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I came across an exercise in Linear Electrical Circuits that says: Find the Thevenin equivalent in the terminals A,B as shown in the picture below.

enter image description here

Regarding the $$R_{Th} = 6||4 = \frac{6*4}{6+4}=\frac{24}{10}$$

But regarding the \$V_{Th}\$ I do not know how to proceed.To knowledge (self studying) in order to do calculate the \$V_{Th}\$ I have to apply the Node Method. But the current flows downwards in the 6 ohm resistor and that confuses me.Any help of can I solve it ?

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  • \$\begingroup\$ What is the current-dependent voltage across the 6 Ohm resistor? It's 6*i1, correct? You know the dependent source is 2*i1. So what's left is 4*i1 for the 4 Ohm resistor. Not so? So how many ohms can be used to replace the dependent source? Would that perhaps be 2 Ohms? (Conventional current has the more positive end where the arrow is directed towards/into the resistor.) \$\endgroup\$ Commented May 14, 2023 at 10:26
  • \$\begingroup\$ @periblepsis now I am confused more.Why the voltage source is dependent from current? \$\endgroup\$ Commented May 14, 2023 at 10:34
  • \$\begingroup\$ Doesn't the picture say "2 i1 volt"? there? This means "2 Ohm * i1" is the voltage across it. (I1 of course is the current through the 6 Ohm resistor.) \$\endgroup\$ Commented May 14, 2023 at 10:43
  • \$\begingroup\$ @ Aaaa so the output voltage of the source is directly proportional to the current i_1 flowing through it, and therefore varies with changes in i_1. That is the reason why it makes the voltage source dependent?.But from there I lost you in the previous comment. \$\endgroup\$ Commented May 14, 2023 at 10:46
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    \$\begingroup\$ No. The I1 is the current in the 6 Ohm resistor. You don't know what that might be. But suppose you applied a 6 volt battery across A (+ side) and B (- side.) Then there would be 1 Amp through the 6 Ohm resistor, right? The dependent voltage source would then be required to present 2*1 A = 2 volts across itself. But there are 6 volts between A and B. This means there must be 4 volts across the 4 Ohm resistor. Otherwise things would not add up right. But it's not clear to me if you are wanting to use nodal to achieve the same result, or not. Or some other method. \$\endgroup\$ Commented May 14, 2023 at 10:51

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You can use super-nodes, if you like, or just regular nodal without the super-nodes. Either way, here's the schematic with some annotations that will be useful:

schematic

simulate this circuit – Schematic created using CircuitLab

I'll start with an explanation of my comment (which uses super-nodes.)

You know that \$V_{_\text{C}}=V_{_\text{A}}-2\:\Omega\cdot I_1=V_{_\text{A}}-2\:\Omega\cdot \frac{V_{_\text{A}}}{R_1}=V_{_\text{A}}\cdot\left(1-\frac{2\:\Omega}{R_1}\right)\$ So treat the two nodes as one, but with different voltages on them. Consider the idea of injecting a current called \$I\$ into node A, which will take on two different values: \$0\:\text{A}\$ and \$1\:\text{A}\$. (You could just as well also consider applying \$0\:\text{V}\$ and \$1\:\text{V}\$, but that wouldn't be using the nodal approach.) KCL using super-nodes follows:

$$\begin{align*}\frac{V_{_\text{A}}}{R_1}+\frac{V_{_\text{C}}}{R_2}=\frac{V_{_\text{A}}}{R_1}+\frac{V_{_\text{A}}}{R_2}\cdot\left(1-\frac{2}{R_1}\right)&=I \\\\ V_{_\text{A}}=3\:\Omega\cdot I \end{align*}$$

So the net resistance of the entire system looks like \$R_{_\text{TH}}=\frac{3\cdot 1\:\text{A}-3\cdot 0\:\text{A}}{ 1\:\text{A}- 0\:\text{A}}=3\:\Omega\$. Since you know that \$\frac{R_1\cdot\left(R_2+R_x\right)}{R_1+R_2+R_x}=3\:\Omega\$ you can solve: \$R_x=\frac{3\:\Omega\cdot\left(R_1+R_2\right)-R_1\cdot R_2}{R_1-3\:\Omega}=2\:\Omega\$. And that is what the current-dependent voltage source looks like, if that helps to see things.

The other approach is to just take things without supernodes. Here, it's three equations (two are KCL, one is an obvious relationship) and three unknowns:

$$\begin{align*} \frac{V_{_\text{A}}}{R_1}+I_2&=I \\\\ \frac{V_{_\text{C}}}{R_2}&=I_2 \\\\ V_{_\text{C}}&=V_{_\text{A}}-2\:\Omega\cdot I_2 \end{align*}$$

Solving those and looking only at the solution for \$V_{_\text{A}}\$ you will find that \$V_{_\text{A}}=3\:\Omega\cdot I\$. Same answer. Just done differently.

And there are many other ways to go.

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