 Hi, I'm Zor. Welcome to a new Zor education. I would like to solve a couple of problems related to circuits containing inductor and capacitor Now this lecture is basically a continuation. It's it's problem-solving, but it's just based on whatever the previous lecture Was about and that's basically about certain formulas which relate inductance, capacitance and voltage and electrical current in the AC circuit So all these lectures are part of the course. The course is called physics for teens. It's presented on Unizor.com I suggest you to watch this lecture from from the website. You just have to follow the menu physics for teens Electromagnetism and then this is part of the alternate current and induction chapter Now the website contains not only the reference to a video presentation, but also Detail explanation and calculations Which I'm not going to do here. I'll just refer you to the written notes on my website Also, the website contains exams for those people who would like to challenge themselves Next is Something very important The physics for teens course has a prerequisite, which is mass for teens and you can't really do physics without mass and This prerequisite course is also in the same site Unizor.com and I do suggest you to familiarize with that course as well or at least with whatever material this course presents Mostly in physics, we will use calculus vector algebra something else Okay, so back to problems So I will heavily rely on the results of the previous lecture And I'll just present certain formula here And if you would like to know how I got it you have to go to the previous lecture and I assume that you did Okay, so first of all, what kind of a circuit are we talking about? We're talking about circuit which has a source a generator of electricity Alternate alternating Alternating current so that's number one. So we have this source of electricity and It has certain electromotive force Which is Variable depends on time Now next what we have here is we have Inductor it has certain inductance L Then we have a capacitor Which has certain capacitance C and This is our circuit so This problem is basically okay known certain parameters we have to Determine some other parameters of this particular circuit and obviously it's all related to Certain laws which were derived in the previous lecture So let's assume that we have Current which is also depending on time It's alternating current now in this particular Problem, but I have basically Presented as as a problem we have given certain things namely Effective voltage Equal to what is it a water? 220 now the frequency is 50 Hertz So that's the characteristic of the generator. So there is some kind of a rotor which Rotates with 50 Rotations per second and the generated voltage has effective voltage is 220 volt now you remember That effective voltage if you have a sinusoidal waves E of t is equal to e0 times sine omega t that's How we describe our? EMF electro motor force Where e0 is a peak Voltage so effective voltage Is e0 divided by square root of 2 and I have derived this as well It's basically you are averaging the square of Of this sinusoidal voltage So that's given effective is given well, which means peak is also given in this particular case It's just square root of 2 greater than this one Now what's next is I Have the effective Current is given It's 5 ampere right right now you remember Now let me go back a little bit Again from the previous lecture. I have derived that The current as a function of time Is also sinusoidal, but it's shifted relative to Voltage its phase is shifted by power two Which means the formula would be Cosine So it's also sinusoidal and the effective Is exactly the same type it's peak divided by square root of 2 again Everything is in the previous a couple of lectures was presented So that's given So we have effective Voltage we have a felt effective current and we also have Capacitance What is it? 10 micro for a Okay, 10 micro farads micro means one millions of for a friend is actually a very big Unit so that's given so what they have to do before the term and he is inductance of this particular Inductor Okay, it's kind of a lot of different variables, etc. In In theory, it's a very simple problem because all these characteristics are connected through Certain equations which were derived in the previous lecture So you remember there is a concept of reactance Reactance for capacitor and reactance for inductor. They are kind of similar to Resistance of the resistor and Basically, there is a similar formulas actually which connect the voltage the amperage and reactance in this particular case So the main formula Which was derived in the previous lecture was the following it's very much like Ohm's law for a direct current. So this is the peak voltage. This is the peak amperage and this is the difference between reactance of the capacitor and reactance of the inductor where inductance are defined as Where omega is the same omega here? This is angular Velocity in in regions By the way, how is angular velocity in radians is related to frequency? Well, frequency is number of rotation if each rotation is 2 pi Right. So if you want to do it in radians My omega is equal to 2 pi times F So using this and This and obviously this We have everything known in this particular thing except XL which is directly related to inductance So if I will just substitute whatever I have in this Formula, I will