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Math Help - Impedance equation

  1. #1
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    Impedance equation

    Hi guys,

    i need some help please with the following equation, I need to make c the subject of z=sqrt(r^2 + (xl - xc)^2).

    My first step was remove the sqrt to become z^2=r^2 + (xl - xc)^2
    Then expand the brackets so z^2=r^2 + (xl - Xc)(Xl -Xc)
    then Multiply out the brackets so z^2 = r^2 + xl^2 - xc^2 - 2xlxc

    then I'm a bit stuck where to go from there. I could move the r^2 to the other side but I'm still none the wiser on getting c as the subject.

    i was also given xl= 2(pi)fl and xc = 1/ 2(pi)FC

    z=104ohms r=100ohms f=50hz, l= 0.1 henry

    any enlightenment would be greatly appreciated!

    thanks

    kris
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  2. #2
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    Re: Impedance equation

    z^2 = r^2 + (x_L - x_C)^2

    z^2 - r^2 = (x_L - x_C)^2

    \sqrt{z^2 - r^2} = |x_L - x_C|


    if x_L > x_C ...

    \sqrt{z^2 - r^2} = x_L - x_C

    x_C = x_L - \sqrt{z^2-r^2}


    if x_L < x_C ...

    \sqrt{z^2 - r^2} = -(x_L - x_C)

    x_C = x_L + \sqrt{z^2-r^2}

    finally ...

    x_C = \frac{1}{2\pi f C}

    C = \frac{1}{2\pi f x_C}
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  3. #3
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    Re: Impedance equation

    Hi skeeter

    Thanks for the input but how do I transpose so C is the subject? Both of the above still with have xc or a c on the both sides. Sorry for not understanding and I'm sure you correct it's just that I still don to now how to use the formula by your solution

    Thanks,

    Kris.
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  4. #4
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    Re: Impedance equation

    Ok I get how c equals that now but the question I've been given is calculated the value of c, but how is the value of c supposed t o be calculated if I'm only given z=104 ohms, R= 100 ohms, f = 50 hz, l = 0.1 Henry. It's just throwing me as I can input the values but I still don't know xc.

    thanks
    Last edited by Krislton; October 10th 2012 at 10:56 AM.
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  5. #5
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    Re: Impedance equation

    Double post sorry
    Last edited by Krislton; October 10th 2012 at 10:56 AM.
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