  ### Author Topic: Resistance across a 1k resistor cube.  (Read 3258 times)

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#### hamster_nz

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• Country:  ##### Resistance across a 1k resistor cube.
« on: April 22, 2018, 12:47:06 am »
While lazing around flicking through the latest Diyode Magazine, I came across this little puzzle.

What is the resistance across a cube of 1k resistors? (see image)

There is a few nice ways to solve it, when insight comes to you.

Gaze not into the abyss, lest you become recognized as an abyss domain expert, and they expect you keep gazing into the damn thing.

#### IanB ##### Re: Resistance across a 1k resistor cube.
« Reply #1 on: April 22, 2018, 12:59:04 am »
I would say 833.3 Ω ?

Edit: I could even say 833⅓ Ω now that the forum has been upgraded « Last Edit: April 22, 2018, 01:50:59 am by IanB »
I'm not an EE--what am I doing here?

#### AlfBaz

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« Reply #2 on: April 22, 2018, 01:27:33 am »
Easy, 12 x 1k resistors multiplied the alfbaz 1k cube constant of 0.069444' #### rs20 ##### Re: Resistance across a 1k resistor cube.
« Reply #3 on: April 22, 2018, 01:45:03 am »
833.33... is correct. hamster_nz, did you deliberately angle the cube in your picture to make one of the insights more obvious, or is that just a coincidence? #### Brumby

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« Reply #4 on: April 22, 2018, 01:47:13 am »
I would say 833.3 Ω ?
So would I.

833.33... is correct. hamster_nz, did you deliberately angle the cube in your picture to make one of the insights more obvious, or is that just a coincidence? I wouldn't have said it was that obvious.  After all, you need some angle to show the cubic structure.  But now that you've said that, there might be a few more who catch on.

#### JDubU

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• • Posts: 241 ##### Re: Resistance across a 1k resistor cube.
« Reply #5 on: April 22, 2018, 01:52:08 am »
LTspice agrees with 833.33... ohms (1 V voltage source has 1.2 mA current through it).

« Last Edit: April 22, 2018, 02:07:46 am by JDubU »

#### IanB ##### Re: Resistance across a 1k resistor cube.
« Reply #6 on: April 22, 2018, 01:57:23 am »
LTspice agrees with 833.33... ohms (1v voltage source has 1.2ma current through it).

It's good that LTspice works then. It would be awful if LTspice failed on such a simple circuit... (Since this is a lighthearted thread and I am a stickler for correctness, I should mention some typography rules for units of measure. There should be a space between the number and the unit of measure, and case is important. Which means: "A 1 V source has 1.2 mA of current through it.")
« Last Edit: April 22, 2018, 02:01:33 am by IanB »
I'm not an EE--what am I doing here?

#### JDubU

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• • Posts: 241 ##### Re: Resistance across a 1k resistor cube.
« Reply #7 on: April 22, 2018, 02:08:37 am »
I stand corrected! #### steverino

• Guest ##### Re: Resistance across a 1k resistor cube.
« Reply #8 on: April 22, 2018, 05:45:58 am »
I get 5/6 * R = 5/6 * 1000 =833.33...

There's also a general solution for 12 arbitrary resistances which works out to be something like

(r1 || r2 || r3) + (r4 || r5 || r6 || r7 || r8 || r9) + (r10 || r11 || r12)

but I don't feel like typing in the derivation.
« Last Edit: April 22, 2018, 06:03:09 am by steverino »

#### rs20 ##### Re: Resistance across a 1k resistor cube.
« Reply #9 on: April 22, 2018, 06:57:20 am »
There's also a general solution for 12 arbitrary resistances which works out to be something like

(r1 || r2 || r3) + (r4 || r5 || r6 || r7 || r8 || r9) + (r10 || r11 || r12)

but I don't feel like typing in the derivation.

