 Hi, I'm Zor. Welcome to Inizor Education. I would like to discuss today a particular circuit, a circuit which includes inductor and a capacitor. The inductor has a characteristic called inductance, which is usually symbolically used as letter L, and the capacitor has a capacity or capacitance, which is usually signified as letter C. So L-C circuit means it's a circuit with inductor, signified by L, and capacitor signified by C. So they are brought into the same circuit. Circuit has the source of energy, which is alternating circuit, and I would like to know what's happening. Now, this lecture is part of the course called Physics 14 on Unisor.com. I suggest you to watch the lecture from the Unisor.com because it has a textual explanation, which is basically like reading a textbook. And it's very important in this particular case for this lecture because it's basically math. And I think that this lecture should really be another evidence, so to speak, that mathematics is especially important in physics. You cannot really study physics without really good knowledge in mathematics. Now, that's why the course Physics 14 has a prerequisite. It's called Math 14 on the same website. I suggest you to basically familiarize with everything whatever is in there. And there is a lot, especially calculus, especially differential equations. Because this lecture is basically less of a physics and more of mathematics including differential equations. And differential equations are everywhere in physics. I mean just starting from the second Newton's law when f is equal to ma, a is actually the second derivative of distance by time. So it's a differential equation. Okay. So, back to physics. So here is my circuit. It has a source of alternating current. Then it has an inductor. It also has a capacitor. Now, this is source of electricity. It's some kind of a generator which has a rotor. And the electromotive force generated, the voltage generated by this particular generator, is sinusoidally changed. Omega is angular velocity. T is time. And E0 is the peak voltage generated by this. So we did talk about this when we were discussing AC alternating current. Now, this is an inductor with inductance L. Now, what L actually means? Well, whenever you have certain current going through this inductor, you know around every wire there is a magnetic field. So whenever you have an inductor which has basically many loops, every loop results in a magnetic field which goes through this. So basically we have a magnetic field going through this inductor. Now, I'm interested in the voltage drop, VL of T, here. So why the voltage drops? Well, the voltage drops as the current goes through the inductor because there is a self-induction, if you remember what it is. So there is a special lecture in this course before this which explained why the self-induction actually exists. It resists if the current goes up, it resists to go up. If the current goes down, it supports it. So again, it resists to go down basically. So it's always against. So it's some kind of equivalent of resistance for a resistor. And here we have this self-induction which resists the electric current to change its value and the ability to do this type of thing obviously depends on what kind of inductor we have, how many loops, for instance, we have, what material, etc. So this voltage drop is actually a result of self-induction. Self-induction depends on how fast the magnetic flux is changing. It's a rate of change. So basically, this voltage drop is proportional to rate of change of magnetic flux. The faster the changing, the more resistance actually occurs. So basically the voltage drop is equal to L, I don't have space. So first of all, we are talking about rate of change. Rate of change means first derivative by time. And we have to multiply it by L and that's what gives me drop of the voltage. Okay, I'm sorry, it's not here, it's just this. The voltage depends only on rate of change. However, the magnetic flux depends on properties of inductor, which is L inductance, and the current which is going through it. That's where it is. So again, we have the current. Current generates magnetic flux which is proportional to inductance of the inductor. And the rate of change of magnetic flux gives me the self-induction which basically resists the change of the current. So that's all about inductor. Now let's talk about capacitor. So capacitor has a different property. The voltage here, if you have certain voltage between these two plates, there is certain amount of electricity accumulated on the plates. And there is a proportionality between the amount of electricity accumulated on the plates and the voltage. The greater the voltage, the greater amount of electricity accumulated. So amount of electricity Q divided by voltage is a constant which is capacity or capacitance of the capacitor constant. This is a constant. Now, what is Q of t? Q of t is amount of electricity. Now, what's important is the current. Now what is the current? Current is rate of change of Q, right? It's basically, again, first derivative. So the current which goes through this capacitor and the current does go, as you remember, because it's an AC. It's all dependent on, basically, it's equal to rate of change of the amount of electricity. This is basically a definition of the current. What is the current? Current is amount of electricity per unit of time, roughly speaking. More precisely, per unit of time means we are differentiating by time. So this is all about capacitor. Now, let's think about it. So we have a change of voltage here, drop of voltage and drop of voltage here, and it's supposed to be equal to original generated voltage. So basically I can say that Q of t should be equal to their sum. So this basically it. So we have completely defined our mathematical universe from which we can derive anything we want. How many? Well, first of all, obviously, I of t is the same. I mean, I didn't specify I l of t and I t and I c of t, because it's the same circuit, which means it's the same current going everywhere. So that's why I i is here and I is here. It's exactly the same I, the same current. The voltage drops are