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Norton's Theorem and Thevenin's theorem

  • Thread starter Zondrina
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  • #1
Zondrina
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Homework Statement



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


Homework Equations




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?
 

Answers and Replies

  • #2
gneill
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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...
 
  • #3
Zondrina
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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}##.
 
  • #4
gneill
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There you go! Much less brain sweat required :)
 

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