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I love rlc circuits : Series

  1. May 9, 2014 #1

    dwn

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

    Image

    2. Relevant equations

    Critical damping : e-αt(At + B)
    R = 1.5

    3. The attempt at a solution
    I'd like to post this before anything else to make sure that I've analyzed the circuit correctly.
    Code (Text):
    t < 0:    v[SUB]c[/SUB](0) = 9V  ,  i[SUB]L[/SUB](0) = 9/4 = 2.25A

    i = e[SUP]-t/3[/SUP](At + B)
    i(0) = B = 2.25A
    di/dt = -1/3e[SUP]-t/3[/SUP](At + 2.25) + Ae[SUP]-t/3[/SUP]
    di(0)/dt = -1/3*2.25+ A = V[SUB]c[/SUB](0) = 9V
     
     

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    Last edited: May 9, 2014
  2. jcsd
  3. May 9, 2014 #2

    rude man

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    Your v_c(0+) is wrong. What is v_c(0-) and what does the capacitor have to say about changing to v_c(0+)?
     
  4. May 9, 2014 #3

    dwn

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    v_c(0-) = 0 V Is the reason for this bc it is not supplying any charge to the circuit? I'm having a difficult time distinguishing when I should have a v_c for 0- and when there is not...I see it as a sort of battery that stores charge in it until there is not longer a source present.
     
  5. May 9, 2014 #4

    rude man

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    Why do you think the charge on C is 9V before the switch is opened?
     
  6. May 9, 2014 #5

    dwn

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    I know that during the steady state the cap is an open circuit, but after reading in a book that cap's are similar to batteries along came the idea of it holding a charge (9V in this case). Essentially the battery will hold the charge that it contains if it is removed from a circuit, until it has somewhere to dissipate the charge (aside from internal resistance of course). Then once the switch is flipped, the voltage in the capacitor decreases by V(0)e^(-t/RC) or using the characteristic equations in RLC circuits.
     
    Last edited: May 9, 2014
  7. May 9, 2014 #6

    rude man

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    You should focus on the inductor. What is the voltage across an inductor when current thru it is constant?
     
  8. May 9, 2014 #7

    dwn

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    0 V --- current is not changing so di/dt = 0. (VL = L di/dt).

    I don't see how that helps me with understanding whether v(c) is 0 or some other constant.
     
  9. May 9, 2014 #8

    rude man

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    Please go back to my post #2 and think about it.
     
  10. May 9, 2014 #9

    dwn

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    So its safe to say that v_c(0-) with any current/voltage source will never "hold" a charge of that source? In which case, why are there instances when v_c(0-) has a value. Just to be clear.
     
  11. May 9, 2014 #10

    rude man

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    Your statement mystifies me.

    Again: what is the voltage across the inductor at t = 0-? What is the voltage across the capacitor at t = 0-? What must therefore be the voltage across the capacitor at t = 0+?

    Hint: you can have all sorts of voltages across a capacitor at t = 0-, but not if there's an inductor across it!
     
  12. May 9, 2014 #11

    dwn

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    It's t(0-) = t(0+) because the cap and inductor cannot change at suddenly at t=0.

    Why does my statement mystify you?
     
  13. May 9, 2014 #12

    rude man

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    Dunno, it just does.
     
  14. May 9, 2014 #13

    dwn

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    well, thats really not helpful.
     
  15. May 9, 2014 #14

    rude man

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    The hint of post #10 was not helpful?

    A capacitor holds any charge sitting on it if there is no load across it. I mean an L or an R or both. In your case what sits across the capacitor?
     
  16. May 9, 2014 #15

    dwn

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    Thank you. That made sense to me. As long as the current has another path to take, the capacitor doesn't hold a charge (hence the open circuit). I had a difficult time understanding how the capacitor operated between t(0-) and t(0+).
     
  17. May 9, 2014 #16

    rude man

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    The basic idea is that the voltage across a capacitor cannot change instantaneously. So if that voltage was zero at t=0- it must also be zero at t=0+.
     
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