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Another RCL Circuit

  1. Apr 27, 2014 #1

    dwn

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    *****Please see 3rd post and ignore calculations on this one******

    1. The problem statement, all variables and given/known data

    Please see image.

    2. Relevant equations

    alpha = 1/2RC
    omega = 1/√(LC)
    iL(t) = e-αt(A1[\SUB]coswdt + A2sinωdt)

    3. The attempt at a solution

    iL(0) = 5 ∴A1 = 5

    Code (Text):
    di/dt = e[SUP]-1.8541t[/SUP](-11sin(2.236t) + 2.236A[SUB]2[/SUB]cos(2.236t)) + 1.8541e[SUP]-1.8541t[/SUP](5cos(2.236t) + A[SUB]2[/SUB]sin(2.236t))
    Evaluating di/dt at t = 0 :
    Code (Text):
    2.236A[SUB]2[/SUB] + (1.8541 * 5) = 0
    A[SUB]2[/SUB] = -4.146
    Code (Text):
    i[SUB]L[/SUB](t) = e[SUP]-1.8541t[/SUP](5cos(2.236t) - 4.146sin(2.236t))  
    I've messed up somewhere...not sure where though. Please help.
     

    Attached Files:

    Last edited: Apr 27, 2014
  2. jcsd
  3. Apr 27, 2014 #2

    dwn

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    I see one mistake right off the bat....α is incorrect, I used ωd. Also, I used ω0 instead of ωd. Aside from that now...anymore errors?

    IDK what I was thinking!
     
  4. Apr 27, 2014 #3

    dwn

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    Okay....my second attempt, rushing the first time around :

    α, ω0, ωd are unchanged.
    A1 = i(0) = 5

    Code (Text):
    i(t) = e[SUP]-1.25t[/SUP](A[SUB]1[/SUB]cos(1.8541t) + A[SUB]2[/SUB]sin(1.8541t))
    Code (Text):
    di/dt = -1.25e[SUP]-1.25t[/SUP](A[SUB]1[/SUB]cos(1.8541t) + A[SUB]2[/SUB]sin(1.8541t)) - 1.8541e[SUP]-1.25t[/SUP](-5sin(1.8541t) + A[SUB]2[/SUB])cos(1.8541t))
    Evaluate di/dt at t = 0:

    Code (Text):
     di/dt = (-1.25*5) - 1.8541A[SUB]2[/SUB]
    A[SUB]2[/SUB] = -3.371  
    :bugeye:
     
  5. Apr 27, 2014 #4

    gneill

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    Staff: Mentor

    How do you figure that iL(0) = 5 A?

    The ##\alpha## for a series circuit is not the same as for a parallel circuit.

    The Wikipedia page for the RLC circuit gives a pretty good overview of the series RLC circuit formulas.
     
  6. Apr 27, 2014 #5

    dwn

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    I just can't hack it with these online courses. I can't begin to explain the benefit of discussing this stuff with a teach or student..:cry:

    I digress...well, recognizing that there is not switch and therefore no sudden changes occurring in the circuit (in other words a t = 0-)...it is safe to say that at i(0) there is no current passing through the circuit which means i(0) = 0 → A1 = 0.

    What exactly do A1 and A2 represent, individually. Is A1 attributed to initial conditions t≥0 and A1 t≤0?
     
    Last edited: Apr 27, 2014
  7. Apr 27, 2014 #6

    gneill

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    Staff: Mentor

    Yup. The initial inductor current is zero. Just a few more things though. Note that the current source is specified as 5u(-t) A. Now u(t) is the unit step function, so there is a "hidden" switching operation that takes place at t = 0. How does the current source behave then? You might also want to consider the voltage on the capacitor at time t = 0. What's the equivalent circuit at t = 0+? Which way will the capacitor voltage want to drive the inductor current?
     
  8. Apr 27, 2014 #7

    dwn

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    I would just like to say that these questions are really helpful in getting me to ask the right questions when I analyze the circuits.

    Okay, I hope I am not missing your point and just stating the (somewhat) obvious, but once the "switch" closes the current will be directed to the capacitor until it is maxed out at which point the current will pass through the resistor and slowly charge the inductor since they like to resist change. Is this the correct flow of current? The current source will drive the capacitor voltage to infinite as t → ∞.

