EEVblog 1472 - Resistor Cube Problem SOLVED
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- Опубліковано 17 лип 2024
- How to solve the resistor cube problem using equipotential nodes, short and open techniques, and circuit simplification.
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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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"You're not an idiot because you're trying to learn"... that's the wisest thing I've heard this week.
I stubbornly assert my right to be an idiot!
Dave is YY, that's for sure
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
Forgot the circuit flattening from my college days, thanks for the informative video Dave!
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
I was hoping you will say "make a cube and 'collapse it'" (lol)
Or, if you are in CAD or handcrafting mood, make the cube out of thin wire and use a flashlight to project it on a piece of paper
In my head, I look down at the cube from above, and take the top face of the cube, "stretch" it, and then flatten it.
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..!
You can also do 4 simple star-delta transformations when you have the redrawn schematic.
And more generally, you can just apply recursively the star-polygon transformation until you've only got one resistor. Mathematically, it's just equivalent to apply gaussian elimination on the Kirchhoff equations, which, obviously, is equivalent to pretty much any method you could dream of. But at this point, we might as well use a computer
even I thought aobut it., star delta transf.
We could also use the Kamen Rider Black RX Solar Transformations recursively, but I don't know what changing the resistors into bugs would help with the solving part 🤔🤔🤔🤔
🤣🤣/s
That's what I thought but anyway this is much simpler
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!
By putting a pair of resistors in parallel, you create an R/2 resistor (you can then repeat the process to make an R/(2^n) resistor) and then by putting resistors in series, you can create a 2R resistor (from two identical resistors) and thus a 2^(n)R resistor.
Then you just write down your target resistance in binary, create that out of the 1R resistors through the processes above, and put them all in series to add them. (And any number can be written in binary to arbitrary precision with enough digits).
Not the most efficient solution, but probably the most mechanical to follow.
Well done, Dave. It was fun to see your analysis and has taught me an additional technique for analysis. Love this Stuff!
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 ❗
At 18:56 you show 121GW has a 100 Ohms in the microAmmeter scale so of course you measure the voltage drop across the resistor and calculate current: I (current) = E (voltage) / R (resistance). But I was confused about your popup text. When I saw 121GW, I thought you were referring to "Back to the future" ... 1.21 Giga Watts... but that was the MODEL of your Digital Multi Meter (DMM). What a coincidence! Great video!
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.
that's the most german thing to have a high school for engineering
The vid at the end, really seals the deal, Dave!
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.
Same here. I think the redrawn version could be better by having A / H on the left/rightmost points. Then the symmetry would be clearer.
@@jameshogge Yep: I agree with all of that. I did it in my head just looking at the original picture of the cube, except that I stupidly got R in parallel with 3R wrong until I re-thought that bit.
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.
Brilliant application of Kirchhoff's law.
Loved the explanation, it was a brilliant set of simple steps.
Great explanation and perfect demo! Thank you!
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.)
Nice practical demo as well as the explanation. We were once shown this as a curiosity at the Electronics course I was attending, but I'm sure the question was for diagonal opposite joints A and G and not A and H.
I learn something new every time I watch your videos - oh and I'm an oldie !
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.
Hello Dave. I don't mean to change your interest in any way, but this video is just design engineers and university students geeking out. This type of analysis has nothing to do with electronics as a hobby... This is one of those "can it be done, yes we did it, let's celebrate", while everyone around you have no idea what you are so happy about...
Keep on learning electronics, just understand this has nothing to do with it!!!
All you need to start with is
Rparallel=(R1*R2)/(R1+R2)
and
1/Total=1/R1+1/R2+1/R3+1/R4...
And basic branch and nodal analysis... (Kirchoff, etc). Very basic!
Far more important than knowing this, as a hobbyist, is when something does not turn ON, ALWAYS check the power supply(ies) first! Power supplies fail much more often that anything else!!! Followed by mechanical problems... Followed by tantalum caps blowing up and short-circuiting, followed by electrolytic aluminum caps drying out and dropping capacitance as they age...
And when you approach or exceed a resistor's rated dissipation rating, they get HOT HOT HOT!!!!
