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Transform circuit

  1. Oct 14, 2011 #1
    The circuit below operates with the switch K closed until steady state is reached. At t = 0 the switch K is opened. Determine the voltage vk(t) across the switch after it is opened. (> 0)
    i3vts2.png


    What I attempted to do first was redraw the circuit when the switch is opened:
    r6ztar.png

    I am unsure of what the next steps should be. But this is my attempt:
    V0 + Vk + LI1 = R1(I1 - I2) + Ls(I1 - I2)
    LI1 + V1/s = R1(I2 - I1) + Ls(I2 - I1) + R2I2 + I2/Cs

    I'm not sure if these equations are correct and do not want to go on if they aren't since everything would be wrong. Thanks in advance.
     
  2. jcsd
  3. Oct 14, 2011 #2

    NascentOxygen

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    Staff: Mentor

    Perhaps it is intended that K impose the initial conditions on the system, then at t=0 K disconnects the battery from the circuit and leaves the circuit to oscillate until its energy is dissipated? The currents in the two branches therefore become equal; the current now being a loop current.
     
  4. Oct 14, 2011 #3
    so i(0) = V0/R1?
     
  5. Oct 14, 2011 #4

    gneill

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    Yes, that would be the initial current in the "new" loop -- the inductor will "insist" that it be so!

    But the problem is going to be that this is a serial RLC circuit. As such it can have three distinct solution forms depending upon the particular values of the components; The circuit can be underdamped, overdamped, or critically damped.

    A straight inverse Laplace transform of the circuit's Laplace domain equation for the voltage at the top node (top horizontal rail), containing all the individual circuit component values, will be a real "dog's breakfast"! Loads of sine and cosine terms, exponentials, and so on.

    It might be worthwhile trying to batter it into a more familiar form by converting the terms that represent the natural frequency ωo, the damping ratio [itex]\alpha[/itex] and any time constants [itex]\tau[/itex] into those symbols.
     
  6. Oct 14, 2011 #5
    unfortunately we have to do Laplace transform and then convert back to the time domain. i just realized that shouldn't it simplify to the circuit below with the switch open?
    vwxzx4.png

    Then I can do I(s) = V(s)/Z(s) where V(s) is the sum of the transform voltages and Z(s) is the sum of the transform impedances so...
    I(s) = [LI1 - V1/s]/[Ls + R1 + R2 + 1/Cs]

    err I suppose I should do V(s) = I(s)/Y(s)
     
  7. Oct 14, 2011 #6

    gneill

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    Staff: Mentor

    Sure, you can do that (Be careful about the polarity of the initial voltage source associated with the initial inductor current. Its polarity should be in the same direction as the initial current).

    Once you have the current in the time domain, presumably you can use it to work out an expression for the voltage at the top. It may involve some calculus.
     
  8. Oct 14, 2011 #7
    is my polarity incorrect? I know usually it's the other way around, but in this case I1 is defined as going up so shouldn't the voltage source be - to + ?
     
  9. Oct 14, 2011 #8

    gneill

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    Disregard the labelled current and determine the actual direction of the initial current flow. The true polarity of the model source will be determined by that.

    Otherwise, calculate the value of your I1 according to the label and you'll find it to be negative. Again, set your model voltage source accordingly. :smile:
     
  10. Oct 16, 2011 #9
    So after combining impedances I have
    2z86iz9.png

    Then I did a source transformation
    smzwp3.png

    is what i have done so far correct?? can i now apply nodal analysis to get Vk?
     
  11. Oct 16, 2011 #10

    gneill

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    Staff: Mentor

    Looks okay. Sure, you can do nodal analysis, or you could continue to simplify the circuit; the current sources and impedances are in parallel...
     
  12. Oct 16, 2011 #11
    awesome. thanks!
     
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