 Hi, I'm Zor. Welcome to Unizor Education. Today I would like to solve one problem related to using capacitors with direct current. This lecture is part of the course. The course is called Physics for Teens and it's presented on Unizor.com. I suggest you to watch this lecture and all other lectures from the website because it contains basically the course, which means everything is logically arranged in chapters, lectures, whatever. There are problems solved, there are exams if you want. Everything is completely voluntarily. There are no ads. And one more thing, there is a prerequisite course for this called physics for physics for teens. It's called mass for teens on the same website. The familiarity with mathematics is very important for physics and this particular lecture, which is about one particular problem, is just an illustration how important mathematics is for studying physics. In this particular case we will have to solve a very simple differential equation. So calculus is a must, vector algebra is a must. I mean there are basically many aspects of mathematics which you really have to know before you seriously approach physics. Even on the level which I am presenting, which is basically like teenagers, high school, maybe first year in college, not higher than that. Okay, so what we're talking today about is, well, it's presented in the form of a problem. So what happens if here is what kind of a setup I have here. First of all we have a battery, direct current battery. Now to this battery we can connect a capacitor. By the way capacitors were introduced already in this course in the previous lectures and I assume that you know something about capacitors from this course or from any other source because I would just use certain properties of capacitors. Now there is a switch here and it can connect this capacitor to the source of energy to a battery. Now then there is another thing here, just a resistor. Now we have the voltage of the battery, we have capacity of the capacitors and we have and we have R here, resistance of the resistor. Now my experiment is the following. First I have this switch in this position. Now what is basically happening in this particular case? Well there is a source of energy so it will basically charge. Now plus and minus, the plus will be here, the minus will be here, so this thing will be again plus and minus, that will be excess of electrons, this will be deficiency of electrons and the capacitor will be charged to a certain degree, right? Okay so what happens then? I will switch this to this position, basically disconnecting the battery. What happens then? Well we have excess of electrons here, deficiency here, so it will go through a resistor and it will and these electrons will compensate the deficiency of electrons here, right? So basically we will have some kind of a current. Now since electrons are moving in one direction only, eventually we will have no difference between the number of electrons on the plates of the capacitor and the torque, the electric current will basically stop. So from certain level in the beginning, electric current will diminish basically to zero and the charge on this capacitor also from some initial value, which depends obviously on the battery and its capacity will also diminish to zero, there will be no voltage at the very end. So I'm interested in this process of discharging electricity from the capacitor through a resistor and how exactly the voltage, actually let's talk about the amount of electricity which is accumulated in the capacitor will change with the time. Obviously it's changing down to zero, same thing with electric current in the resistor. Initially it will be some rather large value but then gradually it will diminish to zero. So I need to express these two functions, the amount of electricity accumulated in the capacitor and the current in the circuit as the functions of time. So what happens? I mean I know it will be diminishing, question is how? I need a mathematical expression of how exactly this is happening. Okay, so this is the problem, how can we address it? The first thing which we do is we have to charge our capacitor, right? So initially we have this switch in this position, now this part of the circuit is not working at all because it's broken and now the capacitor will accumulate a certain amount of electricity. Okay, what is this amount of electricity? When we were talking about capacitors we have introduced this concept of capacity which basically means the following. If you have certain voltage here, you have certain amount of electricity in cool-ons, the basically number of electrons if you wish, accumulated in the capacitor. It depends obviously on the voltage. If I will increase the voltage, the amount of electricity which will be accumulated here will be obviously greater. Now the mechanics of this is simple. Now this is plus, this is minus, right? Plus, minus. So electrons will go here and deficiency of electrons will go here. Now it will go here until it will accumulate sufficiently enough to basically resist any additional electrons which are coming into this plate. So basically it will be some kind of an equilibrium between the number of electrons on the terminal of the battery and on the plate of the capacitor. It will be basically the same. Same sink will be here as far as deficiency of electrons is concerned. So electrons will flow here until there will be so many that they will resist any additional sink. At the same time the deficiency of electrons and extra electrons between these plates, they are attracting to each other, right? So that's another force. So there is a force which resists additional electrons to come here, but there is a force which actually attracts electrons to this plate. And obviously the closer the plates are, the greater that force is which compensates the force against coming electrons from