# How to solve this circuit using KCL

I understand the super node aspect of this circuit, but I'm unsure of what to do as I dont see a direct way to proceed with KCL using all of the nodes. I found the equation for the supernode, but I'm unsure of the rest. simulate this circuit – Schematic created using CircuitLab

Supernode: 12=v1-v2
node 1: 1=v1 + ??? node 2: ???

• How would you approach this if the 12V supply were replaced with a resistor? Mostly, I just want to know if you can apply KCL/KVL for nodal analysis (assuming that's the approach you wanted to take), at all. Or if your situation is more remedial.
– jonk
Dec 5, 2018 at 22:06
• I would do: v1 : 1= (v1 -v2)/R + ???. The thing that I'm unsure about is what the current leaving node 1 to the left would be. Dec 5, 2018 at 22:10
• Well, I don't even bother with supernodes. Some folks are taught to use them. But I think it (1) complicates the situation and makes it look as though one needs to memorize a host of varying situations to make some magic recipe work out -- impairing your ability to think about novel situations; and, (2) it's not necessary and never was. So, why don't you assign a variable for the current in the 12V voltage supply and then perform normal nodal analysis? Have you tried considering that?
– jonk
Dec 5, 2018 at 22:14
• The idea of supernode is that you write KCL for it and not for nodes 1 and 2. You write sum of currents flowing in and out of the supernode in terms of voltages V1 and V2 and combine the equation with V1-V2=12. This works well when the circuit has only one node apart from the supernode. This is the case in your circuit. Feb 11, 2021 at 8:23

Here's a slightly redrawn schematic.

(I'm keeping it mostly as you had it, because nodal analysis isn't about understanding a circuit's design and behavior and more about just applying a systematic mathematical approach. And it's also just an academic exercise -- the schematic isn't supposed to do anything, anyway, except pose a problem to solve using a tool.) simulate this circuit – Schematic created using CircuitLab

I intentionally dropped the ground connections on the other side of your current sources. That connection isn't needed as current sources have $$\\infty\$$ impedance. I also re-labeled your two nodes to avoid confusion with $$\V_2\$$ in your schematic.

I've also introduced a new variable for the current in $$\V_2\$$. This is an unknown variable, but it will be resolved if you solve out the equations.

For each node in the schematic above, I'll list the out-flowing currents from that node on the left side of each equation and I'll list the in-flowing currents into that node on the right side of each equation. The last equation provides the final information relating node 1 and node 2, as you already know.

\begin{align*} \begin{array}{r} {\text{Node } 1:} \vphantom{\frac{V_{N_2}}{R_3}} \\\\ {\text{Node } 2:} \vphantom{\frac{V_{N_2}}{R_3}} \\\\ {\text{Assign}:} \vphantom{V_{N_2}+V_2} \end{array} && \overbrace{ \begin{array}{r} \frac{V_{N_1}}{R_1}+\frac{V_{N_1}}{R_3} \\\\ \frac{V_{N_2}}{R_2}+\frac{V_{N_2}}{R_3}+I_{V_2} \\\\ V_{N_1} \end{array} }^{\text{outflowing currents}} & \begin{array}{c} &\quad{=} \vphantom{\frac{V_{N_2}}{R_3}} \\\\ &\quad{=} \vphantom{\frac{V_{N_2}}{R_3}} \\\\ &\quad{=} \vphantom{V_{N_2}+V_2} \end{array} & \overbrace{ \begin{array}{l} \frac{0\:\text{V}}{R_1}+\frac{V_{N_2}}{R_3}+I_1+I_{V_2} \\\\ \frac{0\:\text{V}}{R_2}+\frac{V_{N_1}}{R_3}+I_2 \\\\ V_{N_2}+V_2 \end{array} }^{\text{inflowing currents}} \end{align*}

Can you solve the above for $$\V_{N_1}\$$, $$\V_{N_2}\$$, and $$\I_{V_2}\$$?

• That makes a lot of sense and is really straight forward. Thank you for the in-depth explanation. Is it just a difference in convention between the method you used and what I am acustomed to? Ex Vn1 - Vn2 / R3 + ... = ... Dec 5, 2018 at 23:11
• @Thomas I've never seen my method documented in any book. It is how I interpreted (made a mental model) of undocumented code I saw when reading Spice source code to see how it approached solving circuit problems. The Spice code was NOT written because of this insight, though. It was written that way for different reasons (having to do with how it moves from node to node to node.) But after seeing it work right, I created my own mental analogy to explain it to myself.
– jonk
Dec 6, 2018 at 1:57
• @jonk Although the system of equations is correct, its explanation is not. VN1/R3 is not an outflowing current, for example. (VN1-VN2)/R3 is the outflowing current. Feb 11, 2021 at 8:12
• @AlexVB You misunderstand the approach. What I show here is exactly how SPICE code develops its solutions. I've been attempting to teach it here. It's not found in traditional textbooks. If you want, you can simply verify by rearranging my equations. You'll find them to be equivalent to anything you develop. I find the technique much better because mistakes in sign are far less likely.
– jonk
Feb 11, 2021 at 10:11
• @jonk I perfectly understand your approach. And I said that equations are correct. But what you call outflowing currents are not them and this may confuse people. Feb 11, 2021 at 11:09