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Maximum and minimum values of a variable capacitor

  1. Feb 25, 2015 #1

    rlc

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    1. The problem statement, all variables and given/known data
    upload_2015-2-25_11-3-1.png
    A parallel circuit like that in the figure forms the tuning circuit for an AM radio. If the inductor has a value of 8.6×10-6 H, what must be the maximum and minimum values of the variable capacitor if the radio receives frequencies from 590k Hz to 1670k Hz?
    a) Maximum C?
    b) Minimum C?

    2. Relevant equations
    w=2*pie*f
    f=1/(2*pie*sqrt(LC))

    sqrt(LC) = 1 / (2 pi f)
    LC = 1 / (2 pi f)^2
    C = 1 / [ (2 pi f}^2 L)

    3. The attempt at a solution
    Is the inductor value 8.6E-6 H the variable f in the equation? Are those equations even the ones I use?
    590k Hz= 5.9E5 Hz
    How do I begin this problem? I'm just really confused
     
  2. jcsd
  3. Feb 25, 2015 #2

    gneill

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    Staff: Mentor

    By convention the variable name used for inductance is L. f is frequency. C is capacitance.
     
  4. Feb 25, 2015 #3

    rlc

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    C = 1 / [(L)(4pi^2)(f^2)]
    Cmax= 1 / [(8.6E-6)(4pi^2)(590E3)^2]
    Cmax=8.461E-9 F

    Cmin=1 / [(8.6E-6)(4pi^2)(1670E3)^2]
    Cmin=1.056E-9 F

    *I kept getting this wrong because the equation in the Relevant Equations part had a 2 instead of a 4*
     
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