Use nodal analysis to find Vo in the circuit of Fig. 3.72

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.

• Should I apply super node for 3v0 and 30ohm resistor Nov 22, 2023 at 6:10
• Hi there! You should add the source to the picture. Thank you. Nov 22, 2023 at 7:43
• @MiNiMe 30V on the left side is source Nov 22, 2023 at 9:18
• @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? Nov 22, 2023 at 9:34
• @periblepsis Okay, source is fundamentals of electric circuits 6th sadiku 6th edition. No, I want the analysis without modification Nov 22, 2023 at 10:02

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.

• Thanks for your answer. I could get the answer but do you mean I don't need to use the supernode here? Nov 22, 2023 at 11:05
• @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. Nov 22, 2023 at 11:26
• Thank you :) for your answer Nov 22, 2023 at 13:44