Doubt about equivalent inductance

Discussion in 'Homework Help' started by jfsas, Sep 1, 2012.

  1. jfsas

    Thread Starter New Member

    Sep 1, 2012
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    Hi, in this part of the circuit i am not sure on how to reduce the inductors to a single one, i need this to solve de ODE for t>=0, i have tried to reduce it using parallel and series asumptions but i have the problem with the 4mH inductor on the center could someone give me a hand? thanks

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    Last edited: Sep 1, 2012
  2. #12

    Expert

    Nov 30, 2010
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  3. jfsas

    Thread Starter New Member

    Sep 1, 2012
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    of course i forgot about delta-star transform, thanks :)
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    And if you find yourself in a similar situation and don't recall that transform (or whatever trick makes it easy), you can always fall back on the fundamentals. You know that inductances add like resistors, so analyze the equivalent resistance (as seen between the top and bottom rails) of the circuit that has the inductors replaced with the same numerically valued resistors.

    Symmetry then dictates that no current flows in the bridging component, and so you have two strings, each with a 4Ω and 0.5Ω component in series, connected in parallel. Hence, by inspection, you have (4.5/2)Ω or 2.25Ω, which translates to an equivalent inductance of 2.25mH. Is that what you got after applying the delta-wye transform?
     
  5. jfsas

    Thread Starter New Member

    Sep 1, 2012
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    Yes i obtain 2.25 mH. That is an interesting approach, i really appreciate it.
     
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