Notron Theorem

Discussion in 'Homework Help' started by Apolonio de Tyana, Jun 13, 2004.

  1. Apolonio de Tyana

    Thread Starter New Member

    Jun 12, 2004
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  2. Mjollnir

    Member

    Apr 22, 2004
    27
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    left side

    VOC = voltage across the 4kohm resistor in parallel (which is in series in the 4kohm resistor next to the source) = 4 / (4+4) * 10 = 5V

    total current delivered by 10V source when short circuited = 10 / (4K + 4K||3K) = 1.75mA

    ISC = 4/(3+4) * 1.75mA = 1mA

    so RTH = 5kohm, ur norton equivalent for the lefthand side is a 1ma current source with a 5 kohm resistor in parallel..

    right hand side

    do the exact same thing, but with 2 sources u can use either superposition or node voltage analysis...
     
  3. Apolonio de Tyana

    Thread Starter New Member

    Jun 12, 2004
    3
    0
    Please, would you try to explane me again? Why you have used "4 / (4+4) * 10"?
     
  4. Mjollnir

    Member

    Apr 22, 2004
    27
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    when it's open circuit, no current flows tru the 3k resistor, hence the voltage across A and B is the same as the voltage across the 4K resistor in parallel..

    but the 4k resistors are in series with voltage source when open circuit, so just use voltage division...
     
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