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Problem with Impedance

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data

    A particular inductor has appreciable resis-
    tance. When the inductor is connected to a
    19.7 V battery, the current in the inductor is
    4.38 A. When it is connected to AC source
    with an rms output of 19.7 V and a frequency
    of 58 Hz, the current drops to 1.65 A.
    Find the impedance at 58 Hz.
    Answer in units of
    .

    2. Relevant equations

    Z=sqrt(R^2 +(XL-XC)^2)

    XL=V/I

    R=V/I



    3. The attempt at a solution

    XL=19.7 V / 4.38 A = 4.49772 ohms

    R = 19.7 V / 1.65 A = 11.9394 ohms

    Z= sqrt( 11.9394^2 + 4.49772^2) = 12.7585 ohms
     
  2. jcsd
  3. Oct 22, 2008 #2
    Re: impedance

    I am afraid you mixed them up.
    The higher current is when you have only resistance. (I1=V/R)
    The lower current is I2 = V/Z and not V/XL.
     
  4. Oct 22, 2008 #3
    Re: impedance

    Ok I understand my mistake now. Thank you for clearing that up for me!
     
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