False. If you set all the resistors lying parallel to the x and y axes (arbitrary choice) to infinity ohms (open circuit), but leave the 4 resistors parallel to the z axis at 1 kiloohm, then:

- Your formula gives a result of 2.5 kiloohm,
- In reality, the such an arrangement breaks the circuit, so the correct solution is infinity ohms.

Therefore, your proposed formula is wrong.

Actually figuring out the correct answer for the general case is a much harder problem than the all-resistances-equal case (because all the resistances being equal is necessary to pull certain tricks); I haven't figured it out yet (not having really tried though.)
« Last Edit: April 22, 2018, 06:59:16 am by rs20 »

#### Kirr ##### Re: Resistance across a 1k resistor cube.
« Reply #10 on: April 22, 2018, 07:37:54 am »
For verifying solutions you can try my online solver: http://kirr.homeunix.org/electronics/resistor-network-solver/

It solves the network by eliminating nodes one by one using star-mesh transform, until only terminals are left. (See examples, which include resistor cube).

A similar quiz: Find resistance across a 4D cube, then also resistance across an edge (singel resistor) of a 4D cube.

#### hamster_nz

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« Reply #11 on: April 22, 2018, 07:47:56 am »
For verifying solutions you can try my online solver: http://kirr.homeunix.org/electronics/resistor-network-solver/

It solves the network by eliminating nodes one by one using star-mesh transform, until only terminals are left. (See examples, which include resistor cube).

A similar quiz: Find resistance across a 4D cube, then also resistance across an edge (singel resistor) of a 4D cube.

I'll have to sketch them out.

I would be interested in hearing who solved this by which means, from LTspice, to changing to an equivalent structure that is obvious to solve..

My first solution was to split all the resistors in half and bisect the design, half way between the test points. Analysis is then a piece of cake.
Gaze not into the abyss, lest you become recognized as an abyss domain expert, and they expect you keep gazing into the damn thing.

#### steverino

• Guest ##### Re: Resistance across a 1k resistor cube.
« Reply #12 on: April 22, 2018, 07:56:36 am »
The solution I gave assumes real values for the resistors - not open circuit.  Perhaps arbitrary was the wrong term.  Correct in this case, no?

#### rs20 ##### Re: Resistance across a 1k resistor cube.
« Reply #13 on: April 22, 2018, 08:05:09 am »
The solution I gave assumes real values for the resistors - not open circuit.  Perhaps arbitrary was the wrong term.  Correct in this case, no?

So we have two facts:
-- Your formula works when all the resistances are completely equal
-- Your formula totally fails as some of the resistances approach infinity

And you're still confidently asserting that it's correct for all arbitrary combinations of finite resistances? That's very very bold! The parallel and series resistance formulae always give sane results even when faced with infinite resistances, and yet it very specially breaks down just for you? Hmm...

Anyway, your formula is still definitely wrong. You can take my previous argument, replace "infinity" by "one gigaohm", and the same  disagreement between common sense and your formula will result. Your formula allows the current to jump freely between the three nodes one step away from the + terminal, and the three nodes one step away from the - terminal. In a real cube, this isn't allowed. This is why your formula gives absurdly low values when very high value resistors are placed in such a way that they provide a unavoidable barrier to current in a real cube.

I've got Maple v11.0 to provide a solution (to the all-twelve-resistors-may-be-different-to-each-other case), and the resulting formula is literally 6 pages long. This doesn't prove much on its own, though (I might have made a mistake in the system of equations I provided, and it's also possible that this 6 page formula could be greatly simplified.)
« Last Edit: April 22, 2018, 08:14:21 am by rs20 »

#### Brumby

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« Reply #14 on: April 22, 2018, 08:10:35 am »
Equipotential shorting method - valid because of symmetry.

