Infinite Grid Resitance

A hexagonal analog of problem number 3.153 from I. E. Irodov's book "Problems in General Physics". Apply the principle of symmetry to get an equation for the P. D. between points $A$ and $B$ :

$V_{AB}=IR_{eq}=I_0R$ , where $I$ is the current flowing through the lead wires, $I_0$ is the current flowing through the section $AB$ , $R_{eq}$ is the equivalent resistance between points $A$ and $B$ to be determined.

Applying the principle of superposition, current enters into any node through three wires. So,

$I_0=2\times \dfrac{I}{3}$ , the factor $2$ arises because of the fact that if current $I$ flowed into $A$ and spread all over the grid, the wire $AB$ would carry a current $\dfrac{I}{3}$ . Similarly, if the current flowed into the grid from infinity and left the grid through $B$ , the wire $AB$ would also carry the current $\dfrac{I}{3}$ .

Hence, $R_{eq}=\dfrac{2}{3}R$ , and $α=\dfrac{2}{3}\approx \boxed {0.6666666666}$ .