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Impedance seen by source

  1. Nov 16, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the impedance as seen by the source.
    upload_2016-11-16_19-55-5.png

    2. Relevant equations


    3. The attempt at a solution
    -j2||j1 = 2i
    Z = 2-j1+j2+(1^2)/(j2+2+j2)
    Ztotal = 2.1+0.8i
    Is this correct?
     
  2. jcsd
  3. Nov 16, 2016 #2

    NascentOxygen

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    Staff: Mentor

    What do you understand is meant by the j2 Ω on each side of the transformer? What does the j1 Ω referring to both sides indicate?
     
  4. Nov 16, 2016 #3
    Mutual inductance?
     
  5. Nov 17, 2016 #4

    cnh1995

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    Homework Helper

    Well, I don't know the direct formula for finding impedance in such coupled circuits, but you can use mesh analysis on both the sides and find the current on the source side. Source voltage/source current will be the impedance seen by the source.
     
  6. Nov 17, 2016 #5

    gneill

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    The final result looks okay, but whether or not your work is correct is hard to say. Since all the values in the circuit are 1's and 2's, they become somewhat anonymous when we can only see digits.

    The Relevant equations part of the template is there for a good reason. Could you not have quoted or referenced the equation that you employed?
     
  7. Nov 17, 2016 #6

    NascentOxygen

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    I asked two questions, and you gave only one answer?
     
  8. Nov 17, 2016 #7
    j1 is mutual inductance
    j2 on each side is the self inductance

    Z(total left) + M^2/Z(total right)
    Z on left side:2-j1+j2
    M=1-> M^2 = 1
    -j2||j1 = 2j
    Ztotalright = (j2+2+j2)
     
  9. Nov 17, 2016 #8

    gneill

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    Okay. Now I can state that you've achieved the correct result in a valid manner.

    For completeness, if the mutual inductance has impedance ##Z_m = 1j## as it was specified on the circuit diagram, and letting ##Z_p## and ##Z_s## be the total primary and secondary circuit impedances, then you can write:

    ##Z_in = Z_p - \frac{Z_m^2}{Z_s}##

    The "j" of the mutual inductance squares out to -1, cancelling with the "-" of the term.
     
  10. Dec 2, 2016 #9

    rude man

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    I did:
    load impedance ZL = (j1 || -j2) + 2 = 2 + j2
    for the input impedance to the xfmr including the load: (-7 + j4)/(2 + j4)
    then add source impedance Zs = 2 - j1 giving total input impedance = (-7 + j4)/(2 + j4) + 2 - j1 = 2.1 + j0.8
    which agrees with your result also. You seem to have come up with a faster way, to your credit!
     
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