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Electrical Resonance and Complex Numbers

  1. Sep 19, 2007 #1
    1. The problem statement, all variables and given/known data

    A circuit is said to be in resonance if it's complex impedance Z (in terms of R, L, and C (being the resistance, inductance, and capacitance)) is real. We are to determine the resonance angular frequency [tex]\omega[/tex] in terms of R, L, and C.

    2. Relevant equations

    The circuit is setup so that the resistor and inductor are in series, and the capacitor in parallel with them. I've solved for he total impedance as,

    [tex]Z_{tot}=\frac{R-i(\omega c R^2 + \omega^3 L^2 C - \omega L)}{(\omega C R)^2+(\omega^2 L C - 1)^2}[/tex]

    And this agrees with the answer supplied in the text. I feel that Euler's relation should be playing a role here as well.

    3. The attempt at a solution

    The question mentions that the circuit would be in resonance if Z is real. I'm thinking that must mean

    [tex]Z_{real}=\frac{R}{(\omega C R)^2+(\omega^2 L C - 1)^2}[/tex]

    But even if that's true, how would I then find [tex]\omega[/tex]? This real component of Z isn't equal to anything. Using Euler's relation seems promising, but for some reason I just don't know what steps to take here. Do I use Euler's relation and set it equal to the real component of Z?

    Also, another question of mine (about a similar, yet different circuit setup) mentions the angle of Z being 45 degrees? That would mean real and imaginary components of Z are equal, but where does that get me? And do I need an angle for my original question.

    This stuff is driving me nuts...any help would be appreciated.
     
  2. jcsd
  3. Sep 19, 2007 #2

    nrqed

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    Just set the imaginary part of Z equal to zero! In other words set
    [tex] \omega c R^2 + \omega^3 L^2 C - \omega L=0 [/tex]
    You may divide by omega throughout (since we assume omega is not zero itself) and that leaves you with a quadratic formula for omega

    It leaves you with an equation. Just set the imaginary part equal to the real part.

    Hope this helps

    Patrick
     
  4. Sep 19, 2007 #3
    ...wow...why did I not see that hours ago?

    Thanks a ton...I'm really sort of awestruck at how simple that was and how I didn't see it sooner.
     
  5. Sep 20, 2007 #4

    nrqed

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    :-) :-)

    Sometimes we miss the obvious.
     
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