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Finding individual impedance given the magnitude.

  1. Aug 12, 2012 #1
    A 10-H inductor, 200-Ω resistor and a capacitor C are in parallel. find C if the magnitude of the impedance is 125Ω at ω=100rad/s




    Now i attempted to solve it by first doing the parallel combination of the inductor and the resistor. Then i would do the parallel combo of these with the capacitor with the variable C. This left me with a massive equation. Im sure there has to be an easier way. please help
     
  2. jcsd
  3. Aug 13, 2012 #2

    tiny-tim

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    Welcome to pf!

    Hi Basher1! Welcome to pf! :smile:

    The total impedance is 1/(1/Z1 +1/Z2 + 1/Z3) :wink:

    Show us what you get :smile:
     
  4. Aug 15, 2012 #3
    hello Tiny tim. I've tried adding the impedance of the inductor and resistior then trying to find the unknown capacitance. but i was given the magnitude of the impedance. I wasn't given anything in rectangular form.
     
  5. Aug 15, 2012 #4

    tiny-tim

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    Show us what you get. :smile:
     
  6. Aug 15, 2012 #5
    ZR = 200Ω,
    ZL = j1000Ω
    ZC = 1/j100CΩ
    ω = 100rad/
    magnitude of impedance = 125Ω

    Zeq = (200 + j1000)/(200.j1000) + j100C = 0.008

    = (200 x 10^6) + (40 x 10^6)j/(4 x 10^10) + j100C = 0.008

    = (5 x 10^-3) + (1 x 10^-3)j + (j100C) = 0.008

    => (j100C) = 0.003 + j(0.001)


    C = (0.003 + j(0.001))/(j100)
    clearly a capacitance cannot have an imaginary value. I have no idea what i'm doing. personally i thought you would require a phase angle to accompany the magnitude of the impedance in order to solve so that we could get a reactance and resitive component and thus i could equate coefficients.
     
    Last edited: Aug 15, 2012
  7. Aug 15, 2012 #6

    cepheid

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    All three of the impedances are in parallel. So why are you adding them as though only two of them are in parallel, and the third is in series with the other two?

    EDIT: Hmm, no sorry, my mistake. I see that your operation is correct after all. I'll reply momentarily with further comments.
     
    Last edited: Aug 15, 2012
  8. Aug 15, 2012 #7
    No thats ok.

    I would like to hear your feedback
     
  9. Aug 15, 2012 #8

    cepheid

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    It's your second line that's wrong. Zeq-1 is NOT equal to 1/125 (0.008), because 125 Ω is just the *magnitude* of the impedance: |Zeq|. The impedance Zeq is in general a complex number.
     
  10. Aug 15, 2012 #9
    so (4 x 10^10)/((200 x 10^6) + (40 x 10^6)j + (4 x 10^12)jC) = Zeq?

    If i rationalise this I end up with a massive value and a c^2. there has to be an easier way?
     
  11. Aug 15, 2012 #10

    cepheid

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    I'm having a lot of trouble following your arithmetic, for instance where the 200e6 and 40e6 come from.

    Personally, I'd prefer to keep things algebraic until the very end. Things become a lot less cluttered, and it's just nicer to end up with a general solution for the equivalent impedance of any parallel LRC circuit that you then plug numbers into for your specific situation.

    The easiest way I've found so far is as follows (I'll just omit the "eq" subscript and call the equivalent impedance "Z"). So we have:

    Z-1 = (1/R) + (1/jωL) + jωC

    Note that 1/j = -j, so you can express this as:

    Z-1 = (1/R) - j(1/ωL) + jωC

    = (1/R) + j[ωC - 1/ωL]

    So now you have Z-1 expressed as a complex number in rectangular form (i.e. with a real part and an imaginary part, as opposed to polar form, which has a magnitude and a phase). How would compute the magnitude of this number using the real part and the imaginary part? I.e. how do you find |Z-1|? How is |Z| (which is what you are trying to solve for) related to |Z-1|?
     
  12. Aug 16, 2012 #11
    argh yes i was complicating things. Square the real part, imaginary part, take the root of the entire thing. equate this to 1/125. thankyou cepheid
     
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