And that all parameter DRIFT with respect to temperature, and to a lesser extent, exotic things like bending of the part...
Then put as much effort into learning how to use an oscilloscope PROPERLY. That everything you touch with the ground lead of the scope probe is automatically shorted to EARTH GROUND. What 10X means with a probe. Aliasing. How to measure high-speed digital signals. And a hundred of other scope-related topics...
Now knowing this, you will be amazed by how many engineers DO NOT know of things I have just told you!
@@nameredacted1242 Which is exactly (I would assume) why you do hobby electronics in the first place.
@@jimmyb1451 "I had no say in the matter". I started in electronics when I was 12. And I was taking mechanical things apart even before then.
@@nameredacted1242 I started when I was 8. Because I was interested in it, and programming, and all things mechanical.
Because I had a father. Who exposed me to fixing things.
R must be your middle initial.
Your best videos are those in which you teach.
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.
that last demonstration was valuable and really nailed it
Thanks for the physical’ information’ at the end. I knew it was right, but, well, verify!
First time complicated electronics! You made it very easy!
Very cool video as always Dave! :)
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.
Removing EF and DC is what I used also.
Great vid! Appreciate the time taken to make this as I also appreciate the youtube algorithm might not be kind to you, but lots and lots and lots of people will thank you all the same!!!
I liked the build at the end. Nice one Dave. I saw this on Twitter and had a panic attack.
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.
I'm afraid it's the very same method, right?
Not really.
I did KCL at the input mode with a "test" voltage source.
No resistor transformations or redrawing circuits multiple times.
It's easier because you just accepted that Ve=Vf and Vd=Vc. While not to hard to understand, if you were to explain it to someone not believing you, you would probably just follow the step in the video.
The rest is just applying KCL instead of parallel and series equivalent resistance, which I would say is not immensely different, just marginaly simpler at the expense of being a less obvious method.
@@jonasdaverio9369 I didn't assume Ve=Vf and Vd=Vc actually, in fact its not even true.
My "trick" is recognizing that all the middle nodes are 1/2 of the input voltage as a result of symmetry.
It just requires less math and is sort of a hybrid approach to what EEVblog guy did.
@@sinank7032 Aren't you contradicting yourself by saying it's not true that Vd=Vc and then saying that all middle nodes are 1/2V? D and C are both middle nodes, and both 1/2V (he even measured it)
But I agree and I've changed my mind, you don't need to know why Vd=Vc if you accept that the symmetry imposes that the drop in voltage from B to C, D, E and F must be the same as the drop in voltage from C, D, E and F to [? can't remember which node, but the output node]
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.
Great explanation!
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.
I do teach, on UA-cam, to vastly more people than I ever could face to face.
As usual, brilliantly explained. 😀
the ability redraw circuits in your head, that's a great skill in electronics 👍👍
Yeh.... My professors would really fail me if I did it this way. But I understand the merit behind equipotential meshes. Well done Dave.
Interested in the problem with all arbitrary resistances. But this vid is awesome. Thanks.
yeah, we use to be given these problems in our first year EE course :) like doing finite element analysis, one item at a time
The demonstration at the end, priceless!
Why did I just watch this...? Maybe it was it just to find out your method of attack... And I just noticed you've relabelled your DMM. NIce 👍🏻
Good educational vid for our next gen engineers!
The key to the redrawing is to think in terms of a polyhedron in which one face is being streched and all the other faces have an image on top of that original single face. The net figure will show all the other faces acting all nice, and the trick is that the space outside of all those other faces is itself a face. This is similar to the idea in geometry of a polyhedron net, which is breaking apart some of the edges between faces so that all faces are in a plane.
A classic circuit used in engineering school used to teach loop (KVL) and node (KCL) analysis utilizing simultaneous equations. You're right, when in doubt, redraw the circuit!
I'm sure that I've seen this problem in an exam, several decades ago. I also recall that we were required to solve this using nodal analysis. :-)
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.
That’s what I saw first as well!
