I have the circuit below drawn on a piece of paper. I am trying to figure out the unknown nodal voltages x1-x9. Actually, from the circuit we already know x8=9 and x9=0.

Here is my thought process: first draw out the direction of current through each resistor. Then use KCL (Kirchoff's Current Law) to write out current equations. Then Ohm's law to convert unknown currents to unknown nodal voltages.


i1 moves from x1->x4
i3 moves from x4->x7
i6 moves from x5->x4

Then my first equation becomes

i1+i6-i3=0 for node x4.  

By similar analysis, I get 4 more equations:

-i1-i8+i2=0 for node x2. 
-i6+i8-i7+i5=0 for node x5
i3-i4+i7=0 for node x8
-i2-i5+i4=0 for node x6

By using Ohm's Law, I can convert all the currents i1-i8 to nodal voltages x1-x9 using the equations:


and so on.

Then I get a linear system of equations with unknown nodal voltages.

My linear system is:

A=  -2     2     0   -10    10     0    -5     0     0
     2    -9     5     0     2     0     0     0     0
     0     2     0    10   -22     2     0     0     0
     0     0     0     0    10     5     5     0     0
     0     5    -5     0     2    -7     0     0     0



The last step is to do A\b in MATLAB, which gives me the wrong answer for nodal voltages. I know it's the wrong answer because when I create this circuit in a circuit simulator online I'm getting different nodal voltage values.

Would appreciate if somebody could point out what is wrong with my thought process here. All I'm doing is using KCL, then transforming unknown currents into unknown nodal voltages. Then solving the linear system.


simulate this circuit – Schematic created using CircuitLab


3 Answers 3


Assume your \$V_1\$ voltage source is called \$V_X\$ and that the current in \$V_X\$ is \$I_X\$, with the arrow pointing from \$x_9\$ to its tip at \$x_8\$. Then:

$$\begin{align*} \frac{V_1}{R_1}+\frac{V_1}{R_3}&=\frac{V_2}{R_1}+\frac{V_4}{R_3}\tag{x1}\\\\ \frac{V_2}{R_1}+\frac{V_2}{R_5}+\frac{V_2}{R_9}&=\frac{V_1}{R_1}+\frac{V_5}{R_5}+\frac{V_3}{R_9}\tag{x2}\\\\ \frac{V_3}{R_9}+\frac{V_3}{R_{10}}&=\frac{V_2}{R_9}+\frac{V_6}{R_{10}}\tag{x3}\\\\ \frac{V_4}{R_3}+\frac{V_4}{R_4}+\frac{V_4}{R_6}&=\frac{V_1}{R_3}+\frac{V_5}{R_4}+\frac{V_7}{R_6}\tag{x4}\\\\ \frac{V_5}{R_4}+\frac{V_5}{R_5}+\frac{V_5}{R_8}+\frac{V_5}{R_{12}}&=\frac{V_4}{R_4}+\frac{V_2}{R_5}+\frac{V_8}{R_8}+\frac{V_6}{R_{12}}\tag{x5}\\\\ \frac{V_6}{R_{10}}+\frac{V_6}{R_{11}}+\frac{V_6}{R_{12}}&=\frac{V_3}{R_{10}}+\frac{V_9}{R_{11}}+\frac{V_5}{R_{12}}\tag{x6}\\\\ \frac{V_7}{R_6}+\frac{V_7}{R_7}&=\frac{V_4}{R_6}+\frac{V_8}{R_7}\tag{x7}\\\\ \frac{V_8}{R_7}+\frac{V_8}{R_8}&=I_X+\frac{V_7}{R_7}+\frac{V_5}{R_8}\tag{x8}\\\\ V_9&=0\:\text{V}\tag{x9}\\\\ V_8&=V_9+V_X\tag{10}\\\\ \end{align*}$$

The above is 10 equations and 10 unknowns. You should be able to populate the matrices from the above. To do that, just move terms around so that your variables are all on the left side with associated factors and there are only constants on the right side. (The factors on the left will then be elements in your left-side matrix.)

There are some obvious simplifications that would allow you to reduce the unknowns and the equations. But I'm sticking with your approach, labeling all those nodes as you did regardless that you could have simplified things a bit.

Since someone has seen fit to mark this answer wrong (-1), I'll take the time to prove that it produces accurate and reliable results. Too bad they were too cowardly to identify themselves and their problem. But I can at least show them their error in thinking about what I wrote.

