 Hi, I'm Zor. Welcome to a new Zor education. I would like to continue talking about alternating current. Today we will talk about something which is called effective voltage in case of alternating current AC. There is another abbreviation which is usually used, RMS. And I will explain what each letter means in this particular case. Today's lecture is in one way it's kind of easy because there is not too much physics in it. But at the same time there is some math. I'll just take an integral to prove whatever the formula you might have been given by your teachers or professors. Without the proof but I will just derive this particular formula. Now this lecture is part of the course called Physics 14. It's presented on Unisor.com website. I suggest you to watch this lecture from the website because each lecture including this one has detailed notes on the site. So you have a video and textbook basically. Plus all the lectures are organized in a course. So you have menu, you can choose a particular topic. And there are exams if you want to take them. And the site is completely free. No strings attached, no advertising, nothing. Okay, so let's go to the concept of effective voltage. Well, we all know that there is some number which we are saying, okay, the voltage in our outlet in our home is such and such. In Europe it's mostly to 20 volts. In the United States it's 110 or 120, maybe some others. Now what does it mean in case of alternating current if we know that whenever we are generating the alternating current by rotating a wire frame in the magnetic field or rotating magnetic field around or inside the wire, whatever it is, you always have, because of this rotation, you always have a variable voltage variable electromotive force generated by this particular process. And it's sinusoidal. So the voltage which is generated is always a function of time, which is equal to basically a sinusoidal kind of value, which means you have some kind of amplitude, maximal value, times sine of omega t, where omega is angular speed of rotation of our wire frame or magnetic field around the permanently positioned wire frame. Whatever it is, there is some kind of a rotation and this is the angular speed of this rotation. And based on this we are describing our voltage generated by this as a function of time. It's changing and we know that there is a good purpose for this, because whenever we have an alternating current, whenever we have a variable flow of electrons, it produces the variable magnetic field which can be used to induce another current of another voltage using transformers, etc. etc. That's how we transmit energy on large distance. Okay, so this is my voltage. And that's exactly the voltage which comes to our outlets at home. Now whatever the values of omega and u max doesn't really matter right now. What matters is it's variable, it depends, sine solidly, it depends on the time. So the question is, what does it mean that we are saying that our voltage is 220 volts or 110 volts? And here we have to actually come to some reasonable approach to this problem. So what is reasonable? Well, we are all practical people and we think it's reasonable to compare whatever the work this variable, sine-osloidal voltage can actually produce during certain amount of time on certain circuit which has certain resistance. We can compare it, we can compare this amount of work which this can do with amount of work which direct current with a fixed voltage can do during the same time on the same circuit with the same resistance. We can actually say, well, it's a definition, that the effective voltage is such a direct current voltage which produces the same amount of work as this one during the same amount of time on the same circuit. So what I'm suggesting to do is, let's just choose some circuit which has certain resistance r, eventually you will see that it doesn't really depend on the value of r, it will cancel. So we choose this one arbitrarily and we choose certain period of time. Now, considering this is a periodic function with a period t equals to 2 pi divided by omega, right? Well, 2 pi is a full circle, omega is, that's 2 pi radians, right? Omega is number of radians per seconds, that's basically speed. This is an angular distance and this is angular speed. If you divide distance by speed you will have the time to cover one full circle of 2 pi, which is basically a period of a sign. So we will choose this period, we will choose any kind of resistance and let's just calculate how much work this particular variable voltage can do. And then we will calculate what's the effective voltage on the direct current which does exactly the same amount of work. Okay, now how can we calculate amount of work if this is variable? Now, we know that if my voltage is constant then my, let's call it u, my current will be i divided by u divided by r, that's the Ohm's law, right? So if we know this, now we have derived a formula for producing heat, electric heat, there is a topic which is dedicated to this particular concept in this course. It's in the direct current part of the course. And the amount of work actually is equal to u times i times time or u squared divided by r time, right? Considering the Ohm's law or i squared r time. So all these are equivalent. This is amount of work which direct current with voltage u and the amperage i during the time i if you have a circuit of resistance r does. Okay, I will use this one for this particular case. However, again, this is variable, u is variable. So how can I basically calculate it? Well, that's the mathematics, it's technicality as we speak. Let's just have a very tiny amount of time from t to t plus dt. dt is differential, it's infinitesimal increment of time. So from the moment t to the moment t plus dt we can consider u voltage to have the value u of t. Variable but on this tiny infinitesimal span of time we can consider it to be constant. So we can calculate a differential of work at that particular moment using this formula. This is u squared of t divided by r times the period now is small. So the period is from t to t plus dt. So the period itself is dt differential, infinitesimal span of time. And during this infinitesimal period we will have infinitesimal amount of work created by the function u of t if the resistance of the circuit is r. Now, what do we have to do next? Well, again, we have to basically stretch it to a period. It doesn't mean to do it greater than a period because function is periodical. We will basically calculate the effective voltage on a period that will be exactly the same on double period, on triple period, etc. So what I will do now, I will integrate this. Integrate means I will summarize, right? Integration is sum on this period of time from 0 to t, where t is divided by omega. Omega is constant so it's okay. And that basically will give me w amount