 Hi, I'm Zor, welcome to Unisor Education. Continue talking about alternating current, and today we will talk about power consumed by a circuit through which we have AC, alternating current. Now this lecture is part of the course, the course is called Physics for Teens presented at www.unisor.com. So I suggest you to watch this lecture from the website because it has for each lecture including this one detailed notes and well sometimes I have to do certain calculations and I might bypass these calculations at the board and refer you to the notes for the lecture where they are presented in all details. Now on the same website you have a prerequisite course called Mass for Teens. Mass is a mandatory topic to know before you start learning physics for any serious kind of a person and there are many problems and exams and the site is completely free, no financial strings attached, no advertising. So it's all for your consumption. All right, so power in the alternating current environment. All right, let me start from a certain reminder about what is the power and how it is approached in the direct current when you have a constant voltage, constant amperage and certain resistor inside the circuit. Now remember what actually the voltage is. Voltage was defined as a certain difference in energy in energetic potential between two ends which we conditionally call plus and minus and the voltage is constant. It means that difference in potential level of potential energy is constant. It's like levels of water above certain level and below certain level so and it can it's constant and the water goes down obviously. Same thing with electricity. So if you have certain constant difference in potentials, let's call it u, then how the difference in potential energy actually is measured. It's measured in the work. So you have certain amount of electricity which is basically going from this level to this level and this certain amount of electricity is called usually measured in coulombs and a letter is usually the Q and what's important is that the rate of flowing electricity, electric current, electric basically charge from one level to another from plus to minus. This level is actually defined as basically a power. So the power is something like this and the work performed by flowing charge flowing electricity from plus to minus is obviously u times Q. That's the work during certain amount of time from 0 to t. So this how it is in the direct current. Now the Q divided by t is constant because it's a DC, it's a direct current, it's constant and it's usually called I. So it's equal to u times I. The difference in electric potential voltage times the direct current, electric current and that's what power actually is. So this is how it is in the direct current and incidentally using the Ohm's law, remember Ohm's law, it's equal to u squared divided by R or I squared times R. So this is something which we know from the direct current. Now we are talking about alternation current but I can definitely say that at any moment of time infinitesimal is small. I can actually do exactly the same thing because during this infinitesimal period of time from t to t plus plus dt where dt is infinitesimal differential of time. So during this interval of time I can consider my voltage and my amperage, my electromotive force if you wish and electric current to be constant and equal to e of t that's voltage that's electromotive force and I of t. They are functions of t, functions of time. So at any given moment of time the work performed from t to t plus, I'll put it as an index, work performed from t to t plus dt. It's equal to this times this times time. This is exactly the same thing as for direct current. The voltage time current amperage times the amount of time during which this happens and correspondingly the power instantaneous power at the moment t is equal to just u of t times I of t. So this is basically follows from consideration of a tiny infinitesimal interval of time as the time during which both e and i are well constant and equal to e as function of time and i as function of time. And this is basically what power actually is. So I'm trying to basically approach my power in alternating current from this perspective. However this is not really very practical because whenever you're saying that okay the power consumed by an electric bulb is a hundred watts. Well it's it's a constant number it's a hundred. I mean what does it mean if my power actually consumption is really changing with the time because my electromotive force is oscillating it's sinusoidal and my current is also oscillating. Now so I will just put exactly what it is equal to. It's e0 times sine of omega t and i is equal to i 0 times sine of omega t plus 5. Now this is something which we have discussed in the previous lecture. This is the sinusoidal electromotive force. This is sinusoidal current and we are talking about RLC circuit which is resistor inductor and capacitor in a series. That's the kind of a standard circuit which we are considering. In this e0 is a peak voltage. i0 is a peak current. Omega is angular speed of oscillations because the alternating current is usually produced by rotating the rotor inside the electric generator and omega is angular speed of rotation. T is obviously time. Now phi is the phase shift introduced by inductor and capacitor. Go to the previous lecture if you don't know what it is and i0 is actually related to e0 in pseudo Ohm's law for AC for alternating current where z square is equal to xc minus xl square plus r square. These are reactants, capacitive reactants and inductive reactants and this is resistor. All this was in the previous couple of lectures. So z is actually a square root of this. So what I meant is that this is an exact power consumption at moment t. Well actually from t to dt and we have to multiply by dt by differential of time. But during the time t as a function of t this is the power consumption. It's a rate of the work which