To see why the D-C and E-F resistors can be ignored, you can swap the source voltages at A and H, and then flip the circuit over lengthwise. The voltage differences across the two resistors will change sign (because you swapped the source voltages), but the resulting circuit is exactly the same as before, so the voltages can't have changed. The only number that is negative itself is zero, so there's no voltage or current across those resistors, so we can remove or short them.
Great video. That is how I first solved it many years ago - but I drew it as one Weatstone Bridge within the other (similar to your more orthogonal approach) and canceled out the two "balanced" resistors to arrive at the 3 parallel branch circuit which is easy to solve: R = 1 / ((1/Branch 1) + (1/Branch 2) + (1/Branch 3)) 0.75 = 1 / ((1/2) + (1/3) + (1/2)). One thing: You should introduce that equation and the two resistors in parallel equation: R = (R1 * R2) / (R1 + R2) for those new to resistance in series and parallel so they know how you got 0.5 Ohms and 0.375 Ohms. Most of us know this, but newbies may not. I teach this puzzle in my STEM class as part of my Weatstone Bridge introduction. Oh, my students each make the cube with 1k resistors and another in the dual bridge flat arrangement on a wooden bread board with test points. They get to keep them both along with many other things they build in class.
Love it. Thank you so much
Nice. Reminded me of the old days.
I remember 3 years ago Dr Heribert Eisele giving us that trick question on first year EEE at university of leeds and 3/4 of the class had trouble solving it for 3 days straight 😂
Lucky freshmen will have it easy next year!
I think everyone who studied EE does know the cube. It is funny.
now, THATs how it should be teaching in school. thanks a lot for the video. aaaa the good old ones. (all your videos are good to watch, but i enjoy that kind) which one did YOU enjoy doing??
o my it has been years since radio/tv class and the resistor cube circuit... 🤥thanks 🥰
great analysis :) love DaveCads :p
By the way Dave the BM786 I got from you is hold tighter tolerance than my 3 hp 3455a ( all in cal) against my voltage and resistor references . So it is a very good value hope every one buys one.
This brings back a lot of memories. Just need to figure out if they are good ones or bad ones...
OMG, at 16:15 I use that "Back to the Future" quote all the time. "Please forgive the crudity of this model; I didn't have time to build it to scale or paint it". It's an engineering joke, I had to point it out to my wife that it was an inside joke for techies.
Great job, Dave!
What would have been fun, is to show the actual full equation for the entire network.
Bonus points, solve the equation in Roman numerals and show all your work.
Great video...cheers.
Absolutely brilliant, so you could physically remove resistors from the cube without altering the resistance measured…
Yep!
Nice reference to Back to the Future :)
nice video. thanks!
Cleaning the tip before inserting is a good idea. I'll do that, maybe using an alcohol wipe. I have resistors about that vintage as well.
I saw a variation solved a bit different. The question was resistance between opposite corners A and G (not B and G) Use a CURRENT source and inject 3A at point A. Being symmetrical, it splits to 1 amp in each resistor connected to point A. From there, at points B, D, and E each of those 1 A currents split to 1/2 amp each. At C, F, and H those 1/2 amp currents sum to 1A again, and those three branches of 1A sum at G to 3A again. Now you can easily find that there is 1V drop where there is 1A flow, and 1/2 V where there is 1/2 A flow. So 3 amps injected causes a 1+1/2+1 volt drop, ergo resistance is (5/2)/ 3 = 5/6 ohms.
Once you see the symmetry and the current law, you can solve it in your head.
The best way is to inject an arbitrary current I into A using a potential V. Then, note that symmetry implies that from A the current splits up into 3 equal parts A/3, and similarly reconstitutes at H back to I. Then, solving for V/I gives the effective resistance between A and H. This is a problem from Resnick and Halliday, a textbook for undergraduate physics.
Great video Dave, now do a dodecahedron instead of a cube to totally amaze us !!!
Interesting Video. I recall we did the same thing in Mech Eng, with water pipe flows. The sum of flows into a node was 0, and the Head Loss and Flow Rate, are analogous to Resistance & Current. So it can be solved the same way, assuming some pipe fitter used the same sized pipes.
