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For a random walk on a finite electrical network,

A real function $h$ on the vertices of a graph of an unweighted locally finite graph is a discrete harmonic function at $u$ when $\mathrm{deg}(u)h(u)={\sum }_{v\sim u}h(v)$, equivalently $Lh(u)=0$ for the Graph Laplacian. For positive degree this says its value is the average over its graph neighbours; at an isolated vertex of a graph the Laplacian condition is vacuous. On an electrical network the corresponding equation uses conductance weights and the Weighted graph Laplacian. Voltages away from sources and sinks, and hitting probabilities away from their prescribed boundaries, are discrete harmonic functions.

For an edge $e=\{a,b\}$ of conductance $c_e$ in a finite electrical network, its inclusion probability in the conductance-weighted uniform spanning tree is $c_e{R}_{\mathrm{eff}}(a,b)$. The same formula holds for the wired limit using the limiting wired effective resistance.

In a finite connected electrical network, run the conductance random walk from $s$ until its first visit to a distinct vertex $t$. Let $G(s,x)$ be the expected visits to $x$ before stopping, and $w_x={\sum }_y{w}_{xy}$. The expected signed number of traversals of $\{x,y\}$ is the displayed expression. Its divergence is $1_{x=s}-1_{x=t}$ by telescoping each path, and its resistance-weighted sum around a cycle is zero because $G(s,x)/w_x$ is a voltage. Uniqueness follows by summing ${\sum }_{\{x,y\}}{w}_{xy}(v_x-v_y{)}^2$: a zero-divergence voltage flow has zero energy. Thus the expected net traversals are exactly the unit source-to-sink current.

An electric potential difference compares electric potentials at two points. For a move from $a$ to $b$, define $\Delta \Phi =\Phi (b)-\Phi (a)$; the terminal convention $V_{AB}=\Phi (A)-\Phi (B)$ instead means the potential at $A$ relative to $B$, so the order must always be stated. In electrostatics, the line integral of the electric field givesHere $\Delta U$ is the change in interaction potential energy of the test electric charge $q$ with fixed sources, while $W_{\mathrm{field}}$ is work done by the electric field. For an external agent moving the charge slowly without a change in kinetic energy, its work is $\Delta U$. A positive electric charge gains kinetic energy when accelerated toward lower electric potential; a negative electric charge gains kinetic energy when accelerated toward higher electric potential. Both motions lower $q\Phi$.^[ref]

For example, fix $Q=+2\mathrm{nC}$ at the origin and compare distances $r_a=0.20\mathrm{m}$ and $r_b=0.10\mathrm{m}$. With the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and zero at infinity,Moving $q=-3\mathrm{nC}$ inward therefore givesIf only the electric force does work, conservation of energy makes the gain in kinetic energy $2.697\times 10^{-7}\mathrm{J}$. Direct integration confirms the signs: ${\int }_{r_a}^{r_b}kQdr/r^2=kQ(1/r_a-1/r_b)=-89.9\mathrm{V}$, which multiplied by the negative test electric charge gives positive work. The electric potential difference is unchanged by choosing a different electrical reference potential.

In an electrical circuit, an electric potential difference is the voltage across a component's terminals. In an electrical network, a vertex voltage is specified relative to a reference vertex; its differences across edges drive current flow. For a uniform electric field, the integral reduces to $\Delta \Phi =-\mathbf{E}\cdot ({\mathbf{r}}_b-{\mathbf{r}}_a)$. A field $\mathbf{E}=300V/m\hat{\mathbf{x}}$ across a displacement $0.04\mathrm{m}\hat{\mathbf{x}}$ therefore gives $\Delta \Phi =-12\mathrm{V}$, as in the interior approximation for a parallel-plate capacitor. The volt, measured in joules per coulomb, expresses energy transfer per unit electric charge, which makes it useful for comparing components without specifying the amount of charge transported.

The Kirchhoff node law says the sum of outward currents at a graph vertex is zero unless that graph vertex is a source or sink. In an electrical network, it is flow conservation; a unit flow has divergences $+1$ and $-1$ at its terminals.