basically have an equation which I should solve to get the L. So let's just do it Well, first of all, you see I0 and E0 which are peak voltage and amperage are related to the effective only by the same factor square root of 2 So if I know E effective and I effective I Can actually divide this by square root of 2 and this by square root of 2 and I will have exactly the same formula I effective equals E effective divided by Xc minus XL Right because it's just divided by square root of 2 both sides So now I have everything actually here Known and the only thing which I have to do is replace effective Now Xc is this one Which is known to pi c right to pi fc Frequency times capacitance Minus This is Xc minus XL, which is 2 pi f Basically, that's it. This is the final equation. Everything is known here except L So how can I basically do this? Well, that's just plain algebra So I will have 1 over 2 pi fc minus 2 pi f L equals E effective divided by I effective From here, this is a linear equation for L So all I have to do is Put this one to here this one to here and divide everything by 2 pi f So it will be 1 over 4 pi square f square c minus E effective divided by 2 pi f I f and this is equal to All right So this thing goes here this thing goes here and I divide everything by 2 pi f Yeah, that seems to be an answer All you have to do now is substitute all these numbers, which I'm not going to do right now But I did do it in notes for this particular lecture. So I do recommend you to read the notes Which basically has the same? kind of calculations Coming up with equation, which you just have to solve and that's all presented in in notes for this lecture Well, that's it for this problem. So again, the problem was to Define one particular to determine one particular characteristic of everything else is known obviously Since it's all related There are other problems which can be imitated exactly from this one. So let's say if L is known and C is unknown Well the same basically approach what if L known and C known and you have to determine The current again, it's all related in one particular equation Which looks very much like the Ohm's law because this is the voltage Divided by something which is kind of equivalent to resistance and you get the current Except that in AC in alternating current. We're talking about either the peak values or the effective values And this is basically equivalent to a resistance of the whole circuit with reactants of capacitor and reactants of Inductor and by the way reactants is actually measured unit of reactants is also Ohm I mean if you will go again back to the lecture which explains the concept of reactants, you will see that the unit of measurement is actually Ohm and the resulting Inductance So I said reactants reactants is it Ohm now inductance is measured in Henry And that's what you will get if you will put all these numbers into this formula You will get the the answer how much Henry is the inductance of This particular inductor Okay, this is all about this problem It seems a little complicated, but in theory it is not because all you have to really know is definition of certain characteristics of The circuit like inductance capacitance reactants of this reactants of this effective voltage effective and fridge So if you know all these concepts and you know this main formula everything goes Now the next two problems actually are very I would say qualitative not quantitative they basically explain the concepts a little bit more of inductors and capacitors So consider the following circuit you have a capacitor and parallel some kind of consumer of electric impulses electric signals In theory it can be something like a radio Now why I'm talking about a radio because on the input I Don't have just a very nice sinus so we do A signal but I have something which is really like a noise Now what is noise? Well consider you have one generator which generates certain variable alternating current of certain Voltage and certain Frequency and then you have another one parallel to this one So you have one and then you have another But this guy has a different frequency and different peak voltage and what if they are Connected together as a source of electricity what you will have here on this Some kind of a Device well you will have a mixture right you have two sinus so it's which are not exactly a sinus so So it'll wait. It's a combination. It's a mixture of different signals of different frequencies What if you have not two but a hundred? Well, you will have a mess you will have a almost like a white noise what they call it So all these signals are coming if you're talking about about for instance radio waves They are from all different sources sources have different Frequencies different amplitudes So you will have a noise so that's basically what here we have so we have some kind of a noisy signal Now what my point is that provided such as noisy signal But I would like to say that If you don't have this capacitor Well, all noise goes to our device right if you do have this capacitor Then the high frequencies of this noise and again noise is a combination of different frequencies. So the high frequencies will actually be Going to this particular device in in less power So certain amount of energy which high frequency are carrying with them will be Going through this chain rather than through this chain So without it Everything called the power goes here with this