Physics Puzzle: Resistor Cube

June 15th, 2009 | Categories: Electricity and Magnetism, Physics Puzzles

Level of Difficulty: Highschool

Here’s a classic puzzle without which I couldn’t call this a puzzle-blog.

Here is a cube with a resistor on each edge. All the resistors have the same resistance R. What is the equivalent resistance between points A and B?

cuberesistor

A Resistor Cube

For dessert, here’s a variant that’s not often seen and requires a bit more work: can you compute the effective resistance between points A and B in the cube below? As before, all edges have the same resistance R:

cuberesistor2

Another Resistor Cube

EDITED, 16/June/2009: a detailed solution has been posted here. Hats off to our reader R4mOn for his solution in the comments section!

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  1. R4m0n
    June 15th, 2009 at 11:58
    Reply | Quote | #1

    Hello there again!
    I think I remember this puzzle from a university physics class, but I didn’t remember how to solve it, so I had to look around a bit (just a tiny bit :-).
    From here: http://www.radioelectronicschool.net/files/downloads/resistor_cube_problem.pdf
    we get 2 ways to solve the first puzzle, which I won’t even bother to put the solution here. Using the first method described there to solve your ‘dessert’, I’ve got to 7/12 R (which sounds about right, but I may have missed some weird current path). I was thinking the second method would not work due to the asymmetry on the paths between A and B, but now I think it would work out quite easily by just ignoring the resistor between A and B and adding it later in parallel with the rest of the circuit… Would have probably been easier than messing with all the current paths like I did, but well… maybe next time :-)

  2. Assaf
    June 15th, 2009 at 16:08
    Reply | Quote | #2

    Yep, you nailed the second one. You could remove the resistor, but there’s an even easier way of doing it which - while not completely hassle free - dramatically simplifies the problem. I’ll publish my solution tomorrow.

  3. Ravi Raj Gupta
    July 19th, 2010 at 13:31
    Reply | Quote | #3

    Solution for case2…..

    case2-
    let us neglect the shortest path between AB(i.e. 1 resistor, and we will consider it in the end in parallel to the circuit.)
    Now, let 2I amount of current pass through A, it gets divided to 2 equal parts of I each, i.e.along -Zaxis and +Yaxis,let I’ be the current which goes along +X axis

    2I-2I’
    _______________
    | |
    I-I’ |I-I’ I-I’|
    ______|_________ |I-I’
    | | I’ | |
    | |I-I’ | |
    I | |________|____|
    | I’ |
    | I |I I
    |______________|
    A B

    BY KIRCHOFF’S LAW
    -RI’+R(I-I’)+R(2I-2I’)+R(I-I’)=0
    5I’=4I
    I’=4/5 I _______________eq 1

    V=2IR’
    I’R+IR=2IR’ ________________eq2

    EQUIVALENT RESISTANCE R’=7R/5

    NOW,
    THIS CIRCUIT AND wire AB are in parallel…
    1/Re=5/7R + 1/R

    therefore Re= 7R/12

  4. Ravi Raj Gupta
    July 19th, 2010 at 13:41
    Reply | Quote | #4

    ……………2I-2I’
    …………________________
    ……… ./|…………../|
    …….I-I’ |I-I’…I-I’../.|
    ………/______|________/..|I-I’
    ……..| ..| I’………|..|
    ……..|.. |I-I’……. |..|
    ……I.|…|____________|..|
    ……..|…/..I’……..|I./
    ……..| I ………….|./ I
    ……..|/_______________|/
    …….A………………B

    DIAGRAM HERE REDRAWN