 Hi, I'm Zor. Welcome to Unisor Education. Today I will continue talking about alternating current and certain devices or certain electrical equipment actually installed in the circuit. Now the devices which we are talking about are resistors, capacitors and inductors. In the previous lectures we were talking about only resistors or resistors and capacitors or resistors and inductors in the AC circuit. Today I will combine them together. So basically it's kind of a continuation of whatever was before discussed in these lectures. There will be a differential equation, there will be a solution to this equation. I do encourage you to, after you've watched this lecture, to read the text which is accompanying this lecture on the Unisor.com. This lecture is part of the course Physics 14's presented on Unisor.com and all the lectures have notes on this website. So if you found it on YouTube, for instance, you will see only the lecture itself. From the website you can see the lecture and you can read the text which is basically like a textbook. The website is absolutely free, there are no advertisements. Also the website contains math proteins, the prerequisite course. You do need math to study physics seriously. Like in this particular case, like today's lecture we will have differential equation which we probably have to at least know what it is. Okay, let's consider we have a circuit which contains the source of alternating current. You have a capacitor, you have inductor and you have resistor. Inductor has inductance L and capacitor has capacitance C. This is given. What else is given? Well, this is a source of alternating current which is generated by rotating some rotor inside the stator and you will have sinusoidal electromotive force generated. So this is electromotive force voltage generated by this AC generator where omega is angular velocity or angular speed and E0 is a peak EMF, peak voltage generated. So it basically goes from minus E0 to plus E0 in a sinusoidal way. So that's given. What I have to determine is what kind of current I will have here. That's my problem. So I will approach this problem in exactly the same fashion I did for RC and RL circuits. RC where resistor and capacitor was, RL was when resistor and inductor was. Very similar, the same logic. We will just have a little bit different differential equation as a result which we will solve etc. Okay, so this is basically is given. I need this space. I will wipe it out. So now as the current goes around the circuit one way and then another way, well it actually meets certain resistance. On a resistor, resistance is based on this value of R. On the inductor the resistance is called actually a reactance. So it's an inductive reactance and inductive reactance is measured by omega times L where omega is the angular speed of voltage and L is inductance of this inductor. So obviously the greater frequency of oscillations we will have more resistance. If my current is not alternating but the direct current which means omega is actually is equal to zero there are no oscillations. The inductor doesn't really present any kind of a resistance. So it's only when you have the variable AC the inductor has variable flux going through it and that's what actually is causing the self-induction which works always against the voltage which comes in. So that's why we have this type of resistance. And finally we will have Xc which is reactance, capacitive reactance of the capacitor and in this case it's just the other dependent set frequency. If there is no frequency the capacitor actually doesn't really let direct current through and the more frequent oscillations are the easier it goes through the capacitor. So we were talking about this when we were discussing capacitors and about this when we were discussing inductors. So these are these resistance inductive reactance and capacitive reactance are the source of resistance to the current. So combined together they actually cause the whole circuit to resist in some way the voltage which is basically running the electrons through this circuit. So we really have to find out how exactly it happens. Now what happens around each element which we have and we have a resistor, we have an inductor and we have a capacitor. So around each we can measure the voltage on both ends. So the difference between the voltage on one end and another end is basically we call it voltage drop and we can definitely localize all the electric behavior of this particular item only using these local voltage drops. Now for instance if you have a resistor then the current which goes through this and a voltage drop around it are related to each other through the Ohm's law. It doesn't really matter that these are dependent on time. At any given time we can consider that during a small interval of time they have the current which is basically constant during the infinitesimal time period. The current doesn't really change and the Ohm's law is supposed to be working. So basically we can say that in this particular case the r of t is equal to r times i of t. This is basically the Ohm's law for resistor. Good. Now let's talk about inductor. Now the inductor inductor has certain quality of resisting through the self-induction resisting the current. Now how is actually happening? So if you have an inductor now this is voltage drop on the inductor right? So what happens with inductor? The resistance this voltage drop is actually the result result of resistance right? So it's a self-induction. We were talking about mechanism of self-induction why it happens etc. What what does it depend on? Well it depends actually on the rate of change of the electrical electromagnetic flux which goes through this inductor. So there is a concept of electromagnetic flux and rate of change actually is causing the self-induction and self-induction is whatever actually the voltage drops from one end to another. So the voltage drop depends basically on the rate of the changing of the electromagnetic flux. Again if you forgot about what electromagnetic flux is go to one of the previous lectures. Now electromagnetic flux in turn depends on the current which goes through this and certain electromagnetic characteristics which are embedded in the inductance L known to us. So we are assuming that it's known. Now how is it actually dependent? Very simply the the flux is actually proportional to current and it's proportional to whatever the inductance actually is. This is the definition of the inductance if you wish. So we can always say instead of this we can put this is equal to L times dI t per dt. So let me write it here that my VL of t is equal to L times I. I will use the little slash and that's all about drop of the voltage around the inductor. Now let's talk about capacitor. Again we have certain voltage drop here. Now when we were talking about capacitors we were saying something like this that if you have certain amount of electricity accumulated on the plates and the voltage on both ends the ratio is