Thevenin & Norton Circuit with Current source?

[Marcin H]

Check your node equation. You've got terms for currents flowing out of the node on both sides of the equation. If you're going to do it that way, one side or the other should describe a currents flowing into the node.

The best way to write a node equation consistently is to write a sum that equals zero. Choose either all currents flowing into the node or all currents flowing out. Then write all the terms on one side of the equation and set it equal to zero. So in this case if we assume that all currents are flowing out of the node:
$$\frac{V_1 -8}{2} + \frac{V_1}{9} + 3 \left(\frac{8 - V_1}{2} \right) = 0$$

Did I draw my arrows at V1 incorrectly? I just look at current in = current out and then write the equation. I'm not too sure why your equation is like that, but using it I get V1= -9V and that would give me IN= -1A

 

[gneill]

Did I draw my arrows at V1 incorrectly? I just look at current in = current out and then write the equation. I'm not too sure why your equation is like that, but using it I get V1= -9V and that would give me IN= -1A

In your equation (post #29) your terms all describe currents leaving the node. That means, for example, if you moved the term on the left side over to the right it would be negated and describe a current entering the node. Then your equation would say that some current entering the node plus all the currents leaving is zero. That's not KCL.

You seem to have made a sign error in solving the equation from post #30. Can you show your work?

 

[Marcin H]

In your equation (post #29) your terms all describe currents leaving the node. That means, for example, if you moved the term on the left side over to the right it would be negated and describe a current entering the node. Then your equation would say that some current entering the node plus all the currents leaving is zero. That's not KCL.

You seem to have made a sign error in solving the equation from post #30. Can you show your work?

I just double checked it. I got + 1A now.

 

[gneill]

Good! So now you have both the open circuit voltage and the short circuit current. What then is the Thevenin resistance?

 

[Marcin H]

Good! So now you have both the open circuit voltage and the short circuit current. What then is the Thevenin resistance?

RT = Vt/Isc = 8V/1A = 8ohms

 

[gneill]

Right.