= Solution
Orient the unique path from $s$ to $t$ in a <uniform spanning tree> $T$, and put $j^T_{xy}=1$ when it uses the <edge> from $x$ to $y$, $-1$ when it uses the reverse direction, and zero otherwise. Then the proposed current is $i_{xy}=\mathbb E j^T_{xy}$; set it to zero on nonedges. It is antisymmetric.
For each <tree> path, every intermediate <graph vertex> has one incoming and one outgoing path <edge>, while the source has one outgoing <edge> and the sink one incoming <edge>. Hence
$$
\sum_y j^T_{xy}=1_{x=s}-1_{x=t}.
$$
Taking <expectations> gives
$$
\boxed{\sum_y i_{xy}=1_{x=s}-1_{x=t},}
$$
the <Kirchhoff node law> for a <unit flow>.
To establish the <Kirchhoff cycle law>, it is not enough to check it separately for each <tree> path: those path currents need not satisfy the voltage law on the original <graph>. Instead, let $\mathcal T$ be all <spanning trees> and let $\mathcal F_{st}$ be all spanning forests with two components, one containing $s$ and the other $t$. For $F\in\mathcal F_{st}$ denote the source component by $C_s(F)$.
For an existing <edge> $xy$, deleting it from a <tree> whose $s$-$t$ path uses $x\to y$ produces precisely a forest $F\in\mathcal F_{st}$ with $x\in C_s(F)$ and $y\notin C_s(F)$. Conversely, adding $xy$ to any such forest produces exactly that <tree> and path orientation. This is a bijection. Applying it to both orientations gives
$$
i_{xy}=\frac1{|\mathcal T|}\sum_{F\in\mathcal F_{st}}\left(1_{x\in C_s(F)}-1_{y\in C_s(F)}\right).
$$
Define $h(x)=|\mathcal T|^{-1}\sum_F1_{x\in C_s(F)}$. Then $i_{xy}=h(x)-h(y)$ on every <edge>. Since the resistances are one, this is <Ohm's law>, and the voltage drops telescope around every oriented cycle:
$$
\boxed{\sum_{xy\text{ along a cycle}}i_{xy}=0.}
$$
Together with the node law, this proves that the mean tree-path current is the electrical unit current. This argument is the <mean spanning-tree path current> identity.
For clarity, these laws determine the current uniquely. The difference of two solutions is a potential gradient $k(x)-k(y)$ with zero divergence. Multiplying its divergence by $k(x)$ and summing over <graph vertices> gives $\sum_{\{x,y\}}(k(x)-k(y))^2=0$. Thus every difference current is zero.
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