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RLC Circuit

  1. Oct 9, 2006 #1
    I have a circuit with a capacitor, resistor, and inductor all in parallel with each other. I have to write the differential equation governing the voltage.
    However, when I attempt to do that [using Kirchoff's Voltage Rule] I will end up with:
    Vc + Vl = 0
    and Vl + Vr = 0
    However, this implies that Vc = Vr.
    If I then take KVR for the entire loop, I will end up with:
    Vc + Vr = 0
    And this implies that Vc = -Vr.
    How can 2 opposite statements be true for one loop?
    Is there a different way to use Kirchoff's Laws for this particular question which I am unaware of?

    Thank you.
     

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    Last edited: Oct 9, 2006
  2. jcsd
  3. Oct 9, 2006 #2

    berkeman

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    Staff: Mentor

    I would use KCL for this one, and then differentiate the resulting equation once to get a strictly differential equation. Then assume a solution, plug it in and solve for the remaining variables. Have you tried it that way?
     
  4. Oct 9, 2006 #3

    SGT

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    Your equations using KVL are wrong! If the three devices are in parallel you should have Vc = Vr = Vl.
    As berkeman said, you should use KCL to set the equations, then replace the currents in the capacitor and in the resistor as functions of their respective voltages.
    But since KVL says that all voltages are equal, you replace Vc and Vr by Vl.
    Finally, you write Vl as a function of Il to get a second order differential equation.
     
    Last edited by a moderator: Oct 9, 2006
  5. Oct 9, 2006 #4

    Andrew Mason

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    As Berkeman suggests, you should use Kirchhoff's current law here. Kirchhoff's voltage law doesn't really apply because the inductor produces an emf. The voltage across the resistor is the sum of the voltages produced by the capacitor and the inductor:

    [tex]V_R = V_C + V_L[/tex]

    AM
     
  6. Oct 9, 2006 #5

    SGT

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    No, if the three elements are in parallel [tex]V_R = V_C = V_L[/tex]
     
  7. Oct 9, 2006 #6
    Does that apply even when there is no voltage source?
     
  8. Oct 9, 2006 #7

    SGT

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    Yes, KVL and KCL apply with or without voltage or current sources.
     
  9. Oct 9, 2006 #8
    Thank you for replying again. Please stay with me a little longer.

    I have made the equation [tex]I_L = I_R + I_C[/tex]
    I then substitute in [tex]I_L = \frac{V_L}{X_L} [/tex], [tex]I_R = \frac{V_R}{R} [/tex], and [tex]I_C = \frac{V_C}{X_C} [/tex].
    However, since the equation [tex]I_R + I_C - I_L[/tex] is equal to 0, I do not know how to simplify any further, in order to get a differential equation for voltage.
     
    Last edited: Oct 9, 2006
  10. Oct 9, 2006 #9

    SGT

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    In reality you have
    [tex]I_C + I_R + I_L = 0[/tex]
    But remember that
    [tex]I_C = C\frac{dV_C}{dt} = C\frac{dV_L}{dt}[/tex]
    [tex]I_R = \frac{V_R}{R} = \frac{V_L}{R}[/tex]
    [tex]V_L = L\frac{dI_L}{dt}[/tex]
    Replace all this in the equation for currents and you have your differential equation.
     
  11. Oct 11, 2006 #10

    Andrew Mason

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    Quite right, of course. I was speaking in terms of
    voltage but thinking in terms of current:

    [itex]I_R = -(I_C + I_L)[/itex]

    where IL and I_C can be plus or minus. Assume that the circuit has no applied voltage and a charged capacitor.

    The emf from the inductor opposes the current flowing from the capacitor through the inductor while the capacitor current is increasing and it adds to the current from the capacitor while the capacitor current is decreasing.

    The differential equation is a little tricky to solve. As SGT has pointed out, it has to be solved in terms of IL.

    [tex]I_C = \frac{d}{dt}Q_C = \frac{d}{dt}CV_C = C\frac{d}{dt}V_L = C\frac{d}{dt}\left(L\frac{dI_L}{dt}\right) [/tex]

    [tex]I_R = V_L/R = \frac{L}{R}\frac{dI_L}{dt}[/tex]

    [tex]I_L = I_L[/tex]

    So you get:

    [tex]\frac{L}{R}\frac{dI_L}{dt} = - (CL\frac{d^2I_L}{dt} + I_L)[/tex]

    AM
     
    Last edited: Oct 11, 2006
  12. Oct 11, 2006 #11

    SGT

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    Yes, these are the equations I proposed to Lisa.
     
  13. Oct 11, 2006 #12
    The next problem in the question is that the capacitor is now charged, and I am closing a switch placed after the capacitor at t=0 when Vc = Vo and i_L = 0.

    I was told to not change the equation for voltage (that I found), but to change the initial conditions. However, no where in my previous solution did I take into account initial conditions.
    Therefore, I was wondering if I should change the capacitor's voltage. It was originally [tex]- (CL\frac{d^2I_L}{dt}[/tex]. Should I change it to some function of (Vo)(e^iwt) or so, to represent the discharging (or in other words: forced, damped motion)?

    If not, what do I do?
    Please answer as soon as possible. Thank you.
     
  14. Oct 11, 2006 #13

    berkeman

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    What does closing the switch do again? Is it shorting out the resistor or something? If you short out a capacitor when it is fully charged, a very large current flows through the shorting element (limited only by the low resistance of the shorting element). I have a feeling that is not what is being asked....
     
  15. Oct 11, 2006 #14

    SGT

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    If you didn't use initial conditions your solution is incomplete.
    The general solution of your DE involves 2 integration constants [tex]k_1[/tex] and [tex]k_2[/tex].
    [tex]i_L(t) = k_1e^{s_1t} + k_2e^{s_2t}[/tex]
    where [tex]s_1[/tex] and [tex]s_2[/tex] are the roots of the characteristic polynomial and can be real or complex.
    In order to eliminate the two constants you must use the initial conditions of your circuit.
    As berkeman asked: what does the closing switch do?
     
  16. Oct 11, 2006 #15
    I would assume that closing the switch discharges the capacitor.
     
  17. Oct 11, 2006 #16

    SGT

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    If your circuit has no power source, the capacitor will discharge through the resistor anyway. There is no need of a switch.
     
  18. Oct 11, 2006 #17
    I think the professor placed the switch there just so that we understand that right at t=0 the capacitor starts discharging.
    Should I assume weak damping for the discharging of the capacitor?
    And then solve the same differential equation as the first half of the problem, except with different initial conditions?

    I also have to find the phase and amplitude of the function.
     
    Last edited: Oct 11, 2006
  19. Oct 12, 2006 #18

    SGT

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    What is the meaning of weak damping?
    Until now I don't know what are the differences between the first and the second half of the problem. Can you state them?
    If your circuit is underdamped, the general solution is:
    [tex]i_L(t) = Ae^{\alpha t}cos(\omega t + \phi)[/tex]
    The amplitude A and the phase [tex]\phi[/tex] are determined from the initial conditions.
     
  20. Oct 20, 2006 #19
    Thanks for your help SGT and Andrew Mason.
     
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