 Hi, I'm Zor. Welcome to Unizor Education. I would like to solve a couple of more problems related to electricity. It's part of physics 14's course presented on Unizor.com. On the same website, by the way, you can find mass 14's, which is a prerequisite for this course. Something like calculus or vector algebra you must actually know before getting any serious about physics. All right, so the problems today, there are some tricks which probably that's exactly why I chose these problems. It's not that trivial. It's not just the plain application of the Ohm's law. So we have three problems. Now the first problem is actually it is a direct application of Ohm's law. So if you have a parallel connection of two different resistors, this is voltage u, then the currents would be inversely proportional to resistances. Why? Well, probably because i times r1, i times r1, that's the difference between electric potential of these points, which is u. Now i2, r2 is exactly the same. It's difference between electric potentials of these two points, which is these two points, which is exactly the same as these two points, which is the same u. So this is equal, so these guys are equal. So i1 times r1 times equals to i2 times r2, from which I want divided by i2 is equal to r2 divided by r1. So the currents are inversely proportional to resistances, which is logically absolutely obvious the greater the resistance, the less number of electrons actually are going through this per unit of time of this. Okay, this is easy. The second one has a little trick, and the third one also. So the second one in the clock, you have a circuit which contains, all these are resistors. Well, I can put something like this, if you wish, on each one of them, and each one is equal to r. All resistors are the same. Now the wires are here. So what I need is resistance of an entire circuit. So basically what kind of a current goes along the common wires. And also I would like to know resistance of, this is A, B, C, D. I would like resistance A, C. Not resistance, sorry, the current A, C. So the current which goes along the diagonal of this assembly of resistors, whatever you call it, circuit. Now you used to solve very simple problems. You have, for instance, series of resistors, then you add them up. If you have parallel resistances, then inverse resistor, resistance of the circuit is equal to sum of inverse resistors of branches, resistances of branches. But this is not like that. That's not exactly serial, not exactly parallel. You don't really know how to deal with this. However, actually it is relatively trivial. It's just a combination of known parallel and series connections. You just have to view it in a slightly separate way. And if I will redraw the picture slightly differently, without changing the currents, without changing any connections, I'll just draw it differently. And you will see how it can be broken into separate pieces, connections which are parallel or series and can be calculated all together. Okay, so here's how I'm going to do it. Let's start from the point A. From point A I have three branches. So let's put three branches. Now, on the bottom branch I have one resistor. On the middle branch I have one resistor, two points C. This goes to point D. And I also have the third branch on the top, which has two in the series, which is easy, which also go to C. So what I missed so far is from C to D I have a resistor here from C. You see, it's exactly the same thing. You have between A and C you have these two resistors, R and R. Then you have the diagonal, R. From the diagonal to point D I have, I stretched it this way, but it's still exactly the same thing. This and this go to point D and D is my source of electricity. Now as you see, it's much better. It's exactly the same, but you view differently and now it's obvious how to do it. Well, first of all you replace these two with one, which is two R, right? Now what happens here? Well, now this part is just two parallel resistors and we can find out their total resistance as one over two R plus one over R is equal to three over two R. So the total resistance is two R over three. So this is two R over three. Now you have two R over three and this one in series. So you have to add them together, which is five R over three. And now you have this five R over three on the top branch and R on the bottom branch. How to add them together? Well, you add inverse of one over R plus three over five R, which is equal to eight over five R. So the total resistance is five R over eight. This is the total resistance of the entire thing. Well, if you know the voltage of the source of electricity, obviously the current E i is equal to U divided by five R eight. So that's my current in the common area, in the common wire. Well, but I'm interested in diagonal, which is this one. Well, I know the resistance. So I need to know the difference in potential between A and C. Okay. Now, how can I do it? Well, the easiest way is I know the the difference in electric potential between A and D, which is U. So from A to D, I have one resistor, which is two R over three, and another, which is R. So the voltage drops first by this and then by this. So first I can determine what's my current in the common area, in the common wire. The common wire is this one, this one. So what's the common wire here? Well, if the total resistance is two R over three plus R, which is five R over three, my I in the common, let's call it I one, is equal to U over five R over three. So this is the current, which is coming here and here. Now, if I have the current, I can find out the drop of the voltage between A and C. Well, my resistance is two R over three. My current is this one in the common area, which comes. So my voltage drop delta U would be equal to my I one times two R over three, which is two U over five R. So that's the voltage drop from here to here. No, I