Physics Puzzle: Infinite Resistor Network

Posted on April 25, 2008
Filed Under Physics Puzzles |

Another weekend has come, and another puzzle with it! This time, we’ll be making good use of Ohm’s law.

Quite a few physics puzzles ask you to find the equivalent resistor of a given network - a particularly famous one is finding the equivalent resistor between two opposite corners of a cube having a resistor R on each of its edges.

This puzzle is even trickier: can you find the equivalent resistance between points A & B in the following (infinite!) network?

Infinite Resistor Network Puzzle

All resistors have the same resistance (say, R), and - as mentioned - the network is infinite in all directions.

Good luck & have fun! :)

ADDED 03/May/2008: I’ve added a solution. To read it, just view the comments (it’s comment #3).

Comments

3 Responses to “Physics Puzzle: Infinite Resistor Network”

  1. wandering.the.cosmos on April 26th, 2008 10:22 am

    Would you know how to solve the problem for the same setup with A and B being diagonally across from each other on one of the squares? The answer I remember involves Pi…

  2. admin on April 27th, 2008 4:00 pm

    Hi wandering,

    Regarding the cube problem:

    No pi in the cube problem. It’s 5R/6, if I remember correctly. There are two tricks: the first is to draw the cube on a 2D plane, simplifying the problem (it makes no difference to the electrical properties whether you have a 3D cube or a 2D drawing). The second trick is that you can “draw lines” between “equivalent” points. e.g., if your initial vertex was A, which split out to B, C and D, then (by symmetry) B, C, D must all have the same potential, and so can be “merged” into a single point. So, we have 3 resistors in parallel between A and the “merged” point (BCD). Then we have 6 resistors in parallel between BCD and the next set of vertices, say (EFG), and then once again 3 resistors in parallel between (EFG) and H, which is the opposite corner. Then the problem becomes a simple problem of adding resistors in parallel & series.

  3. admin on May 2nd, 2008 6:19 pm

    SOLUTION (ADDED 3/May/2008)

    DO NOT CONTINUE READING UNLESS YOU WANT TO ANSWER!

    The trick with the question is not to think in terms of potential difference, but rather in terms of current.

    By symmetry, if we “pump” in a current I into node A, it will flow out of it in all directions equally, (I/4). Similarly, if we “draw” a current I from node B, we will “suck” current equally from all adjacent nodes (I/4 again).

    Thus, assume we connect a battery (i.e. create a potential difference) between A & B in the drawing, pumping a current I into A and drawing a current I from B. This means a total current of I/4+I/4=I/2 will flow between A & B, and hence:

    V = I/2*R

    Or

    V = I*(R/2)

    so R/2 will be the effective resistance between A & B.

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