Yup. We solved this one with a current injection source knowing the current sums to zero at each node. Then add voltage drops across each edge to find total voltage drop divided by current injected to get effective resistance.
Where I worked the design engineers used to calculate pipe flows based on a theoretically perfect nominal bore pipe. I took a real world 1" NB short radius elbow up to the next meeting to show the true bore was well below what they were using in their calculations. I explained that there was no maximum wall for NB parts, only a minimum and that the maximum was based on weight per unit length which didn't apply to elbows that were made thicker on the inside wall and thinner on the outside wall by the bending process. They had to start thicker than nominal to cater for the outer wall thinning. Messed up all their calculations up to that point. As someone who started off as an electrical engineer but ended up as what was referred to as an integrated engineer with both electrical and mechanical engineering training I always looked at fluid flow as though it was an electrical problem.
Also if you have resistors on these paper strips, nearby resistors are a bit more likely to have closer values, some folks say. Makes some sense if they went through the process at about the same time. I think Bob Pease has mentioned that in one of his books.
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.
Thanks for that. I was having trouble understanding why those paths could be ignored.
Thanks! Good way to look at it.
wow, been 20+ years and I got it right; awesome
I learned to solve these problems with matrix algebra. We solved HUGE matrixes with slide rule, because computers were not yet invented. I was quite amazing, you could for example define resistor matrix and divide it with voltage matrix and get current matrix. This example would have been super trivial. Only when you had inductances and capacitors and you did Fourier transforms within matrix, it became more challenging.
My solution using symmetry:
Apply a voltage from A to H, and inspect the currents across the two resistors at CD and EF. From symmetry of these two resistors, the currents cannot be one going up and one going down, but must be both going in the same direction.
Let’s say the currents in the two resistors are going up. Now reverse the voltage applied from A to H. The currents should reverse as well, changed to going down for both resistors. But, the current should still be going up, due to symmetry of the setup. So in fact there is no current flowing through these two resistors at all. The setup would be the same if we removed the resistors and wires joining C to D, and E to F. Then it is a very simple setup to calculate the resistance across A and H.
Wheatstone bridges show up a lot in strain gages, and might be an interesting application to do a video.
They show up EVERYWHERE in instrumentation!!!
They are used in strain gauges to amplify the strain (and at 90 degrees to the direction of strain) to compensate for temperature drift. Same in PT100 temperature sensors to self compensate for ambient temperature stability. Most instrumentation sensors use Wheatstone bridges into an instrumentation op-amp for precision and low drift !
@@simonbaxter8001 I would love @EEVblog to do that full circuit with the instrumentation op-amp. I am familiar with general op-amps and comparators from the other videos and coworkers, but would like to learn more about instrumentation op-amp circuits.
@@andrewpriest9403 It would make a good topic, although I'm sure Dave has done a series on OpAmps and he does cover why instrumentation opamps are used (better CMRR to start with!).
@@simonbaxter8001 Yeah, he has covered opamps already, but I don't recall if he got into instrumentation specifics and CM noise rejection and signal quality.
Nice one!
If only real-world engineering was anything like this...
My most recent adventure with resistors was when I was asked to make a ~1 Giga-ohm measurement.
I found three instruments which we already have in the building, and which could do it:
1) A Source Meter.
2) A bench-grade DMM (the few rare modern models, not the vast majority of older models).
3) An insulation / Hypot tester.
But just to do a sanity measurement and to demonstrate to others that these instruments produce a real reading, and not some noise reading, with load connected, I asked a Project Engineer to order some 10Megaohm, 100Megaohm, and 1Gigaohm resistors from the leading parts distributor. I sent a URL of exact search page for resistors in this range.
Project engineer of course did NOT want to order from my URL, and went ahead, and ordered resistors from another website of his choosing just because he was ordering some other parts from that other website, too. He spent about 3 days going back and forth about some other parts he had to order as well, eventually I stopped paying attention to his e-mails.
Project engineer dropped a package at my desk almost two weeks later (???).
