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RCL Circuit Analysis

  1. Mar 9, 2014 #1
    1. The problem statement, all variables and given/known data

    i(t) = 0 for t<0 and 5*cos(50t) for t>=0
    Vc(0) = 0

    Circuit is current source, 5 ohm resister, 2 Henry inductor, and 3 Farad capacitor in series.

    Need to find Vr(t), Vl(t), and Vc(t), voltage across resistor, inductor, and capacitor.

    2. Relevant equations

    V= IR for resistor
    V= L(di/dt) for inductor
    V = 1/c(di/dt) for cap

    3. The attempt at a solution
    di/dt = -250sin(50t)

    Resistor:

    Vr(t) = 25*cos(50t)

    Inductor:

    Vl(t) = 2*-250sin(50t)
    =-500sin(50t)

    Cap:
    Vc(t) = 1/3*-250sin(50t)
    = (-250/3)sin(50t)

    I ran the simulation on multisim and the numbers do not agree.

    The voltage across the cap shows 66.3mV p-p
    scope.png
    ch1 = node between source and resistor
    ch2 = node between resistor and inductor
    ch3 = node between inductor and cap
    ch4 = current probe 1mV/mA
    traces are from top to bottom
     
  2. jcsd
  3. Mar 9, 2014 #2

    gneill

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    Staff: Mentor

    Something tells me that the simulator is not going to fair very well handling the circuit as described.

    At t=0 the ideal current source is going to want to force the current to be 5A immediately. That's the result of it being a cosine function that kicks in at time t=0. But inductors don't like to change current immediately like that. The simulator might try to generate GV or TV levels of voltages across the inductor for the first instant, causing similarly ridiculous current spikes for the capacitor in the GA or TA range.
     
  4. Mar 9, 2014 #3
    So my results for Vr and Vl seem reasonable but Vc seems like it should be 0.03sin(50t) so i am not sure i am using the right formula.
     
  5. Mar 9, 2014 #4

    gneill

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    Staff: Mentor

    Those look reasonable for the steady-state values for those items, but misses the initial transient response for the reactive components that take place because of the instantaneous forced step in current at t=0.

    If you've been introduced to Laplace transforms, that might be an easier approach to obtaining the response for the inductor voltage.
     
  6. Mar 9, 2014 #5
    Laplace is not for a few more chapters...
     
  7. Mar 9, 2014 #6

    gneill

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    Staff: Mentor

    Aurgh. Then I guess you'll have to make do with a careful investigation of the differential equation for the inductor voltage by more traditional methods. The driving function for the current is not just 5cos(50t), but rather 5cos(50t)U(t), where U(t) is the unit step...
     
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