Does current follow the path(s) of least (total) resistance? Consider Poisson equation $\nabla \cdot (\sigma(x)\nabla u)=0$ in a domain $D$, where $\sigma(x)$ is the spatially dependent conductivity. On the boundary we have $2$ electrodes $E_1$ and $E_2$ (Dirichlet BC $u=0$ on $E_1$ and $u=1$ on $E_2$). And the rest of the boundary is insulating material $du/d\vec n=0$ (Neumann BC). The electrodes do not have any contact impedance. This would be an ordinary Ohmic resistor with resistance R.
To each divergence-free vector field $J(x)$ on $D$ which flow strictly from $E_1$ to $E_2$ we can assign a "resistance" $R_J$ as follows, because we can insert non-conducting walls along the streamlines, we can consider the streamlines from $E_1$ to $E_2$ as resistors of thickness dt, then $R_J$ is defined as the resistance of them all in parallel (so that would be $1/(1/R_1+1/R_2....))$, in the limit as dt goes to 0. Each of the streamline resistors is defined as a series connection of resistors (So thats $R_a+R_b...$) with length dt2, and we take the limit as dt2 goes to zero aswell. The resistance of each resistor of thickness $dt*dt2$ is computed from $\sigma(x)$
I am looking for proof or disproof to the conjecture that $R_J$ is minimized when $J$ is obtained by solving the Poisson equation. I am also looking for some references on this idea or concept.
 A: Perhaps to resolve this issue it helps to work out a simple example.
Take a region $D$ consisting of the strip $|x|<1$, $0<y<1$, and a $y$-independent conductivity profile
$$\sigma(x)=\begin{cases}
1 &\text{for} -1<x<0,\\
2& \text{for}\;\;\;\; 0<x<1.
\end{cases}
$$
The solution of the Poisson equation $\nabla \cdot (\sigma\nabla u)=0$ with boundary conditions $u(x,0)=0$, $u(x,1)=1$, $\partial u/\partial x=0$ for $|x|=1$ is
$$u(x,y)=y\Rightarrow J_y=\sigma(x)\frac{\partial u}{\partial y}=\begin{cases}
1 &\text{for} -1<x<0,\\
2 & \text{for}\;\;\;\; 0<x<1.
\end{cases}$$
The total current $I=\int J_y\,dx=3$. 
Now I am a bit uncertain how $R_J$ is defined. My reading of the OP is that for this simple case of parallel streamlines the definition is
$$\frac{1}{R_J}=\frac{\int \sigma(x)J_y(x)dx}{\int J_y(x) dx}=\frac{5}{3}.$$
The conjecture is that $R_J=3/5$ is minimised by a $J$ which solves the Poisson equation, but one can for example achieve a smaller $R_J=1/2$ for $J_y(x)=0$ for $-1<x<0$ and $J_y(x)=1$ for $0<x<1$.

The correct minimisation principle is not "least resistance" but "least dissipation": At fixed total current $I=\int J_y\,dx$ the dissipation (Joule heat)
$$D_J = \int \frac{[J_y(x)]^{\mathbf{2}}}{\sigma(x)}\, dx$$
is minimised. 
