 Hi, I'm Zor. Welcome to a new Zor education. We continue talking about electricity and it's a good time to solve a few problems. A few problems for the Ohm's law. Actually, I would consider this to be probably the most important law in the electricity. So it really works well to spend some time on problems which are related to Ohm's law. Now, this lecture is part of the course called Physics for Teens, presented on Unisor.com. I suggest you to watch it from Unisor.com because it has detailed notes for each lecture. All lectures are presented in logical sequence as a course and I suggest you to take the whole course. And there are exams for each part of the course. Everything is free, so no financial strings attached, no advertisement. So use the website. It's much better. Okay, so let's talk about the problems. And the first one is probably the most simplest and most classical one. So you have a circuit where you have a combination of parallel and consecutive sequential connections. Okay, these are resistors in a series and these are the resistances of these. This is a resistor which is in parallel to this series. And what's given is the voltage on the battery. And you have to determine the current in this branch and in this branch and in this branch of the circuit. Okay, so how do we solve it? Well, first of all, let's start from the Ohm's Law. Ohm's Law is that U is equal to I times R or I is equal to U divided by R. We know this is Ohm's Law. Now in this case, to find U, we have to find the combined resistance of these three different resistors. Now, it's not one resistor. It's three, so we have to somehow combine. Now, we don't know how to calculate the resistance of this combination of three different resistors. But we do know that we can just consider these three resistors as basically two resistors connected parallel. This is one resistor and this is the combination of these two. Now, all resistors connected in a series are actually the same thing as one bigger resistor which resists more. How much more? Well, the resistance of these two is the sum of their resistances. Remember, if you connect in a series, it's kind of obvious that resistance is added together because the electrons have to drag their way through one and then another. So basically, they're wasting certain amount of energy on one and then another, which is basically to waste for the sum of them. So basically, first what we can do is the resistance R23, which is the combination of these, is the sum of R2 and R3. So this is the resistance of this entire branch of a circuit. Now, we actually convert this scheme into much more simple one. This is R1 and this is R23. Now, this is a regular parallel connection. And again, we know how to calculate the resistance of two resistors connected in parallel. 1 over R is equal to 1 over R1 plus 1 over R23. So if you remember, we have this formula. In a parallel, we are adding inverse of resistors. In a series, we are adding resistances themselves in a parallel inverse. Okay, that's basically enough for us because we know what this is, right? This is R2 plus R3. Now, all we have to do is multiply it by U. Now, what is this? Well, this is the resistance of an entire circuit, right? R. So the I is this. Okay? Now, this is obviously I1 and this is I2 because obviously the current which goes through this one is exactly the same as current going through that one. The current is basically the speed of electrons, right? So we have basically determined everything. So we have I1 which is U divided by R1, we have I2 which is U divided by this and this one would be their sum. Now, obviously, instead of using this approach to calculate the entire resistance and the I as a result of this, we can do it in reverse. Well, let's first determine what's the current which goes through the first resistance. Well, we have the voltage, it's U, right? Because this one is parallel to this one. So on these ends we have the voltage difference in electric potential U. So we have a difference, so we can use the Ohm's law and we can immediately say that I1 is equal to this one. Now, for I2 we do have to calculate the sum of these. So again, we know the voltage in these two ends, so we have to find the resistance of this one which is the sum, so we divide U divided by sum that would be U2, I2. Now, obviously, the current in this common branch of the circuit is equal to sum of these two currents because again, what is the current? It's the amount of electricity, amount of coulombs of electricity per unit of time. So obviously, if we are splitting this flow of electrons, the sum of these is equal to the whole, right? So that's basically it for this very simple, very elementary problem. Next. Okay, next I would like to find out what exactly the amount of work is done by the battery. So this is the regular circuit, this is the Ohm's law or I is equal to U divided by R. Now, so my question is, this battery is producing certain amount of energy and then this energy is basically spent on circulating the electrons in the circuit. So let's consider that we do maintain this type of connection and the battery is generating certain amount of electric energy. It's going around, it probably heats up the resistor and it's all happening during this time T, for instance. So my question is, how much work in JOLS is basically generated during this time, the battery, if you have all these parameters. So you have the parameters of the circuit, you have the resistors, resistance, you have the voltage, I is equal to obviously from the Ohm's law. So you have all the parameters. Now, question is, I would like in terms of these parameters to come up with something from, we know from the mechanical part of this course, energy