Famous question from IE Irodov :For Finding resistance?

Consider a infinite square mesh each side made up of 100 Ohm resistors. What will be the resistance measured across one 100 Ohm resistor somewhere in the middle of the mesh.

Famous question from IE Irodov :For Finding resistance?windows media

The answer: 50 惟 .

Funny. I hadn't heard of I. E. Irodov, ever. Still, I learned about this problem long ago, back in the '70s. Moreover, I knew which the solution was, because it was given along with the problem statement. I quote:

"4.30 A "puzzle problem" that circulated among electrical engineers some years ago was this: An infinite number of 1 惟 resistors are connected together to form a two-dimensional infinite net with square meshes. That is, at every node leads from four resistors join. What is the equivalent resistance between a node and one of its four nearest neighbor nodes? This problem is a startling example of the power of symmetry and superposition. Using superposition shrewdly, you can do it in your head, almost. Try it any other way and you will appreciate the elegance of the first solution. The answer, by the way, is 0.5 惟, but we will not spoil the puzzle by telling you what states of current flow to superpose."

This is taken from Electricity and Magnetism, volume 2, Berkeley Physics Course, published 1965. Written by Edward Mills Purcell, recipient of the Nobel Prize in Physics, 1952.

At any rate, it seems the problem has been around for some time, now. As for the solution, valuable guidance is given within the statement. Heeding the advice, suppose you connect a 1 A current source to any pair of adjacent nodes. Any pair will do, since network extent is infinite. In order to be able to apply superposition, imagine that the positive source terminal is connected first, and somehow current can flow from the source to the network even if the negative terminal is left unconnected. If this bothers you, imagine this negative terminal is connected to a very large square wire frame, as large as you please, which is brought in contact with the network at considerable distance from our pair of nodes.

Since the network is absolutely symmetrical, current will split into four equal parts as it reaches the node. Thus, ? A will flow into each of the four resistors connected to that node. Consider now the other node. The positive source terminal is removed as the negative is connected to the second node. A similar condition exists, only now current is negative, i.e., flows from the four resistors to the negative terminal of the source. By symmetry, all four resistors contribute an equal share to total current: ? A apiece.

Now, according to the superposition principle, current through any resistor will be the sum of the currents originated by each source. Specifically, the current in the only resistor connected to both nodes is ? A. Voltage drop in this resistor can be readily calculated as ? V.

If voltage across the pair of nodes is ? V, while current flow into/out-of the node is 1 A, then the equivalent resistance of the network is, evidently, ? 惟.

If all those 1 惟 resistors are replaced with 100 惟 units, it is inescapable that the equivalent resistance between adjacent nodes is 50 惟.

Famous question from IE Irodov :For Finding resistance?microsoft exchange internet explorer

I don't have the patience to solve it, but I'm guessing you can express the resistance as some sort of geometric series or an inverse thereof.