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Differential equation describing RLC cicuit

  1. Sep 15, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the input/output differential equation for the LRC circuit in the given figure. The figure is shown in the attachment.

    2. Relevant equations

    V(t) = Ri(t)

    For inductor,
    v(t) = Ldi(t)/dt
    I(t) = 1/L∫v(λ)dλ

    For capacitor,
    dv(t)/dt = i(t)/C
    v(t) = 1/C∫i(λ)dλ


    3. The attempt at a solution

    First I use Kirchhoff's voltage law V[itex]_{L}[/itex](t) + V[itex]_{R}[/itex](t) + V[itex]_{R}[/itex](t) = 0. Using a direction of current in the second loop of the circuit, I can see that the branch that has the inductor and capacitor elemens will have opposite signs to the resistor element in the other branch. Because of this, I get the form Ldi[itex]_{L}[/itex](t)/dt + 1/C∫i[itex]_{C}[/itex](λ)dλ = Ri[itex]_{R}[/itex](t). I then take the derivative of the equation to get rid of the integral and get the form Ld[itex]^{2}[/itex]i[itex]_{L}[/itex](t)/dt[itex]^{2}[/itex] + i[itex]_{C}[/itex](t)/C = Rdi[itex]_{R}[/itex](t)/dt.

    This however differs from the correct answer and using Kirchhoff's current law doesn't seem to help much. I'm also not sure why the voltage term for capacitor would depend on the inductor current instead of it's own associated current. I tried looking at my old Electrical Circuits book but none of the examples or problems seemed to have RLC circuits where two elements are on the same branch for me to understand. Could someone please help direct me to the some missing concepts and understanding to this problem?
     

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    Last edited: Sep 15, 2013
  2. jcsd
  3. Sep 16, 2013 #2

    rude man

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    Call the current thru L and C i1 and the current thru R i2.

    Then what is the voltage drop across L and C in terms of i1 and the drop across R in terms of i2?

    And what is the relationship between ig, i1 and i2? So you wind up with one differential equation in i1 only.

    Hint: will be 2nd-order.
     
    Last edited: Sep 16, 2013
  4. Sep 16, 2013 #3

    vela

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    Well, you generally want to get a differential equation for just one quantity, say, ##i_L##. Your equation has ##i_L##, ##i_C##, and ##i_R## in it still. What you did so far was fine. You're just not finished yet.

    The inductor and capacitor are in series, so any current that goes through one has to go through the other. In other words, ##i_L = i_C##.
     
  5. Sep 16, 2013 #4
    Alright, I now understand the relationship between the current in the inductor and the capacitor. For parallel circuits, the total current is related to each branch by i[itex]_{T}[/itex] = i[itex]_{1}[/itex] + i[itex]_{2}[/itex] and since i[itex]_{L}[/itex] = i[itex]_{C}[/itex], i[itex]_{T}[/itex] = i[itex]_{L}[/itex] + i[itex]_{R}[/itex]. I can then substitute for i[itex]_{L}[/itex] in the differential with i[itex]_{T}[/itex] - i[itex]_{R}[/itex]. Is this logic correct?
     
  6. Sep 16, 2013 #5

    vela

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    Yes.
     
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