Yellow lines are connecting equipotential points.  Because they are equipotential, joining them will not affect circuit operation. Section A is 3 resistors in parallel = 333.3 Ω
Section B is 6 resistors in parallel = 166.7 Ω
Section C is 3 resistors in parallel = 333.3 Ω
 A + B + C in series = 333.3 + 166.7 + 333.3 = 833.3 Ω

30 seconds of mental arithmetic (but I admit I wrote down the three intermediate values).
« Last Edit: April 22, 2018, 08:19:55 am by Brumby »

#### IanB ##### Re: Resistance across a 1k resistor cube.
« Reply #15 on: April 22, 2018, 08:17:46 am »
I would be interested in hearing who solved this by which means, from LTspice, to changing to an equivalent structure that is obvious to solve..

My solution arose from observing the symmetry of the system. If all resistors have the same value then all parallel interior nodes of the network have the same voltage, which means they can be treated as if they were connected, which allows the network to be viewed as groups of parallel resistors. Following this observation the overall resistance is (1/3 + 1/6 + 1/3) · 1000 = 833⅓
I'm not an EE--what am I doing here?

#### steverino

• Guest ##### Re: Resistance across a 1k resistor cube.
« Reply #16 on: April 22, 2018, 08:27:59 am »
The solution I gave assumes real values for the resistors - not open circuit.  Perhaps arbitrary was the wrong term.  Correct in this case, no?

So we have two facts:
-- Your formula works when all the resistances are completely equal
-- Your formula totally fails as some of the resistances approach infinity

And you're still confidently asserting that it's correct for all arbitrary combinations of finite resistances? That's very very bold! The parallel and series resistance formulae always give sane results even when faced with infinite resistances, and yet it very specially breaks down just for you? Hmm...

Anyway, your formula is still definitely wrong. You can take my previous argument, replace "infinity" by "one gigaohm", and the same  disagreement between common sense and your formula will result. Your formula allows the current to jump freely between the three nodes one step away from the + terminal, and the three nodes one step away from the - terminal. In a real cube, this isn't allowed. This is why your formula gives absurdly low values when very high value resistors are placed in such a way that they provide a unavoidable barrier to current in a real cube.

I've got Maple v11.0 to provide a solution (to the all-twelve-resistors-may-be-different-to-each-other case), and the resulting formula is literally 6 pages long. This doesn't prove much on its own, though (I might have made a mistake in the system of equations I provided, and it's also possible that this 6 page formula could be greatly simplified.)
Relax my friend.  I simply reduced parallel and series combinations.  If my solution is incorrect, that's ok and I'll double check it in the morning.  If it looks ok to me, I'll post a copy of my notes.  If I find an error, I'll post that also and some kind soul can point out the error.  Now, it's 1:27am and I'm relaxing enjoying a glass of brandy.  Cheers!

BTW, I just looked at Digikey, and I don't seem to be able to find any infinite resistors! « Last Edit: April 22, 2018, 08:33:34 am by steverino »

#### steverino

• Guest ##### Re: Resistance across a 1k resistor cube.
« Reply #17 on: April 22, 2018, 08:48:47 am »
quick request.  Can somebody please post the ltspice file for the cube (i know, i'm lazy).  I'll use it to check my work.  Thanks.

#### rs20 ##### Re: Resistance across a 1k resistor cube.
« Reply #18 on: April 22, 2018, 09:11:37 am »
For those curious, I got Maple (a symbolic algebra solver thing) to solve the general case (resistors in the cube arrangement, but each resistor may have different, arbitrary values). I've attached a printout of the worksheet, but the TL;DR is:

The formula for the resistance is 3 pages long, and has this form (where capital letters are the various resistances, in an weird arrangement explained in the PDF):

total resistance = (A*U*G*F*C*V + K*B*A*E*C*F + ...) / (U*A*B*E*V + K*C*G*V*F + ...)

Which is at least dimensionally consistent!

In the case where four parallel resistors are equal (resistance = Rz) and the remaining 8 resistors are equal (resistance = Ry), then we find:

total resistance = (Ry^2 + 3 * Ry * Rz + Rz ^ 2) / (2 * Ry + 4 * Rz)

Notably, taking Ry or Rz to infinity brings the whole expression to infinity, as Ry and Rz both have higher powers in the numerator.

