[Math] Are resistor-battery networks always uniquely solvable

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Note: if you know the basics of circuits, feel free to skip the brief background; the question is at the bottom, starting below the triple horizontal rule.

Most people with some physics background are familiar with basic circuit analysis. Let's restrict this question just to electrical networks consisting solely of resistors and batteries, and let's set our problem to be calculating the steady-state currents through each wire; take, for example the following network, known as a Wheatstone bridge:

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The simplest problems – with just one battery – can often be attacked using two rules for resistors that every engineer knows. These often allow us to transform a network of resistors into one virtual "equivalent resistor", at which point we can solve for currents using Ohm's law.

The first rule is that resistors in series add, so that the equivalent resistance of the below circuit is 10 ohms

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The second rule resistors in parallel add harmonically, so that the equivalent resistance of the below circuit satisfies $1/R_{eq} = 1/8 + 1/12$, and thus $R_{eq} = 24/5$ ohms. If there were more than two resistors, the sum of all of their reciprocals would appear on the right-hand side.

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Note that these two rules are not sufficient to calculate steady-state currents for every possible resistor-battery network; consider for example the Wheatstone bridge mentioned above.

In these trickier cases, solving for all the currents requires Kirchoff's rules, which state

  1. The sum of the currents entering any node in an electrical network is equal to the sum of currents exiting that node.
  2. The total change in voltage around any loop in an electrical network is zero.

Coupled with Ohm's law, which says that the voltage drop across an ideal resistor of resistance $R$ subject to a current of $I$ amperes is just $\Delta V = IR$, these rules allow us to set up a bunch of linear equations for steady-state currents. There are a bunch of examples of these sorts of problems in the first few chapters of any circuits textbook, or online.

Now let's pass to the underlying graph of an ideal resistor-battery network. Let's go ahead and assume that the electrical network is biconnected, so that it's actually possible to flow a current between any two points. Let's suppose that our graph has $V$ nodes and $E$ edges. Then our system of linear equations has $E$ unknowns, namely the current through each of the wires.

From Kirchoff's first rule, we get $V$ equations of the form $I_j + \cdots = I_k + \cdots$, of which (under the assumption of connectedness) $V – 1$ are independent. (Note that summing all of the equations together gives zero.)

If the network is planar, then Kirchoff's second rule gives $F$ independent equations, where $F$ is the number of faces (including the infinite face), and each equation corresponds to looking at the voltages around the boundary of that face. Each equation looks something like $\pm V_j \pm \cdots + \pm I_kR_k = 0$, where the $V_i$ are the voltages of the batteries and the $R_i$ are the resistances of the resistors.

Again we only end up with $F-1$ independent equations; this is because an appropriately signed linear combination of the equations adds up to the equation corresponding to the infinite face.

Now Euler's formula gives that we have $V + F – 2 = E$ constraints; I doubt it's terribly difficult to show that these equations are all independent. Then we in fact can uniquely solve this resistor network.

On the other hand, what happens if we have a non-planar ideal resistor-battery network, say connected according to a $K_{3,3}$ or a $K_5$? Do we always get enough constraints? Based on what I wrote above, it looks like the number of constraints minus the number of unknowns is going to be $\chi – 2 = -2g$, where $\chi$ is the minimal Euler characteristic of any surface on which the graph can be embedded and then $g$ is the genus of the graph.

In particular, this suggests that resistor-battery networks whose underlying graphs have nonzero genus aren't exactly solvable. This doesn't mesh with physical intuition; how do we resolve these two arguments?

Best Answer

I am assuming that all the resistors are positive. If you have negative resistors, all bets are off!

Having said that, you can show more. Even if the resistors are non-linear, as long as their v-i curve is in the first and third quadrant, under fairly weak conditions such as $i(v)$ has strictly positive slope for all $v$, there is a unique solution.

Easiest way to prove it is to use the following result:

For each resistive element, let $v_k$ be the voltage across it. Define $W_k$ as $$ W_k(v_k)=\int_0^{v_k} i_k(v) d v $$ Now consider the optimization problem $$ \text{minimize} \sum_k W_k(v_k) $$ subject to Kirchoff's voltage law.

Using Lagrange multiplier, it is possible to see that the solution, if it exists has to solve the circuit.

Now use convexity argument to show that there is a unique solution.

Note: For the linear case, $W_k$ is the power consumed in the resistor. The above theorem is often stated as: Solution is where the power loss is minimized subject to one of the Kirchoff's laws. If the loss is written in terms of element voltages, then we need the Krichoff's voltage law. If the loss is written in terms of element currents, then we need Kirchoff's current law.

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