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: ???

  • 1
    \$\begingroup\$ 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. \$\endgroup\$ – jonk Dec 5 '18 at 22:06
  • \$\begingroup\$ 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. \$\endgroup\$ – Thomas Dec 5 '18 at 22:10
  • \$\begingroup\$ 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? \$\endgroup\$ – jonk Dec 5 '18 at 22:14

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.

$$\begin{align*} \frac{V_{N_1}}{R_1}+\frac{V_{N_1}}{R_3}&=\frac{0\:\text{V}}{R_1}+\frac{V_{N_2}}{R_3}+I_1+I_{V_2}\\\\ \frac{V_{N_2}}{R_2}+\frac{V_{N_2}}{R_3}+I_{V_2}&=\frac{0\:\text{V}}{R_2}+\frac{V_{N_1}}{R_3}+I_2\\\\ V_{N_1} &= V_{N_2}+V_2 \end{align*}$$

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

| improve this answer | |
  • \$\begingroup\$ 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 + ... = ... \$\endgroup\$ – Thomas Dec 5 '18 at 23:11
  • \$\begingroup\$ @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. \$\endgroup\$ – jonk Dec 6 '18 at 1:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.