certain amount of high frequencies Will go here and will be built will be weakened here So it's actually like a filter for high frequencies and this filter is Taking upon itself Certain amount of energy was high frequency and that beacons the high frequencies from this white noise which are going to This particular Device now question is why? Well, very easily It's it's actually a more explanation than the reality here again remember reactance of the Capacitor which is equal to one over omega C C is a capacitance of the capacitor It's it's a characteristic basically which depends on the area of the of these two plates and the distance between them and what's in between them what kind of Insulating material is in them. It's a this is a characteristic of the it's given it's constant Now this omega is very important. You see the greater omega omega is angular Freak angular velocity, which is actually proportional to the frequency of mega is equal to 2 pi f So the greater the frequency the smaller um reactance of this capacitor so this thing is Relatively functionally equivalent to resistor. So you have a resistor here a resistor here now if you have this resistor Being smaller and smaller it means that the greater and greater current will go this way Then this way because if you remember in this parallel Circuit the currents are inversely proportional to Resistance and it's obvious from the you know pure logical standpoint If this has certain resistance and this has certain resistance So the the current goes here and here, but if this resistance is gauging smaller and smaller When the frequencies increasing This resistance is gauging smaller and smaller. Well, I shouldn't say resistance. I should I should say reactance because they're talking about Capacitor, but if functionally resist the resistance and reactants are the same so the more Current of these high frequencies will go here and this high frequency will be weakened on this particular device Okay, that's hopefully it's understood the same thing basically if you have let's say flow of water and that's flow is branching like a Y letter now if Resistance in in some ways equivalent to the width right, so if you have This thing and then it splits now the resistance is Equivalent to the widths the greater the widths the less resistance right so if you will increase the widths of this Then more water will go here and less water will go here, right? So that's exactly the same thing with currents with electrons because you know what is current current is the flow of electrons So same thing they they basically obey the same kind of a logic. So the smaller reactants of the capacitor the greater energy will go here and Obviously the high frequencies give you smaller reactants and then more so in the higher frequencies You will have more Energy going through this then through this it beacons the high frequency it filters out the high frequencies from this white noise As they are coming into this particular device Great, that's my second kind of a qualitative problem And now my third qualitative problem is related to this What if instead of capacitor I have an inductor? Inductor also has its own reactants, right? And that's a mega L so in this case as you see the lower Frequency corresponds to lower reactants and again lower reactants means that these Low frequency Oscillation of the current will go more to this device and will be filtered out from this device So this is a low frequency filter so to speak So capacitor will filter out high frequencies Inductor will filter out Low frequencies by combination of these two filters and properly Calculating what exactly capacitor and what exactly inductor should be used we can actually narrow The Oscillations the frequency of oscillations which are coming to this particular device very very Narrow actually interval and that's how the tuning actually is obtained in let's say radios or whatever else whatever devices is using So this is very important kind of Connection so to speak and it's a very important quality of capacitors and inductors and On a similar note, I can also mention that we all know the dimming Switches you're just moving certain Slide and the lights go up and down, right? So how can that be accomplished? Well again the same thing you can actually do something like an inductor inductor is easier to change The inductance because you can just shift the contact this contact Can be floating right and the shifting it you will change the number of Loops of this particular inductor and obviously the greater the number of loops the greater inductance the smaller the smaller And that's how you can actually change the inductance. It's easy and Changing conductance you can actually change something which is an equivalent of Resistance so in this particular case you can include into the loop An electric lamp Let's say so it's your electric lamp Incadescent lamp so changing this Inductance will basically change the effective Current which goes to this particular lamp. So that's another property of Inductance which you can use here Well, basically, that's it. These are Kind of three problems, which I wanted to present And after analyzing what LC circuits actually are I do suggest you to read the notes for this particular lecture and there are Maybe a little bit more detailed explanations of these two qualitative problems so to speak And well, that's it for today. Thank you very much and good luck