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RLC Circuit - 2nd order

  1. May 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the voltage across the resistor.

    2. Relevant equations

    V = L*di/dt
    I = C*dv/dt

    3. The attempt at a solution

    I'm not too worried about the differential equation part but I need some help setting up the circuit for me to start the process.

    Since the current through the capacitor/inductor block will equal the resistor's current can I put the resistor *before* the capacitor/inductor block which I can then setup a source transformation with the voltage source and resistor?

    I'm feeling that I can't do this because the capacitor/inductor will cause some oscillations to go through the resistor but wanted to check.

    Thanks.
     

    Attached Files:

  2. jcsd
  3. May 3, 2012 #2

    gneill

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    Staff: Mentor

    Is that a DC source or an AC source?
     
  4. May 3, 2012 #3
    DC and I know what the answer to my question is now, I can't move that resistor. I'm really having trouble setting this one up.. I know VC = VL and the current iS= iC + iL = iR. I keep getting things in terms of VC for the differential equation.
     
  5. May 3, 2012 #4

    gneill

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    Staff: Mentor

    You CAN move the resistor to the top rail if you wish, just take note of the direction of current flow to determine the direction of potential drop across it.

    Perhaps you can show one of your attempts to set up the differential equation?
    You don't happen to be familiar with Laplace Transforms I suppose...?
     
  6. May 3, 2012 #5
    Unfortunately, we haven't learned Laplace yet. Here was one attempt:

    Starting with VS = VC + VR and i = iC + iL = iR

    Which after a little work leads to:

    VS = VC + RC*d2VC/dt2 + R/L*VC.

    Now I know this is in terms of VC so I plugged it back into VS = VC + VR which then gave:

    VR = LC*d2VC/dt2 + L/R*dVC/dt - VS

    Does that seem right? Thanks for helping me by the way, I really appreciated it.
     
  7. May 4, 2012 #6

    gneill

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    Staff: Mentor

    It looks like you have a correct equation for VR.

    My own choice might have been to just solve for Vc(t) and then afterwards subtract Vs which is a constant. Then the differential equation to be solved will be in the more compact form:
    $$ \frac{d^2 V_C}{dt^2} + \frac{1}{R C}\frac{dV_C}{dt} + \frac{1}{L C} V_C = 0$$
     
  8. May 4, 2012 #7
    I finally solved it correctly late last night. Here's what I did:

    Starting with VS = VC + VR and i = iC + iL = iR

    Which after a little work leads to:

    VS = 1/C[itex]\int[/itex]iC + i*R

    VS = 1/C[itex]\int[/itex]iC + (iC+ iL)R

    [itex] V_S = \frac{RCd^2V_C}{dt^2} + \frac{dV_C}{dt}+ \frac{RV_C}{L}[/itex]

    Then I used the characteristic equation which showed that it's an underdamped solution. After that I subtracted the solution from VS to get the voltage over the resistor.

    Again, thanks for your help.
     
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