Solution to Infinite Grid of Resistors Challenge Problem

Solution #1 (my solution I posted on the chalkboard)

First I stood there looking at the picture, and thought that if it is infinite in extent, all the square grid is essentially a repeating pattern of squares, then the input impedance looking into a side of one of the squares must also be periodic and symmetric in all directions. Therefore I said that if inside one of the squares, looking toward any of the sides of the square must be the same impedance, which I called Zin. As a result, the input impedace looking out of the box on any side is the same as looking into any one of the sides. Finally, I said if I look into the box from any one side, I can say the other 3 sides look like Zin in series all in parallel with a resistor “R”, as shown below:

So from the last sketch on the right, I believe this first step to be the trick — because from here on out is simple circuit analysis theory. Another way to think of it is that since it’s periodic, addind or subtracting a box should not affect things, so remove the box to the left of the box and then look along the Zin path.

So next, redraw the circuit like a normal circuit and solve for Zin as shown below:
This is not the answer, just yet. This is what the resistance looks like looking out of or into one side of the box.

In order to get the actual thevenin resistnace, now solve for the inpedance looking into the terminal set a-b, as shown below:
SO, the answer turns out to be that Rth is equal to one half of R, as shown above. This was the mathematical solution.

Solution #2 – Here’s the cool, clever solution! (that I did not think of but that I found on google — what can’t you find on google?!)

This solution is so much simpler. It requires not thought about periodic impedances or input impedances looking this way and that way. It relies on symmetry, Ohm’s law and Kirchoff’s Current Law.

The first step here is to look at node “a” and “b” separately:
The next step is to apply a current source of 1A to terminal “a”, leaving terminal “b” alone. The current can be setup flowing into node “a”, as shown. The consequence is that since this huge grid is symmetric and infinite in extent, the current branches off into 4 equal currents, or 1/4 of an amp. Also, the other side of the current source can be said to be placed at infinity.
So, from the source at “a”, 1/4A will flow into the branch between node “a” and “b”. Now take away that source and apply another 1A source to node “b”, only this time flowing out of the node:
Once again 1/4A flows through each branch. Using the superposition theorem, the total current when both sources are applied will be the sum of the current from the source at “a” and the source at “b”, giving 1/2A flowing between node “a” and “b”.
Since one end of each current at infinity is moving 1A through the node at infinity (at node “a” source 1A is extracted whereas at node “b” source 1A is injected) then the two ends of the current sources can be thought of as connected, and therefore replaced by a single 1A current source.
Now, by definition of the thevenin equivalent circuit theorem, driving two terminals with 1A of current and then dividing the voltage that developes across the terminals by 1A, you get the equivalent resistance. This is shown above. The voltage is equal to the resistance “R” times the total current flowing through it, and diving this voltage by 1A gives the final equivalent resistance.