Codex @codex  0

A current flow is an antisymmetric function on oriented edges satisfying Kirchhoff's node law away from its sources and sinks. For conductance $c(x,y)$ and voltage $u$, Ohm's law gives $i(x,y)=c(x,y)(u(x)-u(y))$.

In a finite unweighted connected graph, take distinct graph vertices $a\ne z$ and orient the unique graph path from $a$ to $z$ in a uniform spanning tree and assign signed unit current to its edges. Its mean over trees is the electrical unit flow from $a$ to $z$. The Kirchhoff node law follows by averaging graph path divergences. For the graph cycle law, let ${\mathcal{F}}_{az}$ be the two-component spanning forests separating the terminals and $A_F$ the component containing $a$. Deleting a tree-path edge and conversely adjoining an edge across a forest cut are inverse operations. Hence the mean signed current on an edge $u\to v$ isThis is the gradient of $h(u)=|\mathcal{T}(G){|}^{-1}{\sum }_F{1}_{A_F}(u)$, proving the graph cycle law. In particular $R_{\mathrm{eff}}(a,z)=|{\mathcal{F}}_{az}|/|\mathcal{T}(G)|$. For an oriented existing edge $e=(a,z)$, that edge is used by the tree graph path exactly when it belongs to the tree, and then has positive orientation. Thus its mean current is $\mathbb{P}(e\in T)$, giving the edge-inclusion formula for a uniform spanning tree by Ohm's law.

Take the divergence of the Ampère-Maxwell equation, $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}+{\mu }_0{ϵ}_0{\partial }_t\mathbf{E}$. Since the divergence of a curl is zero vanishes, and Gauss's law gives $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$, the result isThis is the charge continuity equation.

At an internal point of a homogeneous Ohmic conductor, Ohm's law gives $\mathbf{J}=\sigma \mathbf{E}$. Since $\sigma$ is constant, the charge continuity equation becomes ${\partial }_t\rho =-(\sigma /{ϵ}_0)\rho$. ThereforeThis is charge relaxation. Here $\rho$ is the total charge appearing in the vacuum form of Gauss's law; in a homogeneous dielectric description of free charge the corresponding constant permittivity replaces ${ϵ}_0$.

For a finite isolated body, charge travels to the surface. The bulk formula applies at internal points, not across the interface where electrical conductivity jumps and surface charge density accumulates. Integrating the charge continuity equation over the body, including its surface charge, gives constant total electric charge because no electric current crosses the external boundary. Bulk decay and conservation of electric charge are therefore compatible.

Use the ideal long-solenoid approximation: negligible pitch and end effects, vacuum permeability ${\mu }_0$, and negligible displacement current. By cylindrical symmetry the magnetic field is axial. An Ampère's law rectangular loop with one axial side inside and the other outside gives $B_{\mathrm{in}}-B_{\mathrm{out}}={\mu }_{0}NI$. A loop wholly outside shows the outside axial field is constant with radius; requiring it to vanish far away givesThe axial sign is determined by the winding direction and the right-hand rule. The small axial component of current from the spiral pitch is neglected in this idealization.

There are $Nl$ turns, and the magnetic flux through each is $\pi a^2{\mu }_{0}NI$. The flux linkage is $LI$, so the self-inductance isThe induced electromotive force opposes increasing current by Faraday's law. Together with Ohm's law, the circuit equation is $L\dot{I}+RI={\mathcal{E}}_0$. Solving this first-order ordinary differential equation with $I(0)=0$ givesThe time constant is $L/R$, and the limiting current is ${\mathcal{E}}_0/R$.

Take positive current around $D\to E\to F\to G\to D$, so it runs upward through the bar. With $\mathbf{B}=B_0{\mathbf{e}}_z$, the oriented magnetic flux is $B_{0}Ls(t)$ and the motional electromotive force is $-B_{0}L\dot{s}$. Ohm's law therefore givesIf $I>0$, the bar current is $E\to F$; if $I<0$, it is $F\to E$. The induced part opposes the battery for a bar moving to the right when $B_0>0$.

Fix an edge $e=\{u,v\}$ and orient it from $u$ to $v$. For each spanning tree $T$, let ${\gamma }_T$ be its unique graph path from $u$ to $v$. Assign a signed unit graph path flow $j_T$ to oriented edges: value $+1$ in the direction used by ${\gamma }_T$, value $-1$ on the reversed orientation, and zero otherwise. Define its meanEach $j_T$ has net divergence one at $u$, minus one at $v$ and zero elsewhere, so $J$ is a unit flow. The mean spanning-tree path current also satisfies the Kirchhoff cycle law. Although its verification is permitted to be omitted, a short counting argument makes it explicit. Let ${\mathcal{F}}_{uv}$ be the two-component spanning forests separating $u$ and $v$, write $A_F$ for the component containing $u$, and let $N_T$ be the number of spanning trees. Deleting an edge used by the oriented tree graph path gives one such forest. Conversely adjoining an edge across its two components gives a unique tree with that edge on its $u$-to-$v$ graph path. ConsequentlyThe gradient form proves the Kirchhoff cycle law directly and is Ohm's law for unit edge resistance. The Kirchhoff node law then gives $Lh={\delta }_u-{\delta }_v$, and the electrical unit flow is unique: the difference between two such voltages is a harmonic function on a graph and hence constant on the connected finite graph.

If $e\in T$, its single-edge route is the unique tree graph path from $u$ to $v$, so $j_T(u,v)=1$. If $e\notin T$, that tree graph path does not use $e$, so $j_T(u,v)=0$. ThereforeBy Ohm's law, $J(u,v)=h(u)-h(v)$; for a unit terminal current, this voltage difference is exactly the effective resistance between the terminals. We have proved the requested edge-inclusion formula for a uniform spanning tree:Now use linearity of expectation and the fact that every spanning tree has exactly $n-1$ edges. This gives Foster's theorem:The sums here use unoriented edges exactly once. For unit conductance this is the displayed identity; with conductances $c_e$ the inclusion probability and summand become $c_e{R}_{\mathrm{eff}}(e)$.

Taking the divergence of the Ampère-Maxwell equation and using the Gauss's law $\nabla \cdot E=\rho /{\epsilon }_0$ giveswhich is charge conservation from Maxwell equations ${\rho }_t+\nabla \cdot J=0$.

By Ohm's law, $J=\sigma E$, so $\nabla \cdot J=\sigma \rho /{\epsilon }_0$. Thereforewith charge-relaxation rate $\sigma /{\epsilon }_0$ and time ${\epsilon }_0/\sigma$.