 Hi, I'm Zor. Welcome to a new Zor education. Today we will continue talking about electronic oscillators, and we will expand whatever we had on the previous lecture. The previous lecture was slightly simplified, very slightly, so we will just add another component into our electronic oscillator, the resistor, which obviously exists everywhere, because every electronic circuit has some resistance, and it will slightly change our calculations, but this will be a very practical kind of RLC circuit. R stands for resistor, L stands for inductor, and C stands for capacitor. So, now this lecture is a continuation of the course Physics for Teens, presented on Unisor.com. I suggest you to watch this lecture from the website. The website contains the course, which means there is a menu, there is a sequence, there is a logical connection between the lectures. There is a prerequisite course called Mass for Teens, which definitely contains a necessary information for studying physics. And everything, the whole website, is completely free. There are no advertisements, no strings attached, etc. So, okay, so we will start with the same electronic circuit as in the previous lecture, which contains inductor and capacitor, and just add one more element, the resistor. Okay, so let me just remind you. First we had inductor, we had capacitor, and now we will add a resistor. Now, if you recall, we had a battery to charge the capacitor and a switch, a-b-switch. So, in this position, this thing is not connected to anything and the battery is charging the capacitor. Then we flip the switch to A position. Now, the battery is no longer involved. Now we have a circuit and magic begins. Now, what's the magic? Okay, let me just remind you. We had certain amount of charge accumulated on the capacitor from the batteries. So, let's say this is plus and this is minus. Okay, what happens in this case? Well, electrons will start moving from axis of electrons plate, which is minus, to the positive plate, through resistor, through inductor. The problem is, gradually, we are discharging basically our capacitor. So, the amount of charge is diminishing because the electrons are moving from here to here. So, that means that our electric current in this circuit is changing. It's variable, it's diminishing, and so far it's diminishing, right? Now, the problem is that this is an inductor and we know that the electric current which goes through a wire has magnetic field around it. Now, Faraday's self-induction principle says that whenever the magnetic field is changing, it actually produces the electromotive force. So, in this particular case, the magnetic field is changing because the current is changing. But since it's changing, it produces electromotive force and it supports the current. So, the current not only goes all the way to discharge completely, but since it supports it, it continues actually going. And as it continues because of electromotive force produced by the inductor, this minus becomes eventually plus. Electrons are moving in axis to this part and this becomes minus. What happens next? Well, the next thing is just a completely reversed situation. Electrons from here, whenever the charge will be too much and the force of the inductor will no longer... the electromotive force which is produced by inductor is no longer supporting this current. The current dies, but when it dies, it's already charged with a minus and plus and an opposite charge. And then the whole thing, the whole cycle starts back. Now, without this resistor, if there is absolutely no resistance, that was the last lecture and this oscillation would be infinite. So, it would be a sinusoidal current. Current is I of t. Current is the same everywhere, obviously. It's a closed circuit. So, the I of t was, in this particular case, sinusoidally changing. Now, what happens if there is a resistor? Or, more precisely, the whole circuit has some resistance somewhere. But right now, I just concentrate it in this particular case. I mean, this particular object in the circuit considering that the rest doesn't have any resistance. It doesn't really matter. So, what happens then? And that's the subject of this lecture. So, let's assume that the inductance of the inductor is L. The capacitance of the capacitor is C and resistance of the resistor is R. Now, we will do exactly the same as in the previous lecture, which means I will talk about voltage on each of these things. So, this voltage on this would be VL of t. Voltage on this will be VR of t. And voltage on capacitor will be VC of t. All of them are changing, obviously, all the time. But what's important is there is no source of electricity here, which means that the sum of the voltage drops on each of these things should be equal to zero. Otherwise, we would have some kind of an infinite source of energy, right? So, this is equal to zero. Okay, fine. What I'm doing now is basically repeating the same thing which was in a previous lecture, and then I will add something related to resistor. Now, first of all, let's talk about capacitor. Now, what is C? What is capacitance? What is C? Well, basically, it's a connection between amount of charge, which is accumulated on the capacitor and the voltage between the plates. So, it's Q of t and VC of t. They are related. Now, how are they related? Well, they are related this way, constant. And that's what capacitance is. So, whenever you have any kind of a capacitor, the more charge you put in, the more voltage you have between the plates, and they are proportional, always. And that's the capacitance, it's a constant. Well, from which I will do two things. Now, the same number two, I will put that VC of t is equal to 1 over c times Q of t. And then, you see, I'm going to use current. However, the current is very much related to the charge. So, what is the current? Well, if this is the charge between the plates, and it's discharging, basically, electrons going from negative to positive, well, the electric current, by definition, is a rate of change of the charge, well, rate