constant and this is basically a capacitance of the capacitor. So the more voltage you apply the more electricity is concentrated on the plates of this capacitor and that's kind of obvious. And how much more it all depends on the capacitance. For instance bigger plates can accumulate more electricity on its plates than smaller plates. Smaller distance between the plates also contributes to greater capacitance. But whatever capacitance is the amount of electricity accumulated on the plates is always divided by the voltage. Voltage drop in both ends is always a constant and that's the constant which defines the capacitor. Now there is a very interesting thing here. How this amount of electricity accumulated on the plates relate to current in the whole circuit? Well very simply what is the current? Current is rate of change of the amount of electricity. Right? So if you have two plates and there is certain accumulated charge on these plates if we will take the derivative by time we will have actually the current which goes around it. So what I would like to say here that the dL no not dL we see the voltage drop around the capacitor is equal to qc divided by c and q is my i of t. This is basically all we need from the physics. Now we start mathematics. I didn't mention it before but I can assume it's obvious that i of t this is the current which goes in the whole circuit. It's the same. This current is the same for capacitor for inductor and for resistance. Now it's the same circuit so the same electrons are going around. So whatever goes through my resistor it's exactly the same amount of electricity the same amount of electricity per unit of time if you wish which goes through the inductor and exactly the same which is goes through capacitor. So that's why this i t and this i t and this i t they're all the same i of t. This is the same amount of current as a function of time. Now what now happens here is one very simple equation and equation actually is if you have a closed circuit and you have an emf generated by a source of electricity and then you have one two three different drops voltage drops on the resistor on the inductor and the capacitor then obviously the generated amount of electricity the voltage basically generated is sum of these three voltage drops. This is our equation which will allow us to determine the value of current in the in the circuit. So this is actually a differential equation so let me right now switch to pure mathematics in this particular case. What's a little bit unfortunate is you see this is a charge q is a charge and these are the currents i of t is a current q of t is a charge but they are related so what I will do I will express everything in terms of q so I will have v r of t is equal to r times i of t which is first derivative of c the l of t is equal to l times first derivative of i is a second derivative of q right so it's a q second derivative of c of t and finally vc of t is equal to qc of t divided by c now everything is expressed in terms of q q first derivative of q and the second derivative of q and we can substitute it into this equation we know the expression for e of t so it's e zero times sin of omega t and it's equal to now let me just make it a little bit shorter I will use y of t instead of qc of t just easier to to write so what happens is some of these three well let me start from the highest derivative so it's l times second derivative of r now the first derivative is r the plain one without any derivative is this one okay as you see we have very normal I would say linear differential equation and this sum is equal to some trigonometric function now all the details how to solve this equation are actually presented in the text for this lecture in the notes for this lecture so you have to go to this website you have to choose the the part which is called electromagnetism then you should find the alternating current on the next screen and this is one of the lectures related to the ohm's law for ac now and this is the name of the lecture rlc circuits in ohm's law now what I would like to basically explain right now how people solve these this type of equation with as less as small amount of details actually so how can we solve this equation now I'm not talking about any kind of a general approach to solve equations this is a very particular one you see this is a trigonometric function and what actually people do they solving trigonometric equations and sometimes it's an art it's not a skill now as far as the art is concerned well basically it depends on experience etc etc and people have done it many times before they see that if there is a trigonometric function on the left of this and this is just a linear equation then it's convenient and most likely it will bring some good results if we will use this form of potential solution to this equation well let me see can I find f and g such that this particular function will satisfy this equation again based on experience intuition whatever it is so that's exactly what we're going to do we will try to find f and g to satisfy this is it possible well yes it is actually because you know that derivative of sine is a cosine derivative of cosine is a sine so we're not actually going out from the sine cosine if we will start differentiating functions so this is this has sine and cosine this has sine and cosine this has sine and cosine with different coefficients obviously and this is sine so we can basically combine this into one equation and equate all the sines and all the cosines to whatever is necessary separately and that's how we will get two equations with two variables okay very very quickly the first derivative is it's a omega f cosine omega t minus a derivative of cosine is a minus sine and there is an inner function so it's omega g cosine omega t sine sorry sine now the second derivative would be from cosine is a minus sine and another omega so it's minus omega square f sine omega t from sine is a cosine minus two minus omega square g cosine omega t incidentally this is equal to minus omega square times y of t you see if i will take minus omega square factor out i will have f sine plus g cosine exactly the same as this one so it's easier because right now if i will substitute all of these into this equation i will not even have the second derivative i will have basically only one particular equation and what happens is what are my what are my coefficients actually separately for sine and cosine well let's see on the left i have only sine so for the sine i have only e zero on the right i have l times the second derivative and sine is equal this so it's minus omega square l f now from this i have r times this so it's plus r omega f no sorry we need sine so it's not this so it's minus omega r g and from this i have f again with one over c plus one over c times f okay this is equation for f and g i just added this expression l multiplied by this r multiplied by this and one over c multiplied by this and added them together and took only sine sine