Main question

  • In the context of resistor networks, and particularly purely from a graph theory point of view, is there a consistent way of assigning the two electrode nodes in order to compare the conductivity or equivalently the resistance of two networks?

The intuitive choices might be:

  • Choosing the furthest (graph distance) two nodes in the network and computing the equivalent resistance between those.

  • Randomly choosing the two nodes and averaging over a certain number of realisations.

  • Fixing the distance between the measuring (electrode) nodes, therefore, measuring resistance as a function of graph distance.

Either of these definitions have an arbitrary aspect to them. For example, if we choose the first definition, then the two networks may have very different diameters, and therefore, we'd be comparing resistances at different distances.

For simplicity, we can assume the individual resistors (edge values) to be constant and uniformly distributed.


Generally, for this question one can assume any connected graph of $n$ nodes and $m$ edges where each edge is a resistor. Then with the generalized Kirchhoff's laws and assuming Ohm's relation holds, the circuit theory equations in matrix form can be written as follows:

The nodal voltages and currents:

$$ \psi = (\psi_1, \cdots, \psi_n)^T $$ which together with the incidence matrix $D$[*] of the graph allows to write the voltages across edges as:

$$ V= D^T \psi $$ Similarly, for the nodal currents ($J$) and currents across edges (I): $$ J = DI. $$

With Ohm's law, we have a linear relation between the current and voltage, e.g. current at a given edge $x$ is given by:

$$ I_x = G_k V_x $$ where $G$ is an $m$ by $m$ diagonal matrix of conductances (inverse of each value corresponding to resistances) across each edge, which we can assume to be uniformly assigned to $1,$ so $G$ is for now an identity matrix.

Then given two electrode nodes $i$ and $j,$ in order to measure the resistance between them, we can insert a current of $1$ and $-1$ at each node respectively with other nodal currents set to zero, and we solve the linear system of form $Ax=b$ given by $DGD^T \psi = J,$ which yields $\psi.$ Having the nodal voltages and currents, we can compute the resistance with $R_{i,j}=\frac{\psi_i - \psi_j}{1}.$

[*]: Each column of $D$ corresponds to an edge of the graph and contains exactly one $1$ at the tail-node of the edge and one $-1$ at the head-node of the edge.

Re-iterated questions:

  1. Given two graphs $G_1$ and $G_2$ (each assumed connected), one might e.g. be a transformed version of the other, how should the electrode nodes be assigned in order to consistently compare the conductivity of the two graphs? Are any of my suggestions at the start valid? In other words, is there any way of making statements about the conductivity of two graphs independently of the choice of electrode nodes?

  2. In general, in the context of conductivity of random graphs and random resistor networks, is the dependence of conductivity/resistance on the choice of electrodes a well studied problem?


In order to better clarify the specific context of interest and in view of the discussions so far, I write here additional details:

I am primarily interested in the case of graphs treated as resistor networks, i.e. simple connected graphs (can be random graphs), given by a vertex set and edge set, where each vertex is circuit junction and each graph edge $(i,j)$ is a resistor of $r_{ij}=1$ Ohm.

In this context, we can for instance define a metric in terms of the effective resistance between two arbitrary vertices, often also called resistance distance between the nodes. And at the level of graphs, we do not have a dimensionality problem, of 2D or 3D that we'd have in the context of uniform continuous bodies (generally simple connected spaces) for which methods such as the vdP as mentioend in the answer below exist for computing their resistivity.

Instead, focusing on graphs, we can compute the resistance tensor $R_{ij},$ by e.g. assuming a homogeneous edge-wise resistance distribution $r_{ij}=1$ Ohm for all edges $(i,j)$, and insert a current $I=1 $A between two arbitrary nodes and compute the voltage drop across all pairs of vertices yielding the effective resistance $R_{ij}$ between them (aka the resistance distance between nodes $i,j$), as summarised in the context section above. But this (i.e. the $R_{ij}$'s) is always dependent on the choice of vertices, and I am trying to learn if there exists a measure of resistivity/conductivity for the entire graph, quantifying how conductive/resistive a graph is based on its underlying connectivity structure, and thus providing a means of comparison between different graphs, and irrespective of where on the graph one would place the electrodes.

In this regard, I have e.g. found out about the Kirchhoff index of graphs, which is the sum of the effective resistances across all pairs of vertices, and the Cheegar constant of graphs, which if I understood correctly, characterizes how "bottlenecked" flow across the graph can become, but given the close analogy to random walks, and confusing terminology (e.g. Cheegar constant is also called conductance of a graph), I don't know which of these, if any, represent a valid measure of electrical conductivity of the graphs, or if other definitions of graph resistivity exist.


1 Answer 1


Best practice to determine the resistivity tensor is to use the van der Pauw method. For resistor networks this method is used for example in Nonlinearity of resistive impurity effects on van der Pauw measurements (2006). This is appropriate for a two-dimensional system. For a three-dimensional resistor network, see Electrical resistance between pairs of vertices of a conducting cube and continuum limit for a cubic resistor network (2017).

  • $\begingroup$ Many many thanks for this suggestion, I hadn't heard of the van der Pauw method yet, really interesting! In case of a 2D-embedded graph where coordinates are assigned to nodes, I can imagine one could still define nodes belonging to the perimeter of the graph, but in general, for a graph irrespective of an embedding, we simply have the node set, edge set, and unit resistors assigned to each edge, can we still apply the VDP method? I ask because most articles on the subject tackle samples of arbitrary shapes,and do not discuss the applicability of the method at the underlying graph level. $\endgroup$
    – user929304
    Oct 1, 2019 at 8:30
  • $\begingroup$ Dear Carlo, to phrase it more clearly, what I meant to ask is: Is there a measure of conductivity or resistivity for arbitrary resistor networks? i.e. not in the context of continuously connected bodies, but simply at the graph level where nodes are circuit junctions and edges $(i,j)$ correspond to resistors (e.g. $r_{ij}= 1$ Ohm). If I understood correctly, an approach such as vdP only works for geometries in 2D and simply-connected, then a sheet resistance can be defined. But how about the case of resistor networks? Is e.g. the Kirchhoff index a valid measure of conductivity? Thanks again! $\endgroup$
    – user929304
    Oct 3, 2019 at 12:42
  • $\begingroup$ it all depends on whether the network is spatially uniform; if it is, then you can use any method that will give you the conductivity tensor (van der Pauw in 2D, or a 3D generalization); from the conductivity tensor you can then find the current distribution also if it is multiply connected; if the network is not spatially uniform you will have to work with the conductance rather than the conductivity; the conductance $G_{12,34}=I_{12}/V_{34}$ is the ratio of current $I_{12}$ between nodes 1,2 and the voltage difference $V_{34}$ between two nodes 3,4; it depends on where you place the nodes. $\endgroup$ Oct 3, 2019 at 13:32
  • $\begingroup$ Thanks for the clarifications! I think now I understand better what you meant. On the other hand, I think our discussions are still partially cross-purpose. At the risk of embarrassing myself, if I understood correctly, you are still discussing the case of conductivity of samples being continuous bodies. To better explain my original question, I have written an addendum covering more details. Please let me know if anything remains vague/unclear. $\endgroup$
    – user929304
    Oct 3, 2019 at 14:52

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