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2-position switch circuit

  1. Oct 16, 2011 #1
    In the circuit below, the switch reaches a steady state while in position 1 then at t = 0 gets thrown to position 2. I am supposed to find the voltage (s-domain) across the capacitor under these conditions using thev. theorem
    63soew.png

    I just want to make sure I am approaching the problem with the correct methods:
    1. Find initial conditions using the properties that an inductor is shorted in steady state and a capacitor is open circuit
    I got i(0) = 5 A
    v(0) = 5 V
    2. Transform the circuit to the s-domain, adding an independent voltage source with voltage = 10 (from inductor). Remove the capacitor to make it an open circuit. Find VTH via voltage divider equation.
    3. Find Zeq by shutting off all independent sources.
    Zeq = 2 + (1 || (1 + 2s))
    4. Recreate the circuit with VTH and Zeq in a loop with the 1/2 F capacitor and 5/s V independent voltage source from the initial conditions (steady state).
    5. KCL to get current
    6. voltage across the cap = current * capacitor value in s domain

    mainly i want to know if my initial conditions are correct and if i'm using them correctly.
     
    Last edited: Oct 16, 2011
  2. jcsd
  3. Oct 17, 2011 #2

    gneill

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    Staff: Mentor

    Your initial conditions look fine.
     
  4. Oct 17, 2011 #3
    when doing the last step, "6. voltage across the cap = current * capacitor value in s domain", do i have to subtract the initial condition so it would be voltage = (capacitor value in s domain)*(current - initial voltage)? i thought i had already taken that into account in step 5 when i combine the voltage source and initial voltage, but i am unsure if i am using the correct formula or not
     
  5. Oct 17, 2011 #4
    I am getting different answers when i do it in thevenin vs. norton
    I'm not quite sure why. My formula for I in thevenin is I = (Vth - 5/s)/(Zeq + 2/s). Then I do the last step to get V.
    Vth = ((2s+1)(10s+11))/(2(s+1)^2)
    Zeq = (6s+5)/(2(s+1))
    ** my Zeq does not include the capacitor and Vth does not include the initial condition voltage for the capacitor

    Below, I used V = 1/Cs[I - v(0)]... i am not sure if i was supposed to use the initial condition or not

    Thevenin:

    V = (20 s^3+22 s^2-9 s-10)/(2 s (s+1)^2)
    Z = (6 s^2+9 s+4)/(2 s (s+1))

    I = V/Z = (20 s^3+22 s^2-9 s-10)/((s+1) (6 s^2+9 s+4))

    V = IZc = (2 (20 s^3+22 s^2-9 s-10))/(s (s+1) (6 s^2+9 s+4))

    V = (10 s^3-31 s^2-83 s-40)/(s (s+1) (6 s^2+9 s+4))
    ---
    Norton:

    I = (70 s^2+119 s+47)/(12 s^2+22 s+10)
    Z = (10 + 12 s)/(4 + 9 s + 6 s^2)

    V = IZ = ((12 s+10) (70 s^2+119 s+47))/((6 s^2+9 s+4) (12 s^2+22 s+10))

    V = (40 s^3+44 s^2-18 s-20)/(s (s+1) (6 s^2+9 s+4))

    Since my denominators match, I am thinking I may have used an incorrect sign somewhere?
     
  6. Oct 17, 2011 #5

    gneill

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    No, you should include the initial voltage in the total for the capacitor; the real capacitor was replaced by a model that consists of a freshly-minted, uncharged capacitor and a separate voltage source. That model, in its entirety, represents the real capacitor, and the voltage on the real capacitor will be the sum of the voltage on the "new" capacitor and that voltage source. To that end, you might want to leave the capacitor voltage source out of the Thevenin equivalent.
     
  7. Oct 17, 2011 #6

    so i disregard the initial condition voltage and then add it back in after finding the voltage across the capacitor with Zeq and Vth? i am still getting different results for my thevenin and norton methods; it differs by an extra s in the denominator
     
  8. Oct 17, 2011 #7

    gneill

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    Staff: Mentor

    The voltage source should be part of the circuit that you analyze hen you find the voltage on the capacitor; it will affect the current that flows. You can either keep it separate from the Thevenin equivalent so that the capacitor model remains intact, or you can if you with bundle it into the Thevenin equivalent. In the latter case you'll have to remember to add it to whatever voltage you find on the "new" capacitor when you report the capacitor voltage.
     
  9. Oct 17, 2011 #8
    So if I simplify the circuit like I did below, when I do KVL, I will get the current. Then is it right that V across the capacitor is simply V = I*(2/s)
    jhbvvs.png
     
  10. Oct 17, 2011 #9

    gneill

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    Staff: Mentor

    That would be the voltage across the capacitor in the model of the "real" capacitor, yes. The voltage across the "real" capacitor includes the voltage source that represented its initial charge.
     

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