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I'm solving this high-school level problem, where I have to find the current through each resistor.

schematic

simulate this circuit – Schematic created using CircuitLab

There aren't any resistors in parallel or series, which could be combined, so I tried using Kirchoff's law for junctions and loops. The resulting system of linear equations doesn't have any solution according to Mathematica. \$I_0\$ is the current through the source, \$I_{15}\$ is an auxiliary current through the empty wire in the middle of the circuit. My resulting system of linear equations:

System of equations

My questions are:

  • Is my approach to this problem right? Maybe I chose wrong loops. Is there any rule for which loops I should choose?
  • There are 16 unknown currents, I have 10 junction equations. Does it mean I have to create at least 6 loops, to find all the unknowns?
  • If my approach is wrong, what is the first step or direction I should use to solve this problem?

EDIT: There are a few mistakes in the equations:

  • equation 3 is \$I_3=I_5+I_7\$
  • equation 15 and 16 don't include \$I_{15}\$ (there is no voltage change over the wire without any resistor)
  • With these changes, the system can be solved by satisfying \$\mathbf{A}^\intercal \mathbf{A}\vec{x}=\mathbf{A}^\intercal\vec{b}\$. However, the accepted answer shows a much better approach to this problem (the results are the same).
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  • \$\begingroup\$ Current isn't "on" a resistor. \$\endgroup\$ – Olin Lathrop Dec 29 '17 at 14:35
  • \$\begingroup\$ @OlinLathrop sorry, I'm not native \$\endgroup\$ – Sorashi Dec 29 '17 at 14:36
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    \$\begingroup\$ For starters I3=I5+I7, not I5+I9. \$\endgroup\$ – Finbarr Dec 29 '17 at 14:43
  • \$\begingroup\$ @Finbarr thank you so much, this solves it. Do you want to post it as an answer? \$\endgroup\$ – Sorashi Dec 29 '17 at 14:46
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    \$\begingroup\$ Now that is some highschool. \$\endgroup\$ – StainlessSteelRat Dec 29 '17 at 19:48
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Extending Olin's answer for those that want to see how much of a difference the art makes.

schematic

simulate this circuit – Schematic created using CircuitLab

Once you see it that way it's pretty easy to tell, or should be, what voltage those nodes in the middle are at. After that it's simple Ohm's Law stuff.

You should also be able to see that the nodes I marked as \$VT\$ have the same voltage, as do those I marked with \$VB\$.

Since there is no voltage across \$R6\$ and \$R9\$,

\$I6 = I9 = 0A\$.

Because of that, \$R6\$ and \$R9\$ can be considered as shorts, so \$R4\$ is effectively in parallel with \$R2\$ (\$R24 = 50R\$). Which means \$VT\$ is \$50/150\$ or \$1/3\$ of \$12.5V\$ above the mid-point, that is,

\$VT = 12.5 + 12.5/3 = 16.667V\$

Similarly \$R8\$ is effectively in parallel with \$R10\$, making

\$VB = 2/3 * 12.5V = 8.333V\$.

So,

\$I1 = I3 = I12 = I14 = 8.333/100 = 83.33mA\$ \$I2 = I4 = I5 = I7 = I8 = I13 = I11 = I10 = I1/2 = 41.667mA\$

\$Io = I1 + I2 = 166.67mA\$

The puzzle is actually delightful in it's symmetry. Of course this only falls out like this because all the resistor values are identical. Had they not been, your original method, or some other analysis method, would be required.

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    \$\begingroup\$ I think you've made one slip. You say "Since there is no voltage across R6 or R9, I6=I9=0A. Because of that, R6 and R9 can be considered as shorts." No, they can be considered as open (and so removed from the circuit entirely). I haven't worked this through, but it probably makes little difference to the rest of your answer. \$\endgroup\$ – Dreamer Dec 30 '17 at 1:04
  • \$\begingroup\$ What does node voltage mean? Is it the voltage between the node and the grounded node? \$\endgroup\$ – Sorashi Dec 30 '17 at 11:13
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    \$\begingroup\$ @Dreamer actually, no. Although they can be considered both shorts or open - because there's no voltage difference and no current, it really doesn't matter if you envision them as one or the other - it won't change neither the equations or the working logic a bit. Still, what Trevor says is true - if two points have the very same voltage regardless of operating conditions, considering them shorted is IMVHO much more sensible than considering them open - since for two open points, there no guarantee of 0A and deltaV = 0, while a short essentially guarantees it. \$\endgroup\$ – vaxquis Dec 30 '17 at 12:29
  • \$\begingroup\$ @Dreamer as vaxquis mentioned then can be considered as either open or short, but I did not mention that in it's entirety. Since the voltage is the same at the ends of R4 and R2, it changes nothing if you short out or remove R6. \$\endgroup\$ – Trevor_G Dec 30 '17 at 12:47
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The first step in solving these kinds of problems is to redraw the schematic in a more logical layout. A main component to these problems is the deliberately obfuscated circuit. Often the problem becomes easier to solve, and certainly easier to see, when its schematic is drawn logically.

