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Finding a Unknown Inductor

Discussion in 'Homework Help' started by Trav, Dec 12, 2011.

  1. ljcox

    ljcox Well-Known Member

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    This is correct.
     
  2. mvs sarma

    mvs sarma Well-Known Member

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    Cant be ljcox ,
    because against an equation F and L have to be inverse square
    Fr =1 /(2*pi*Sqrt(L*C))
    Sqrt(L*C) = 1 /(2*pi*Fr)
    L*C = 1/ (2*pi*Fr)^2
    L = 1 / {((2*pi*Fr)^2)*C}
    L= 1 / {4*3.141*3.141 *Fr*Fr*C)
     
    Last edited: Dec 14, 2011
  3. ljcox

    ljcox Well-Known Member

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    Yes, you're right. I should have looked at it more carefully.

    It should be L = 1/{C * (2pi x fr)^2}
     
  4. dave

    Dave New Member

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  5. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    hi Len,
    We now agree
    Knowing that Fr= 1/(2Π*√LC)

    I suggested to the OP that he moved the '2Π' to the left side of the equation, so that it became 2Π*F and substituted ω for 2ΠF.

    This would give ω = 1/(√LC)

    Then its a simple matter of squaring both sides of the equation to, ω^2 = 1/LC....

    Transposing gives, L = 1/(ω^2 * C) ..... (1)

    E
     
  6. meowth08

    meowth08 Member

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    If you are not sure of your answer, all you need to do is substitute any number to your variables in the original equation say 2 for L, 3 for C. From here, you can get fr. Now, from your derived equation, substitute the value of fr that you got, and 3 for C. If you get L equal to 2, your derivation might be right.

    Note: These are not always true all the time. You have to use different values for your variables to verify your equation.
     

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