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Electrical Resonance Problem

  1. Apr 10, 2009 #1
    A constant voltage at a frequency of 1MHz is maintained across a circut consisting of the inductor in series with a variable capacitor. When the capacitor is set to 300pF the current has its max value. When the capacitor is reduced to 284pF the current is 0.707 of its ma value. Find

    The inductance and the resistance of the inductor





    Im not sure if im using the correct formula.

    f=[1/(2.pi)] x [1/(square root of (L).(C)]

    f=frequency
    L=Inductance
    C=capacitance

    L = 1/[(300x10^-12) x (2,000,000.pi)^2]

    =337micro H

    What formula could I use to find the resistance of the inductor???
     
  2. jcsd
  3. Apr 10, 2009 #2
    use the second pair of info, and the formula for the current in an RLC circuit...
     
  4. Apr 10, 2009 #3
    What do you mean by use the second pair of info?

    Thanks for your help
     
  5. Apr 10, 2009 #4
    well you know that when the capacity is 284pF then the current is 0.707 times the max...

    So use the general formula for the current (containing R) and make it equal to 0.707 times the max current...
     
  6. Apr 10, 2009 #5
    oh thanks,

    so is the first part correct??
     
  7. Apr 10, 2009 #6
    yes thats cool...
     
  8. Apr 10, 2009 #7

    rl.bhat

    User Avatar
    Homework Helper

    I = 0.707Imax when XL - XC = R
     
  9. Apr 10, 2009 #8

    So i use this formula

    I = V.R.C/L

    I(max) = [V. R. (300x10^-12)] / (337x10^-6)

    and then I = 0.707 I(max)

    But i dont know V???
     
  10. Apr 10, 2009 #9

    rl.bhat

    User Avatar
    Homework Helper

    Just calculate 2*pi*f*L -1/(2*pi*f*C). That is the value of R.
     
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