electricity is making inside the circuit. Okay so as I was saying it's a variable and it's not very convenient to talk about. Now let's recall we were introducing things called e effective and i effective. Effective voltage and effective amperage. Incidentally each one of them are equal to peak divided by square root of 2. Peak divided by square root of 2. And again we have derived y. It's all kind of an averaging of the current and voltage during let's say one period of oscillation. Now here we also can do very very similar thing because we can always find work which is basically performed by the electricity during one period of oscillation and then divide by the length of this period. And that would be average work within one particular period of oscillation which is exactly the same as average on any number of periods because the functions are periodic as you see. And the same period by the way here and here the period is t is equal to 2 pi divided by omega. Omega is a speed angular speed so one oscillation which is 2 pi 360 degree if we divide by speed we will get the time. So we will find out how much work is done during this period t one period of oscillation. Now as soon as we will find out what's amount of work if we'll divide it by t we will have an average power per unit of time if you wish. Which is basically not exactly right. The correct terminology is basically we have to have the first derivative of it but doesn't really matter because we were talking about one period we divide by the period so that's the average per period. And that's what conveniently called p effective. Effective power and that's where and this is the constant because it's the same for each period and that's what actually is meant when people are saying that okay this electric bulb is consuming 100 watt. That means that during one particular period of oscillation the average power consumption is 1 watt and one period of oscillation is a very small amount of time it's 150 or 160 of a second because that's what usually commercial AC oscillations are. So for any reasonable amount of time practical amount of time like minute or an hour or something like this we can very safely use this particular power consumption. So our purpose right now is to find the work performed by electricity during one period of time and then we will divide it by time. Alright so let's do that. So what is amount of work which is performed by electricity during the time from zero to certainty any t actually but we will use t as a period obviously. Well it's integral p of t dt right. This is power consumed at the moment t. We are assuming that during the infinitesimal period of time dt from lowercase t to lowercase t plus dt this power remains constant so we will multiply power by time get the work performed from t to dt to t plus dt and then we integrate from zero to capital t and that's how we get the whole work. Well let's just calculate this integral no we did right what's equal to we know what pt is so we will take e0 and i0 outside and we will have integral from zero to t of multiplication of two sine. Now from this is a very easy integral by the way from trigonometry we should convert product of sines into sum or difference actually here it is cosine of alpha plus beta is equal to what cosine alpha cosine beta minus sine alpha sine beta and cosine of alpha minus beta is equal to plus beta plus sine alpha sine beta so what do we do from this from the second we will subtract the first we will have cosine cancelled we will have two sine by sine so basically sine alpha times sine beta is equal to the second one cosine alpha minus beta minus the first one alpha plus beta divided by two right subtract from this this we will have double multiplication of sines so that's why we divided by two this is a difference between two cosines now in this case we will do exactly the same so we will have e0 pi0 integral from zero to t so sine times sine is cosine of minus so if we will take minus it will be phi minus phi actually but doesn't really matter because the cosine is even function so cosine of minus phi is equal to cosine of phi so we will have here cosine of phi minus cosine of the sum which is 2 omega t plus phi and we have to divide everything by two so we will put one second to two and dt so that's our integral now obviously we will make it as two integrals because integral is additive function so it will be integral of cosine and cosine is a constant right phi is an angle which is measured based on difference between xc xl remember tangent phi is equal to xc minus xl divided by r but this is reactance of capacity capacitor this is the reactance of inductor so it's a constant it's basically determined by the characteristics of the circuit right just in case you forgot about what is reactance xc is equal to 1 over omega c and xl is equal to omega l where c is a capacitance and l is inductance of the inductor okay so this is all from the previous couple of lectures so I'm just using the cosine is a constant so it goes out from the integral and all we have is integral of dt which is actually from 0 to t which is actually t so we have equal to 1 over 2 is 0 i is 0 cosine phi times t that's my first one now minus now what is the minus now let's think about integral of of cosine well even if you don't think about anything if t is a period and the periodicity of this function is what it's t divided by 2 omega right which is you know which is multiple of periods so we should expect actually is to be equal to 0 right because if you take a plane function let's say cosine now this is the period obviously if you integrate it that's sum of this plus sum of this it's supposed to be equal to 0 and it is equal to 0 now if you want we can prove it analytically so our intuition is right so we have integral from 0 to t we have minus cosine of 2 omega t plus phi dt right that's what we have now cosine is a derivative of the sine right so we can have the indefinite integral should be minus sine 2 