delta-wye configuration of resistors

Discussion in 'Homework Help' started by PG1995, May 23, 2011.

  1. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Hi :)

    I was having a look on the "Example 2.15" which you can see in the following link: http://img23.imageshack.us/img23/5218/deltawye.jpg

    1: I don't understand how "abn" form a delta configuration. For example, take the case of "can" configuration. We start at 'c' and traverse through 12.5Ω resistor toward 'a', and then from 'a' we traverse through 10Ω and 5Ω resistors and end up at 'c' where we started the traversing. It is a "can" delta configuration. But what is happening in case "abn" deta - which in my view is not a delta? Please help.

    2: Now consider the statement which I have highlighted: Another approach would be to solve for the equivalent resistance by injecting one amp... Chap 4.

    In Chap 4, which we haven't covered so far, the author introduces these topics: Linearity Property, Superposition, Source Transformation, Thevenin's Theorem, Norton's Theorem, Maximum Power Transfer.

    Which of the above theory/topic the author would have used to solve the problem in "Example 2.15" instead of detal-wye appraoch? Please tell me.

    Thank you for your help and time.
     
  2. jegues

    Well-Known Member

    Sep 13, 2010
    735
    43
    Loop abn certainly forms a delta configuration, I've attached a figure to help you visualize, but after doing so many of these problems, they should become easier to spot.

    As for the second question. Pull off the 120V voltage source for now and set it aside.

    We've be asked to find the equivalent resistance as seen by the a-b terminals.

    The approach the author is hinting at is that you should wire in a 1A current source at that a-b terminals, and find the voltage across the a-b terminals.

    With this we can determine Rab,

    R_{ab} = \frac{V_{ab}}{I_{source}} = \frac{V_{ab}}{1A}

    After we've got Rab, our resistor network has essentially been reduced down to once resistor, Rab, now we can reattach the intial 120V voltage source and find the desired current, i, is simply,

    i = 120V \times R_{ab}
     
    Last edited: May 23, 2011
  3. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Thank you very much, jegues. It was very helpful.

    How should I include a current source between the terminals a-b? Should it point upward or downward?

    Are you sure that we can find i this way? "i" is total current in the circuit delivered by 120V source but I see you are only finding the current "i" through Rab. What am I missing? Please guide me.

    Thanks a lot.
     
  4. jegues

    Well-Known Member

    Sep 13, 2010
    735
    43
    It doesn't matter which direction you choose to inject the current in, you have to understand that we are only doing so to determine the resistance as seen by the a-b terminals.

    After we've determined Rab we can hook our 120V current source up to this one resistor, which models the previous resistor network we had before, thus pulling the same current, I, from the voltage source.


    Is that clear?
     
  5. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Many, many thanks, jegues. It is clear now, I think.:)

    Regards
    PG
     
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