Power Dissipated Into Wire

Discussion in 'Homework Help' started by hitmen, Jan 6, 2010.

  1. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    The impedance of the wire is (0.05 + j0.15)
    The current through it is (97.6<-41.6688)(phasor)

    Why cant we use the equation P = ZI^2 and expand everything out?
    I got a garbage value.
     
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Does your text explain the difference between real power and reactive power?

    The key to this question is the word "dissipated"; that means that they want the real power.

    Try again, and show your calculations.
     
  3. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    I am supposed to determine the "active power loss" so is it real, imaginery or apparent power?
     
  4. The Electrician

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    Oct 9, 2007
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    That should be real power.

    Do you know the correct answer?

    Is 97.6^2*.05 anywhere near the correct result?
     
  5. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    Yes.

    But my answer is P = Z * I^2
    = (0.05 + 0.15j)Ω (97.6<-41.6688)A^2
    = (0.15811<71.565)(9525.76<-83.3376)
    = 1506.11794<-11.7726
    = 1474.436 - j307.290 :confused:

    This is nowhere near your answer. Is there something wrong with my concept?

    Thanks!
     
  6. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    Hi electrician:

    Why are you using the magnitude of the phasor for the current and the real value for the impedance?

    When is the time to use which?

    Thanks!
     
  7. The Electrician

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    Oct 9, 2007
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    Only the real part of the impedance is dissipative. When current passes through the impedance, the value of the real part is what determines how much heat will be produced by the current.

    The RMS value of the current is what you use in the formula P = R * I^2, and the RMS value is just the magnitude of the current phasor. Notice that I've used R * I^2 instead of Z * I^2 to remind us that it's the real part of the impedance that dissipates power.
     
  8. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    ok. In that case what is the imaginary part of the impedance used for?

    Why does multiplying everything out like I have done give me the wrong answer?
     
  9. The Electrician

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    Oct 9, 2007
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    It is used in determining the total voltage drop along the wire when the current in the wire is known; it's used for determining the reactive power, which is not dissipative.

    Because when you do that you are including the reactive part of the line impedance in the calculation, and the reactive part doesn't contribute to the dissipative power loss. Also, you need to use the RMS value of the current, which is the magnitude of the current phasor neglecting the phase angle.
     
  10. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    Okay I get it. :)
    Thanks.

    But why does my lecturer use the formula P = ( Vmag)(Imag)cos(θv - θi)
    And this time the real part is NOT taken :confused:
     
  11. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    It's just a different formula. It gives the same result as the method I gave you : 476.288 watts.

    It's an example of what the reactive part of the line impedance can be used for.
     
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