omega t plus phi with some kind of a multiplier so the derivative of this is cosine times 2 omega so I have to divide it by 2 omega I have to do it from 0 to t right now t is one period right one period is 2 omega divided by 2 pi divided by omega so if I will substitute this t for this what I will have I will have sine of t is equal lowercase t is equal to capital T so what will be it will be 4 pi right omega will cancel out 4 pi plus pi divided by 2 omega that will be for t and for 0 it will be sine of 5 but again sine is a periodic function so this value and this value give the same value of sine that's why this will be 0 so we will have 0 so this part disappears that's the answer basically this is an answer of this is my work performed during this time where t is a period of oscillation now from this we can obviously say that the power the average power or we can call it effective power is equal to work during this period of time divided by time so it's equal to one half E0 I0 cosine pi let me make it a little more practical again I0 and E0 are peak voltage and voltage and amperage and again we are not really used to these peak kind of characteristics in practice but we're talking about voltage in the AC outlet to be like 220 or 120 or 100 whatever volts it means effective voltage and again if you forgot about what effective is go to one of the previous lectures it's all explained there so effective voltage is by square root of 2 smaller than the peak and it's defined basically through the work I mean what is my effective voltage and effective amperage which if it was a direct current it perform exactly the same work so that's what effective is so in this particular case we know that effective voltage is smaller by square root of 2 of effective of peak voltage and effective current is also by square root of 2 smaller than peak so the effective ones are constant which give exactly the same work if we have a direct so direct current with this voltage in this amperage give exactly the same work as alternating current with sinusoidal voltage and sinusoidal amperage so if this is true you see this 2 this 2 is 1 square root of 2 times another square root of 2 so I can actually use them and I will put that P effective is equal to E effective times I effective times cosine phi so easy root divided by square root of 2 is effective I 0 divided by square root is I effective and that's why we have this formula which is very simple and what's very important is this cosine okay now let's talk about this cosine if we are talking about RLC circuit it contains resistor capacitor and inductor so let's consider different cases what if I have only our circuit if I have our circuit it means that capacitive reactants and inductive reactants are absent they are 0 no capacitor no inductor just a resistor tangent phi is equal to xc minus xl divided by r which in this particular case phi is equal to what if tangent is equal to 0 my angle is equal to 0 so the cosine is equal to 1 and I have this formula P effective is equal to E effective times I effective because the cosine 0 is 1 this is exactly the same as if it's a direct current you see that's a very important demonstration of why effective voltage and defective current are very important because the law which basically expresses the power consumption is exactly but I'm not exactly it looks exactly the same as the corresponding law for direct current where the P is just equal to E times I where all of them are constant okay so that's very important so there is no capacitor and no inductor similarly if there is a capacitor and there is an inductor but they're equal to each other and we call this a resonance then we will have exactly the same we will have the same formula because then my tangent is equal to 0 my angle is equal to 0 my cosine is equal to 1 so if I don't have capacitor and inductor or I have both of them and they have the same reactances capacitive reactance is equal to inductive reactance which means that 1 over omega c is equal to omega l or omega square is equal to 1 over l times c if this is given so my inductance times capacitance give me the inverse give me a square of angular velocity or if you wish l times c is equal to 1 over omega square same thing so if this condition is satisfied then we still have exactly the same formula because the cosine will be equal to 1 if however I have something like RC which means reactant resistor and capacitor with certain reactance so XL is equal to 0 what do I have in this case well I have certain tangent positive by the way so in my formula the phi will be with a plus sign actually and I will have you know certain correction to the product of effective voltage times effective and the correction factor will be the cosine of x of the angle 10 tangent of which would be equal to xc divided by by r same thing with rl in this case the tangent would be negative right x c would be 0 and this would be minus something so we will have a negative angle and we will have corresponding with the negative angle but the cosine of negative angle is the same as cosine of the positive angle so that doesn't really affect the power so basically that's the final formula I wanted to basically convey to you and certain cases when when we can apply this formula and okay basically that's it read the notes for this lecture they contain maybe some other considerations like one of the considerations what we don't have are I mean it's not practical we don't have any circuit without any kind of resistance right without any kind of resistors everything even just plain wire consumes certain it has certain resistance and it consumes energy but it's interesting that l by itself or c by itself inductor or conductor or combination if there is no resistance my denominator is 0 which means my tangent is equal to 90 degree the cosine will be equal to 0 and the power consumption is 0 so if there is no resistor there is no power consumption well that's it thank you very much and good luck