Thevenin and Norton Equivalent Circuits problem

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Hi, I found this question from a textbook:

Suppose there are two black boxes, Box A has a 1V voltage source and 1ohm resistance (Thevein Circuit); and Box B has a 1A current source and 1ohm resistance (Norton Circuit). You only have access to the two terminals. How do you know which is which by using only a shorting wire?

When I look at the problem, I know that the Norton circuit is running with the terminals open, while the Thevenin one is not. Shorting the terminals of the Thevein circuit would make it start running, while shorting the terminals of the Norton circuit would not stop it running, but would greatly reduce the current through the 1ohm resistance. But how would I tell whether the circuit is running with just the wire? Perhaps the wire is not put between the two terminals?

I look online and find an answer about how the Norton circuit would be hotter. But if that is the answer, why would I even need a shorting wire?
 

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  • #2
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When open, the Norton is hotter. When shorted, the Thevenin is hotter. You don't really need the wire. But the wire when used, illustrates the duality of nature.

Claude
 
  • #3
Hi if your dealing with thevenin and norton circuits then i suggest you memorize the effects on wire that is short circuited (if i remember right, voltage and resistence = 0, and i = unknown), and on open circuited (if i remember right, i = 0, voltage and resistence = unknown).

This can be of use in problems with dependent voltage sources, where they are dependent on the current in some wire that got shorted when you followed the shorting rules, etc.

See: http://wolfsfiles.googlepages.com/norton.jpg
And my thread: https://www.physicsforums.com/showthread.php?t=255090

Also theirs something i used in this problem: Rth = Vth/Isc When i used it in an exam i had to mathematically and logically prove that formula to be correct in order to get the points... in short i had to fight for the points... but ya know it.,. it might be of use.
 

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