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I have been doing this sum with nodal analysis. As nodal analysis says when a voltage source is between to make a super node. But I cannot do KVL for the super node. Please help me with this question.

(30-3v0/60)+3 = (v0/30)+(v1/60)

I applied super node.

  • \$\begingroup\$ Should I apply super node for 3v0 and 30ohm resistor \$\endgroup\$ Nov 22, 2023 at 6:10
  • \$\begingroup\$ Hi there! You should add the source to the picture. Thank you. \$\endgroup\$
    – MiNiMe
    Nov 22, 2023 at 7:43
  • \$\begingroup\$ @MiNiMe 30V on the left side is source \$\endgroup\$ Nov 22, 2023 at 9:18
  • \$\begingroup\$ @nithyamanish I think MiNiMe meant "From what book did that schematic picture come from?" Also, are you able to first consider converting the right-side Norton source into a Thevenin source before nodal analysis? Or do you want to do the analysis as it is shown and without modification? \$\endgroup\$ Nov 22, 2023 at 9:34
  • \$\begingroup\$ @periblepsis Okay, source is fundamentals of electric circuits 6th sadiku 6th edition. No, I want the analysis without modification \$\endgroup\$ Nov 22, 2023 at 10:02

1 Answer 1


Call the bottom node ground. Then the node at the intersection of those three resistors is \$v_o\$. Call the node with the \$3\:\text{A}\$ current source entering it as \$v_x\$. Then in super-node form, you may write it as:

$$\begin{align*} \frac{v_o-0\:\text{V}}{R_1}+\frac{v_o-V_1}{R_2}+\frac{v_o-\left(v_x+3\cdot v_o\right)}{R_3}&=0\:\text{A} \\\\ \frac{\left(v_x+3\cdot v_o\right)-v_o}{R_3}+\frac{v_x-0\:\text{V}}{R_4}&=I_1 \end{align*}$$

But you could just as well have called the (+) node of the dependent source as \$v_y\$ and then, recognizing that \$v_y=v_x+3\cdot v_o\$, may have instead written:

$$\begin{align*} \frac{v_o-0\:\text{V}}{R_1}+\frac{v_o-V_1}{R_2}+\frac{v_o-v_y}{R_3}&=0\:\text{A} \\\\ \frac{v_y-v_o}{R_3}+\frac{v_x-0\:\text{V}}{R_4}&=I_1 \end{align*}$$

Exact same thing.

Or you could swap the series resistor and the dependent source, so that there would be a dependent source of \$v_o-3\cdot v_o=-2\cdot v_o\$ for the super-node. Then you'd write:

$$\begin{align*} \frac{v_o-0\:\text{V}}{R_1}+\frac{v_o-V_1}{R_2}+\frac{\left(-2\cdot v_o\right)-v_x}{R_3}&=0\:\text{A} \\\\ \frac{v_x-\left(-2\cdot v_o\right)}{R_3}+\frac{v_x-0\:\text{V}}{R_4}&=I_1 \end{align*}$$

Again, same thing.

Just different ways to see the problem.

Note: I've used \$R_1=30\:\Omega\$, \$R_2=60\:\Omega\$, \$R_3=30\:\Omega\$, and \$R_4=60\:\Omega\$, with \$R_1\$ being the left-most resistor and \$R_4\$ being the right-most resistor. \$I_1=3\:\text{A}\$ is your current source.

  • \$\begingroup\$ Thanks for your answer. I could get the answer but do you mean I don't need to use the supernode here? \$\endgroup\$ Nov 22, 2023 at 11:05
  • \$\begingroup\$ @nithyamanish All the methods above show the use of a super-node. I could have avoided the use of super-node methods entirely. I would need to create a new variable for the current of the dependent source (and its series resistor), though. But then the nodal equations would not use super-nodes, at all. It would be just a basic nodal analysis but with more equations. So you can either choose to use super-nodes, or not. Both work fine. I just kept the super-node idea because I thought that is what you wanted to know. \$\endgroup\$ Nov 22, 2023 at 11:26
  • \$\begingroup\$ Thank you :) for your answer \$\endgroup\$ Nov 22, 2023 at 13:44

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