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Complex current along a resistor

  1. Jul 21, 2010 #1
    If a current of .587e^j1.12 is passing along a 200ohms resistor, how would I go about calculating the power dissipated in the resistor?
     
  2. jcsd
  3. Jul 21, 2010 #2
    [tex]P = (V)(I^*)=(IZ)(I^*)=Z(II^*)[/tex]

    Now, I times I conjugate leaves the magnitude of I squared. Z is not complex, only resistive.
     
  4. Jul 21, 2010 #3
    the complex conugate of i multiplied by i is .0653 - .5487j isn't it?
    Then do I just multiply the real part of this by the resistance?
     
  5. Jul 21, 2010 #4
    No. Complex conjugate means you take the vector, keep its magnitude, and add a negative to the angle. You can also think of it as adding a negative to the imaginary component of the vector.

    [tex]I = .587e^{j1.12}[/tex]
    [tex]I^* = .587e^{-j1.12}[/tex]
    [tex]II^*=.587e^{j1.12}.587e^{-j1.12}=(.587)^2e^{j1.12-j1.12}= (.587)^2e^{0}=(.587)^2[/tex]
     
  6. Jul 21, 2010 #5
    does Z represent the resistance?
    .587^2 x 200 = 68 Watts which is wrong for some reason.
     
  7. Jul 21, 2010 #6

    Zryn

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    Gold Member

    Z = Impedance (complex resistance)

    What does your book say is the correct value?
     
  8. Jul 21, 2010 #7
    It says the correct answer is 34.5 watts
     
  9. Jul 21, 2010 #8
    It's because you gave us peak current without saying! Almost always will you be given RMS and when it's not stated, RMS is assumed. Everything I said above is the same for peak values of voltage/current except you need to divide by 2.

    [tex]\frac{I_p}{\sqrt{2}}=I_{rms}[/tex]
    So when you have I^2 (in rms) you have 1/sqrt(2) squared = 1/2
     
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