the battery. So as a result, depending on certain physical properties, like how big these plates are and how close they are to each other, and actually what else is what's exactly in between these plates, all these physical characteristics, they play a very important role to amount, to total amount of electricity which capacitor can hold. So there is a characteristic called capacity. And what happens is that given a particular capacitor with a particular characteristic of capacity, it's basically a ratio between the amount of electricity accumulated on the capacitor and the voltage of the battery which is attached to it. So it's a constant. If we double the voltage, the amount of electricity accumulated between these plates will also double. Obviously it cannot go to infinity because at some moment the forces between these two plates will be so big that there will be an electric discharge between them. So we're not talking about this borderline kind of behavior. But in general, you increase the voltage by certain factor and the amount of electricity accumulated here will also increase by the same factor. So the ratio will be the same. It's constant. And this is a characteristic of capacity, a capacitor. It's called capacity. So it's given. Now this voltage is also given. Well, if that's true, then we can actually find out how much electricity is accumulated in the very beginning of our process before we switch, before we flip the switch, right? So obviously it will be Q which is V times C. V is a constant, C is a constant. Q. Well, again, that would be in the very beginning of the process discharging. So now I switch it and I start the timing. Okay, this is as soon as I switch, this is time is equal to zero. From this time on, I would like to know what happens. Now, at this moment, so that's why I will put zero here. So at the moment zero, at the moment T equals to zero, Q amount of electricity on this capacitor is equal to V times C. Both V and C are given. What happens then? Well, as soon as it happens, now my battery is completely out of the picture and the plus and minus start working basically. There is a voltage between plus and minus, right? So the capacitor will start discharging. The electrons will flow from wherever they were into the opposite plate, basically causing the electric current. Okay, that's good. We definitely have some kind of, since we have a certain amount of electrons in one plate and certain deficiency of electrons in another plate of the capacitor, we definitely have certain voltage between them, right? It's a difference between potential difference in number of electrons. So there is certain amount of, there is certain voltage here and it's obviously a function of time because as soon as we start the process of discharging, the amount of electricity amount, number of electrons, so to speak, is diminishing. So the voltage between these plates will be diminishing. But we would like to introduce this concept of voltage because it will help us to find out what exactly is the current in this particular resistor, right? So what is the current? Well, we know there is Ohm's law. So the current, which is also variable, multiplied by resistance, which is constant, will give us the voltage between these two plates, right? Now this V and this V of T have nothing to do with each other. This V is given, this is the voltage of the battery. It's out of the pictures right now, completely. So this is the voltage between these two plates. Maybe I should use another letter, but it doesn't really matter. You understand? So this is the Ohm's law. Now, what else is important? Well, we know that the voltage between these plates is related to amount of electricity accumulated and the capacity of the capacitor. This is exactly the same as before, but before we had the constant voltage connected to this. So we had certain constant amount of Q accumulated, but as soon as we switch, now the voltage between these two obviously is related by the same equation. But now both of these are variable, okay? So we don't really need this voltage. So I will substitute this into this. And what do I have? Well, this can be expressed as C times V of T is equal to Q of T. So if I will substitute V, it will be I of T times R times C equals Q of T. Now, this is an equation, a functional equation, where two unknown functions, which are supposed to be actually found, that's the purpose of this problem, have to find I of T and Q of T. The current, as it changes with the time and the amount of electricity accumulated in the capacitor as it discharges. So this is my equation, but now we have two functions and one equation. We need another one. Well, another one is very simple. You just have to remember the definition of the electric current. Now, what is the definition of electric current? It's a rate of electricity by time, right? Which, mathematically speaking, is just first derivative. So if you have the amount of electric charge Q of T here, and you have the current here, so the current is actually the derivative of the amount of electricity. So as this is changing, this is also changing, but this is actually a derivative by time, well, rate of change basically in a simple language, but in mathematical language, it's a first derivative of Q by time. So I of T is a derivative of Q by time. It's basically definition of the current. The only little thing I would like to add here, I would like I of T to be a positive value, certain amount of amperage, right? Regardless of the direction, because it all depends on where is plus, where is minus, I don't want to worry about this. I know right now that Q of T is changing how. It's diminishing, right? If it's diminishing, so its first derivative is negative. So I have to put the minus sign here that would solve all my problems. So now, I of T will be positive, because this is diminishing function, derivative