Put the power supply at left. Draw the + power as a line across the top, and the - power as a line across the bottom of the schematic. Now draw any obvious + to - resistors or strings of resistors vertically. Higher voltages go higher on the page than lower voltages. Show any resistor connected directly to + power vertically, going down from the + horizontal line. Similarly any resistor connected directly to - goes up vertically from the - line. Try to visually simplify the remaining resistors as much as possible.

We can get into your actual problem after you've posted the de-obfuscated schematic as described above.

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    \$\begingroup\$ thanks for these tips for drawing circuits, we don't learn that at school \$\endgroup\$ – Sorashi Dec 29 '17 at 14:48
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You've made a mistake with your equations.

You have shown

\$I_3=I_5+I_9\$

but it should read

\$I_3=I_5+I_7\$

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  • \$\begingroup\$ This makes my system of linear equations solvable: oi68.tinypic.com/f3tez4.jpg \$\endgroup\$ – Sorashi Dec 29 '17 at 14:57
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    \$\begingroup\$ I suggest you try following the instructions from @OlinLathrop anyway. You will see that this diagram has a very neat symmetry to it when drawn better, and having all the resistors identical makes it easy to solve without relying on clever tools. \$\endgroup\$ – Finbarr Dec 29 '17 at 15:06
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The key principle of this problem is symmetry.

[The other key principle I have demonstrated is not being lazy about a problem that looks easy! I previously got sloppy and tried to make it TOO simple. I believe I've corrected the errors.]

I would further say that the purpose of this problem is precisely to make you think clearly about the circuit, rather than try to solve it mathematically. I'm sad to hear that you didn't learn techniques for drawing clear circuit diagrams in school, because that's maybe the most useful practical skill in solving this problem. (When I saw your question, and decided to try to solve it for myself before reading the answers, redrawing it clearly was the first thing I did.)

Others have already given some useful information about the solution, but I'd like to try to give you my intuition for how to approach it.

When I first saw the diagram, I was immediately suspicious that it was probably more symmetrical than it looked. You described it as a 'high school' problem; sometimes school exercises will involve doing gruntwork, but more often they will involve clever tricks you have to spot. And though the drawing obscures this, it's pretty easy to see that each terminal of the power supply is connected to exactly two resistors, each of which is connected to exactly two more resistors; and all the resistors in the circuit have the same value.

Symmetry is a critical principle to keep in mind, across all kinds of problems in all kinds of classes, and beyond that into the working world. You may never have to solve the current flow through a resistor network once you graduate, but you can amaze your coworkers all your life by making complex problems trivial using hidden symmetry. It often lets you prove that dramatic simplifications of a problem are possible with very little work.

In this case, as you can see in Trevor's beautiful drawing, the resistor network in this problem is extremely symmetrical. Trevor uses this to prove that the voltages in the middle nodes must be 12.5 V -- can you see why? Because, whatever the total resistance comes out to, between those nodes and V+/V-, it is obvious that it's the SAME above and below. It's then easy to see that no current can flow in resistors r6 and r9, since as Trevor has shown, the voltages across them are 0 by symmetry.

Now we can continue using the left-right and top-bottom symmetry of the whole diagram to see that the voltages must also be the same at all the mirror points on the left and the right, and the currents must be the same through the paired resistors r1-r3-r12-r14, r2-r7-r8-r13, and r4-r5-r10-r11. That alone reduces this to a fairly simple set of equations to solve.

But we can go a little further. Given two points with the same voltage, it won't change the current flows to connect them; and given a wire through which no current flows, it won't change anything to disconnect it. So we can make the diagram even more symmetric in a couple of ways, the easiest of which is this: Since we know no current flows left-right or right-left across the center (by symmetry), we can disconnect r4-r6-r10 from r5-r9-r11. This reduces the problem to a pair of parallel circuits, so you can solve one and then apply the result to both (and within each one, there are parallel strings of identical resistors where you can do the same thing again.)

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  • \$\begingroup\$ I don't think this is correct, There is no potential across R6 and R9, so no current flow, but there is potential across R4, R5, R10, and R11 \$\endgroup\$ – Mr.Mindor Dec 29 '17 at 19:13
  • \$\begingroup\$ Hmmm, argh, I think you're right. I will think about whether my answer can be salvaged or not. Thanks. \$\endgroup\$ – Glenn Willen Dec 29 '17 at 19:16

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