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The expressions

\$N-1=\# KCL \$

and

\$E-N+1=\# KVL \$

(where \$N\$=number of nodes; \$E\$=number of elements on the circuits) gives us the linear independent equations on a circuit.

Per example:

Here I have 8 nodes and 10 elements, so I got 7 independent KCL equations and 3 independent KVL equations, but I only want to know I,I1,I2,I3,I4,I5, so I just have 6 unknowns. So my doubts are:

  • Why I have more equations that unknowns? I only have 6 unknowns.

    What are the 7 KCL independent equations?

    How can I select the 7 KCL equations in order to don't get dependent equations?

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    \$\begingroup\$ You are persistent. I think you've been asking the same or very similar questions perhaps four times, already in the last day or so. You can identify a total of 8 nodes. Agreed. But most of them are trivial: one entrance and one exit. So you really only have two nodes: the top-left and the bottom nodes. Some people call them super-nodes. I just think of the other nodes as being degenerate. You can also identify a total of 10 elements. Agreed. But most can be chained up in series to create four, still rather simple, elements (or branches.) It's actually a lot simpler than you say. \$\endgroup\$
    – jonk
    Apr 12, 2017 at 5:22
  • \$\begingroup\$ The voltage at the upper left node (relative to the bottom node called "ground") is: \$V=\left[2- \frac{V_1}{R_2}+ \frac{V_2}{R_3+R_6}+ \frac{V_3}{R_4+R_5} \right]\cdot\big[R_2\vert \vert \left(R_3+R_6 \right)\vert\vert\left(R_4+R_5 \right)\big]\$. Do you see how I can just quickly dash that out? The left side is the sum of currents into the upper-left node. The right side has the voltage sources shorted and the current source opened and the rest taken as the impedance between the upper-left node and the bottom node. Current times resistance equals voltage. \$\endgroup\$
    – jonk
    Apr 12, 2017 at 5:28
  • \$\begingroup\$ If you want me to explain, I'd appreciate seeing more specifics in your work. Some attempts, for example, to actually develop the equations you imagine exist. It's okay to be wrong. But engage yourself in the details. That would help direct a better answer, I think. (No promises, of course.) \$\endgroup\$
    – jonk
    Apr 12, 2017 at 5:32
  • \$\begingroup\$ @jonk firstly, It was different questions based on the same circuit, because it was a good example to show my points; I had to be persistent because I had none useful answers and I keep having doubts; I know very well how to solve this problem, I'm just testing those expressions. I can easily see that I have 2 supernodes, and 3 meshes, so 4 equations will be enough to find all currents. But I have to know how to use these expressions because in harder circuits with more unknowns I'll have to choose the correct equations (linear independent); How could I show you my work? I got 3 examples more. \$\endgroup\$
    – Vitor Aguiar
    Apr 12, 2017 at 11:02
  • \$\begingroup\$ If you were this persistent with your instructors (I assume they exist), you'd have all your answers. I have to therefore assume you are involved in a bad educational system which isolates students in some unfortunate way from tutors and instructors who could help them; or from forming into knowledgeable student groups who help each other. \$\endgroup\$
    – jonk
    Apr 12, 2017 at 18:20

1 Answer 1

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I'll encourage you a little by showing how I might approach your problem and to show you the use of LaTex on this site, as well. Since you are aware of the supernode concept (a term I don't like, but live with), let me approach your problem from that perspective and see if you follow it fine. I'm going to "ground" the bottom supernode (call it \$0\:\textrm{V}\$) and label the upper-left supernode as simply \$V\$. Then it follows that:

$$\begin{align*} \frac{V+V_1}{R_2}+\frac{V-V_2}{R_3+R_6}+\frac{V-V_3}{R_4+R_5}&=2\\\\ \frac{V}{R_2}+\frac{V}{R_3+R_6}+\frac{V}{R_4+R_5}&=2-\frac{V_1}{R_2}+\frac{V_2}{R_3+R_6}+\frac{V_3}{R_4+R_5}\\\\ V\cdot\left[\frac{1}{R_2}+\frac{1}{R_3+R_6}+\frac{1}{R_4+R_5}\right]&=2-\frac{V_1}{R_2}+\frac{V_2}{R_3+R_6}+\frac{V_3}{R_4+R_5} \end{align*}$$

Solving for \$V\$ is easy, now. Just divide the right side by the left side's factor. But once inverted onto the right side, this is the same as putting those resistor groups in parallel, so the resulting equation is:

$$V=\left[2-\frac{V_1}{R_2}+\frac{V_2}{R_3+R_6}+\frac{V_3}{R_4+R_5}\right]\cdot\bigg[R_2\vert\vert\left(R_3+R_6\right)\vert\vert\left(R_4+R_5\right)\bigg]$$

That's it.

This is the same as dismantling the upper-left node and bottom node (disconnecting all "feeders" into them) and then placing an ammeter between the now-isolated ends to measure individual loop currents, then summing these various measurements into the left side of the above equation. And then returning to the original circuit, but now replacing the voltage sources with their impedance (zero) and the current sources with their impedance (infinite), and then analyzing the resistance between the two nodes in order to form the right side of the above equation. (\$R_1\$, of course, disappears because of the infinite impedance of the current source there.)

Now. I've shown you my work. Show me yours. It's more algebra involved to approach the problem with all the nodes and elements you suggested in your question. But the important results will be the same.

In any case, I see nothing from you regarding the development of your KVL and KCL equations. Sure, you might worry about over-specification (there may be reason for worry there.) But I don't see any attempts to develop any of those equations. Let's see the attempt, at least. You need to expose your way of thinking about things.

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