Home hobbyist here. I've been doing a lot of theoretical resistor, voltage divider, Ohm's law, and Kirchhoff law type problems just for practice, and ran across this older video on YouTube, which presented this problem:

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I didn't have a problem solving it, but it got me to thinking: Is it possible to come up with a resistor network type problem like this which isn't solvable?

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    \$\begingroup\$ You can always build it, apply a voltage across it and measure the current. \$\endgroup\$ Commented Mar 14, 2023 at 0:45
  • \$\begingroup\$ Yes, I understand it can be measured, I'm just wondering if there's one which could be mathematically unsolvable. \$\endgroup\$
    – LarryBud
    Commented Mar 14, 2023 at 0:57
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    \$\begingroup\$ All finite resistor networks are solvable. If you want a rigorous proof, you will have to look elsewhere. \$\endgroup\$
    – Mattman944
    Commented Mar 14, 2023 at 1:25
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    \$\begingroup\$ Computers do math. Spice is building up a system of equations and then solving them, so a spice simulation is a mathematical solution to the problem. I think in the case of a linear network made of ideal components like this, it is even doing the same math you would, or at least something pretty close. \$\endgroup\$ Commented Mar 14, 2023 at 1:39
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    \$\begingroup\$ ITT: xkcd.com/356 \$\endgroup\$ Commented Mar 14, 2023 at 12:12

3 Answers 3


If you have two terminals to a network of resistors, the resistance between the two terminals will always be a single value.

This is a special case of Thevenin's Theorem, without the complicating voltage and current sources. The wikipedia article linked has a proof of this theorem.

If the network is infinite and general, then there will be a computation impossibility for a general circuit, not surprising as the network cannot be represented in a finite way. In order to define it, the connectivity and resistor values will usually employ some symmetries, for instance ..., which allow a computation to be made.

For a finite network, application of series, parallel and delta-Y transformations allow any network to be reduced down to a single equivalent resistance in O(n) time.

Although any finite network can be reduced to an equivalent resistance, the network cannot necessarily be fully solved in the sense of determining all branch currents when a voltage is applied. If the network contains some zero resistance elements, then the branch currents may be ambiguous. This is what happens in SPICE if you have loops of zero impedance, you get an uninvertible matrix.

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    \$\begingroup\$ @psmears I think I mean O(n) time. However, an ohmmeter can do it in O(1)! \$\endgroup\$
    – Neil_UK
    Commented Mar 14, 2023 at 10:53
  • \$\begingroup\$ Haha, fair point :) \$\endgroup\$
    – psmears
    Commented Mar 14, 2023 at 10:54

As long as you only allow a nonnegative resistance, then you can solve it. If you also include negative resistances then you can have a problem, for example: What is equivalent resistance of a 10 Ω and a -10 Ω resistor connected in parallel?

While negative resistors dont exist, you can build active device that behave as one. Also, some passive device can exhibit negative resistance region, for example tunnel diode.

  • \$\begingroup\$ For 1V, you will have 0.1A in one resistor and 0.1A out the other resistor, so 0A total, so why can't I just say the two resistors in parallel are an open circuit (equivalent R is therefore infinity)? \$\endgroup\$
    – bobuhito
    Commented Mar 14, 2023 at 14:08
  • \$\begingroup\$ It isnt so easy, problem is that R isnt only infinity, but have some extra properties, for example can you actually change voltage over it? With ∞ Ω there is zero current from outside, so it should keep voltage over them at 0 V no matter outside voltage and this sound like a paradox to me. \$\endgroup\$
    – Rokta
    Commented Mar 14, 2023 at 15:46
  • \$\begingroup\$ @Rotka why should it keep voltage at 0? In your example with 1 V, there is 1 V over both resistors, no current from outside, and 0.1 A of current looping through the two resistors. I don't see a paradox. A negative resistor generates power, so it is not a paradox that current flows only through the two resistors and not through the 1 V generator. \$\endgroup\$
    – wimi
    Commented Mar 14, 2023 at 17:06
  • \$\begingroup\$ I think it is like the resonance in a parallel LC circuit , the inductive reactance is jXl and the capacitive reactance is -jXc so at resonance when Xl = Xc no current flows because the currents are 180 degrees out of phase. \$\endgroup\$ Commented Mar 14, 2023 at 19:49
  • \$\begingroup\$ @Rokta The combination of +10Ω, -10Ω and a voltage source all in parallel is fine. No current flows in the source, and all current flows in the resistor loop. Those resistors can have any voltage across them. It sure looks weird, but the maths works out fine in an application of KCL, KVL and Ohm's law, nodal analysis yields solvable equations. It's only conversion from a parallel pair to a single resistance that's problematic. \$\endgroup\$ Commented Mar 16, 2023 at 2:53

All finite resistor networks are solvable because you can write a finite number of linear equations for such a network and solve the system of equations.

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    \$\begingroup\$ This is an incomplete proof because not all finite systems of linear equations are solvable. (For example x+y=3, x+2y=5, x+3y=8) \$\endgroup\$ Commented Mar 14, 2023 at 1:56
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    \$\begingroup\$ @user253751 your example equations cannot have been derived from the same system, because they aren't consistent with each other. They effectively say y=2 and y=3. If we apply KCL, KVL and Ohm's law consistently in a single system, none of the equations will contradict any other. I agree, though that for certain values of resistance, current or voltage, that one or more things may remain undefined, but not algebraically inconsistent. \$\endgroup\$ Commented Mar 14, 2023 at 8:13
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    \$\begingroup\$ @SimonFitch: What you say is true, but the point is that the answer just says "you can write a finite number of linear equations and solve" - but that's incomplete, because you actually need (and get, in the case of applying circuit laws to a circuit) stronger conditions in order to guarantee consistency. \$\endgroup\$
    – psmears
    Commented Mar 14, 2023 at 16:37
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    \$\begingroup\$ @appliedSciences the answer as written is circular. "it is solvable because you can write a system of equations and solve it". Ok... why? Why is it guaranteed to be solvable? \$\endgroup\$
    – mbrig
    Commented Mar 14, 2023 at 17:07
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    \$\begingroup\$ @appliedSciences: No, it's still circular, even if it's true! You need to give a justification for such a network not existing, or at least (if that's hard to do) acknowledge that fact. Otherwise your answer amounts to "No such network exists, because no such network exists", which isn't really an answer :) \$\endgroup\$
    – psmears
    Commented Mar 14, 2023 at 23:03

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