EEVblog 1472 - Resistor Cube Problem SOLVED

How to solve the resistor cube problem using equipotential nodes, short and open techniques, and circuit simplification.

00:00 - The Resistor Cube problem
04:28 - The SECRET to circuit simplification using equipotential nodes
06:04 - Analysis "by inspection"
14:22 - Summary of the solution
16:14 - Let's BUILD it!
19:28 - Two resistors are entirely redundant!

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[OzRetrocomp]

"You're not an idiot because you're trying to learn"... that's the wisest thing I've heard this week.

[wickedxe]

Dave, I just wanted you to know that even though I rarely comment on your videos I get massive value from your fundamental basic electrical/electronic theory. Even if your other non theoretical content gets more views, your basic this is how it works and why videos are imo amongst most important content on the internet because you explain theory really well and it makes sense to everyone no matter how limited their theoretical knowledge might be, and I think in this day and age where most people have a very poor understanding of science it is vital

[felipemakara]

If you still cannot see the 3D to 2D transformation, take a cube and look at it directly from top side, putting the A and H to left-right sides (cube should be "diagonal"). It makes the same drawing as the schematic, with the closest side forming the "outer" loop and the far side forming the "inner" loop. The only difference being the angle the wires make

[AlexKrieger]

Forgot the circuit flattening from my college days, thanks for the informative video Dave!

[Geek_Chef]

Snipping the resistors at the end was the best part of the video...! The physical proof that the math is valid was a great moment. Thanks Dave..!

[Robin-kylc]

You can also do 4 simple star-delta transformations when you have the redrawn schematic.

[sigmaxi]

It really shows the power of symmetry as a tool! Back in high school* I learned to solve these kind of problem sets by applying star-to-delta conversions or vice versa, but symmetry is much better.

Another great advice for beginners: If you have trouble identifying series or parallel connections, think of resistors as toilet paper rolls on a string (=wires). If you can slide them towards each other until they touch, it's series; if not it's something else (not necessarily parallel). Nodes are knots in the wire that prevent the rolls from moving.

* in Austria there are high school level (actually ISCED level 5B) engineering schools, called HTL, that's what I am referring to.

[ZomB]

7:27 It's was very obvious to me that all four points are the same potential, almost right from the start. Just look at the cube itself. If you apply -10V to A and +10V at H, then due to the symmetry, every node exactly half-way (so, those four points) must be 0V.

[tubastuff]

Many years ago, there was a story in one of the electronics mags called "A Jar of Resistors" (or something like it). The basic plot was that a young fellow picked up a large jar full of identically-valued resistors and wondered what could be done with them. The gist of the article was showing how resistances of any value could be fashioned, given enough of the little bugs. Wish someone would refresh my recollection!

[evensgrey]

Always nice to have multiple paths to the answer, as that gives you a quick way to check the answer you got.

And this isn't an unusual thing to need to design into a product. You may need a specific value that isn't one of the standard resistor values you can get. Even if you can get it as a standard part, you also have to be aware of the set-up costs of the assembly house you're using and how much they will charge to have another part in the BOM versus how much more the BOM costs for using multiple resistors to eliminate that additional part from the BOM. At some point, you hit the limit for how many parts the assembly house can put on the board in one pass, and they either won't put on any more parts or they'll charge you a hefty price for a whole second run through the pick-and-place system. Never forget, the effects on the cost of manufacture are ALSO an engineering calculation. Engineering is about balancing all the factors in a design to produce the optimal result. (Famous example of this in some circles: If you've ever used Cisco profession-grade networking gear, it all has RS-232 serial ports for local device access. These ports use RJ-45 jacks instead of BD9 or DB25 connectors because that choice eliminated the extra assembly cost of those different parts and instead used a part that the device already had to have in the BOM.)

[davedave]

I’m relatively new in my journey through learning electronics. I didn’t quite understand that explanation but I’ll watch it again until I do. I went ahead and built a resistor cube out of 5.6k resistors and it measured 4.2k, exactly .75 the value of one. My train of thought had me doubling the 2 resistors on the closest path and assuming that would be the answer.

[IAdryan]

My method : After noticing the simetry, i've removed resistors (EF) and (DC). The 4 resistors on a cube side are 1R on diagonal so i left with 1R (ADHE) paralel with series of 1R (AB)+ 1R (BFGC) + 1R (HG). Then 1R paralel with 3R. Then 3/4 R.

[Dwarg]

About halfway through the analysis by inspection part I recognized the technique you were using as being factorization, and things just mathematically clicked for how to solve this problem.
Amazing as to where you can end up finding applications for these algebraic methods.

[johnbhancock]

I first solved this in about 1970 when I was in 3rd year high school doing the subject applied electricity. I used the current method and symmetry. Assuming all resisters were 1 ohm and 1 V applied across the cube, I calculated the current in each resister then converted that back to a resistance using Ohm's law. Later on using simultaneous equations I generated an equation to solve for any value resistor on any node, which I still have somewhere. An interesting problem whichever way you look at it.

[evensgrey]

6:40 another way of looking at this is there logically can be no current flowing between F and C due to a voltage applied between B and G. Yes, there are current paths, but if you add them up, they cancel out to exactly zero between equipotential points, as they must because the current paths, and thus the currents they carry, are also symmetrical.

This problem was part of my exam some fifty years ago.
I had completely forgotten about it until I saw this video.
It brought back good memories of my student days.
Thanks Dave, I enjoyed your explanation very much ❗

[sinank]

I've got an even EASIER way to do this.

1. Hook up a voltage source V so node B.
because of symmetry voltage of F, E, D, C are all V/2.
2. 3 currents flow from voltage source to nodes F, A, C.
assuming all resistors are 1.
3. If = Ic = V/2 /1 = 1/2V
4. IA = V/2 / (1+ 1/2) = V/3

INPUT resistance then is just V/I = V/(4/3 V) = 3/4

much less math and conceptualizing short circuits.

[marcus_w]

Great! Not a problem, I'm ever going to face, not a problem for me at all - but the explanation is top notch. You really should teach.

[OnboardG]

I always like finding these problems. I've been doing electronics professionally for ten years and you end up moving away from the basic mathematics like this as you get into more whiz-bang stuff. It's nice to go back to "Intro To Electonics" problems and see if you can still remember how to do them.

[malpa]

Simpler solution: According to symmetry current from A to D is equal to current from D to H. In that case there is no current from D to C as well as from E to F (we can remove these resistors as infinite resistance). In that case we will get your last graph.