Solving Potential Differences on a Network

From Edge Currents to Vertex Potentials

In a resistive network, the unknowns are often the electric potentials at the vertices. The current along an edge depends on the potential difference between its endpoints and on that edge’s conductance. A matrix formulation gathers all of these edge relations and vertex balance equations into one linear system. Choosing a reference potential then removes the freedom to shift every potential by the same amount.

Consider a finite undirected graph with \(n\) vertices and \(m\) edges. Give each edge an arbitrary orientation: one endpoint is its tail and the other is its head. The orientation is bookkeeping; it does not turn the physical edge into a one-way connection. Let every edge \(e\) have positive conductance \(c_e\). If its resistance is \(R_e>0\), then \(c_e=1/R_e\). Let \(v_i\) be the potential at vertex \(i\), and let \(s_i\) be the external current injected into the network at that vertex. Positive \(s_i\) means current is supplied to the network; negative \(s_i\) means current is withdrawn.

Write the vertex potentials as the column vector \(\mathbf{v}\in\mathbb{R}^n\), and the oriented edge currents as \(\mathbf{i}\in\mathbb{R}^m\). For an edge oriented from \(i\) to \(j\), its potential drop in the oriented direction is \(v_i-v_j\). The corresponding current, positive in that direction, is \(c_e(v_i-v_j)\). In matrix form, all edge drops are \(\mathbf{B}^{\top}\mathbf{v}\), so Ohm’s law gives \(\mathbf{i}=\mathbf{C}\mathbf{B}^{\top}\mathbf{v}\).

With our incidence convention, the \(i\)th entry of \(\mathbf{B}\mathbf{i}\) is the current leaving vertex \(i\) along its outgoing edges minus the current entering it along its incoming edges. At a vertex, the current leaving through network edges must balance the external current injected there. This is the vertex current-balance equation \(\mathbf{B}\mathbf{i}=\mathbf{s}\), where \(\mathbf{s}\) is the column of external injections.

Proof. Ohm’s law gives \(\mathbf{i}=\mathbf{C}\mathbf{B}^{\top}\mathbf{v}\). Substituting this expression into the balance equation \(\mathbf{B}\mathbf{i}=\mathbf{s}\) yields \(\mathbf{B}\mathbf{C}\mathbf{B}^{\top}\mathbf{v}=\mathbf{s}\), which is exactly \(\mathbf{L}\mathbf{v}=\mathbf{s}\). Each column of \(\mathbf{B}\) has one \(1\) and one \(-1\), so the entries in each column sum to zero; therefore \(\mathbf{1}^{\top}\mathbf{B}=\mathbf{0}^{\top}\), where \(\mathbf{1}\) is the column whose entries are all \(1\). Consequently, \(\sum_i s_i=\mathbf{1}^{\top}\mathbf{s}=\mathbf{1}^{\top}\mathbf{B}\mathbf{i}=0\). \(\square\)

The zero-sum condition says that total current supplied must equal total current withdrawn. Also, adding the same constant to every vertex potential leaves every edge potential difference unchanged. In matrix terms, \(\mathbf{B}^{\top}\mathbf{1}=\mathbf{0}\), so \(\mathbf{L}\mathbf{1}=\mathbf{0}\). A specified reference potential fixes this otherwise irrelevant common shift.

Grounding a Reference Vertex

Choose a reference vertex \(r\) and prescribe its potential \(V_{\mathrm{ref}}\). The reduced Laplacian \(\mathbf{L}_{gg}\) is obtained by deleting row \(r\) and column \(r\) from \(\mathbf{L}\); the subscript \(g\) denotes the remaining, unknown vertices. If \(\mathbf{L}_{gr}\) is column \(r\) with its \(r\)th entry removed, the equations at those vertices are

It is often simplest to solve for potentials relative to the reference. Define \(u_i=v_i-V_{\mathrm{ref}}\). Since subtracting a constant from every potential changes no edge difference, the relative potentials satisfy the same Laplacian equations, with \(u_r=0\). Thus the grounded system is \(\mathbf{L}_{gg}\mathbf{u}_g=\mathbf{s}_g\), and the absolute potentials are recovered by adding \(V_{\mathrm{ref}}\).

Proof. Extend any column \(\mathbf{x}_g\) on the non-reference vertices to a column \(\widetilde{\mathbf{x}}\in\mathbb{R}^n\) by setting \(\widetilde{x}_r=0\). From \(\mathbf{L}=\mathbf{B}\mathbf{C}\mathbf{B}^{\top}\),

Every term in the sum is nonnegative. If \(\mathbf{L}_{gg}\mathbf{x}_g=\mathbf{0}\), then the left side is zero, so the sum is zero. Since every \(c_e>0\), each edge must have \(\widetilde{x}_i=\widetilde{x}_j\). Connectivity means there is a path from every vertex to \(r\); equality along every edge of such a path gives \(\widetilde{x}_i=\widetilde{x}_r=0\) at every vertex. Thus \(\mathbf{x}_g=\mathbf{0}\), and the grounded matrix has trivial null space. It is square, so row reduction has a pivot in every column and hence in every row; its RREF is the identity. Therefore the system has exactly one solution for every right-hand side \(\mathbf{s}_g\). \(\square\)

This argument also gives a useful energy identity: the quadratic expression on the left is a sum of conductance-weighted squared potential differences. It explains why a connected network with positive conductances has no ambiguity once one reference potential is fixed.

Worked Network Calculations

Common Pitfall: Ignoring Conductance Weights

A Laplacian uses conductances, not just the unweighted pattern of which vertices are connected. In the weighted triangle example, replacing the edge conductance \(2\) by \(1\) would incorrectly give the reduced matrix \(\begin{bmatrix}2&-1\\-1&2\end{bmatrix}\). Solving that incorrect system with right-hand side \((3,0)^{\top}\) gives \(v_1=2\) and \(v_2=1\). But in the actual network these potentials produce current \(2(2-1)=2\) on edge \(1\to2\) and current \(1(2-0)=2\) on edge \(1\to3\). The total current leaving vertex \(1\) is \(2+2=4\), not the specified \(3\). The correction is to include every conductance in \(\mathbf{C}\): with the correct reduced matrix, \(v_1=\tfrac95\) and \(v_2=\tfrac65\), and the checked currents sum to \(3\).

Another common error is to mix sign conventions. With the incidence convention used here, positive entries mark tails, so \(\mathbf{B}\mathbf{i}\) is net current leaving each vertex. If instead one chooses negative entries at tails, both the edge-current and balance conventions must be adjusted consistently. State the convention first, use it in every edge equation, and verify the final vertex outflows against the injection vector.

In practice, the procedure is concise: orient the edges, construct \(\mathbf{B}\) and \(\mathbf{C}\), form \(\mathbf{L}=\mathbf{B}\mathbf{C}\mathbf{B}^{\top}\), prescribe a reference potential, and solve the reduced equations. Check the answer by calculating each oriented edge current and summing the net outflow at each vertex. The orientation can be chosen for convenience: reversing an edge reverses its incidence column and the corresponding current sign, while leaving the physical potential differences and vertex equations unchanged.