EEVblog 1472 - Resistor Cube Problem SOLVED

hi it's twitter question time thank you very much aditya ah you behindy i'm butchering that sorry dave care to make a quick and easy video on this one for idiots like me trying to learn electronics well you're not an idiot because you are trying to learn which is excellent anyway world of engineering i follow them uh basic electrical knowledge test so let's have a look what is the resistance between a point a and point h here for a cube of resistors like this three dimensional cube uh with so 12 identical resistors of resistance they happen to be 10k here are 0.05 percent but we'll just assume they're like one ohm it doesn't matter they're all identical r values and this is actually a very common uh exam question you'll get in any electronics course and i've done of course way way back which is my the infinite resistor problem and i didn't solve that mathematically i actually physically built it up and solved it and you can physically uh build one of these yourself go and get some resistors measure them so they're all reasonably close you know choose your you know pick and choose and then solder them together and actually build it up and verify the answer might do this at the end actually so how do we solve this well there's many ways to skin this um engineering cat and here's the way i would approach it i think it's the way probably majority of people would actually approach this now i'll leave the network up here now the first thing you have to do is to redraw it because this is very common in electronics if you redraw circuits uh to be like familiar to you or easier to understand then often boom you'll go oh that's easy right um take any uh bob pease circuit for example your eyes just rolling in the back of your head going how does that work well if you redraw it in a sensible manner it it works just fine that was that was his style okay so what i've done here i've i've taken these four top resistors here and i've redrawn them in the middle here this is the first thing i did so i've got points a here a and h and h and hopefully right you can see that that's obvious if you if you have problem actually translating this into a three-dimensional shape into a two-dimensional diagram not sure i can help you any further but um yeah try and work through it it's it's pretty easy so e here obviously goes has a resistor going down to point f so i drew a resistor here going off to point f and likewise point d going down to point c here like that and likewise for point h and points a you've got a resistor going off here in this case it goes from h to g like that and it goes from a to b like that and then obviously you can see that there's a resistor between f and b so i've added that and another one from b and c etc so at this point the question asks what is the resistance between a and h and well we've got this point in here and this point in here and we've got resistors going out this way and we've got resistors going out this way and they fold back on each other and they're in series and parallel and all higgledy piggledy but by looking at this cube here you can see that the question is exactly the same between a and h here this point here and this point here it'll be exactly the same between b and g it'll be exactly the same between f and c etc like two opposite points what's the difference because these are all identical resistors so this cube has symmetry okay so a and h here there's no difference as i said between b and g so if you solve for b and g it's exactly the same as solving for a and h it makes no difference so our circuit conveniently already has b on the outside here and g on the outside here so we'll just solve for b and g so we'll forget a and h now i won't bother i could re-label them and everything but i won't bother i'll just leave it so you've got a more familiar resistor network problem um with you know what is the point between here and here so it's starting to feel a bit more familiar hopefully but we still have this problem of like oh look we've got paths going up here and down here you know it's like these things are all over the shop right it's a dog's breakfast how do we like possibly solve for any of this now here comes the neat trick okay you remember how we talked about symmetry before obviously this circuit is symmetrical but hopefully you can visualize this okay it's obvious to me and hopefully it'll be obvious to you now because this is the key trick you need to actually solve this thing otherwise it's really ugly like you could solve it without simplifying the circuit even further but the equations is just going to be horrible they're all the same value so hopefully you can see if i split this down the middle like this all this side is identical mirror image to what's on left and right if i split this like this it's a mirror image top and bottom right you can see that so this is a completely symmetrical circuit so we can use a technique called equipotential nodes or equi potential it means equal potential same potential nodes right you remember kirchhoff's uh current laws the sum of the currents exiting the junction equals the sum of the currency going into the junction well if all of these are the same value okay then the current going up here is going to be identical to the current going down here if you apply a voltage between b and g okay so you can try and solve it in terms of like ohm's law and kirchhoff's current laws and everything else right but we can do it simpler than this because we know the current up here is the same as the current down here and interestingly the current through here is going to be the same as the current through here but you might be thinking oh what about look there's a sneaky current path up here stick with me we'll deal with this in a minute now i'm sure i've mentioned this term in a previous video by inspection due to the symmetry of this circuit top and bottom left and right we can determine that if we applied a voltage between b and g here then the voltage at point f would be identical to the voltage at point c we're inspecting this circuit and we're deeming that the voltage here must be equal to the voltage here because this is a symmetrical circuit everything's symmetrical why wouldn't those voltages be the same and it's true that volt if you build this up