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Circuits: impedance = undefined? (j8ohms-j8ohms)

  1. Jan 17, 2007 #1
    1. The problem statement, all variables and given/known data
    I'm working a problem where I have a j8ohm and a -j8ohm in parallel, and I need to reduce the circuit to get the current in one of these branches. Being these numbers, I get a fraction with zero on the bottom. How should I treat this?

    3. The attempt at a solution

    I'm thinking it shorts it, but that's a guess
     
  2. jcsd
  3. Jan 17, 2007 #2
    well I change my answer on the basis that I think the impedance is infinity...

    treat as an open?
     
  4. Jan 17, 2007 #3
    Okay, i'm convinced it's the open circuit, but I'm still curious: if you actually did this in a lab, and set the frequency to make these impedances, would it actually not let current get through? or is that more of the "ideal" vs. "real"
     
  5. Jan 18, 2007 #4

    berkeman

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    Staff: Mentor

    The +j and -j matching impedances represent a parallel LC circuit, correct? What is the impedance of that kind of circuit at resonance (where the +j and -j reactances match)? And what is the thing in a real circuit that keeps the input Z of a parallel LC circuit from being infinite?
     
  6. Jan 18, 2007 #5
    Well...it's either the fact that you can't reach infinite impedance, so it will appear as very large, but finite.

    Or is the fact that practical capacitors will have a small amount of leakage current
     
  7. Jan 18, 2007 #6

    berkeman

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    Staff: Mentor

    I suppose the leakage current might be a factor, but it's a different property of both the inductor and capacitor that keeps it from being infinite. The thing that makes the Zin of a parallel LC look infinite is that you can get a current going back and forth, with the energy going back and forth between the charge stored on the cap and the magnetic field stored by the inductor, all with very little energy being input from outside.

    Once you get the oscillation going, you can put almost no energy in to keep it going. But that's when you have an ideal cap and ideal inductor. Now what happens if you add in the DCR of the inductor and ESR of the cap? What would happen to the LC oscillation if you didn't put in any energy?
     
  8. Jan 18, 2007 #7
    hmm...

    I do believe energy associated with a resistor is turned into heat/friction, so since the practical models of C and L has resistance, and it's oscillating back and forth...

    W = IVt, So to keep the current and voltage oscillating, you have to add power (w) because it's being converted (lost as far as circuits go)

    am I in the ballpark here?
     
  9. Jan 18, 2007 #8

    berkeman

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    Staff: Mentor

    Yep, exactly. Now to finish figuring out your OP question about the parallel combination of the reactive phasors and what happens. Go ahead and draw the parallel LC circuit now with a resistor Rdcr ("DC Resistance") in series with the L, and a resistor Resr ("Equivalent Series Resistance) in series with the capacitor. Solve for the input impedance now, with those real-world resistances accounted for.

    And then put in some real-world numbers, either by hand, or run them as SPICE simulations, to see what the input impedance is at resonance. Do you have any guesses about how the input Z at resonance is affected by Resr and Rdcr?

    BTW, you can use these values for practical starting values:

    C = 10uF, Resr = 1 Ohm (and step it up by 0.5 Ohms at a step)
    L = 1mH, Rdcr = 1 Ohm (and step it up by 0.5 Ohms at a step)
     
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