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Delta-Y transformation of resistors

  1. Feb 1, 2017 #1
    1. The problem statement, all variables and given/known data
    Starting from the expression of the Delta-Y resistor transformation work out the conductance transformation equation.
    2. Relevant equations
    3. The attempt at a solution

    I will just be using one equation as others are done analogically. My Δ has ##(R_{12},R_{23},R_{13})## and my
    γ has ##(R_1,R_2,R_3)##.
    The first equation of transformation goes:
    ##R_1=\frac{R_{12}*R_{31}}{R_{12}+R_{23}+R_{13}}##
    When i use that ##R=\frac{1}{G}## i get that
    ##G_1=\frac{G_{12}+G_{23}+G_{31}}{G_{23}}## which is not what they get. They get the same equation you get for Y-Delta transformation except for R u write G. ow did they get to that? I used a logical move and it cant get that way. What am i missing?
     
  2. jcsd
  3. Feb 1, 2017 #2

    gneill

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    Staff: Mentor

    Must be an algebra issue somewhere... but we can't tell you what's missing since it's almost all missing --- we can't check what we can't see :smile:

    Can you show your work in detail?
     
  4. Feb 1, 2017 #3
    Sure, :).
    ##R_1=\frac{R_{12}*R_{31}}{R_{12}+R_{23}+R_{31}}##
    ##G_1=\frac{1/G_{12}+1/G_{23}+1/G_{31}}{1/G_{12}*1/G_{31}}##
    ##G_1=\frac{(G_{12}+G_{23}+G_{31})/(G_{12}*G_{23}*G_{31})}{1/(G_{12}*G_{31})}##
    ##G_1=\frac{G_{12}+G_{23}+G_{31}}{G_{23}}##
    See it? :)
     
  5. Feb 1, 2017 #4

    gneill

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    Staff: Mentor

    I don't see how you went from your second line to your third. You haven't involved he denominator yet, so everything must be happening in the numerator. There you should end up with a sum of products divided by a product since all the terms are different.
     
  6. Feb 1, 2017 #5
    Ohhh, i see now, i had this all wrong. Turns out algebra is the killer here. Thanks man! :D
     
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