[Circuits] Finding the Thevenin and Norton Equivalents #3

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  • Thread starter ainster31
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Homework Statement



vtG9YnW.png


Homework Equations





The Attempt at a Solution



Attached to this post. I had to take the absolute value of Rth. What did I do wrong to get a negative value for Rth?
 

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  • #2
gneill
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The Attempt at a Solution



Attached to this post. I had to take the absolute value of Rth. What did I do wrong to get a negative value for Rth?

No doubt it has to do with the direction of the current that you're "injecting" into the circuit versus the polarity of the resulting voltage across the terminals. The resistance of a component (or circuit for that matter) is defined as the current into the port divided by the resulting potential across that port. The polarity of the voltage "measurement" is such that the terminal where the current is injected is taken to be positive:

attachment.php?attachmentid=66261&stc=1&d=1391463145.gif
 

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  • #3
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So my mistake was that the 1A current source was pointing down when it should've been pointing up?
 
  • #4
gneill
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So my mistake was that the 1A current source was pointing down when it should've been pointing up?

Not a mistake per se, not if you take it into account when you assign the polarity of the voltage. You are looking for the potential at terminal a with respect to terminal b, so automatically that implies that the a terminal is the where you'd put the + lead of your meter (if you were measuring it). You are driving the current into the b terminal, so it becomes the "+" terminal for the resistance determination. That's contrary to the desired arrangement, and will result in a reversed sign for the voltage Vab.
 

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