and sine and this is sine on the left now what about cosine cosine on the on the left is equal to zero right now on the right what do i have with the cosine i have l times the second derivative which is minus l omega square cosine then i have plus r this so it's plus r plus omega i forgot g by the way here right yes i forgot g minus l g omega square omega r f from this and the cosine from this is g times one over c g right so we have system of two equations with two variables f and g it's a linear system which is we know how to solve system of two linear equations with two variables and i will just write the result again all the details you can find in the notes for this lecture but i will just write down the result so y of t is equal to f sine omega t plus g cosine omega t where f is equal to e zero divided by omega xc minus xl divided by z square g is equal to minus e zero divided by omega times r divided by z square now what is this well first of all my coefficients were l and c and r right so instead of l i used xl divided by a mega instead of c i was using uh what was what was using uh mega there yes uh one over xc one over xc omega so i used these and i have basically simplified all these formulas and the z square is equal to xc minus xl square plus r square so i skip all this arithmetic and this is basically the result of solving that system of two equations with two variables this is easy but tedious unfortunately and this is the result so i have my answer i have the a solution to differential equation and this y of t which is actually amount of charge accumulated on the plates of capacitor and the derivative of this is equal to current which i actually need before going into derivative let me just mention something here you see what's interesting here is xc minus xl r and z square is equal to this now what it means is that xc minus xl divided by z and r divided by z are two things which are which has which have the following property well first of all the absolute value of this and absolute value of this is between zero and one right because you know because of this so xc minus xl square is less than this so the absolute value is from zero to one same thing here and some of their squares is equal to z square in this case some of these squares is equal to one right square of this plus square of this is equal to one now as you understand we can always find an angle phi which has the following property and cosine of phi is equal to r over z and tangent by the way of phi is equal to x c minus minus xl divided by r so we can always find angle if we have these two values some of their squares is equal to one each one of them is by absolute value from zero to one then we can find the angle basically the angle phi is arc sign of this and the same thing is arc cosine of this or arc tangent of this so if we will just take this particular angle we will get this situation why did i do it here is why because now instead of f i can use e zero divided by omega z times times this so one z goes here and another i will use replacement for sine sine phi and this one is equal to minus e zero divided by omega z r times z is cosine phi and what do i have as a result as a result my y of t is equal to um e zero divided by omega z times what sine times sine right sine omega t times sine phi minus cosine omega t cosine phi correct and what is this this is minus cosine now this yeah minus cosine of omega t plus phi so my whole formula actually is much easier to express in this particular fashion f t is equal to e zero divided by um omega z minus minus cosine of omega t plus phi now what do i have to do next y t is charge i need to take the first derivative to get the current first derivative of this is what minus cosine sorry would be derivative is a sine inner function omega goes out uh cancels with this one and my i of t is equal to e zero divided by z times sine of omega t plus phi this is my final formula where z is basically z is square root of this okay that's it basically we have come up with the solution this is my uh current you see that the current is also sinusoidal as emf as initial voltage generated um it has exactly the same frequency of oscillation angular speed therefore uh frequency by the way omega is equal to 2 pi f where f is a frequency right you know that um okay now this thing has a special name z is called impadence of the circuit and actually this impadence impadence plays exactly the same role as resistance you see the formula is very much like peak i zero peak of i is well i zero you can say that i is equal to i zero the peak value times sine of omega t plus phi and i zero is equal to e zero divided by z this looks like the ohms law this is the voltage this is the current and z is measured in exactly the same ohms as resistance you see xc xl r this is all ohms square and z is the square root of this so z is also measured in ohms so this is actually an equivalent of ac um in ac equivalent of the ohms law now from this obviously follows that the same thing for effective voltage and effective current because effective is less than peak by square root of 2 as we know so what else is important here well basically um if you will take a look at this formula it corresponds to whatever we were um talking before about rl and rc circuits because rl circuit is when your xc is equal to zero so you will have square root of xl square plus r square which is exactly what we have derived with in the previous lecture same thing if my xl is equal to zero we will have xc square plus r square which is also corresponding to the one of the previous lectures where we were talking about rc circuit so basically this is just a more general formula where all three components are involved resistor inductor and capacitor and um it gives you basically the picture of how the oscillation of the electric current actually is it's exactly the same frequency of oscillations but it shifted so there is a phase shift and what's interesting is that this phase shift phi that's what we were talking about phi is equal to tangent phi is equal to xc minus x l divided by r well what it means is it's positive or negative depending on how my reactances are related if my um capacitor's reactance is greater then the phase shift would be positive this will be positive if my um inductive reactance is greater than capacitive reactance it will be negative so it will be shift to another side and if they are equal to each other which happens this is a resonance kind of situation when there is no shift when the phi would be equal to zero you see if they are equal to each other tangent is zero so angle is zero and my um current would be um not will not be shifted in phase relative to the uh generated voltage well that's it i do recommend you to read all the notes for this lecture there are much more detail um uh solution which i i just jumped to a solution but in in theory there are some more details and more formulas if you are interested other than that that's it good luck thank you very much