don't need R anymore. Cancels, three cancels. So I have two fifths. So I have two fifths of U and I have the resistance R. So between this and this, the voltage drops by this, and I have the resistance. So it's two U over five divided by resistance. That's my Iac. That's the problem solved. Okay. And the third one also with a little trick, but again, it depends on how you view the whole thing. You have a cube made of resistors. So each edge is a resistor, R, all identical. Now, this is the main wire. It goes, if this is A, B, C, D, E, F, G, H. So from A and G points are connected to source of electricity. The main diagonal points. And I have to find out what is my resistance of an entire circuit. So for instance, I'm interested in what kind of a current will be in this common wire. So I need the resistance of the whole thing. Okay, how can we do it? Well, if you will just look at this like this, again, you don't see any kind of a series connection, parallel connection. It's all kind of messed up, quite frankly, because it branches all the time. So we have to view it somehow differently. And here is what I suggest. Now, consider all the resistors are identical. What you have here you have from this point you split your current into three branches. They are equivalent because each branch splits into two, this splits into two, and that splits into two. And it meets from another side with exactly the same thing. They have three branches, one, two, three. Each one is splits into two, this splits into two, and this splits into two. So it's all symmetrical from both sides. And if you will consider points E, H, and D, they are coming from point A. And they're going into the same thing. And they're supposed to be dropping the voltage in exactly the same fashion, because it's all symmetrical. Relatively to this main diagonal, all these three branches are exactly the same, which means they have to drop the voltage in exactly the same way. Okay. But what does it mean? It means that if I will bend the wire and connect point D, H, and E together, between these points, if I will just connect them, there is no current between these points. Because the voltage is exactly the same and the difference is electric potential is zero. So there is basically no voltage and no current. If there is no current between E and H, if I will connect them together. So let me just again bend the wire and what happens? From A, I have three branches, and I bend them together and merge them. I can do it without any problem, because there is no difference in electric potential between these points. So there is no current between them if I connect them. So I just merge them together. That's okay. Now from another side, I do exactly the same thing. This is my G point. And this point is X, which is equal to E, D, and H. And this point Y is equal to F, G, and C. Now, if I merge E, H, and D, now actually, that's not H, wait a minute. A, B, it's B. It's B, H is on the top. Okay. H is on the top. Sorry. So this is B. E, D, B. Right, B. And this one is F, H, not G. F, H, C. Sorry about that. So again, E, B, and D are connected into X. F, H, and C are connected into Y. Now what's between X and Y? Well, each point E, B, and D has two connections to this group. So basically, I have two connections from E, two connections from B, and two connections from D. Now, each line is a resistor which has a resistance R. So here I have three different resistors parallel to each other, and they're combined is 1 over R plus 1 over R plus 1 over R. It's 3.R over R, so it's R divided by 3. This is the total resistance of this. The total resistance of this will be R divided by 6, and this will be R divided by 3, and they have to connect them together sequentially, right? It's in series, so I have to add them together, and their combined would be what? 2 and 1, and this is 5R over 6. And that's my combined resistance of the entire cube of resisting wires. Well, here is what's very interesting about this. If you have points where electric potential is exactly the same, or if you wish the voltage is dropped in exactly the same fashion from here into all these three points, then you can connect them without any problem. In the previous lecture with the problems, I had similar problem, but there I had points which are also the same electric potential, it was something like this. R, 2R, 2R, so far so simple, right? But then I had this one, 3R, which really distorts the picture. It's no longer like parallel sequential connection, whatever. But again, from the consideration of symmetry, you see R and R, 2R and 2R. So in these two points, there is no difference in electric potential, because the drop is supposed to be exactly the same. The current is supposed to be the same, so the voltage drop should be the same, which means the potential in these points is the same. And if it's the same, I can just completely take it out from the circuit without changing anything. It has zero current. So in this case, I removed a piece of the circuit along which there is no current. Here I added, I basically merged whatever, I added certain connection, because I know that there is no current going through these connections. In both cases, I was using the same principle. If the potential is the same between two points, then it doesn't really matter whether they're connected, not connected, there is a resistance, no resistance, et cetera, between them, you can safely consider both ways, whatever is convenient for you. And in this case, it was convenient to connect them. In that case, it was convenient just to remove piece of the circuit along which there is no current. Okay, that's it. I suggest you to go to the website, to this lecture, you have notes, you have all the problems and solutions, read the problems, solve it yourself, then check the solution. It's a very, very useful exercise. So thank you very much and good luck.