of work on the period from 0 to t. So let me calculate it and then I will calculate the same period from 0 to t amount of work which direct current, which effective voltage does. And I will compare them, I will equalize them and that's how I will derive the effective voltage. Okay, fine. So all we have to do is, as I said, it's technicality. So it's equal to integral of 0 to t. So instead of u of t I will put this u max squared divided by r times sine squared of omega t dt. Alright. Now this is obviously constant so I will put it aside, outside. Now integral of 0 to t. Now we need a little trigonometry. I will use the formula sine squared x is equal to 1 minus cosine 2x divided by 2. And again, if you don't remember this formula and I don't actually, I just wrote it here, but I do remember another formula. Cosine 2x is equal to cosine squared x minus sine squared. That I do remember. But I also know that sine squared plus cosine is equal to 1. So instead of cosine I will put 1 minus sine squared x and minus another sine squared would be 2. From which sine squared is equal to 1 minus cosine of 2x divided by 2. So that's how I, without remembering this formula, remember only this one, but I can very easily derive this. So there are certain basic formulas in trigonometry from which you can derive everything else. So that's why I have my sine of 2x and cosine of 2x. Remember, everything else I derive. Okay. So what else do I do? I think what makes sense now is, so I can use this formula to get rid of the square. Square definitely is something which is complication and that would give me this chance. I would also like to simplify this and I will put x is equal to omega t. Omega is a constant, so it's okay just plain replacement from which dx is equal to omega times dt. Now if my time belongs to the interval 0 t, my x would be from interval, so I would like to replace these. This is for time from 0 to capital T. Now this x would be from 0 to omega times t, which is from 0 to omega times t is 2 pi. So actually I can put it here in terms of x to pi. Instead of sine, I will change to this. So it would be 1 minus cosine of 2x divided by 2. And instead of dt, I will have dx divided by omega. So dx and omega would be here. So I just replaced the variable and came to a simpler integral. Why is it simpler? Because this is a square and this is not a square. Equals. Okay, 2 goes outside, so let me just do it here. Now integral of difference is difference of integrals, right? So if I have 1 here, this is integral from 0 to 2 pi of dx. Integral of 1 is x, so you have to substitute 4 million Newton-Blake means. So it will be 2 pi and minus 0, so it's 2 pi. So I will have 2 pi, I will have u max square divided by 2r omega 2 pi minus integral of 0 to 2 pi. Of cosine 2x dx. Now if you wish, again I can always replace it with... Instead of 2x, I would like to have y, so it would be cosine of y. Instead of 0 to 2 pi, it would be from 0 to... For y it would be 4 pi, right? And instead of dx, if my 2x is y, it would be dy divided by 2, right? If 2x is equal to y, 2dx equals to dy. So instead of dx, I will have dy divided by 2. Now what is this? Well, this is obviously 0, because the derivative of cosine is sin. And then I have to substitute upper and lower limits of integration, sin of 4 pi, sin of 0, 0 minus 0 is 0. So the whole thing is equal to just this part. And 2 and 2 will cancel out, so I will have this. That's my W, 0T, for variable voltage, for alternating current, sinusoidal voltage. Fine, great. So that's basically the most important piece of this lecture to come up with this formula, because to equalize it with direct current is easy. Now direct current is constant voltage. And the constant voltage is something u effective square divided by r times t. So I would like to find the voltage of the direct current which produces exactly the same amount of work as the variable one. That's why I put this equal sign. u effective is unknown. Okay, t is 2 pi divided by omega, right? So what's left here? r omega r omega pi and pi. What's left is u max square is equal to 2 u effective. That's my formula. And I would like to emphasize that in many cases you have complex calculations, but the final result is relatively simple. This is as simple as it can be, quite frankly. So from which follows that u max is equal to square root of 2 u effective. So if we are talking about our voltage in the alternating current circuit is let's say 220, then it means that my variable voltage is changing from maximum equals to 220 times square root of 2 to minus 220 times square root of 2. And these are the peak voltage, u max or peak voltage or amplitude, whatever the terminology is. That's the maximum voltage. But effective voltage is something which you are used to know about your own outlet. By the way, if it's 220 volts effective voltage as we are saying in our outlet, it means effective. It means that u maximum, the amplitude, the peak is equal to 220 times square root of 2. I have calculated it somewhere. It's 311 volts. So this is the maximum voltage. So it goes from plus 311 down to 0 to minus 311 up to 0 plus 311 etc. But considering the amount of work which this variable voltage is producing, it's equivalent to having a constant voltage of 220. Now if it's 110 volts, then u maximum is equal to 156. If it's 120 volts, it's 170. So that's basically it. That's all I wanted to talk about right now. I would like to explain what exactly is when we are saying that the voltage is such and such in our outlet. What does it mean if we have an alternating current? Now let's talk about these three letters. R stands for root. This is a root. M stands for mean and S stands for square, mean square. Because what we actually have done, we have squared our voltage and integrated, which means we are averaging it on an interval from 0 to t to period. So that's why it's a root mean square. We are basically, what actually root mean square is if you have a and b, it's a square plus b square divided by 2 square root. That's the mean between two different things. But if you have an infinite number of these points, if you have a variable function, integration means what? You are dividing into infinite number of infinitesimal pieces and then you summarize it and then you divide it by period basically because you have considered only one period with any amount of time. So that's equivalent of dividing by time. So that's average. So again, root square, square root, I mean root mean square. That's what it is. Well, that's it. I do suggest you to read the text accompanying this particular lecture on Unisor.com. So you go to this website, you choose the Physics 14 course. It's in the chapter called Electromagnetism. And then in the AC you have properties of alternating current. This is one of those lectures in this topic. That's it. Thank you very much and good luck.