Calculating Thevenin equivalent circuit with series current source

Question

Wikipedia writes this about calculating the output current \$I_{AB}\$ when finding the Thevenin equivalent:

2a. Replace voltage sources with short circuits, and current sources with open circuits.

2b. Calculate the resistance between terminals A and B. This is \$R_{Th}\$.

But! What if the circuit contains the following (independent) current source and resistance:

+--R---o
|
I
|
+------o

I think they could be replaced with a voltage source with the potential difference \$V=I*R\$.

If we calculate the output current before replacing and after replacing the above, we would get different answers! What am I misunderstanding?

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Answer 1

That could only be replaced by an infinite voltage source.

Current sources operate on the basis that they will output their current no matter the load. When you leave it open circuit your open circuit voltage would be infinity. IE, this circuit cannot exist.

The standard replacement will be a current source with a parallel resistance to a voltage source with series resistance. As shown in this wikipedia image.

http://en.wikipedia.org/wiki/Th%C3%A9venin's_theorem">

Answer 2

There is nothing wrong with your answer - you just have to work through it and then see the implications.

You come up with something similar to what Kortuk said except that you don't need to change what is asked IF that was what was asked.

Following your rules you get Rth = infinite.
The resistance of a perfect current source is infinite.
It's voltage will rise to infinity if necessary to cause its define current to flow. Open circuiting a perfect current source is on the dangerous side compared to sticking a bent nail into a near-perfect voltage source such as the AC mains.

So your instructions work.

ie Thevenin's laws should work for anything linear, but a current source is a bit of a strange beast and needs an apparently strange answer to make sense of it.