Suppose we have an electrical network consisting of $n$ nodes and the graph is connected. Let $a$ and $b$ be two of the nodes. Now we put an external source at the network such that the voltage at $a$ is 1 and the voltage at $b$ is 0. Then is the maximum voltage difference between any 2 nodes in the network 1?
We can assume the degree of each node is roughly the same.
Consider the case where the maximum of voltage $V_{max}$ is greater than $1$. Let $c$ be a vertex where this maximum is achieved. It is clear $c \ne a, b$.
Since voltage at $c$ is no less than all its adjacent vertices, net current flowing out from it unless all vertices adjacent to it has same voltage. By conservation of charge, this is impossible. This implies all vertices connected to $c$ has same voltage $V_{max}$.
Since the graph is connected, there is a path connecting $c$ to either $a$ or $b$ without visiting the other. Let's say we have such a path to $a$.
$$c \to d_1 \to d_2 \to \cdots \to d_m \to a$$
Apply above arguments to $d_1$, then $d_2$ and so one, we can deduce $d_1, d_2, \ldots$ and finally $a$ has voltage $V_{max}$. Contradicting with the fact the voltage at $a$ is $1 < V_{max}$.
From these, we can conclude $V_{max} \le 1$ and hence $= 1$ (because voltage at $a$ is $1$). By a similar argument, we can deduce the minimum of votage $V_{min}$ is $0$.