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Frequency domain and mesh currents

  1. Feb 24, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-2-24_11-1-52.png

    2. Relevant equations


    3. The attempt at a solution
    So far I was able to transform each element into the frequency domain:
    upload_2017-2-24_11-2-46.png

    I am just unsure how to work with the dots to find the mesh currents. Can someone explain? Thanks.
     
  2. jcsd
  3. Feb 24, 2017 #2

    cnh1995

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    Homework Helper

    Have you studied dependent sources?
    In circuits with mutual inductance, you represent mutually induced emf using current controlled voltage source(s).
     
  4. Feb 24, 2017 #3
    I believe we have studied this but I am definitely not clear on the subject.
     
  5. Feb 24, 2017 #4

    cnh1995

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    Ok. Think about what the dot convention means here.
    When the current "enters" a coil at the dot, the dotted terminal of the other coil becomes positive i.e.both the dots become positive (and negative) simultaneously.

    You have assumed i2 "leaving" the dot of the 6H coil. What will be the polarity of the dot?
     
  6. Feb 24, 2017 #5
    Negative since it is leaving ?
     
  7. Feb 24, 2017 #6

    cnh1995

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    Yes. So what is the polarity of the dot of the 8H coil? That would be the polarity of the mutual voltage induced in the 8H coil.i.e. the emf induced in 8H coil due to the current in the 6H coil. Add a current controlled voltage source in series with the 8H coil with the polarity you just found. Controlling current would be i2 for mutual emf in the 8H coil and i1-i2 for mutual emf in the 6H coil.
     
  8. Feb 24, 2017 #7
    Would these be the two equations?
    upload_2017-2-24_18-41-25.png
     
  9. Feb 24, 2017 #8

    gneill

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    Staff: Mentor

    Your calculation of Zc is not correct. Zc should be of comparable magnitude to the other impedances in the circuit.

    Your first loop equation does not include a mutual inductance term, and the signs for the currents flowing through the j200 inductor are not correct.

    For your second loop you didn't account for I1's contribution to a mutual inductance term, and I2 should make two contributions (I2 flows through both inductors, hence though both dots).

    I find that it can be helpful to re-draw the circuit with the mutual inductance inspired voltage sources inserted. Start out with them having their negative ends anchored to the dot of their inductor. Figure out their values by using labelled currents and the dots to assign appropriate signs (after which you can adjust the voltage source directions to eliminate negative terms if you wish), then write the loop equations. This way you can't drop terms if you do your KVL walks methodically around the loops.

    upload_2017-2-24_19-8-49.png
     
  10. Feb 24, 2017 #9
    Thanks for such a complete answer. One more question. Can you always attach the mutual inductance voltage source by the negative end to the dot? For example in this circuit:
    upload_2017-2-24_22-48-23.png
    Would I attach it to the dot, or do I need to take into account the sign of V1 and V2 ?
     
  11. Feb 24, 2017 #10

    gneill

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    Staff: Mentor

    Attaching the negative end to the dot is just a convenient starting point. It follows the convention that a current flowing into one dot results in a current flowing out of the other dot, and really is just a matter of getting the sources in place without considering any details of the circuit currents. Once this is done it's up to you to define the mesh currents and then check whether they are flowing into or out of the dots. You assign signs to the voltage values on the sources accordingly.

    upload_2017-2-24_23-24-52.png

    You are of course free to reorient the polarities of the voltage sources as you like, provided that you change the signs of their values accordingly. In the end it's the currents flowing into or out of the dots that dictate the induced voltage polarities.

    In your example i1 is flowing into the dot on the 4H inductor, and so the voltage source value attached to the other inductor gets a positive sign. i2, on the other hand is flowing out of the dot of the 6H inductor, so the voltage value assigned to the source on the 4H inductor gets a negative value.

    The v1 and v2 shown in the diagram appear to be related to the potential drops caused by the mesh currents across the inductors (inductor impedances), or they're just labels for the net voltage across the inductor that you need to solve for with the indicated polarity being defined.
     
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