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 $z$.

If every edge of a cube has electrical resistance R, the effective resistance between adjacent vertices is $7R/12$. Apply a unit voltage between the terminals. Reflection symmetries leave two interior potentials x,z and their complements. Kirchhoff's current law gives $4x-z=2$, $4z-2x=1$, so $x=9/14$, $z=4/7$. The terminal current is $[1+2(1-x)]/R=12/(7R)$, proving the formula. No symmetry identifies vertices at different potentials across the source edge.

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.

For a finite connected undirected electrical network with positive edge electrical conductances $c_e$, the Thomson principle stateswhere $a\ne z$, each undirected edge is counted once with an arbitrary orientation, and $\theta$ ranges over unit flows with divergence $+1$ at $a$, $-1$ at $z$, and zero elsewhere. The expression is the dissipated energy for unit electric current.

To prove the minimum, let $i$ be the current flow arising from a voltage $u$ that drives one unit of electric current. Write another unit flow as $i+h$, so $h$ has zero divergence. Since $i(e)/c_e$ is the endpoint voltage difference, summation by parts givesHence the energy of $i+h$ is the energy of $i$ plus the nonnegative energy of $h$. The energy of $i$ is $u(a)-u(z)=R_{\mathrm{eff}}(a,z)$ by Ohm's law. For two resistors of electrical resistances $2$ and $3$ in series, every unit flow uses both, giving effective resistance $5$.

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

For a finite connected loopless undirected electrical network without killing, give each edge a positive electrical conductance $c_e$, put $W={\sum }_e{c}_e$, and use transition probabilities $P(x,y)=c_{xy}/c_x$, where $c_x={\sum }_y{c}_{xy}$. Then the commute time identity isThe factor is total electrical conductance, not total electrical resistance. A killed random walk can fail to reach its target and is outside this statement.

Here is a direct proof. Let $h_z(x)={\mathbb{E}}_x{T}_z$ and define the network operator $Lh(x)={\sum }_y{c}_{xy}(h(x)-h(y))$. First-step analysis gives $L{h}_z(x)=c_x$ off $z$. Summing $L{h}_z$ over all graph vertices gives zero, so $L{h}_z(z)=c_z-2W$. Therefore $u=(h_z-h_a)/(2W)$ has divergence $+1$ at $a$, $-1$ at $z$, and zero elsewhere. Its voltage drop is $(h_z(a)+h_a(z))/(2W)$, which is the effective resistance for unit electric current.

For an unweighted path with $n$ edges, $W=n$ and endpoint effective resistance is $n$, giving commute time $2n^2$. Uniformly multiplying all electrical conductances changes $W$ and effective resistance inversely and leaves the Markov chain unchanged. ^[ref]

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.