    V = 1/C ∫(i * dt) + v(0)
    Vc(0) = 0 V

    The equivalent circuit would be a series RLC circuit not including the source.

    Is this good brain function on my part? haha
     
  9. Apr 27, 2014 #8

    dwn

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    If this is the case, which I hope it is, then when it comes to solving diL/dt at t=0, I will not get an answer for A2.

    Code (Text):
    di/dt|t = A[SUB]2[/SUB] = V[SUB]c[/SUB](0)/L = V[SUB]c[/SUB](0) - Ri(0)/L = 0
     
  10. Apr 27, 2014 #9

    gneill

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    Staff: Mentor

    No, I suspect that you're missing the import of the unit step function controlling the current source: 5u(- t) A (note the negative sign on the t!). What will be the value of the current source at all time before t = 0? And after t = 0?
     
  11. Apr 27, 2014 #10

    dwn

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    t < 0 current will be positive 5
    t > 0 current will be negative -5
     
    Last edited: Apr 27, 2014
  12. Apr 27, 2014 #11

    gneill

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    Staff: Mentor

    Nope. The sign of the argument of the function doesn't change the sign of the function itself. The value of the unit step function can only be 0 or +1. So what might be the function of the negative sign on the argument t?
     
  13. Apr 27, 2014 #12

    dwn

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    The current source is turned off at t = 0. The step function is only applicable during -t...?

    t < 0 equates to 1

    t > 0 equates to 0

    guess, I missed all this from previous chapter.
     
    Last edited: Apr 27, 2014
  14. Apr 27, 2014 #13

    gneill

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    Staff: Mentor

    Right. For all time t < 0 the current source is producing +5 A. At time t = 0 it "switches" or "steps" to zero amps. And, something that passes zero current essentially disappears from the circuit...

    So you should determine what the voltage will be on the capacitor at time t = 0, since that will be the source of energy for the circuit after the "switch".

    A useful thing to remember is that if you have a capacitor with an initial voltage on it, an equivalent circuit model for that capacitor is an uncharged capacitor in series with a voltage source of the same voltage. This can sometimes help to clarify your overall equivalent circuit and make writing the equations for the circuit easier.
     
  15. Apr 27, 2014 #14

    dwn

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    I think have got it now, at least the step functions. That last bit is good to know, I will try to remember that and see if it helps make things a little clearer. I think I need to consider a different major.

    Given everything, I really hope this is correct:

    Code (Text):
     
    i[SUB]L[/SUB](t) = e[SUP]-t[/SUP](A[SUB]2[/SUB]sin(2t))
    di[SUB]L[/SUB](t)/dt = -e[SUP]-t[/SUP](A[SUB]2[/SUB]sin(2t)) + e[SUP]-t[/SUP](2A[SUB]2[/SUB]cos(2t))

    if V[SUB]c[/SUB] = 1/C∫i*dt  evaluated at t=0  → 5A/0.2F = 25 V
    2A[SUB]2[/SUB] = 25    A[SUB]2[/SUB] = 12.5  
    Final Answer is...?
    iL(t) = e-t(12.5sin(2t))
     
    Last edited: Apr 27, 2014
  16. Apr 27, 2014 #15

    gneill

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    Staff: Mentor

    You don't need an integral to evaluate Vc for t = 0. It will have achieved steady state long before t = 0, so use the usual methods to evaluate it: The current through the capacitor and inductor is zero, the capacitor looks like an open circuit and the inductor a short circuit. All the current from the source is passing though the resistor. What's the potential that must appear across the "open" capacitor.

    If it helps, you might think of replacing the current source and resistor with a Thevenin equivalent (voltage source in series with a resistor).
     
  17. Apr 28, 2014 #16

    dwn

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    Hey gneill, I think I got it figured out...I was up until about 3 a.m. trying to figure all of this stuff out. I realize I need to work on first figuring out how the circuit is functioning before jumping into the calculations--I think thats how I get so lost, in addition to staring at this stuff for 8 hours straight! :)

    Thanks a lot for all your help!
     
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