In the bag were PRECISION, MEDIUM POWER resistors, of values 10 milliohm, 100 milliohm, 50 milliohm, and a pack of 1 Gigaohm resistors. The precision sub-ohm resistors were 2W and 4W current-sense types (but thru-hole radials). Turns out when he was typing in, manually, in the search field (INSTEAD OF USING THE FILTER CRITERIA LIKE I WAS DOING FOR HIM), "Resistor 10Mohm", what you get is a match for a 10 milli-ohm resistor (it literally is spelled out all in uppercase in the leading distributor's part number as RESISTOR 10MOHM PRECISION 1WATT, etc.............
When they allow me to do adjunct professoring, I do my best to tell my students that all of this BS math they will most likely never encounter once they graduate...
Oh don't worry, I did not even yell at him.
1) The 1G resistor was what I really needed, and it demonstrated that all three instruments were roughly in agreement.
2) I already made a 1G+ potentiometer from things lying around in the lab, well before I even asked him to source some parts for me. How, you might ask? I wetted a thin tissue sheet used for cleaning safety glasses with said liquid lens cleaner, and hooked up the instruments with J hook micrograbber leads, to the wetted tissue paper. Now, the critical part is poaitioning of leads. If one or both leads are on the dry section of tissue paper, resistance measurement exceeds capability of my instrumentation (probably 10-100G for measured value).
If you place both leads over wetted region of paper, you only read ~100Mohm. Only by precisely putting one lead in the wetted region, and another lead RIGHT ON THE EDGE of the wet to dry line was I able to dial in a reading of about 1Gohm. This was a "pot" because very small movements around the moving and evaporating wetted line, I was able to adjust readings around 1Gohm. SO, this proved to me that the meter would respond and produce some display readings that tracked real conditions, and not just show me a bullcrap noise reading that did not change until something like a 100Meg load was connected... The 1Gigaohm purchased part would then "calibrate" / verify the instrument with more certainty once I had the part on hand...
@GsaUce Rug Oh, they will still try to do "substitutions" or second-guess you the engineer. You would think that a project engineer, who used to be a real engineer way back when, would know better than to think they know better than a "design" engineer. Or they did not like DigiKey and went with Mouser (in my case), to combine shipping charges... And these days, it's likely by the time you research to the time you request the part, it becomes out of stock...
I'd analyze the circuit and look for potentials... Elimination of resistors is a very cool method here, and remember: if it doesn't have a voltage drop across it, current won't flow, so basically you can short it or erase it, to your preference :)
I think I had an A-G problem back in 2005 when I was studying EE.
Glue on "ammo tape" components should be banned by the law... even if we're just talking Ohm's and Kirchhoff's laws :D
Brilliant!
For the axial lead components on an adhesive band, just do what the machines that install these do - clip the leads at the adhesive band. No need to clean up sticky residue that way.
I could do this but I'm too tired today, so I'm just going to nerd out and enjoy the show. Besides, it's been a couple of decades.
At 11:28 it is not straightforward why symmetry means equipotential. Dave also recognizes this with the added on-screen text. I think a good explanation is the following. Suppose first, that the vertical R/2 resistor is not there. Then, clearly the voltage between the two resistors on the upper arc is the half of the input voltage (U/2) and the same way the voltage between the two resistors on the lower half is U/2 as well. Now, put back the vertical resistor. As it has the same voltage at both of its leads, no current flows through it, so you can really omit it.
I think the problem is the same at 12:55 and also the explanation is very similar.
I was expecting the question to be for the full 3D diagonal, from A to G. But maybe that was too easy compared to the A-H face diagonal:-
By symmetry, nodes B D E from A are equipotential, and thus can be wired together giving R/3. Likewise for nodes C F H to G.
The remaining six resistors are now all in parallel between wired nodes BDE and CFH.
This makes three sets in series: R/3 for A-BDE, R/6 for BDE-CFH, and R/3 for CFH-G.
The total resistance is then:-
R/3 + R/6 + R/3
= 2R/6 +R/6 + 2R/6
= 5R/6
= 0.83333...R
(confirmed using LTSpice)
I did it with DY transform which is somewhat harder, but works at the end
I wonder how hard a 4-dimensional version of this would be
You could still flatten it into 2D. The only difference would be that a 4D-hypercube (actual name) has more edges etc and hence more resistors.