of electron flow, if you wish. Now, what is this mathematically speaking? Well, mathematically speaking, A of t is just the derivative of the function Q. So, if Q is changing, if charge is changing, its derivative is a rate of change, and the rate of change is called electric current. So, what I will do here, I will differentiate this to get rid of the Q and deal with the VC. First, the derivative is equal to 1 over c i of t. Great, done. Next, we have an inductor. All right, now, what about this inductor? We know that the electromotive force, which we call DL, which is developed by changing the electric current, is related to this change, and let's talk about how. Electric current is producing magnetic field. Magnetic field has such a concept called flux. Okay, now, what is flux? Magnetic flux is actually proportional to the current. So, if you have, let's say, a wire loop, and you have a current, then around this current, the magnetic field is proportional to magnetic flux. This flux is proportional to inductance of this inductor. That's characteristic of this inductor. So, the magnetic field which goes through this thing, through this inductor, is magnetic flux which goes through this inductor, is proportional to electric current. Now, the electromotive force is related to change of magnetic flux. So, the electromotive force is actually a rate of change of magnetic flux of this inductor. Now, all this was started in details in electromagnetism part of this course. So, if you do not remember this, and you would like actually to refresh it, go to the electromagnetism part of the course. There is a Faraday's law, which basically explains all these things. I'm just using whatever the material was presented in electromagnetism part of this course. So, electromotive force, which is produced is proportional, is actually equal to a rate of change of the magnetic flux, which in turn, magnetic flux, is dependent on the current. So, in this particular case, what we can say is this is equal to L times I derivative. L is a constant, L is inductance of the inductor. So, that's how we have derived the value for VL. This is the electromotive force, which is developed by the inductor. Finally, we are the drop of the voltage on the resistor. Well, this is the Ohm's law, as we know, right? It's R times I of T, where R is resistance. So, we have VR drop on the resistor, we have VL drop on, well, it's not drop in this case, it's actually electromotive force, which is produced by inductor, and we have derivative of the voltage on the capacitor. And now, I would like them to somehow come up with this particular equation. Well, unfortunately, I have the derivative here, and if I will put integral, it would look ugly, quite frankly. So, instead, what I will do, I will do this. I will differentiate this, and I will put second derivative, I will differentiate this, I will put first derivative, and then I can differentiate this, too, because if some of them is equal to zero, some of their derivatives is also equal to zero. And that's my equation. And let's do this right now. So, what's my equation? L times second derivative of T plus R, first derivative of T plus one over C function without derivative is equal to zero. That's my differential equation, which basically delivers how current is changing. So, any solution to this equation gives me basically behavior. Okay, now, if you will take a look at this equation, and if you recall the mechanical oscillations in viscose environment, it looks exactly the same. So, let me just remind you. So, M times X of T plus C X C plus K X of T equals to zero. Okay, so, this is an equation which was derived when we were studying mechanical oscillations in the viscose environment, damping. So, let's say you have your object in the spring in water. Water is basically damping the oscillations. It was exactly the same equation, just different coefficients, different letters basically. So, from now on, everything, I mean, I can just stop here and just give you the final formula. But I will just do some explanation, obviously, but it's exactly the same, the same logic. Okay, so, now, this is M is mass, C is viscosity coefficient, and this is the spring's elasticity. And X is displacement from the neutral position. So, exactly the same equation, absolutely. Okay, so, let's forget about this. Again, there are maybe more details when I was explaining solution to this than whatever I will do right now. But since every lecture on the Unisor.com has detailed explanation, so you can go to basically notes for this lecture which are quite detailed. All right, so, how can we solve this equation? Okay, first of all, this is a very interesting equation. It's homogeneous, homo-genius, which means if function i of t is a solution, function i of t times constant, any constant would be a solution as well. Why? Because the constant is going out from the derivative. So, if instead of i, I will put k times i of t. Well, the k will be here, the k will be here, and the k will be here, the k will cancel out. It will be exactly the same solution, right? So, obviously, it's a homogeneous. It's also linear. Now, what linear means? Well, if you have two solutions, i1 and i2, then any linear combination of those is also a solution. And again, it's obvious because the derivative of some would be some of derivative, or multiplied by a constant would be, et cetera. So, and it's second order because it's a second derivative. So, the linear and homo-genius is very simple thing to solve. If you would like to find some solution, there is obvious function which would satisfy very easily. We can find this function. The thing is that for these homo-genius linear equations, differential equations of the second order, it is sufficient to find just two partial solutions, two concrete solutions. I just find two functions which correspond. And then their linear combination with any coefficients describes all the general solutions. Okay. This is the fact. You can just take it as it is because there are some, again, explanations