We were asked to form the node-voltage equations of this circuit.

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But how can I form the equation that specifies the constraint imposed by the controlled voltage source on the nodes it is connected to?

All the equations must be in the form of v1, v2, and v3, and other unknown variables. All must be simplest form

The constraint that I solved so far was this:

$$ V_3 - V_1 = 2V_y $$ Where $$ V_y = V_2 - V_3 $$ $$ V_3 - V_1 = 2V_2 - 2V_3 $$ $$ 3V_3 - 2V_2 - V_1 = 0 $$

Is it like this? Or do I need to substitute this in the nodal equation?

Also, what is the equation of \$I_x\$ ?

  • \$\begingroup\$ You know that the current in the 2 Ohm resistor must be 2-ix. So the voltage across it must be 4-2*ix, The current in the 3 Ohm resistor must be (2-ix)-3*ix or 2-4*ix. So the voltage across it must be Vy=6-12*ix. Adding these two drops you find 10-14*ix. But this must equal -2*Vy, or -12+24*ix. Knowing 10-14*ix=-12+24*ix, you should be able to easily find ix. Of course, since asked for nodal you've still got that to do. But at least you can test your results, now. \$\endgroup\$ Jun 28, 2023 at 21:53

2 Answers 2


Well, we are trying to analyze the following circuit:


simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \begin{alignat*}{1} \text{I}_0&=\text{I}_1+\text{I}_4\\ \\ \text{I}_1&=\text{k}\cdot\text{I}_4+\text{I}_2\\ \\ \text{I}_3&=\text{I}_2+\text{I}_4\\ \\ \text{I}_0&=\text{k}\cdot\text{I}_4+\text{I}_3 \end{alignat*} \end{cases}\tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \begin{alignat*}{1} \text{I}_1&=\frac{\displaystyle\text{V}_1-\text{V}_2}{\displaystyle\text{R}_1}\\ \\ \text{I}_2&=\frac{\displaystyle\text{V}_2-\text{V}_3}{\displaystyle\text{R}_2}\\ \\ \text{I}_3&=\frac{\displaystyle\text{V}_3}{\displaystyle\text{R}_3} \end{alignat*} \end{cases}\tag2 $$

We also know that \$\displaystyle\text{V}_3-\text{V}_1=\text{n}\cdot\left(\text{V}_2-\text{V}_3\right)\$.

Using \$(2)\$ we can rewrite \$(1)\$ as follows:

$$ \begin{cases} \begin{alignat*}{1} \text{I}_0&=\frac{\displaystyle\text{V}_1-\text{V}_2}{\displaystyle\text{R}_1}+\text{I}_4\\ \\ \frac{\displaystyle\text{V}_1-\text{V}_2}{\displaystyle\text{R}_1}&=\text{k}\cdot\text{I}_4+\frac{\displaystyle\text{V}_2-\text{V}_3}{\displaystyle\text{R}_2}\\ \\ \frac{\displaystyle\text{V}_3}{\displaystyle\text{R}_3}&=\frac{\displaystyle\text{V}_2-\text{V}_3}{\displaystyle\text{R}_2}+\text{I}_4\\ \\ \text{I}_0&=\text{k}\cdot\text{I}_4+\frac{\displaystyle\text{V}_3}{\displaystyle\text{R}_3} \end{alignat*} \end{cases}\tag3 $$


You have four nodes (three nodes labeled + ground), therefore you have to write three independent equations—one for each labelled node:

$$V_1 = V_3 - 2v_y$$ $$\frac{V_1 - V_2}{2} = 3 i_x + \frac{v_y}{3}$$ $$V_3 = V_2 - v_y$$

Now you find some way to substitute \$v_y\$ and \$i_x\$, since they are dependent variables. In this case use KCL:

$$i_x = 2 - \frac{V_1 - V_2}{2}$$ $$\frac{v_y}{3} + i_x = \frac{v_3}{4} $$

Now solve it.


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