 Hi, I'm Zor. Welcome to Inizor Education. I would like to solve a problem which I consider to be a general problem related to Ohm's law and now before the problems which we were solving even the tricky ones actually had certain logical consideration, I would say which prompt us to make the solution actually quite easy. This one is not easy but very much straightforward because this is a universal approach to how to calculate the circuits which I'm going to present you, which does not have any kind of a symmetry. Now this lecture is part of the course called Physics 14 presented in Unizor.com. There is a prerequisite course called Math for Teens on the same website, which I consider to be a mandatory as a pre-course. Okay, so let's just solve this particular problem. Now we have a circuit. So plus and minus here it branches into 2, r and 3r. r is some kind of a number, so this is three times as more resistant than this one. Here I have 3r and r and here I have 2r and we are connected back to this. So there is no symmetry here, which we were using in a very kind of a similarly looking problem before. So we need some more general approach and that's exactly what I'm going to explain right now. How to approach any kind of a circuit which doesn't really have inherent symmetry in it. Now the first thing which you should do, you should put some errors which basically define direction of electrical current. Well, in general, you know that this is plus, this is minus. Electrons are going from negative to positive, but the electric current by definition goes in an opposite direction and don't ask why, it's just by definition people determine that before they knew about electrons. So this is arrow which is supposed to be this way and this is arrow supposed to be this way. So electrical current goes this way, here it splits into two, so it's reasonable to put arrows in this direction. Now at the exit it goes this way obviously and this way and this way. Now here we really don't know which one is stronger. I mean we can guess, but it doesn't really matter what the reality is. What I'm suggesting you is to put any kind of a direction and then we will actually use this particular direction depending on we will go along or against it. And for instance, here and here and here and here, that's our directions. So basically what I'm saying is that these errors can be really any, whatever you want in any direction because what is important is how you take the currents which are going along these lines in into consideration into equations which you will have to create. Now first you have to consider all connections like this one, this is the connection, this is the connection, this is the connection, this is the connection, this and this. Now once you put all the errors you can always consider every connection as some electrical current going in it and some going out. For instance, this particular thing, we have one connection going into this. You can consider this vertex of the graph using the mathematical terminology. So into this vertex connection you have one current which goes in and two different currents which are going out. So this is I current, this is I1, this is I2. Well, this is also I1 obviously and this is also I2. Now this would be let's say I3, this would be I4, I4 and this would be I5, I5. So what you can say about this particular connection that certain amount of electricity coming in and certain amount of electricity coming out and therefore you can consider as a first equation this one. Now let's consider this particular vertex or not or whatever you want to call it. You have one connection coming in and two connections coming out. So I1 equals to I3 plus I4. I3 goes down, out from the node from the vertex and I4 goes out but I1 goes in. Here we have I3 goes in and I2 goes in and I5 goes out. So we can have I2 plus I3 equals I5. Next this, so it's 1, 2, 3, 4. So the fourth equation is I4, I5 goes in and this is by the way it's the same I as here, right? This is like a one giant resistor so it goes, whatever goes in should go out back into this. So this is I, this is I4 plus I5. So that's about currents. So every node can be converted into equation of this type. By the way how many unknowns do we have? We have one and five we have six unknowns. This is only four equations for six unknowns, not enough. Let's go further. Further is that whenever you're going through any kind of a circuit on every resistor you have certain loss of the voltage. It's called voltage drop if you remember. So for instance we go this way along this particular circuit. On this line, on this particular circuit voltage drops here by I1r, here by I4 times 3r. And that's it. So the sum of these two voltage drops should be equal to initial voltage between the terminals of the battery. So if this is you, the voltage, then you can say that I1 times r plus I4 times 3r. So it's 3i, this is I1, 3I4r is equal to you. So this is where the voltage drops by I1 times r. This is voltage drops by I4 times 3r. And the sum of these two should be equal to voltage between these two points, which is the same as between these two points, right? They're connected, so it's you. But we can consider another circuit, for instance this one. So here we have I2 times 3r. Here