to this in the course, Mass for Teens, when I was talking about differential equations. But anyway, just take it for granted that it's a theorem, actually, which can be proven. So, what my problem right now is just to find two solutions, two particular partial, I would say, solutions to this equation. And basically, the combination of these two solutions with any coefficients would be a solution. Okay. So, the obvious way to find the solution and, again, it's not... I just came up with this. I'm not that smart. Very smart people, actually, were doing this before me many times and they have suggested to look for a solution among function which looks like this. Were gamma is some coefficient? Why? Because the first derivative of this would be this. The second derivative of this would be this. And if I will substitute it to this, you see that e to the gamma t can be just cancelled out because it never equals to zero, right? So, it will be L times gamma squared times e to the gamma t, but it will cancel out plus r gamma plus 1 over c equals to zero. It's a quadratic equation and they can obviously find two roots for this quadratic equation and two roots give me two independent partial solutions. One will be e to the gamma first t another will be e to the gamma second t and their combination would be a solution, a general solution. Okay? So, what are gamma? Gamma 1, 2 is equal to minus r divided by 2L plus minus square root r squared r divided by 2L squared minus 1 over Lc, right? So, I've got two gammas. Now, the problem is I don't know if this is positive or negative or zero because there are definitely the problems here. First of all, if it's zero, I will not have two gammas. I will have only one gamma and that represents only one partial solution and we need two to have their linear combination as a general solution. Now, if it's positive, I can basically do whatever is necessary and if it's negative, that would be an imaginary number and I have to really think about how to deal with imaginary number. Now, again, all these aspects were presented when I was talking about mechanical oscillations with damping. There are three different cases of damping and it depends on the same thing basically using different letters in mechanical oscillation but we had the overdamping underdamping in critical damping, critical level of damping when it's equal to zero. We'll do exactly the same here and I will try to be very brief. Well, first of all, let me just make, okay, what is if r to o l square minus 1 over lc is greater than zero and I will therefore call it omega square. So my gamma 1, 2 is equal to minus r to l plus minus omega. Right? So for the positive expression under the root, under the square root, I can put it is equal to omega square since it's positive and the square root would be omega and my two solutions are i is equal to a some coefficient e to the power minus r to l plus omega t plus v e to the power minus r to l minus omega t. Now, what's interesting is that this is r over 2l. This is r over 2l square minus something. Now, this something is positive, this is positive, this is positive, but this thing is smaller now than r over 2l square, which means omega is smaller than r over 2l. Right? So this is omega square root of this. Now, since we are diminishing the r over 2l square, we have less, still positive but less. So the square root of this would be by absolute value less than r over 2l, which means that both of those guys would be negative. This is minus r over t and we're trying to plus something which is by absolute value smaller. So the gamma is negative. So this thing and this thing both are negative. So what we have here, we have e to the sum negative number and as the t is increasing the graph of this thing would be this and graph of this thing would be also kind of this. It's all diminishing and their sum will be diminishing as well. Maybe a little wavy in the beginning but again it will be asymptotically going to zero and after certain amount of time it will be only positive or only negative. It will not be both because this is positive and this is positive. So no matter what our coefficients are eventually all these coefficients will be brought down not to zero but it will be basically staying on the same side plus or minus so there will be no oscillations back and forth. The current will not go back and forth it will go this way maybe a little bit back depending on these coefficients but in a couple of oscillations it will stay on the same positive or negative side so it will be either this way or that way depending on certain initial conditions but there will be no oscillations. Second Second is this is equal to zero. Let's think about what happens in this case. Okay. If omega is equal to zero this is omega. Okay. The problem is we have only one root. We have only one solution. We have only one gamma. Where is the second one? We need two to make a general solution. Well, we have to look somehow differently. Okay. The first we were trying to find among these type of functions. Well for this thing we can find another we can try t times e to the gamma t. Is there a solution to this? Well if I will substitute this into the equation I put first derivative, second derivative put the equation I will also get a solution. It will be a very easy kind of calculations and I will get exactly a solution gamma 2. And it will be equal to exactly the same thing. But now instead of two functions of this kind I will have one function of this kind and one function of this kind with the same gamma. Which means my general solution would be a to the minus 2 minus r divided by 2Lt plus b times t. Because this is a different function by e to the same or you can just put e to the minus r over 2Lt out of the brackets and here I have something like this. So this is a different function it's a general solution in case of critical damping critical when this is equal to 0. But this is also not permanent oscillations around some kind of neutral slide. Because this is a linear function and this