How does a current source work in Norton's Therom

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Hi.i know that a ideal current source has infinite resistance and it supply's constant current to aload connected across it. The current source doesn't have the infinite resistance in series to it but in pparalle. All is well until I think of this Norton's Therom. When we transform a voltage source to current source we connect the same thevinin resistance in parallel to current source (now it's name is Norton's resistance). So for an ideal voltage source the internal residence is zero and all the voltage across it is dropped across the load only. But when we transform it we connect zero resistance across a current source (as per Norton) no current passes through the load as the current source is shorted?? How is it possible that Norton's Therom it's true this case? Correct me if I am wrong.http://www.zen22142.zen.co.uk/Theory/images/norton_th.png
 

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  • #2
NascentOxygen
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Hi dushyanth. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

Don't overlook the need to set VTh = R × IN

▻ What is the open circuit voltage across AB in each of the above?
▻ What is the short circuit current when you place a short across AB?

Your answers in each part should be identical.
http://thumbnails112.imagebam.com/37333/0363e9373324851.jpg [Broken]
 
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Hi dushyanth. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

Don't overlook the need to set VTh = R × IN

▻ What is the open circuit voltage across AB in each of the above?
▻ What is the short circuit current when you place a short across AB?

Your answers in each part should be identical.
http://thumbnails112.imagebam.com/37333/0363e9373324851.jpg [Broken]
Can you please elaborate???
 
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NascentOxygen
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NascentOxygen
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So for an ideal voltage source the internal residence is zero and all the voltage across it is dropped across the load only. But when we transform it we connect zero resistance across a current source (as per Norton) no current passes through the load as the current source is shorted?? How is it possible that Norton's Therom it's true this case? Correct me if I am wrong.
I seem to have missed the significance of what you were asking in your OP.

Rather than look at the extreme case, it is more enlightening to consider what happens as the internal resistance gets progressively smaller. Take, for example, a source having V = 10 volts and r = 10 ohms, and determine the Norton equivalent of this voltage source. Now, recalculate with smaller values of r, say, 1 ohm, 0.1 ohms, 0.0001 ohms, and 0.00000001 ohms.

Comment on the trend you observe as r approaches 0.
 

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