Would the test points be across the same 2-d plane or would they get a dimensional upgrade to A and G?
Probably not much harder. More vertices, but also more symmetry to exploit.
We used to have similar questions for Physics MCQ questions (2 mins per question) in school, I think we solved these quickly by considering symmetry.
I was shown this in high school and then learned than because of symmetry, there are points that are at equal potential, so you could connect them together and simplify and it becomes easy.
I was asked about the resistance across the diagonals. Same idea.
In all my 50 years of Electrical engineering this is the most impractical this I have ever seen. Yes is a great practice exercise but nobody would actually need on in a circuit.
A quick way to think about this to ignore all the resistors except the top four and only calculate the top 4 parallel resistors and you will be within 10%. The time and headache you save can be better spent elsewhere.
Hello. I calculated I lot similar schematics with different resistance, when I was at university, we simply use first and second Kirchhoff's circuit laws, it can be solved without 3d to 2d transform even with all different resistance. And If you completely lazy like I am, you can use mathcad it can solve system of linear equations for you.
Nice, but what is the tolerance of the 0.75*R resistance?
Related problems solved in a similar way a) across a single resistor A to B is 7R/12 and b) across the cube large diagonal A to G is 5R/6.
Hi Dave, at 11:41, it seems to me that if you somehow had potential at the node below that vertical resistor, the less resistant path would be to go up and around to reach G. So I'm having trouble seeing that there would be no current across that vertical resistor. Is it that the two paths have already balanced out (appropriate current per resistance) by the time any potential reaches that center vertical wire? So would that be a voltage difference measured across that center vertical line?
I'm a 44 years old electronic engineer and still learn. Nice video!
Using the same approach, you could have removed the two vertical resistors F-E and D-C and make it more simple to solve. Actually, you can ever do the math in your head. ((2R//2R)+R+R)//(2R//2R) = 3R//R = (3/4)R.
In my first year, the very first exam, had one problem, a cube. It had a voltage source and a dependent voltage source and resistor on the other edges. Only one class mate unfolded it correctly, and that alone was half the points. I did not, but thought me to see problems in a very different way.
I got that question on job interview, but it was the resistance between A and G.
I have a problem with Bob Pease schematic. How do you mount three transistors and 7(???) resistors of various power dissipation ratings "on a heatsink"??? And what's the schematic of a power supply that produces 50V 36V 22V outputs?
You also can transform "stars" to "triangles", and then solve for simple serial-parallel circuit.
An old colleague of mine, a technician, handed me, many many years ago, such a construction with the words: "calculate the resistance between point x and y". Both points were opposite of each other (A and G in your drawing). The only other difference in the construction was 'cross members': resistors from A to H, E to D, and that on all sides, you get the jest, and finally two cross the inside: A to G and B to H. I was dabbling a bit in electronics then and was really a novice. Needles to say I never managed to work it out. Still a bit of a poser though...
EXCELLENT MAN!
In school this was an end of chapter question asking you to find the values at different places. My solution was to assume a 3 amp input between points A and G and all resistors to be one ohm. The amperage divides equally to one Amp between the three input resistors after point A. The next division is to one half Amp then they are recombined to one amp at points C, F and H. With this approach you can calculate any resistance at any point. Example A to D and A to E are one Amp x one Ohm = one volt, and D to H and E to H are one half Amp = one half Volt. A, D, H = 1.5 and A, E, H also = 1.5, in parallel they = .75. Kirchhoff's current law.
The ADH=1.5 is a voltage. 1.5V in parallel with 1.5V is 1.5V, you hit the correct answer by using totally wrong logic and calculation. If you applied 1.5 volts across AH you will not get 3 amps flowing in the circuit from A to G and this circuit will not result in 2.5V across AG.
Actually I shouldn’t have used Volts it’s about the ratios between the two paths the units have little meaning.
thank u dave . if there was no symmetry! . i would use star delta transformation
Brilliant.
I'm not an electrical engineer (physics major) so I'm not sure exactly how these trikcs work, but would is it also possible to find out the voltages at each node and deduce the R some how from those values?