unknown and the current is unknown. So we have three variables. Okay, we have three variables and how many equations do we have? Well, these two make one equation where I of t and Vl of t is related. Now, these two make another equation if I will substitute Q. It will be also depending on I and Vc of t. And this is the third equation. We have three equations with three unknown functions. Now, a couple of these equations are differential, but still, I mean, it's still an equation. So from now on, we will basically devote our attention to mathematics rather than to physics. This is the end of the physical part of this lecture. Now let's talk about math. Okay, so what I would like to do is I would like to really reduce it to three equations. Now, first of all, this one. So if I will differentiate this 5t, I will get this. So let's just differentiate L times derivative of I should be equal to V, right? Okay, L times derivative of I should be equal to Vl of t. That's my first equation. Now, second equation from here. Now, if we will multiply here, it will be Q is equal to C times V. And now we're differentiating and we will get I. So I of t is equal to derivative of C times V. So C times Vc derivative. Okay? And third equation is as it is. Okay, this is my system of three equations. Two of them are differential, as I said. That's about it. Now, I'll name this. Now, E is a known function, right? So I assume that E0 and omega are known. So E is a known function. Unknown are I, Vl and Vc. Okay, so what do we do first? How do we solve it? Well, how do we solve any equations? We're trying to simplify it as much as possible, right? So first, we will find Vc from here, from the third one, and differentiate it and put it into the second one. So what I have is Vc of t is equal to E of t minus Vl of t. Now, I'm differentiating and multiplying by C. I of t is equal to C times derivative of this. Derivative of E of t is E0 cosine omega t and omega. Okay? Minus Vl of t. This is my equation. Now, together with this, the first one, I have only two unknowns, I of t and the L of t. So I will use the L of t from here and substitute here. All I need is differentiated. So I will have I of t equals C times omega times E0 times cosine omega t minus derivative of this, which is L times second derivative of t. This is the first and I'm differentiating. I will have second derivative. Now, I have one equation but of the second order because I have a second derivative here. But that's what it is. So let's just think about this. Let's just simplify it a little bit. I will move everything, move all unknown to the left. All right? So what I will have is... Oh, I think I've, by the moment, I of t is equal to C times this minus C. I forgot to multiply it by C. Sorry, I forgot the fact. So I of t equals C times this, derivative of this and derivative of this should be multiplied by C, right? And that's why C also should be here. Okay, let me check. C equals sine minus C L. Right, that's right, now it's right. Okay, well, as you see, sometimes we do make mathematical mistakes. Okay, so let's move everything unknown to the left. And what we will have is C times L times second derivative of t plus first plus I of t. It's equal to C omega E0 cosine omega t. Okay, so this is my differential equation of the second order, which I have to solve. Now, I'm not going to go into depths of how to solve differential equations in general. I'll do what physics usually are doing, the guess. Well, let's just think about it. See, this is the cosine. Now, this is the function and this is the second derivative of a function. If you recall the derivative of cosine, the first derivative is minus sine and derivative of sine is again cosine. So, doing the main function is second derivative. If main function contains cosine of omega t, my second derivative will also contain second derivative of the cosine, which is, again, contains cosine, right? First derivative is sine, second is cosine, with the coefficients. So, what I'm saying right now, that it makes sense for me to look for an equation of this type. I don't know what K is, but I think I can come up with certain K in such a way that this would be satisfied. Yes, it's a little trick and yes, it's only because this specific differential equation has this specific format. I have the function and its second derivative and I have a cosine here. I just recall that cosine, second derivative is also cosine, with some coefficients, so that's why if I will have this, most likely my cosine will just be reduced because the cosine will be here and the cosine will be here and all I need to do is find out what's my K. Let's try. So, first derivative is the derivative of K is a constant, so the derivative of cosine is minus sin and then the derivative of the inner function, which is omega, so it's minus K omega sin is omega, right? Now, my second derivative is derivative of sin is a cosine and another omega, so it's minus K omega square cosine omega t. So, if I will substitute the second derivative here and the function here, all of them will have cosine omega t and the cosine will go out, right? I can reduce everything by cosine. So, what will I have? I will have C L and the second derivative. So, it's minus C L K omega square and the cosine, which will be reduced. Now, plus function, this is also just K and this part, again, cosine goes out so I have C omega E0. That's my equation for K. It's a linear equation from which I can just find the K. So, what is the K? Okay, if I will take it out, it will be one minus this thing, right? So, it will be Cw, C omega, sorry, E0 divided by one minus C L omega square. Now, let me just reduce it in such a way that it will be much more simple and understandable, the physical part of it. So, let's talk about reducing numerator and denominator by C omega but will be E0 here. Here will be one over C omega minus L omega equals two. Now, let's recall our lectures on capacitors and inductor. In the capacitor lecture, one over C omega was actually designated as Xc, which is capacitive inductance. And L omega was, again, that was described in the inductor. This is inductive, how is it called? Inductive capacitive