we have I5 times r. And the sum of these two voltage drops should also be equal to you. Well, we have two more equations, right? Well, we can have more. We might have more equations than we have unknowns. Okay, let's consider this circuit. From here we go up and then down and then to the right and go this way. This is another way the electrons are going, right? Which means in every particular resistor they lose certain amount of voltage, the voltage drops. And eventually whenever we go from here to here, it drops by I1r, I1 times r. Then from here to here it's I3 times 2r, 2I3r. And then from here we go to I5r. And this is also you. Now we can go a different way. We can go from here, here, here, here and here. I'm sorry. It was a spam. So we go from here to here. It's also the circuit. So let's consider what here will have some kind of a voltage drop. Here we have I2 times 3r. So it's 3I2r. Here we have I3 times 2r. But, and this is very important. Remember, we were putting this arrow in this particular way. We don't really know how the electricity is really going through. What kind of electrons flow will be. But we have put it this way and we counted this as something which goes out from this node for instance, right? Now we go against, so we have to put it with a minus sign. So it's a minus 2I3r. And then plus I4, 3I4r. And it's also equal to you. So we have more equations than we have unknowns. So we have what, 4, 5, 6, 7, 8 equations. Which means what? Well, it means that our equations are somehow linearly dependent. However, just what's very important is they are linear equations. They are simple linear equations. Which means this system can be resolved in a straight way just substitution. First, you can substitute, let's say, I5 with I minus I4. Put it into all other 7 equations. So you will have 7 equations with 7 variables. Then you can find I4 from others and do that, etc., etc. This is a straightforward substitution method. It's long in this case because we have 8 equations, so it's long. And I actually spent a certain amount of time and put in writing all these steps, how you solve these equations. And I put it in the notes for this lecture. So I do recommend you to read the notes. And there is a complete set of equations which basically solve this problem. I would rather suggest you to do it yourself and see if you will have exactly the same answer as I did. In any case, my point was that any kind of a circuit, however complex it is, it can be solved as far as the currents and resistance of the entire thing. For instance, after you find I, you can divide U by I and you will have a resistance of the entire thing. So all kinds of problems like this. After you put all these equations, well, first you put the errors. And again, it's absolutely up to you. You don't really have to put it like this. You can put it any way you want, as long as, whenever you are going along the circuits to calculate voltage drop, if you go along the arrow, you should do the plus. If you go against the arrow, like in this particular case, against this arrow, use the minus. That's a very, very simple thing, right? So as long as you do it correctly, you will have as many equations as necessary, and you will be able to calculate everything about this particular circuit. And that was actually my very important point about problems related to Ohm's law. By the way, I didn't specify it as the problems number three. Okay, in any case, so it's very important to understand that there is a general principle how to solve the problems of that kind, how to basically make all the calculations for the circuit. You have resistors and the voltage in the very, very beginning. Then you can calculate all the currents and all the voltage drops and resistance of the entire circuit, etc. And this is the way how you do it. Okay, let me just repeat. Errors number one, every vertex or a node in and out gives you these equations and every circuit gives you drop of the voltage. Now, by the way, there are some other circuits. You can consider this circuit, for instance. The voltage of this circuit is supposed to be drop of voltage, supposed to be equals to zero, right? If you start from this point, then you go along this, you go along this, and it gains this. And it's supposed to basically give you the right solution. Whatever you have, in which case the whole circuit will be at zero as a result, will be i times r plus i3 to r minus i2 to 3r. And it's supposed to be equal to zero because there is no difference in potential, right? There is no extra battery like in this case. So all kinds of loops you can make in the circuit are good enough. And again, you will have a lot of duplicates, a lot of, I would say, linearly dependent equations. But that's okay, I mean. So it will be eliminated in the process of solution. Alright, so try to solve it yourself in this particular schema. For this circuit, go to the website Unizor.com to this lecture and see if you will have exactly the same results as I did. I spent a lot of time actually to solve these equations and they made a lot of errors because all this arithmetic is really kind of cumbersome. But whatever you do, you do up to you basically. I do suggest you to try it yourself. So thanks very much and good luck.