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Norton's Theorem and Thevenin's theorem

  1. Dec 11, 2014 #1

    Zondrina

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    Homework Helper

    1. The problem statement, all variables and given/known data

    For some reason I don't use this theorem since Thevenin's theorem is more common. I'm asked to find the Norton equivalent source between the terminals ##A## and ##B## for the following:

    Screen Shot 2014-12-11 at 4.23.33 PM.png

    2. Relevant equations


    3. The attempt at a solution

    The first step is to short ##R_2##, and to find the short circuit current across that wiring.

    Screen Shot 2014-12-11 at 4.23.33 PM.png

    I believe KCL at the node above the ##3 mA## would be useful:

    ##\sum I = 0 \Rightarrow 3mA + I_{sc} - I_R = 0 \Rightarrow I_{sc} = I_R - 3mA = \frac{V_{Node} + 20V}{4k} - 3mA##

    Now ##V_{Node}## is the voltage across the ##3 mA##. Applying KVL around the left loop using the current source as a constraint:

    ##\sum V = 0 \Rightarrow 20V + V_R + V_{Node} = 0 \Rightarrow V_{Node} = -(20V + V_R) = -(20V + (3mA)(4k)) = - 32V##

    This implies that ##I_{sc} = \frac{-32V + 20V}{4k} - 3mA = - 6mA##.

    Is this reasonable?
     
  2. jcsd
  3. Dec 11, 2014 #2

    gneill

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    Staff: Mentor

    You went astray when you wrote equations with ##V_{Node}##. The short circuit forces ##V_{Node}## to be zero volts. If you think about the Superposition Principle you can take advantage of that, since it makes calculating the current contribution of the 20 V source quite simple.

    If I may suggest another approach? Start by converting the 20V source and R1 to their Norton equivalent. Then take advantage of parallel connections to simplify...
     
  4. Dec 11, 2014 #3

    Zondrina

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    Homework Helper

    Ah source transforms, those are awfully useful and I hardly ever use them.

    First, ##V = -IR \Rightarrow I = \frac{-20V}{4k} = -5mA## so ##-5mA## points up or ##5mA## points down. Also, the ##5 mA## is in parallel with the ##4k## now.

    So the combined current in parallel is ##-5mA + 3mA = -2mA##. This is the only current in this new circuit (comprised of the current source, 4k resistor and shorted 6k resistor). Hence ##I_{sc} = -2mA##.

    Now going back to the original circuit, short circuiting the voltage and open circuiting the current source, we find:

    ##R_{th} = (\frac{1}{4k} + \frac{1}{6k})^{-1} = \frac{12k}{5}##

    Hence the norton equivalent would be ##-2mA## pointing up in parallel with ##R_{th}##.
     
  5. Dec 11, 2014 #4

    gneill

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    Staff: Mentor

    There you go! Much less brain sweat required :)
     
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