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Star Delta Transformation Proof

  1. Aug 23, 2008 #1
    I've come across a proof for star-delta transformation which goes like this.(Refer to the diagram for notation. Pardon me for my bad drawing skills.)

    In the delta, he found the effective resistance between two vertices ( say a and c, which can found easily). Then he found the effective resistance between the same two vertices (a and c)
    in the star. According to him Rac in the star(which denotes the resistance between a and c) is equal to Ra + Rc. What about Rb? It might not be connected here but it may in a circuit. It is like this(refer to second attachment). The resistance about ac in the left diagram of second attachment is 2[tex]\Omega[/tex](drawing analogy from the "faulty" proof) but link it to another circuit (shown in right diagram), the resistance about ac is different. So just because it isn't connected, it doesn't mean that you can ignore it. Here's the surprise. The proof gives the CORRECT EQUATIONS. Work it out. You'll end up with the right thing!Can anybody make things clear for me?
     

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  2. jcsd
  3. Aug 23, 2008 #2
    Did I not follow any guidelines? How come nobody is replying?
     
  4. Aug 23, 2008 #3
    I wonder why the third line in your second diagram is connected directly to 'a' and not between 'a' and 'c'.
     
  5. Aug 23, 2008 #4

    rbj

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    you need to be more patient. i hadn't seen this post before this moment. give it a couple of days and if no one responds by 48 hours, then you have good reason to inquire.


    okay, first of all, this is normally called a "delta-Y transformation" or, less often, a "[itex]\pi[/itex]-T transformation". never heard of the "Y" called a "star" in this context.

    if it's not connected to anything, no current flows through it. if no current flows through it, the behavior is no different than an open circuit (a.k.a. nothing).

    if the impedance (or any two-terminal part) isn't connected on one end, how does it make any difference if it's connected (or not) on the other end?

    this "proof" you see is standard in any first-semester electrical engineering text (and i can go through it in detail if you want). the result is true, but i have never been satisfied with the proof because usually it requires assuming at the outset that the transformation of the circuit with all three terminals connected is equivalent to the transformation of the circuit when any arbitrary pair of terminals are connected. it turns out to be true for linear components (resistors or complex impedances) where the super-position property is valid (and that fact has to be established), but is not true for non-linear devices.
     
  6. Aug 23, 2008 #5

    berkeman

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    Staff: Mentor

    Since this is homework/coursework, I moved it from General Physics to Homework Help.
     
  7. Aug 24, 2008 #6
    Can it be established using the principles we learn in high school physics? (Actually this "Y-delta" transformation is not in our syllabus, but I picked it up so that I could get a convincing solution for some problems like a cube of resistors-It turned out to be one heck of a workout though. :smile:)

    I haven't received a convincing reply for the contradiction we are facing in circuit2.
     
  8. Aug 25, 2008 #7
    In the second image, Rac will still be given by Ra+Rc.

    The right hand circuit has an extra resistor (R1) and as such the overall resistance of the circuit will be calculated using a different equation (this doesn't have any effect on Rac only on the TOTAL resistance of the circuit). As the resistor R1 in in parallel with the Rac this would be calculated either using the product over sum rule:

    [tex]R_{T}=\frac{R_{ac}\times R_{1}}{R_{ac}+R_{1}}[/tex]

    or the equation:

    [tex]\frac{1}{R_{T}}=\frac{1}{R_{ac}}+\frac{1}{R_{1}}[/tex]

    I have attached a picture that will hopefully clarify my point a little further.

    I think it is important to note that the circuits shown in the second image are in face connected in Star formation, if it were in Delta then there would only be a single Resistor connecting node A to node C (using the notation from your first diagram this would be R2.)

    In addition to what rjb was saying in relation to Rb when finding the value for Rac, Rb has no effect but obviously when you come to find the value for Rab and Rbc then Rb comes into effect.

    again using notation from your first diagram:

    [tex]R_{1}=R_{ab}=R_{a}+R_{b}[/tex]

    [tex]R_{3}=R_{bc}=R_{b}+R_{c}[/tex]

    I hope this helps

    Ram
     

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