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Driven RLC circiut in parallel

  1. Sep 17, 2006 #1
    I need some help setting up the DE for this system. I have a voltage source pumping at [tex]V_0cos(wt)[/tex]

    Connected to a capacitor, inductor and resistor all three in parrellel. I am guessing then, that the voltage drop among all the parts are equal.

    [tex]
    V_0 cos(wt) = \frac{Q}{C} = L \ddot{Q} = \dot{Q} R
    [/tex]

    Then, The total charge on all parts would be equal to the sum of charges, right?

    [tex]
    Q_{Total} = \\Q_C + Q_L +Q_R
    [/tex]

    But what is the Q total? I have these two equations, but I am having a hard time understanding how to combine them into one single DE that I can solve.
     
    Last edited: Sep 17, 2006
  2. jcsd
  3. Sep 17, 2006 #2

    Andrew Mason

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    Homework Helper

    If you just give it a one time zap and remove the voltage, you will have:

    [tex]L\frac{dI}{dt} + \frac{Q}{C} = IR[/tex] or

    [tex]L\frac{dI}{dt} + \frac{Q}{C} - IR = 0[/tex]

    This is because the current through the resistor has to come from the emf produced by the capacitor and the inductor.

    Now take the same circuit and apply a forcing voltage.

    [tex]L\frac{dI_L}{dt} + \frac{Q_c}{C} + V_0\cos\omega t = IR[/tex] or

    [tex]IR - (L\frac{dI_L}{dt} + \frac{Q_c}{C}) = V_0\cos\omega t[/tex]

    Of course: [tex]I_L = \dot{Q} \text{ and } \frac{dI_L}{dt} = \ddot{Q_L}[/tex]

    and [itex]I_L + I_C = I[/itex]

    AM
     
    Last edited: Sep 17, 2006
  4. Sep 17, 2006 #3
    okay, I appriciate the help. It seems that this is the same as setting up the circuit in series. (V1+V2+V3=V, Q1=Q2=Q3=Q) Is there no difference when setting up the circuit in parallel? I was thinking V1=V2=V3=V, and Q1+Q2+Q3=Q.
     
  5. Sep 18, 2006 #4

    Andrew Mason

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    The difference is that the currents are not the same. [itex]I_L, I_C \text{ and } I_R[/itex] will be different. They are necessarily the same when in series.

    AM
     
  6. Sep 20, 2006 #5

    SGT

    User Avatar

    Since all elements are in parallel they have the same voltage drop.
    You have three currents:
    [tex]i_R(t) = \frac{V_s(t)}{R}[/tex]
    [tex]i_C(t) = C\frac{dV_s(t)}{dt}[/tex]
    [tex]i_L(t) = i_L(0) + \frac{1}{L}\int_{0}^{t}V_s(\tau)d\tau[/tex]
    Where [tex]V_s(t) = V_0 cos(\omega t)[/tex]
     
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