 Solution

Orient each edge both ways. A unit flow $j$ from $x$ to $y$ is antisymmetric, $j(u,v)=-j(v,u)$, with divergence of a flow equal to $+1$ at $x$, $-1$ at $y$ and zero elsewhere. Its energy iswhere each unoriented edge is counted once. Put $c_e=1/r_e$. The electrical unit current is $i(u,v)=c_{uv}(V(u)-V(v))$, where the potential $V$ solves the Kirchhoff node law with those source and sink divergences. Because $G$ is a connected graph, this potential is unique up to an additive constant. The effective resistance is the electric potential difference needed for unit current,Discrete summation by parts gives ${\mathcal{E}}_r(i)={\sum }_{u}V(u)\mathrm{div}i(u)=R(x,y)$.

The Thomson principle says that this current minimizes energy among all unit flows. Indeed write any other unit flow as $j=i+k$, where $\mathrm{div}k=0$. ThenConsequentlyThis also proves uniqueness of the minimizing current.

If $r_e^{\prime }\ge r_e$ for every edge, then ${\mathcal{E}}_{r^{\prime }}(j)\ge {\mathcal{E}}_r(j)$ for every admissible flow. The admissible set is unchanged, so taking its minimum proves the Rayleigh monotonicity principle:Deletion is the limiting operation $r_e\to \infty$: finite-energy flows must then carry zero current on that edge. If the terminals become disconnected, the effective resistance is infinite.