physically in this configuration like this which will be no different to the cube over here it's just in a physical flatter two-dimensional format if you build this and put a voltage across here here's some homework for you go and physically build this measure the voltage here and here relative to a reference point you can choose a reference point anyway these voltages will be the same and likewise voltage at point e node e and node d will be the same as well bingo if these voltages are the same then we can treat them as a short circuit so what we can do now is we can actually treat these points as a short circuit and i can draw a short circuit in here like this okay but it's a bit how you do it now because we have to like jump over some things like this we'll simplify this further in a minute stick around so we've got two points now shorted out in our circuit now we can actually go further and we kind of like have to at this point because these two resistors here are still confusing us right is current flowing this way is it like what's what's going on here well hopefully you can see that we've still got symmetry right like this now you remember these two resistors are now in parallel these two resistors are now in parallel so technically we could redraw that so well let's do that okay so i've redrawn that here these two resistors because you remember these were in parallel these are now r on two these are their value this resistor remains r or whatever that is you know it could be 1 ohm 10k doesn't matter and these ones up here these all everything else remains resistance r so the only ones we've sold for now are these two here but once again you can see we've got symmetry like this so this point up here has to equal this point here and that's what we determined before okay short those out now here's where you can go in two different directions to solve this and you'll get exactly the same answer okay so we know that these two f and c here are you know equi-potential nodes right they're the same potential because of the symmetry we have here now we can actually short these again and we can do it that way if we want but also what you can do is you can simply say that no current flows down here or here in either direction no current at all flows so you can actually eliminate these two resistors from uh in fact let's choose two different parts and we'll solve it two different ways and by the way we kind of have a wheat stone bridge kind of uh thing happening here which means that like we're sort of because these are all balanced right this side is balanced with this side here and this half it's right it's all balanced oh might i don't think i've ever done a video on wheatstone bridge go check out wheatstone bridges right so you can actually once again by inspection and knowledge of wheatstone bridges you can say that the current down here and down there is no current flowing down here and down here once again build it up and put your ammeter in there and measure the current for yourself there'll be no current flowing and you can do all this in the newfangled simulators as well so you could just put this into your simulator and you could actually measure the current so it will measure the simulated current through there and through here and at zero anyway let's solve two different ways okay so the first way we're going to do this is we're going to actually physically short f to c here and that's the f to c point here so this resistor here and this resistor here become these two resistors and they're both in parallel because we're shorted physically short at this point up to this point up here and then this resistor down here and this resistor here become these two in parallel okay and then we've got our existing resistors in here they don't change and that becomes our new network so we put these in parallel so the this r on r becomes r on two like this you can keep it in fraction form or you can do like 0.5 r i've kind of mixed it here sorry about that for those who don't like it like i've put 1.5 r and and are on fraction and decimal here and whatever anyway hopefully you're still with me so these become r on two r on two and this becomes r on two so this is getting much simpler but once again we've got these current paths like this okay so this isn't your traditional series parallel problem once again you simplify it again using equipotential nodes symmetry right down here like well just imagine b and c are in the middle right symmetry like that so once again you can short out this point to this point here so let's do that okay so there's no current flowing through this r on two at all here there's no current flowing through these two here so we can short those out and bingo now we've simplified it to one if you were given this in an exam you'd solve that easy peasy lemon squeezy because we've just got two resistors in parallel and then the total in series easy so i've converted back to decimal here r on 2 becomes 0.5 r now just to keep it consistent in this diagram here so half an ohm in parallel with 1.5 ohms and that's 0.375 r and likewise here 0.375r you add those up because they're in series bingo your answer is 0.75 r for the resistance between b and g or as we said before a and h or e and d or c and f or whatever doesn't matter now i'm going to show you an easier method to actually do this you notice that there's like this one had one two three and like four kind of steps well this one only really has two steps and you get exactly the same answer let's see how we did it okay before path number one we actually are treated this as a short circuit if you remember that okay but i also said remember you can think in terms of this as like a balanced like wheat stone bridge so there's no current flowing through here so we don't have to actually short these out what we can do is just actually eliminate these entirely from the circuit we can scrub them out so that's what i'm going to do in uh path two here okay so we've got these four resistors in series like this there they are they're identical we've eliminated the resistor in here we've got it out of the circuit because if there's no current flowing through there is it shorted or is it open ah doesn't matter could be either you can treat it either way okay and so therefore we're left with the two series resistors