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Complex impedance problem

  1. Feb 11, 2014 #1
    parallell.png

    For the problem on the left, the 2 capacitors are parallel to each other so the 1/z = 1/-2j + 1/-2j = 2/-2j

    so 1/z = 1/-j = -1/j

    so the total impedance of the 2 capacitors, z = -j

    Now if you add this to 5j + 4 (the total impedance of the resistor and inductor) you get 4j+4. However this doesn't give the right answer..
     
  2. jcsd
  3. Feb 11, 2014 #2

    berkeman

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    The capacitors are not in parallel. They have an inductor between them. You need to write the KCL equations and solve them simultaneously...
     
  4. Feb 11, 2014 #3

    gneill

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    No they are not. One pair of their leads (the bottom ones) are connected together, but there's an inductor in the way of the other connection. So, not parallel.


    EDIT: Oops. Berkeman beat me to the punch!
     
  5. Feb 11, 2014 #4

    berkeman

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    Actually, it's easier than using KCL equations. Do series-parallel combinations of the complex impedances to get the overall input impedance...
     
  6. Feb 11, 2014 #5
    So if they are not in parallel they must be in series? In which case we just add them? -2j +-2j =-4j?
     
  7. Feb 11, 2014 #6

    berkeman

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    For the left circuit, you first make a series combination, then a parallel combination, then a final series combination. You fold the circuit up from right to left in this case...
     
  8. Feb 11, 2014 #7
    Ah I see..

    For the one on the right, I done the following but I don't get the answer correct?

    the 4 ohm resistor and the 1j ohm inductor are in series so you just add them up to get 4+1j, then this combined total is in parallel with the 2j ohm inductor so

    1/(4+1j) + 1/(2j) = 1/Z

    so Z = 16/25 + 38j/25 (once simplified and rationalised)

    then you add -1j ohm to the answer of the above (as the -1j capacitor is in series with our above answer) so

    Z = 16/25 + 13j/25 (2)

    Then the remaining 1 ohm resistor (the one to the left) is calculated in parallel with (2)

    so 1/Z = 1/1 + 1/(16+13j/25)

    But simplifying the above and finding it in the form r<theta, gives me a different answer to the given answer. Where have I gone wrong?

    Thanks
     
  9. Feb 11, 2014 #8

    berkeman

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    It looks like you are using the correct steps. Can you post your work in detail so we can check it?
     
  10. Feb 14, 2014 #9
    I just made a small algebraic mistake!

    Thanks anyway.. :)
     
  11. Feb 14, 2014 #10

    berkeman

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    Sweet! :smile:
     
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