in JOLS basically, the work. Okay, so what is the work? Let's just think about it. What is the voltage? Let's just go back to the definition of voltage. Voltage is amount of energy which we have spent or is needed for to separate certain amount of electricity in Coulons basically, right? And that's why basically it's per unit of electricity, per Coulon. So how much energy we need to produce one Coulon of electrons, so to speak, right? So what is the one volt? One volt is amount of energy we need to separate one Coulon of electricity. So amount of work therefore it's from definition of electric potential is equal to the voltage times number of Coulons which this battery actually is performing. Now, this is done for the purpose of circulating these Coulons. Whatever the Coulons of electricity is produced is circulating and what is I? I is amount of electricity per unit of time, right? That's the current. What is the current? Current is basically speed of electricity, amount of Coulons of electricity per unit of time. Now from these two we conclude that A is equal to U times I times T. So my very important point is that work which is performed by the battery during the time T is equal to U times I times T. The voltage times the amperage current times time. Now from this obviously follows that the power if you remember is amount of work per unit of time is equal to U times I. Okay? Amount of work per unit of time. Now we can express this differently using the Ohm's law. For instance I can use it, power is equal to, so let's say instead of U I can use I R so it will be I square R and amount of work would be equal to I square R times T. Or I can put it in terms of voltage so instead of I I will use U divided by R so I will have U squared divided by R and amount of work is equal to U squared divided by R times T. So these are different formulas for basically the same thing. How to express something which we know from mechanics, the work. Now we are actually talking about electricity so the concept of work also exists. And again eventually we will understand that the work actually is to heat up the resistor but that's the next problem. And these are formulas which express the amount of work which is done by the battery or any kind of source of electricity to basically force the electrons to go through the circuit and to do whatever is necessary they are doing. Because as they are going through the resistor if it's something like a lamp then the temperature of the filament of the spiral whatever in the lamp is increasing. That's why we have the light, right? If it goes through the electric motor for instance in electric car all these batteries which are connected in some kind of a configuration are spending their energy to basically force the car to go forward. There is an electric motor which is a consumption, a consumer of electricity and converts into mechanical energy. So whatever the work is done to heat up the element or to force electric motor to spin the wheels of the car whatever it is the amount of work is always calculated using and power these formulas. So it's very important. I would like to actually make a very small comment that I'm sometimes using the problems to introduce certain theoretical concepts. This is obviously the problem but at the same time it's a theory actually which allows you to basically understand from the concept of energy and work what electricity is all about these voltage and amperage how they are connected to energy. So work and energy and power are connected to voltage and amperage in exactly this fashion. Next. So now the third problem is basically a continuation of this one and we will talk about the heating. So this is an element and this is our bulb and it produces certain amount of electricity. There is a voltage here and let's say this is the bulb like a regular bulb we are using in incandescent lamp which we are using at home. So it has this tungsten filament inside and I would like to find out actually how high the temperature will go inside this lamp. How far up the temperature of the tungsten will go if we will start pushing the electricity through it. And let's think what exactly we need for this. Well obviously we need the resistance which gives us the all elements we need to calculate the amount of work which is done by the battery. And from the previous problem it is u squared divided by r times t. So this is amount of work. So we know the work. Now where is it going? It actually goes to produce heat right? So this is exactly the same as amount of heat which is consumed by this lamp. So it consumes amount of heat. Now if it consumes certain amount of heat during this time, if we know the resistance and we know the voltage. Now my next question is how high the temperature will rise because of that. Because since we are applying heat then the temperature is supposed to rise. How much? Well there is a specific heat for number for every material for tungsten for instance from which the filament is made as well. That's amount of heat which is needed to heat up unit of mass by unit of temperature. So if we will multiply it by mass and multiply it by difference in temperature we will get exactly the amount of heat which is consumed by this lamp. That's basically it because from this we know delta t. That's the temperature. It's q which is the same as this one, u squared times t divided by r and divided by specific heat. It's a constant for metal and mass. So if we know the mass of this tungsten filament inside the lamp we know that this is tungsten for instance which means we know it's a specific heat capacity. We know the resistance, we know the time, let's say second or two seconds and we know the voltage. We know what is