Finally, setting all the resistances to 1000 gives an answer of 833.33..., which is reassuring Btw, hamster_nz, the above is not how I initially solved the simpler problem: I initially solved in exactly the way that Brumby described. Unfortunately, that method can't be applied in the case where the resistors have different values, as the underlying symmetry breaks down I'd be fascinated to hear if anyone has a non-maching-requiring method of solving the arbitrary case; I'll be the first to admit that I have no idea how to do it by hand.

#### RoGeorge ##### Re: Resistance across a 1k resistor cube.
« Reply #19 on: April 22, 2018, 09:28:58 am »
OK, then, the cube was easy, try this one: What is the resistance between 2 nodes A and B, in an infinite 2 dimensional square grid of resistors, each resistor having the same value, R? « Last Edit: April 22, 2018, 09:32:09 am by RoGeorge »

#### JacquesBBB ##### Re: Resistance across a 1k resistor cube.
« Reply #20 on: April 22, 2018, 10:28:03 am »
I would be interested in hearing who solved this by which means, from LTspice, to changing to an equivalent structure that is obvious to solve..

My solution arose from observing the symmetry of the system. If all resistors have the same value then all parallel interior nodes of the network have the same voltage, which means they can be treated as if they were connected, which allows the network to be viewed as groups of parallel resistors. Following this observation the overall resistance is (1/3 + 1/6 + 1/3) · 1000 = 833⅓

This is a great solution Ian and Brumby !

I  worked out nearly the same way, considering the symmetries of the system, and  thus three
different voltages,  V, V1, V2, 0 ,  with by symmetry also I/3 or I/6 current, and got  the solution,
but  although this was easy to solve, I  did not use your brilliant shortcut.

which is to say that all summit with same voltage can be considered as being connected, which
then simplifies the solution as one can use the already know rules.

This equipotential trick  is powerful !

« Last Edit: April 22, 2018, 10:31:30 am by JacquesBBB »

#### Wimberleytech ##### Re: Resistance across a 1k resistor cube.
« Reply #21 on: April 22, 2018, 01:22:38 pm »
Here it is.

#### Wimberleytech ##### Re: Resistance across a 1k resistor cube.
« Reply #22 on: April 22, 2018, 01:26:31 pm »
This was my method to solve years ago...still works!

A couple of months ago I decided that I wanted to build one of these cubes as a piece of desk art.  I toyed with some 3D printer corner designs and such, ordered some 1/2 watt precision resistors...ADD as I am, the project is still incomplete along with the other hundred that are incomplete...lol...but I am always busy!!

#### Brumby

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« Reply #23 on: April 22, 2018, 01:57:55 pm »
A lot of us know that song only too well.

#### Wimberleytech ##### Re: Resistance across a 1k resistor cube.
« Reply #24 on: April 22, 2018, 03:41:42 pm »
OK, then, the cube was easy, try this one: What is the resistance between 2 nodes A and B, in an infinite 2 dimensional square grid of resistors, each resistor having the same value, R? Using superpostion, go out to infinity and ground the last (LOL) nodes.
Then insert a current of 1 amp somewhere in the center of infinity (LOL²).
That current will split into fourths (1/4, 1/4, 1/4, 1/4).
Now remove that current source and connected a different current source of 1 amp LEAVING an adjacent node.
The current leaving will split into fourths.
Now, for a linear system, we can add these result yielding 1/2 amp flowing in the resistor between the adjacent points.
1/2 amp flowing through R gives V = R/2
That voltage (R/2) resulted from a 1 amp source, so the equivalent resistance is V/I = R/2/1 = R/2

Edit:
After pondering further, I do not have to ground the array at infinity...just insert the first test current and then the second.
« Last Edit: April 22, 2018, 04:06:23 pm by Wimberleytech »

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