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.