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Kirchhoff's law in LC circuit

  1. Oct 15, 2014 #1
    I am confused with explanation about Kirchhoff's law in LC circuit.
    Please refer to the file I attached for the LC circuit.
    First, the switch was connected to the emf and the capacitor was charged to its full capacitance.
    Then the switch is connected to the capacitor, the capacitor will start to discharge clockwise. Let me trace the closed loop clockwise and sum all of the voltage differences. Then I get (q/c) - L(dI/dt) = 0 for the Kirchhoff's second law, because I trace the capacitor from its cathode to anode and the inductor will opposite the current. But common textbooks says that (q/c) + L(dI/dt) = 0 with no thorough explanation. What am I missing for this issue?

    Attached Files:

    • LC.jpg
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    Last edited: Oct 15, 2014
  2. jcsd
  3. Oct 15, 2014 #2


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    Welcome to PF!

    Both equations are right if you have the correct relationship between q and I. What is it in your case?

  4. Oct 16, 2014 #3
    Thank you ehild.
    What do you mean by the correct relationship between q and I?
  5. Oct 16, 2014 #4


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    Q (the charge on the capacitor) and the current I are related. Taking this relation into account you get a differential equation for Q that you can solve and you get the time dependence of the charge and from that, the time dependence of the current.
    How are Q and I related?
  6. Apr 12, 2015 #5

    I have been agonizing over the exact same problem. I spent several hours today thinking about this and trying to figure out the stupid sign. I slightly edited the image posted by OP to add the capacitor charges and to specify my starting point. I am considering a situation where at time t = 0 the capacitor has +q on the top plate and -q on the bottom. Current will thus flow in the direction of the arrow, initially. If I start at point a in the attached image and go around clockwise, then, just as the OP wrote, I get:
    q/C - L(di/dt) = 0

    The inductor will experience a voltage drop when following the direction of current; thus I get the negative sign in front of the L term since it's a "drop". Qualitatively, the inductor will initially oppose the increasing current, so it creates a potential opposite in direction to the potential across the capacitor, similarly to how it would be if this were an RC circuit.

    To answer ehild's question, i = dq/dt. But I still don't see how this allows me to get the correct differential equation with the correct signs.

    I read what Young & Freedman wrote in their book about this, but it made no sense (to me, at least). I would really really appreciate help here. This has been driving me absolutely crazy.

    Attached Files:

  7. Apr 13, 2015 #6
    Hmmmm when you close the switch does q increase or decrease?
  8. Apr 13, 2015 #7
    After the switch is closed, dq/dt < 0 for sure. I am treating q(t) as the charge on the positive plate of the capacitor. So, as the switch is closed, charge begins draining away from the top plate as the capacitor discharges -- i.e. q is decreasing.

    I believe what Young & Freedman wrote in their book has to do with this, but I don't see why it matters. How does the fact that dq/dt is initially negative have to do with d2q/dt2? That's the term that really matters.

    In the analogous mechanical system, 1D oscillator on a spring, if I stretch the mass to the right of the origin (positive x), then the initial dx/dt is also < 0 because the mass moves to the left. But that doesn't change change the sign you put in front of the m*dx2/dt2 term in the differential equation. If I instead were to stretch the spring to the left of the origin, then dx/dt would be > 0, but I would still have the same differential equation. Put another way, dx2/dt2 doesn't care whether dx/dt < 0 or > 0.
  9. Apr 16, 2015 #8
    Bump. No one knows how to resolve this?
  10. Apr 17, 2015 #9
    dq/dt is the rate of increase of the charge on the upper plate and your diagram defines "i" as charge leaving it.
    So the equation relating the two is . . . . .
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