up here and two series resistors down the bottom so we're left with a simple parallel circuit of three resistors because these are all in series okay so you solve your series ones first so these two resistors become two r here likewise that becomes two r and this one becomes r plus r plus a half an r plus half an r which becomes 3r so now you've got three resistors in parallel solve that using whatever method of the parallel resistance equation you want to get and bingo you get the answer 0.75 r so they match regardless of the method that you use so there's there's two different methods one is using equipotential nodes shorted one is using equipotential modes nodes open so there you go and there's other techniques please leave it in the comments down below how you would actually solve this this is how i would solve this problem and i just did and hopefully that makes sense to you so when you ever see questions like this just go right i i can't think in three no one thinks in 3d calculations like this so you redraw it in two dimensions once you redraw it in two dimensions you look for any symmetries have you got any equipotential nodes that nodes at the same potential um and then can i short them together and or can i um open them and you know which choose the path of least resistance i'm here all week um and choose whichever method you want and you get to the same answer so there you go i hope you found that useful i hope i've answered uh that question he said simple um but this is i think it's a very simple uh concept to do once you know you have these tools available in your mathematical and analysis circuit analysis toolkit to actually do this using equipotential nodes and just redrawing things and symmetry then yeah you can really simplify these circuits but hey you can do it using complex equations and everything i'm sure and i wouldn't even bother for me this is like the easiest way to do it just you know step by step reducing it so in this case i think probably you know that method there would be like the simplest way to do it so anyway thoughts and comments down below if you found that interesting if you did please give it a big a thumbs up and as always discuss down below and yes i'm on the twitters i've got 60 000 followers now on twitter and it's a way to directly interact uh with me and you can also do it on patreon of course and the um supporters section of the uh forum as well you can interact directly with me and ask me questions like this and hopefully okay i'll do a video answering them because i think you know this has nice broad appeal anyway waffled on long enough catch you next time all right please excuse the cruddy of the model i didn't have time to build it disguise or to paint it i have flatified the cube here as we uh saw before uh one percent resistors and if you follow me on twitter you'll know that i could not find damn it my box of several boxes of thousand resistors um several thousand resistors per box um so i could have matched them uh like actually hand picked them out to match them a bit better but anyway uh this is all i could scrounge 4.7 k resistors have not matched them at all so just one percent tolerance like crusty old ones from like 30 years ago you get what you're getting you don't get upset let's measure it what does the confuser say we should get well 4 700 ohms times 0.75 ah is 3.525 k do we get it oh that's good enough for australia three point five one seven that's well within spec and if we work that out we get point seven four eight three um as the scale factor so that's well within uh the one percent tolerance of the resistors that we're using and for those curious and you should be minus one percent on 0.75 is 0.7425 which would be 3.489 k and we're only 0.23 off so yeah where where balls and that one in and this is a good time to give you a trap for young players when you're using axial resistors like this or bandoliered uh components in these bands as they're called when you pull them out you'll notice look at that there's glue left on the end of it so yeah don't go sticking these directly into breadboards because you're going to come a gutser and they're not great to uh solder either so you want to get a scalpel in there and get rid of that glue okay so what i've done is i've put a 10 volt voltage source across this i've trimmed it to exactly 10 volts we'll use uh the this point over here as a ground reference but you know you can do it anywhere anyway what we want to see is that these two points here are exactly the same and these two points here are exactly the same you know that 5 volts perfect oh look at that 5 volts a smidgen out because remember we're just using stock 1 resistors here five volts and you guessed it five volts down there now as for the uh current flowing through these resistors remember how i said it should be zero well we can measure that uh we can't just use the ammeter on the multimeter because the burden voltage of the multimeter even though this is a low burden voltage multimeter um it still could be a problem because we're only dealing with a 4.7 k there so you know if you whack a couple hundred ohms in there or whatever even for a low burn voltage one it uh it will unbalance these resistors and remember when i said it's a wheatstone uh bridge and a wheatstone bridge you adjust the two of the resistors until you actually null out uh the current or null out the voltage anyway we can measure the voltage across there 0.93 millivolts and that works out too with the 4.7 k resistor 197 nano amps nano amps that's pretty low it's essentially zero and this one's even lower works out to 51 nano amps so yeah there's no current flowing through those resistors so if you actually uh trimmed all these absolutely perfectly or you put them in the simulator and they're all ideal um yeah there's actually zero current in those two resistors there so this means that i can physically snip these two resistors and it makes absolutely no difference to the circuit resistance at all cool huh catch you next time [Music]

Channel: EEVblog

Views: 102,118

Keywords: eevblog, video, resistor, resistor cube, resistor puzzle, solution, resistor cube problem, resistor cube equivalent resistance, equipotential, ohms law, electronics problem

Id: Eoh-JKVQZwg

Length: 19min 56sec (1196 seconds)

Published: Wed May 04 2022