is negative, plus this is another minus, it will be positive. And this is my second equation, which connects my two functions, I of T and Q of T. So I have two equations with two unknowns, unknown functions. One of them happened to be differential equation. Well, so be it. I mean, we're still, as I was saying, mathematics is a must to know before you go into physics. Now, it's a very simple actually differential equation, and I will solve it right now. But again, for those of you who have any kind of a problem with mathematics, with calculus, you can go to Mass 14's prerequisite course on the same website, or any other source of information. But I assume that we know what differential equation is, because again, mass for T is a prerequisite course, and I assume you have to know it. So how can we solve this? Well, it's actually kind of simple thing. So what I will do, I will resolve this I T for I T, I of T, and substitute to this. So I will have only one differential equation for one function Q of T, right? So I of T is equal to Q of T divided by RC. So I will put it here. And my equation would be Q of T divided by RC, right? Is equal to minus DQ of T by DT. So this is my very important differential equation, which I have to solve. And I will. It's very easy. And I don't think I need this anymore. So how can I solve this? It's very easy. This is very easy to transform into DT divided by RC minus is equal to DQ of T by Q of T, right? I change that. That's simple. Now, this, what is this? Well, this is differential of logarithm of Q of T. So if you remember the first derivative of logarithm is one over the function, and then times the derivative of the inner function, which is this. So that's what it is. And on the left is differential of what? T divided by RC with a minus sign. Okay? So now, well, yeah, I will do even better. I will put this minus inside. Okay. So we have equality of two differentials of two different functions. One function is logarithm of Q of T. Another is minus T over RC. If differentials are equal, then the functions themselves differ only by a constant, right? So you can say that logarithm Q of T is equal to minus T divided by RC plus some kind of a constant, which I don't know. Well, from here, I can do even further simplification. I will use exponent for both sides. If I will do e to the power of logarithm, that's the definition of logarithm, obviously. So e to the power of logarithm Q of T is equal to Q of T, right? That's the definition of logarithm, which is another thing you have to know from the math. So if I will use e to the power on the left, I will have Q of T. If I will have it on the right, I will have this. Now, exponent of some of these is a product of exponents. So it will be exponent of constant, which is some kind of a constant, okay? I don't know. And exponent of this, I have already received my function Q of T. So this is amount of electricity as it's changing during the discharge of my capacitor. So the only thing I know is K, but I know this. This is my initial condition. It's always for differential equation. You need initial conditions. Otherwise, you cannot obviously resolve these constants. So if I will substitute T is equal to zero, what happens here? e to the power of zero is one. So on the left, I will have Q of zero, which is V times C. And on the right, I will have K. So K is equal to Vc. So I can put it here. And this is the answer to the question, what is my amount of electricity accumulated by a capacitor? Now I have to find out the electric current. I will do it from here. Q of T divided by RC. If I divided by RC, it will be divided by R times, and that actually completes the problem. By the way, let me check if this is actually true. Let me have a first derivative of this. Will I get this? Okay, derivative of this function. So I need minus derivative of Q of T divided by DT is equal to minus. Now V times C is just a constant coefficient factor. Derivative of e to the power is e to the power times derivative of the inner function. Inner function is basically T times constant. So it will be just a constant minus one over RC. And what happens here? C and C minus and minus. I will have V divided by R and exactly what is here. So that's kind of a checking. Well, that's it. I think it's a little bit more than people at school usually go through as far as the problem solving. But I think it's a very useful problem. And primarily I think its usage is towards mathematics. As a mathematician, I appreciate actually the usage of math in all aspects of physics. I mean, I respect experimental physics and it gives certain food to the theoretical physicists. But then there is an opposite kind of influence. And you can obviously arrange this type of experiment and really see how gradually changing, exponentially changing the current, you can measure the current using ammeter. And you will see how it will change from something which is initial where T is equal to zero. So which is V divided by R as if there is no capacitor at all. That's what in the very beginning you switch the flip. You flip the switch and the capacitor has basically the same voltage as the battery used to have. So that's why in the very beginning you would have V divided by R which is the Ohm's law. But then this V is changing. It's not the initial value V which is the same as in battery. But it goes down. And you have exponentially, the graph of this function is obviously this. So the current is exponentially going less and less and less because you have less and less charge accumulated in the capacitor. Well, that's it. I suggest you to read the notes for this lecture because this website contains not only video lectures but also notes for each lecture which is very important to read. It's basically the same thing which I was presenting here but it's like a textbook. And well, that's it. Thank you very much and good luck.