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Star Delta Transformation

  1. Dec 29, 2014 #1

    M P

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    1. The problem statement, all variables and given/known data

    zad n.png
    2. Relevant equations


    3. The attempt at a solution

    a) the question is to transform star to delta ?
     
  2. jcsd
  3. Dec 29, 2014 #2

    gneill

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    Staff: Mentor

    The question asks you to find a single Δ-connected load that is equivalent to total of the two loads shown.
     
  4. Dec 29, 2014 #3

    M P

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    attempt:
    90+90j for single delta ?
     
  5. Dec 29, 2014 #4

    gneill

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    Staff: Mentor

    We won't confirm or deny a guess. You'll have to show how you got there.
     
  6. Dec 29, 2014 #5

    M P

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    I was not guessing I just used the T to π converter supplied with course and added result to another ZΔ =(45+j45) doubling them given 90+90j
     
  7. Dec 29, 2014 #6

    gneill

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    Staff: Mentor

    Can you justify adding them? How do the voltage sources "see" them when you combine them?
     
  8. Dec 29, 2014 #7

    M P

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    angle is changing ? on the other connection V ∠0 ∠-120 ∠-240 thank you for the interest.
     
  9. Dec 29, 2014 #8

    gneill

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    Staff: Mentor

    Ignore the phase angles of the sources for now. How do the individual impedances of the two loads combine as seen by the sources? You have two Δ loads. What connections are in common? Can you see serial or parallel reduction possibilities?
     
  10. Dec 31, 2014 #9

    M P

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    attempt: ZΔ x ZΔ /ZΔ+ZΔ and then x3 ?
     
  11. Dec 31, 2014 #10

    gneill

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    Can you elaborate?
     
  12. Dec 31, 2014 #11

    M P

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    I have Y and Δ from the start. I transform Y to Δ and obtain 2 x Δ with ZΔ = 45 + j45 and to get single Δ after your hints I thought I need to combine each 2 impedances in parallel so ZΔ x ZΔ /ZΔ+ZΔ and I need to do that 3x ?
     
  13. Dec 31, 2014 #12

    M P

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    An thank you for the prompt reply :cool:
     
  14. Dec 31, 2014 #13

    gneill

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    Staff: Mentor

    Okay, that is correct. But it is important that you can see why the individual impedances are in parallel by considering the circuit diagram. If you can't see it, be sure to ask.
     
  15. Dec 31, 2014 #14

    M P

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    I will improve I promise.
     
  16. Dec 31, 2014 #15

    M P

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    So similar to b) with 2 x Y at 15+j15 and ZYx ZY/ ZY+ZY
     
  17. Dec 31, 2014 #16

    gneill

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    How are we to interpret your coded messages? o_O A few words of explanation and justification would help.
     
  18. Dec 31, 2014 #17

    M P

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    Sorry. after Δ to Y transformation I have 15+j15 for each impedance 2xY and then combine each impedance ZYx ZY/ ZY+ZY as in a)
     
  19. Dec 31, 2014 #18

    gneill

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    Staff: Mentor

    Okay, what leads you conclude that you can combine the Y's in that fashion? I'm not saying it's incorrect for this very special case, but you should be able to justify it. That way you won't inadvertently make the error of trying to do the same thing when the scenario is different and combining them in this fashion would be incorrect.
     
  20. Dec 31, 2014 #19

    M P

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    as per drawing??
     

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  21. Dec 31, 2014 #20

    M P

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    now I am thinking what happens when I add 0 to it? I wonder if you have any hints...:)
     
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