Effective resistance ($R_{\mathrm{eff}}(a,z)$)

The effective resistance between vertices $a$ and $z$ is the voltage difference required to send one unit of current from $a$ to $z$.

For the conductance-weighted random walk on a graph, set $d(x)={\sum }_v{c}_{xv}$. The voltage $h(u)=P_u(T_x<T_y)$ has boundary values one and zero. The total electric current is ${\sum }_v{c}_{xv}(1-h(v))=1/R(x,y)$. The first-step analysis therefore gives the displayed positive-return-time formula. Wiring the exterior of finite sets and taking limits gives $P_x(T_x^{+}=\infty )=1/(d(x)R(x,\infty ))$.

For a finite connected unweighted loopless graph with $n$ graph vertices,The edge-inclusion formula for a uniform spanning tree gives the left-hand side as $\mathbb{E}|T|$, and every spanning tree has $n-1$ edges. On the same finite connected graph with positive edge conductances $c_e>0$, the corresponding identity is ${\sum }_e{c}_e{R}_{\mathrm{eff}}(e)=n-1$.

The effective resistance to infinity is the limit of the effective resistance from a fixed vertex to the wired complement of increasing finite sets. It is finite exactly when the finite-energy flow criterion for transience holds.

Increasing edge resistances, including deleting edges, cannot decrease effective resistance. Decreasing resistances or identifying vertices cannot increase effective resistance. These statements also follow from the Thomson principle.

The effective conductance between two terminals is the reciprocal of their effective resistance. Under a unit voltage difference it equals the total current flowing from one terminal to the other.

Thomson principle says that effective resistance is the minimum energy among unit flows from $a$ to $z$:

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

For a random walk on a finite electrical network,

Let simple random walk travel from $a\ne z$ to its first visit to $z$, then stop on its subsequent first hit of $a$ in a finite connected unweighted loopless graph. Every directed edge $u\to v$ has expected traversal countThe Strong Markov property splits the count into the two killed legs. By killed-walk occupation voltage, their contributions are $v_{az}(u)$ and $v_{za}(u)$. Their Laplacians cancel, so the harmonic maximum principle on a finite graph makes the sum constant, with value $v_{az}(a)+v_{za}(a)=R_{\mathrm{eff}}(a,z)$. Summing over all directed edges recovers the commute time identity.