In Sage, this is how you might formulate the above:

var('v1 v2 v3 v4 v5 v6 v7 v8 v9 ix vx')
var('r1 r2 r3 r4 r5 r6 r7 r8 r9 r10 r11 r12')

eq1= Eq( v1 *                (1/r1 + 1/r3), v2/r1 + v4/r3 )
eq2= Eq( v2 *         (1/r1 + 1/r5 + 1/r9), v1/r1 + v5/r5 + v3/r9 )
eq3= Eq( v3 *               (1/r9 + 1/r10), v2/r9 + v9/r10 )
eq4= Eq( v4 *         (1/r3 + 1/r4 + 1/r6), v1/r3 + v5/r4 + v7/r6 )
eq5= Eq( v5 * (1/r4 + 1/r5 + 1/r8 + 1/r12), v4/r4 + v2/r5 + v8/r8 + v6/r12 )
eq6= Eq( v6 *      (1/r10 + 1/r11 + 1/r12), v3/r10 + v9/r11 + v5/r12 )
eq7= Eq( v7 *                (1/r6 + 1/r7), v4/r6 + v8/r7 )
eq8= Eq( v8 *                (1/r7 + 1/r8), ix + v7/r7 + v5/r8 )
eq9= Eq( v9                               , 0 )
eq10=Eq( v8                               , v9 + vx )


[(i, ans[i].subs({r1:.5,r3:.1,r4:.1,r5:.5,r6:.2,r7:.2,r8:.1,r9:.2,r10:.1,r11:.2,r12:.5,vx:9})) for i in ans]
[(v9, 0),
 (v4, 7.67171239356670),
 (ix, 16.4545884578997),
 (v6, 3.29091769157994),
 (v3, 4.05723746452223),
 (v8, 9),
 (v7, 8.33585619678335),
 (v2, 5.58987701040681),
 (v1, 7.32473982970672),
 (v5, 7.68661305581835)]

The above uses a simple loop to walk through the list and emit the values, but in no particular order.

If you want them sorted, you could instead write:

 [(eval(key), ans2[eval(key)].subs({r1:.5,r3:.1,r4:.1,r5:.5,r6:.2,r7:.2,r8:.1,r9:.2,r10:.1,r11:.2,r12:.5,vx:9})) for key in sorted([str(key) for key in ans2])]
[(ix, 16.4545884578997),
 (v1, 7.32473982970672),
 (v2, 5.58987701040681),
 (v3, 4.05723746452223),
 (v4, 7.67171239356670),
 (v5, 7.68661305581835),
 (v6, 3.29091769157994),
 (v7, 8.33585619678335),
 (v8, 9),
 (v9, 0)]

You can also hand-request them like this:











You will note that these values (printed using any method) will be precisely matched by any Spice simulation you composite up. As I wrote earlier, this approach does work. Though of course there are simplified approaches that would get the more important results from which you could easily get the rest.

My personal opinion for folks caring about symbolic matrix solutions to electronic problems is to become very familiar with SageMath and Python. These are incredibly powerful and free tools available to anyone and well worth a small investment into their learning curves.

  • \$\begingroup\$ what software do you use to build these calculation codes? \$\endgroup\$ Sep 26, 2018 at 3:39
  • \$\begingroup\$ @IronMaiden SageMath. \$\endgroup\$
    – jonk
    Sep 26, 2018 at 4:23
  • \$\begingroup\$ Thank you very much. I´ve been programming in C since I used to develop some microcontroller projects, but now i´m getting into more math codes, and started studying Python. And as you are the beloved jonk from EE.SE, I´m really going deep into Python in order to learn curves better along with properly graph some electromagnetism equations. Take this as a praise, I really like your answers and have learned a lot from them. \$\endgroup\$ Sep 27, 2018 at 2:26
  • \$\begingroup\$ @IronMaiden The technique I've shown here shows some C code that can EASILY be adapted to help when you are trying to work out the distribution of charges on ANY shape of antenna or conductive object. It's a very powerful tool (checkerboard technique.) I'd recommend playing with it enough until you get the ideas solid. Because it is very useful when equations are very, very hard. Just use partial diff eq and solve. And thanks for the kind words! \$\endgroup\$
    – jonk
    Sep 27, 2018 at 2:43
  • \$\begingroup\$ Thank you once again for sharing this great knowledge jonk. I´m certainly going to play with it to get a good understanding. I´ve once heard of this checkerboard technique, maybe used in image resolution or something like that. I´ll stick to it for sure! \$\endgroup\$ Sep 27, 2018 at 2:55

Node $7$,$1$ and $3$ may be forgotten and you only have to rearrange resistances in series. Done that you may also forget node $8$ since you already know its voltage which is $9V$

Now you are left with 4 unknown node voltages: $V_4$, $V_5$, $V_2$ and $V_6$. Write $4$ node equations for them. You can think of current leaving the nodes in all situations, thus you don´t need to bother with current directions since the overall result will always be the same.

Also, are you taking the inverse matrix of A ($A^{-1}$) to calculate the resulting matrix? Remember that you´ll have $AX=B$, and that can be rearranged to $X=A^{-1}B$


Treat it as a Thevenin problem: remove \$R_1\$ and \$R_3\$, and call this the load; then the circuit becomes simple Ohm's law: \$ (R_5+R_9+R_{10})//R_{12}\$, and \$ (R_4+R_6+R_{7})//R_{8}\$, easily gives the voltage at x5, and the rest follow.


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