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Determine source frequency and impedance

  1. Apr 6, 2015 #1
    1. The problem statement, all variables and given/known data
    The current in a series circuit of R=5 ohms and L=30 mH lags the applied voltage by 80 degrees. Determine the source frequency and impedance.
    (Ans: Z=5+j28.4 , f=150.4 Hz)

    2. Relevant equations


    3. The attempt at a solution

    Current in the inductor,
    IL = Vm/wL*sin(wt-80)

    Current in the resistor,
    Ir = Vm/R * sin(wt-80)

    Current in the resistor equals current in conductor
    Equating them you get,
    1/wL = 1/R
    1/w(30*10-3) = 1/5
    w = 166.67 rad/s

    w = 2*pi*f
    f = w/2*pi
    = 166.67/2*pi
    = 26.53Hz

    But they get a frequency of 150.4Hz
     
  2. jcsd
  3. Apr 6, 2015 #2

    donpacino

    User Avatar
    Gold Member

    have you learned phasors yet? if you have there is an easy way to do this problem.

    That being said, your problem is voltage Vm is equel to the voltage drop across the inductor AND the resistor. They are is series. Therefore the voltage drop across the inductor is not vm. You also wrote your time domain equation wrong.
     
  4. Apr 6, 2015 #3
    Yes, I have done phasors

    I think I found a way using phasors

    if z = x∠80

    Then, x*cos(80) = 5 (The real part, which is the value of the resistor)

    Therefore x = 28.79

    So, z = 28.79∠80

    Therefore the impedance of the inductor is equal to 28.79sin(80) = j28.4

    Angular frequency for inductor
    z = jWL
    W = 28.4/30*10-3
    = 946.67 rad/s

    Therefore the frequency is
    w/2*pi
    = 946.67/2*pi
    = 150.67Hz

    I think that's correct, yes ?

    Thanks
     
  5. Apr 6, 2015 #4

    donpacino

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    Gold Member

    yes! here is how i did it

    i=v/z
    z=R+LS

    we know I=mag(I)∠-80
    so z=mag(z)∠80=sqrt(R2+(Lw)2)∠tan-1(Lw/R)

    80=tan-1(Lw/R)
    solve for w

    in other words, the same thing, but used algebra until the end. I recommend you do that, with more complicated problems avoiding calculations until the end can prevent mistakes.
     
    Last edited: Apr 6, 2015
  6. Apr 6, 2015 #5
    Yeah, I see my final answer was a bit off there

    Thanks for you help :smile:
     
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