is exponential with a negative exponent. So the graph would be again it might actually be something like this but then it will stay on the same on the same side. After initial fluctuations it will be gradually diminishing because this is very strong. As t goes as t increasing the time is increasing this is a major factor and whatever it is it will just go straight down without changing the sign without anything. And the third I would say more interesting case when this is negative I will call it minus omega square. So the result would be gamma12 is equal to minus r2L plus minus omega i where i square is equal to minus 1 it's an imaginary unit. So now you have to know the complex numbers. And again if you don't go back to mass proteins there is a chapter for complex numbers so you can learn it over there. But you see how many different mathematical pieces you have to really know to learn physics. Okay. Now what does it give me? Well it gives me two solutions but the problem is gamma is this. The function is with complex values and that's not exactly what I'm looking for. I'm looking for i is electric current it's obviously a real valued function right? But here is an interesting case you see this is as I was saying before linear, homogeneous, etc. Now what if you have a complex function let's say i of t is equal to x of t plus i y of t. Complex function obviously can be represented in this way right? Complex values. Well if you substitute this to this now again derivative is a linear function so derivative of sum would be sum of derivatives. So your real parts will be l times x the second derivative plus r times first derivative of x and 1 over c x itself. And then i times the same for y. And if I have a complex number a plus bi is equal to 0 it means a is equal to 0 which means x should be a solution and y should be a solution because x should be if x in this particular expression plus i multiplied by y of this expression is equal to 0 then this expression for x should be equal to 0 and this expression for y should be equal to 0. So my point right now is that if I will be able to express this function with this exponent as real part and separately imaginary part I will have two different solutions real part would be a solution and this coefficient at i would be another. What I have to do is I have to represent function e to the gamma t for this gamma in real plus imaginary part and this is very easy to do using Euler's formula e to the ix is equal to cosine x plus i sine x Euler's formula great formula. So let's do it this out and I will put it here so e to the power minus r over 2l plus omega i t this is what this is e to the power minus r to l t multiplied by e to the power omega i which is plus minus plus minus equals to e to the power minus r to l t times and I will use this formula so it will be cosine omega i plus sine i sine omega t now where is minus in theory plus or minus should be here plus minus but cosine is an even function so I can drop plus or minus sine is an odd function so plus or minus can go out so that's what I have and as I was saying if I would take only the real part it would be a solution which means which means e to the power r over 2l t times cosine omega t is a solution and e to the power minus r over 2l times sine omega t is another solution and if I will do this I will have complete solution plus or minus doesn't really matter because a and b are as I was saying for a general solution I just have two partial and any coefficients including plus or minus doesn't really matter so these two partial solutions give me the general solution for this particular equation now how this thing is looking well now you have the oscillations because this is obviously something which goes down to zero but this is amplitude but if this amplitude now the cosine for instance is something like this right but if you will multiply by this it will just diminish the amplitude so it will be this and then this and then this it will be lower and lower amplitude will be smaller and same thing with this so basically the total solution general I would rather say solution to this problem is e to the power minus r over 2l t a cosine omega t plus b sine omega t so that's a general solution to this particular equation and now we have oscillations and oscillations are in case r to l square minus 1 over lc less than zero so r is not really very big r is resistance so the bigger the r the greater this and it will be greater than zero that's it basically there will be no oscillations but while resistance is not very big everything will be fine now if resistance is zero you will have this and this is omega square is equal to 1 over lc if you remember this is from the previous lecture so that's why it's a mega it's basically the angular speed of oscillations well basically that's it now just as a trigonometric another part of mathematics trigonometric exercise I did it a few times this can be converted into sine or cosine with phase shift if you remember so what I can do with this equation I can multiply it and divide it by this a over square root cosine omega t plus b divided by the same square root sine omega t now this and this can be cosine of some kind of phi and sine of some kind of phi and now this is a trigonometric formula cosine plus sine sine that's actually a cosine of omega t minus phi so it's phase shift so instead of this you can put this and that would be some kind of d this square root so the whole thing depends also now it depends on two parameters a and b now this thing depends on two parameters g and phi where g is defined like this and phi is defined like cosine phi is this and sine is that so everything is defined and this is just looks a little bit better now we really see that this is an oscillations with gradually diminishing because of this multiplier amplitude well that's it for today I do suggest you to read the notes for this lecture they are much more detailed than whatever I was just talking about with references to previous lectures etc etc alright good luck thank you