the rise of temperature will be during this time. But obviously as the temperature is rising it emits the heat as well. So it's not like we are accumulating heat and the longer we are under this condition the higher the temperature will dissipate obviously. But we are not talking about this, right now we are talking only about amount of heat which is consumed. Now obviously because the heat dissipates it's not going up to infinity as the time is increasing. It goes to certain level and then amount of heat dissipated will be equal to amount of heat consumed by the filament of the lamp. So this is the formula and if you will put some normal values which I have in the notes for this lecture. So the u let's say will be 110 volts which is normal for the United States. Let's say T would be two seconds. Let's say resistance would be 200 ohm. And specific heat for tungsten would be I think it's 134 J per kilogram degree of and the mass I don't know what the mass 0.001 kilogram. Let's say so if you will do all this calculation I have come up with 900 degrees temperature rise in two seconds. Well I hope I don't make any mistakes but it doesn't really matter. Right now we are talking about the theory and the theory is basically here. That's the whole thing. So that's my third problem and now I have the fourth one which is I consider very interesting because it's very simple but there is a trick. I love the problems with the tricks. So let's say we have the following connections. So far it's simple right? We calculate the resistance of these two which is the sum of these. The resistance of these is the sum of these so it's 3R and 3R and then these are parallel. We calculate the total resistance without any problems. So if you have u, you have I, you have I1 and you have I2. Now let me introduce the bad thing. What if I have another resistor which is connected this way? Now the whole circuit is neither in a series nor in parallel and it cannot be split into pieces and calculate each piece separately and then combine. Like without this I have two separate pieces and calculate this series and this series and then parallel to each other. Here I can do it. Question is what can I do? And this is basically one of the tricks which can be used in problems like this. Let's just think about it. The whole thing is very much symmetrical. So the top part and the bottom part are completely symmetrical and this one doesn't distort the symmetry. Symmetry is still there. Now if symmetry is still there it's very reasonable to basically state that I1 is equal to I2. Because from the perspective of these two currents there is absolutely no difference and if there is no difference then this is obviously true. And it's equal to I divided by 2. Because they are together we are splitting one flow into two absolutely symmetrical ones. But if it's symmetrical remember what is the voltage drop? The voltage drop from here to here is equal to I1 times R. And what is the voltage drop here? Let's just talk about I1 in both cases. Because they are equal to each other. It's also I1 times R. These Rs are the same. So the voltage drop is the same. So what is the voltage between this and this? Well it's whatever comes here minus voltage drop you come here. Now here you have exactly the same voltage and exactly the same drop. So you have the difference between these two electric potentials are zero. Because we have the same here and we dropped it by the same voltage. And we have exactly the same in these two pieces. These two pieces whatever it doesn't really matter. Now what does it mean that the voltage potential is the same in these two points? Well it means that there is no current through it. That's what it means. If there is no difference in electrical, why is there a current here? Because there is a difference in potential between minus and plus. There is an axis of electrons here and a deficiency there. This deficiency is exactly the same here. And this axis is exactly the same here. So from here to here the voltage drops. Here it drops by this one. And from here it drops by this one. The same thing. So these are exactly the same electric potentials. There is no difference and there is no current. And since there is no current to make all our calculations we can do just this. As it does not exist. That's the trick. So you have to understand that because of the symmetry of this particular problem there is no difference between electric potential in these two points. And therefore we can just completely ignore it in all our calculations. And therefore we have r plus 2r which is equal to 3r on the top. The same thing on the bottom. And they are parallel to each other. So it's 1 over 3r plus 1 over 3r is 1 over r total. Which is what? 2 third 3r. So r total r total is equal to 3 second of r. And that's why u divided by this r total will give me i. i is equal to u divided by 3r2. And i1 is equal to i2 is equal to half of this which is u divided by 3r. And that's the end. So I like this particular problem because it actually needs some kind of a logical trick. And there are some other problems which I will present similar to this one. I do suggest you to go to the Unisor.com to this particular lecture. Take a look at the notes and solve the problems yourself right now. One thing is to listen to whatever I said. And another thing is to basically do the problem yourself. This is the most important part of any kind of an education. You can do it yourself based on whatever you have learned from the lectures or text books or whatever. Don't pay attention to the solution which I present. Just read the problem itself, solve it and then check your result with the solution. That's it for today. Thank you very much and good luck.