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Delta to Y method

  1. Jul 26, 2012 #1
    Hello, I have this question that I've been working on. The question requires me to find the current across an inductor in a Delta circuit, so I want to solve this question using Delta-to-Wye method. so I transformed it using the method and I found the total impedance of the whole circuit, then I divided the phasor voltage by the impedance and I found the current, but I still have no idea how I can find the current across the inductor. I attached a picture of the question so you guys can see my problem.
    The result of the total impedance was around ( 82 - 64.07i)
    and the phasor current after dividing by the phasor voltage is 0.6731 < 38

    Thanks a lot, your help is really appreciated.
     

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  3. Jul 26, 2012 #2

    uart

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    Say that you transform R1 R2 and ZL1 from delta to star. As you've found, you now have a series - parallel circuit that is amenable to "ladder style" collapsing. You do lose track of L1 in the process (it gets tangled up with R1 and R2 in the transform), more importantly however you do still have access to nodes A and B.

    So you can calculate the voltages at A and B from the transformed circuit, then using this knowledge of [itex]V_{AB}[/itex] you can then go back to the original circuit to solve for Ix
     
  4. Jul 26, 2012 #3
    Okay. I got your idea, by that Va should the voltage across the capacitor and Vb should the voltage across R3 but the thing how could I find the voltage or apply the voltage divider rule in case of a sinosoidal voltage source.
    just to tell you
    Z1= 16 - 21.33i
    Z2= 51.2 + 38.4i
    Z3 = 64 - 48i

    when I apply the voltage divider rule to Z1, VZ1 = Vin * Z1/ Ztot
    all of Z1, Z total and Vin should be in phasor form?

    Thanks a lot man.
     
  5. Jul 26, 2012 #4

    uart

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    Yeah that's right. Just use complex numbers for all the impedances and use the voltage divider rule with complex numbers. For the source voltage use 70 + j0 (70 angle 0).
     
  6. Jul 26, 2012 #5

    gneill

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    Re: Delta To Y method in solving this Q

    If you apply a Δ-Y transform to the circuit you will "transform away" the component through which you want to find the current (L1). However, nodes A and B still exist in the "new" layout, so you could calculate their node voltages. With the two node voltages in hand you can go back to the original layout to determine the potential difference across L1 and hence its current...

    A more straightforward approach might be to write KVL mesh or KCL node equations for the original circuit.
     
  7. Jul 26, 2012 #6
    Okay that's great, I think I got the picture. I'll solve the whole thing and I'll scan and attach it to show it to you so I can know your remarks, of course if you don't mind?
     
  8. Jul 26, 2012 #7
    Re: Delta To Y method in solving this Q

    Okay. The thing I'm supposed to do it using delta to wye, however; If I ever have time, I'll try the node analysis. And if you don't mind, can I solve the whole thing, and attach it so I can show and you could tell me what are my mistake, if I have any?

    And thanks a lot for your help;.
     
  9. Jul 27, 2012 #8

    gneill

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    Re: Delta To Y method in solving this Q

    Yes, you are encouraged to show your work :smile:
     
  10. Jul 27, 2012 #9
    There is this one weird thing that I came across when solving the question, as you know to find the voltage of node A, you have to find the voltage of the capacitor, therefore; firsr you need to find the voltage of the whole parallel part by applying voltage divider rule like this (Zparallel/ Zparallel + Z1)* 70<0 ... The answer would be 52.92<5.076. Then you use this in voltage divider rule again to find the capacitor voltage Like this, (Xc/Xc + Z2)* 52.92<076. However the weird thing that the voltage of node A (capacitor) is actually higher than the voltage of whole parallel part, and this should be impossible. I tried to check all of my results, but I'm quite sure I haven't made any calculation mistakes?
     
  11. Jul 27, 2012 #10
    Here you go, this is my work. BTW I tried also to find the current through the capacitor and then find the voltage but I got a different result as well :S
     

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  12. Jul 27, 2012 #11

    gneill

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    You'll want to recheck your calculation for Z3. Otherwise, your method of work looks fine.
     
  13. Jul 27, 2012 #12
    You're right Z3 here is wrong, but when I calculated the total Z I used the right calculation which is 64 + 48i , and the answer of the Z of Parallel would still be 66 - 42.74. And thus the total Z is 82 - 64.07. And when we calculate Va we just need the Z parallel and Z tot so the mistake in calculation Z3 won't have any effect on the result of Va.
     
  14. Jul 28, 2012 #13

    gneill

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    It does, however, impact Vb which in turn effects Va-Vb for the calculation of Ix.
     
  15. Jul 28, 2012 #14
    Anyway I was concerned bcuz when we find the magnitude of Va in polar form it's greater than the magnitude of the parallel voltage in polar form. However; I decided to check the total voltage of Vz2 + Va in rectangular form and it was equal to the parallel voltage which I guess my answer is correct. Anyway I resolved everything and I corrected my mistakes and here is my final work.
    Thanks a lot for your help, could you please check it for the last time. I really appreciate your help.
     

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  16. Jul 28, 2012 #15

    gneill

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    It all looks fine except for your final calculation of Ix; the impedance of the inductor should be an imaginary value (240j Ω), but it seems that you've treated it as a real value.
     
  17. Jul 28, 2012 #16
    oh missed that. SO final answer should be (28.93 - 22.27j)/240j = -0.093 +0.120i which equals to 0.152 < -127.59 in phasor form. Thanks a lot gneil, and sorry for that coin in my last solution sheet :D
     
  18. Jul 28, 2012 #17

    gneill

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    Yes, that looks better :approve:

    Next time post a cheque; paper coins are hard to cash :smile:
     
  19. Jul 28, 2012 #18
    Hahaha, you got it :biggrin: take care
     
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