26-09-09 Resistance of an infinite grid

http://www.targetiit.com/iit-jee-forum/posts/resistance-9716.html

plz chek the link given by tushar...there are quite a few unanswered doubts in dat post

For those who are interested.....
If we consider the two dimensional square grid and map it into the Cartesian plane with each node occupying a lattice point, then the equivalent resistance between the node (0,0) and the node (a,b) is given by
R_{a,b}=\dfrac{R}{4\pi^2}\int_{-\pi}^\pi\int_{-\pi}^\pi\dfrac{1-\cos(ax+by)}{2-\cos x-\cos y}\ \mathrm dx\, \mathrm dy
where R denotes the resistance of each unit of wire.

As an example, the above formula yields (correctly) the equivalent resistance between two adjacent nodes (0,0) and (1,0) (which of course would be obtained easily using superposition) as
R_1,0 = R/2.

another solution which i found while surfing :
principle of superposition:

first consider a battery whose one end is a at and another at infinity:

Force the circuit at point A while holding voltage zero at infinity. By symmetry, 0.25 current flows through the resistor between A and B (going from A toward B).

now :
Take current out of circuit at point B while holding voltage zero at infinity

By symmetry, 0.25 currentflows through the resistor between A and B (going from A toward B).

Superimposing these two situations gives 0.5 i flowing through the resistor between A and B
hence equivalent resistance R/2