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LCR circuit and frequency equation

  1. Dec 9, 2011 #1
    1. The problem statement, all variables and given/known data
    Im doing a lab experiment on LCR circuits and have gotten a graph of 1/capacatance against frequency^2. With my results I got the slope to be 0.89±0.01 and the intercept is 358708±43189


    2. Relevant equations
    the equation that was given was
    frequency^2=(1/4*pi*L)[(1/C) - (R^2/2L)]
    we then get
    f^2 = 1/(4*pi*L*C) -R^2/(8*pi*L^2)

    3. The attempt at a solution

    since y =mx + c

    the slope is
    m=1/(4*pi*L) where m is 1/C
    then c = -R^2/(8*pi*L^2)
    we rearrrange the first equation to get L= 1/(4*pi*m) then when we know L we can get R
    R^2 = -c*8*pi**l^2

    When I put in my values for m for the first equation I get L = 2.85E-2 Henry
    Then when to get R^2 I can as you cant the square of a negative number.

    Would I get the positive of the R^2 value and then say the resistance is (plus/minus) 151 ohms?
    Or can someone tell me what I am doing wrong
    my lecturer does not know what is wrong so I kinda want to get it right
     
  2. jcsd
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