reactance and inductive reactance. Okay, yes. So, they are actually serving as equivalent in the capacitor and inductor's world equivalent to resistance. So, it's a capacitive reactance and inductive reactance. So, these are two reactances. But look at this thing, the difference between these two reactances, it looks like it actually can be viewed as a resistance, so to speak. Because, you see what happens here? What is I? Now, I would be equal to... I would be equal to... I would be equal to... that's my current. Which is this one, times cosine omega g. So, this is our first and most important part of the solution. We have found what is exactly the current which is circulating here. This current is equal to some multiplier times cosine. So, let's just compare it with our EMF generated by the generator of electricity. Now, both of them are sinusoidal. Both of them have the same frequency, the same angular velocity omega. Frequency is 2 pi times frequency. So, omega is 2 pi times frequency, right? Remember this. And this is the peak amperage. So, the peak amperage is equal to peak voltage divided by something. Well, remember the Ohm's law. The amperage is equal to voltage divided by resistance in a simple direct current circuit. So, basically, this difference between xc and xl, the reactants of the capacitor and reactants of inductor play exactly the same role as a resistance in direct current with a resistance in a circuit. So, this is our solution and that's very important. What's another important thing is, this is cosine and this is sine. Well, if you remember, cosine x is equal to sine of x plus pi over 2. Am I right? Sine x cosine of pi over 2, which is 0 plus cosine x and sine, yes. So, it looks like the oscillations of my current are shifted by pi over 2 to the left, right? This is a graph of this shifted to the left. Sine of x plus pi over 2 is shifted to the left by pi over 2 relative to sine of x. So, these oscillations and these oscillations, they're all oscillations with the same frequency at the same period, different template adopted, obviously, but the cosine is shifted to the left relative to the sine. So, that's very important. It's not in sync. I mean, it's shifted by, it's called phase shift. So, the current is shifted by pi over 2 to the left, which means it's prior in time than our oscillations of our EMF, voltage generated on the generator. Okay, now from here, we can very easily find the voltage drops here and here. So, let's do that. Okay. What can we use? Well, we definitely don't need this. And we don't need this. And we don't need this. Okay. So, first of all, we can very easily find VL. It's L times this. VL of T. That's the voltage drop on the inductor. It's equal to L times derivative of this. Okay, we have this multiplier times derivative of cosine. It's a minus sine of omega T times omega inner function, which is equal to... Look at this. L and omega are XL. So, it's minus E0 XL divided by XC minus XL divided by sine of omega. Okay, let's talk about this first. This is minus sine, which means it's opposite to this. It's anti-phase. I mean, sine of something with a minus sine is sine plus pi. So, this function is this. This function is this. So, it's opposite. Or if you wish it's shifted by pi to the left or to the right, it doesn't really matter because it's pi and it's periodicity. Okay, so that's what we have gotten for this I of T. It's equal to E0 divided by XC minus XL cosine of omega. Okay, now let's talk about VC. Now, VC derivative is 1 over CI of T. So, how can I find... If I know the derivative, I have to integrate this, right? So, let's just integrate it. 1 over C, integrate of this. So, integrate from 0 to T, I believe. E0 divided by XC minus XL cosine of omega T equals... Now, what's integral of cosine? Well, what's the function derivative of which is cosine? Sine, right? So, we have to start from sine omega T. But if we will derive... If we will take a derivative from sine of omega T, we will have extra omega. So, I will put this, and then these are just multipliers and sine of omega T. Right? The derivative of sine is a cosine and omega will reduce with this one and 1 over C is still here, which is equal to... Again, 1 over C omega is XC by definition, right? So, we have E0 XC divided by XC minus XL sine omega. Well, first of all, let's just have a quick check. If you will add this one, which is VC. Well, now this is... It's not equal. I mean, I've just taken an integral. This is VC. So, if you will add VC and VL, you see XC and minus XL. So, it will be just E0 times sine, which is our original AMF. So, that's kind of checking, okay? Now, as far as what exactly these two things are, as you see, this is shifted by pi relative to this AMF. But this is the same in-phase. So, this one is in-phase. The voltage drop on the capacitor is in-phase with AMF. Voltage drop on inductor is in anti-phase. And my current is shifted by pi over 2, because cosine is shifted by pi over 2 to the left. So, which means this voltage on the inductor, which is shifted by pi to the left. This is shifted by pi over 2. So, this one precedes by pi over 2, my current. And this one follows by pi over 2, the current. So, that's how interesting it is. Well, that's basically it. What's important about this lecture, it kind of comprises everything you know about the AC, about inductors, inductance, capacitor, capacitors, capacity, capacitance, about inductive reactance and capacitive reactness of the inductor and reactness of the capacitor. It's kind of simpler. Which basically act as a resistance, as resistors. That's basically it. I do suggest you to read the text for this particular lecture. It basically has exactly the same thing. Maybe a little bit more concise and without little problems, which I have. But, you know, it's like a textbook. I do recommend you to basically take the whole course from munisor.com. It's logically arranged and I'm referencing certain other lectures. Before this, like for instance when I'm talking about reactance, it's all explained in the previous lectures and that's why it's very important to take the whole course. And, well, that's basically it. So, good luck. Thank you.