Norton's theorem

Discussion in 'General Electronics Chat' started by KCHARROIS, Mar 27, 2014.

  1. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
    292
    1
    So there's a question in a book asking to solve a circuit using Norton's theorem and I've only been able to solve it using superposition. (Circuit attached)

    So to solve using superposition here's what I did, I opened the current source and calculated 12V across R1 and 8V across R2, simple series circuit. Next I shorted the voltage source and calculated -6V across both resistors, simple parallel circuit. Then adding and subtracting the values I get 18V across R1 and 2V across R2 and this makes sense.

    But how can I calculate using Norton's theorem, It may not be the best method but I'd like to learn.

    Thanks

    Edit: Forgot to mention that the 1mA source is actually an unknown resistor but has 1mA running through it.
     
  2. daviddeakin

    Active Member

    Aug 6, 2009
    207
    27
    You have a 20V source in series with R1. Convert that source into its Norton equivalent instead:
    20/15k=0.1333mA

    You can now re-draw the circuit as two Norton sources in parallel:
    I4 in parallel with R2
    and a 0.1333mA source in parallel with R1.

    The total resistance becomes R1||R2 = 6k
    And the total combined current is -1mA + 1.333mA= 0.333mA

    Ohm does the rest.
     
    KCHARROIS likes this.
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