[/itex]

    If so, I'm not too sure what I should do with it.

    I'm sorry I'm not able to make any more progress than this, but I'm trying. It's frustrating not being able to do more when you're obviously trying hard to help me. Sadly, I can't just force myself to understand it...
  14. Mar 24, 2014 #13


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    Staff: Mentor

    I1 and I2 are on the left, and I3 is on the right.

    The KCL equations are the sum of all currents out of each node = 0.

    So the currents leaving the left node are I1 and I2 and the current flowing left-to-right through the inductor.

    The currents leaving the right node are I3 and the current flowing right-to-left through the inductor.

    The current flowing through the inductor depends on the voltage across it (via the differential equation). Call the left node's voltage with respect to ground V1 and the right node's voltage V3.

    Now try to write the 2 KCL equations for the left and right node...
  15. Mar 24, 2014 #14


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    Staff: Mentor

    BTW, your differential equation has a typo in it -- it should be di(t)/dt, not dt(t)/dt. :smile:
  16. Mar 24, 2014 #15
    You're absolutely correct, of course.

    It's approaching 2am here, and I need to get some sleep before my lecture tomorrow. I also think, or hope, that I will do better at this tomorrow, rested and motivated.

    Thanks so much for helping me out - I really appreciate it - and I guarantee I will reply, or edit this post, by tomorrow afternoon.:smile: :zzz:
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