# Thevenin & Norton Circuit with Current source???

• Engineering
gneill
Mentor
8V is the potential across so would the In just be I=V/R = 8V/9ohm = 8/9A?
Are you sure that it's 9 V 8 V ? Remember, V1 in this circuit is not guaranteed to be the same as in the previous one where the output was open. Show your solution for V1.

[Edit: Fixed the voltage value to match the quote]

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Are you sure that it's 9 V? Remember, V1 in this circuit is not guaranteed to be the same as in the previous one where the output was open. Show your solution for V1.
Oh, right. It won't be 9V. But it should be equal to V1 (the voltage across the 3ix current source) but I am not sure how to find that now after shorting ab. The way i did it in the picture of my solution is the only way I can think of it. I didn't learn mesh/super mesh yet so I have to do it using KCL and node voltage I think. Or loop analysis. I don't know how to find the potential across the 3ix source once the circuit is shorted.

gneill
Mentor
Write a node equation for the junction of the 2 and 9 Ohm resistors. The auxiliary equation remains $i_x = \frac{8 - V_1}{2}$. Solve for $V_1$. The current through the short ab is the same as the current through the 9 Ω resistor, and you'll have the potential across it. Write a node equation for the junction of the 2 and 9 Ohm resistors. The auxiliary equation remains $i_x = \frac{8 - V_1}{2}$. Solve for $V_1$. The current through the short ab is the same as the current through the 9 Ω resistor, and you'll have the potential across it.

View attachment 105734
Ok so my equation would be (V1-8)/2 = 3ix + (V1-0)/9

and plugging the ix in I get a value for V1 of 144/17

Then I=V/R = (16/17)A

gneill
Mentor
Ok so my equation would be (V1-8)/2 = 3ix + (V1-0)/9

and plugging the ix in I get a value for V1 of 144/17

Then I=V/R = (16/17)A
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:
$$\frac{V_1 -8}{2} + \frac{V_1}{9} + 3 \left(\frac{8 - V_1}{2} \right) = 0$$

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
Mentor
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?

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

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
Mentor
Right.

• Marcin H