Phasor circuit question

Discussion in 'Homework Help' started by lurikeen, Apr 13, 2011.

  1. lurikeen

    Thread Starter New Member

    Apr 13, 2011
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    I've been working on the following phasor circuit problem, but the results seem incorrect to me, because I get one of the currents, I2, to be to be 0A when it appears that it shouldn't be. I am solving using the mesh current method.

    [​IMG]

    My equations, where x=I1, y=I2, and z=I3:

    [​IMG]

    This outputs I3=-3i, I2=0. I can't see how this is correct.



    Link to the equations on wolfram alpha: http://www.wolframalpha.com/input/?i=solve+x%3D5%2C+0+%3D+-5i+%2B+%28y-x%292i+%2B+%5By-z%5D5%2C+0%3D%5Bz-y%5D5+%2B+%5Bz-x%5D%5B-3i%5D+%2B+%5Bz%5D3i+for+z

    Can anyone help me find my error? I don't have the solutions but I think its wrong. Thanks.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Your solution is correct. It's certainly possible that the circuit conditions allow for zero current draw from the voltage source. In effect it doesn't matter if the voltage source is there or not.
     
  3. steveb

    Senior Member

    Jul 3, 2008
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    Why do you feel your answer is wrong?
     
  4. lurikeen

    Thread Starter New Member

    Apr 13, 2011
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    Well, since its a constant voltage source, I just felt that the chance that it would have no current running through it was very low. I've just started with complex numbers, so I really don't understand exactly what is MEANT by the solution -3i, or the voltage on the source of -5i.

    Thanks btw.
     
  5. steveb

    Senior Member

    Jul 3, 2008
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    Well, actually you are correct about that, if you consider a real world example. However, mathematically, a zero current is no problem. Basically, what is happening is that the current source is supplying the power to the circuit and just happens to produce a voltage (at the terminals of the voltage supply) perfectly equal to the supply voltage. Hence, the voltage source doesn't need to supply current to force the constraint that the voltage across the terminals is -5j. Note that in the real world, such perfect balancing does not happen. There would always be a small current coming from (or going into) the voltage supply.
     
    Last edited: Apr 15, 2011
  6. lurikeen

    Thread Starter New Member

    Apr 13, 2011
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    Cool, thank you very much!
     
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