If ${\Pi }_1,\dots ,{\Pi }_m$ are pairwise edge-disjoint cutsets separating $a$ from $z$ in an electrical network, then

Write $w_{xy}=w_{yx}>0$ for an edge conductance and $w_x={\sum }_{y\sim x}{w}_{xy}$. Take $s\ne t$, as implicit in a unit source-to-sink electrical network. An oriented current flow is antisymmetric, $i_{yx}=-i_{xy}$. Kirchhoff's first law is conservation of current: ${\sum }_y{i}_{xy}=0$ at every vertex other than the source and sink, with the divergences there equal to the prescribed total flow and its negative. Kirchhoff's second law says that the sum of oriented voltage drops around any closed cycle is zero. Ohm's law isThus the resistance of an edge is $1/w_{xy}$, and the cycle law reads ${\sum }_{j=0}^{m-1}{i}_{x_j{x}_{j+1}}/w_{x_j{x}_{j+1}}=0$ for $x_m=x_0$.

Let $\tau =\inf \{n\ge 0:X_n=t\}$. Its expectation is finite. Indeed, from each state there is a path to $t$ of length at most $|V|-1$ with positive probability; finiteness gives a common lower bound $a>0$ on reaching $t$ within that many steps. The Markov property then gives $\mathbb{P}(\tau >k(|V|-1))\le (1-a{)}^k$, proving the assertion. Define the expected occupation count before absorptionConditional expectation at every pre-stopping visit to $x$ gives ${\mathbb{E}}_s{N}_{xy}=G(s,x)p_{xy}$ for the number $N_{xy}$ of directed traversals. Hence the expected signed count isThis proves Ohm's law with voltage $v_x=G(s,x)/w_x$, grounded at $v_t=0$. Summing its voltage drops along a cycle telescopes, proving Kirchhoff's second law.

For Kirchhoff's first law, count entrances and exits on each stopped path. Their net difference at $x$ is the initial indicator minus the final indicator. Taking expectations,Thus the source emits exactly one unit, the sink receives one unit, and all other vertices conserve current.

We also establish the uniqueness required for the conclusion. The cycle law implies that edge voltage drops are differences of vertex potentials: define a potential by summing drops along a path to $t$, and cycle sums make this independent of the path. If two unit flows obey both laws, their difference has the form $j_{xy}=w_{xy}(h_x-h_y)$ with zero divergence at every vertex. ThereforeStrict positivity of the conductances and connectedness force $h$ to be constant, so $j=0$. We conclude the electrical current from stopped random-walk traversals identity

Put $A=\{a,b\}$ and let ${\tau }_x^{+}=\min \{t\ge 1:X_t=x\}$. The conductance-hitting identity for an electrical network iswhere the vertices of $D$ are wired together. It follows by taking the hitting probability of $D$ before returning to $x$ as a voltage and computing its total current out of $x$.

Split the walk into successive excursions from $x$. The first excursion that hits $A$ determines whether $a$ or $b$ is hit first. Its conditional probability of hitting $a$ is at most the probability that an arbitrary excursion hits $a$, divided by the probability that it hits $A$. Hence

A voltage is a scalar potential on the vertices of an electrical network. Away from prescribed boundary vertices it is harmonic, and its difference across an edge drives current through that edge.

For a finite undirected graph without self-loops, give each edge a symmetric nonnegative weight $w_{ij}=w_{ji}$. In the weighted graph, set $(A_w{)}_{ij}=w_{ij}$ for an edge and zero otherwise, and $(D_w{)}_{ii}={\sum }_j{w}_{ij}$ with all other entries zero. The Weighted graph Laplacian isEach edge appears once in the quadratic form. For the path graph $A-B-C$ with weights $w_{AB}=2$ and $w_{BC}=3$,The Weighted graph Laplacian is still a positive semidefinite matrix. Its kernel consists of vectors constant on each connected component of a graph after zero-weight edges are removed: an edge of weight zero imposes no constraint. In an electrical network, taking the weights to be electrical conductances makes $x^T{L}_{w}x$ the total power dissipated by Joule heating when $x$ gives the voltages at the vertices. This physical interpretation explains why stronger connections penalize disagreement more heavily.^[ref]