Take a Cayley graph $\Gamma$ (thought of as an electrical network with all edges having equal resistance) and break one edge $e$ and put a battery there. (Assume the graph has only one end* so that this operation does not disconnect the graph.) Are there any estimates on the resulting current?

In more mathematical terms, write $e=(s,t)$, then define the voltage at a vertex $x$ to be [Edit: Formula incorrect see Ori's comment after Victor's answer] $$ V_x = \mathbb{P}(\textrm{a simple random walk starting at } x \textrm{ hits } s \textrm{ before it hits } t) = \mathbb{P}^x(X_{\tau_{ \lbrace s,t \rbrace }}=s) $$

$V_x$ is (up to a constant**) a harmonic function, and the current on the edge $(x,y)$ is just $V_y - V_x$. It turns out $V_x$ may be chosen so that $(x,y) \mapsto V_y-V_x$ is in $\ell^2(E)$.

Are there any finer estimates on the decay of these functions (on the edges)? For example, might it even belong to $\ell^1(E)$?

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*[Edit] Having one end implies being infinite, not X where X stands for cyclic, free, amalgamated product or HNN extension over a finite subgroup (a theorem of Stallings). One could add virtually before these adjectives but it is redundant (see HW's comment)

**[Edit] it could happen that the function is even less uniquely defined. For example, if there is a harmonic function with $\ell^2$ gradient on the graph $\Gamma \setminus e$. However, the element in $\overline{\nabla \ell^2(X)}^{\ell^2(E)}$ (see PPS below) is unique up to a constant. As far as my little knowledge of the topic is concerned, it might happen that this element is also not the best in terms of decay.

PS: it does not matter whether the edge $e$ is removed or not when defining the probability $V_x$.

PPS: this is classical, but a way to see that the current belongs in $\ell^2(E)$, is that $\ell^2(E)$ decomposes as $\overline{\nabla \ell^2(X)} + \ker \textrm{Div}$, projecting the Dirac at $e$ on the first factor gives the desired current (except on the edge $e$). [Edit: another way is given in Victor's answer below]

a finite subgroup. For instance, $\mathbb{Z}^2=\mathbb{Z}∗_{\mathbb{Z}}$ is of course, one-ended. You don't need the 'virtually': it's true that you can't be virtually cyclic, but virtually cyclic groups do in fact split as HNN extensions over a finite subgroup. You might like to mention that this is a theorem of Stallings. $\endgroup$ – HJRW Feb 18 '13 at 11:19