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Paradox of star - mesh transform ?

  1. Sep 18, 2012 #1
    Dear All,

    I have encountered the following situation:
    Given a node that is connected through n resistors to n ports.
    One if the resistors is a negative resistance and equals to minus of all other resistors in parallel.
    [itex]\frac{1}{R_n} =- \sum_{i=1}^{n-1} \frac{1}{R_i}[/itex]
    Using the well known star - mesh transformation, I get
    [itex]R_{k,j} = R_k R_j ( \sum_{k=1}^{n} \frac{1}{R_i} )[/itex]
    that equals to 0, due to the definition of [itex]R_n[/itex].
    [itex]R_{k,j} = 0 [/itex]

    Therefore, the equivalent mesh representation of this circuit is a shortcut of all nodes.
    It makes sense from one side, because the total circuit has zero impedance, but, from the other side, this transform says that all nodes should be of equal voltage. The last one is not true, because in the original network the nodes may have different potentials.

    What do I miss ?
    Any ideas ?
    Are the equivalence transforms applicable for negative resistances ?
     
  2. jcsd
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