 Hi, I'm Zor. Welcome to Inizor Education. I would like to continue talking about three-phase alternating current and solve a very simple problem. The problem has actually practical implementation and practical need because sometimes in our households we can use the electricity of let's say 120 volts and sometimes 220. Certain appliances like electric stove for instance or powerful air conditioner require 220 volts to reduce because they consume a lot of energy so if you increase the voltage you can decrease proportionally decrease the amperage so to prevent certain accidents whatever. In any case we do need both. The question is how it's done. Well you call an electrician an electrician probably does whatever is necessary with your connections and here is your outlet with 220. So how is it done? Well I will answer this question after I will finish talking about this little problem. So here is the problem. Let's say you have three-phase generator. Well it has four different wires zero which is basically like ground. It has phase number one which is some maximum times sine of negative. You have phase two which is shifted by 120 degrees by 2 pi over 3 relative to and you have a phase three which is shifted by yet another 120 degree which is if you wish it's 4 pi over 3 or the same thing as minus 2 pi over 3 because 2 pi is a period of function sine so that's more convenient to do this. So my question is we do know these voltages relative to to the ground to the zero okay. My question is what is the difference in voltage between pair of phases? All right well the only thing is you can calculate the pair simply. Well here it is. If this is let's say x y z and this is let's say a you have a you have x minus a you have y minus a and you have z minus a these are differences. If you need x minus y this is x minus a minus y minus a right so by subtracting one from another so to find the difference between these two we can take the difference between these two which is difference between this and this and this and this right so let's just do the calculations and this is just a simple trigonometry and that's why on this website I have a course which is called mass for teens which is prerequisite for this one and I do recommend you to take it okay so let's do let's say x minus y so this is e12 as a function of time so e is a peak voltage now if you remember effective voltage is square root of 2 less than the peak voltage we did it in some other lecture okay so the difference between this and this all right so this is e sine of omega t minus e sine omega t plus 2 pi over 3 it's more convenient for me to do it differently it's sign let's substitute phi is equal omega t plus pi over 3 then omega t would be phi minus pi over 3 minus e obviously goes factor out sine of phi plus pi over 3 so 2 pi over 3 is 5 plus pi over 3 and plane omega t is just phi minus pi why do I need it well you will see that if I will open all these parentheses and use the function of sine of difference or sum of two angles you will see how things basically just reduce cancel each other equals to so e okay sine phi cosine pi over 3 minus cosine phi sine pi over 3 right that's this one minus this is the sum of three angles well you remember I will remind sine alpha minus beta is equal to sine alpha cosine beta minus cosine alpha sine beta and this is plus this is plus well again go to trigonometry so minus some sign of some of this angle so it's sine phi cosine phi not I divided by 3 and minus again because this is minus so and this is sum of two angles so it's plus and plus so minus and minus cosine phi sine pi over 3 what happens here that's what happens so it's minus I e cosine phi sine pi over 3 equals okay now it's minus e cosine omega t plus pi over 3 oh I'm sorry that's supposed to be 2 right it's 1 and 2 so it's 2 cosine now what is the sine of pi over 3 pi over 3 16 degrees sine is square root of 3 divided by 2 so it's 2 here square root of 3 divided by 2 is equal to minus square root of 3 e cosine so what do we have here that's what's very important this is what very important these guys all are sinusoidal now doesn't really matter whether it's plus 2 pi over 3 minus it's just a shift in time but the behavior is sinusoidal with e being the peak here we have also sinusoidal oscillation because cosine and sine are basically the same graphs again shifted in time by pi over 2 so basically it's also sinusoidal but what's important is the peak voltage is not just e it's but in magnitude it goes from e to minus e to e to minus e but this goes from square root of 3 times e so it's by square root of 3 greater the difference in voltage between these two guys at any moment is greater the peak voltage is greater by square root of 3 which means that effective voltage which is square root of 2 less than the peak is also square root of 3 greater between two phases now as an example let's just take two other phases let's say phase 2 and phase 3 it will be the same so phase 2 and 3 with the e 2 3 of t difference between this and this so again it's e times sine of sum which is sine of omega t cosine 2 pi over 3 plus cosine omega t sine 2 pi over 3 minus this one minus so this is sine of difference between angles so it's cosine 2 pi over 3 minus but this is minus here so it's plus cosine omega t sine 2 pi over 3 so what do we have here we cancel this and we have 2 cosine omega t times sine of 2 pi over 3 what is sine of 2 pi over 3 it's 2 pi over 3 is 120 degree and sine is an opposite angle which is one half is it one half sine of 2 pi over 3 sine is no that's 120 degrees sorry it's not correct it's this way it's 90 plus 30 here so sine is square root of 3 over 2 so this is square root of 3 over 2 and this is one half minus one okay so what we have here again square root of 3 by the way I forgot e obviously times e times cosine omega t so what we have here exactly the same thing as before the peak voltage is 100 and square root of 3 greater than the peak voltage of the phase 2 relative to 0 so from 2 to 3 is square root of 3 greater the peak voltage from 1 to 2 and if you calculate the same thing from 1 to 3 it will be the same thing well very quickly let me just do e from 3 to 1 so from 1 to 2 from 2 to 3 from 3 to 1 okay so it's e times sine of difference which is okay here it's also convenient to do this phi is equal to omega t minus pi over 3 and I have sine of phi minus pi over 3 right phi is omega t omega t minus pi over 3 and another pi over 3 that would be this one minus sine of phi plus pi over 3 because if I will add pi over 3 will have omega t here that's what I have equals to equals sine or e obviously again I forgot e sine phi cosine pi over 3 minus cosine phi sine pi over 3 minus this is sine of sum of two angles so it's minus and minus sine phi cosine pi over 3 minus cosine phi sine pi over 3 so what do we have canceling sine cosine minus sine cosine so it's equal to 2 e minus minus and minus cosine phi and sine pi over 3 which is square root of 3 divided by 2 so we have the same thing again square root of 3 greater so as you see the difference in voltage between two phases is by square root of 3 greater than the difference between the phase and the zero so if you are in house which needs both voltages let's say 120 and 220 127 actually 220 what you do is you have you need to have at least three wires one zero and any two out of three phases and that's how this distribution actually is happening so the three phases and zero obviously go to well some kind of district let's say a building a big building gets three phases and then every apartment has zero and two out of three phases distributed evenly more or less heading let's say this in your apartment for a regular like regular lamp or something like this you can use two wires between two wires zero and and one of the phases so you have certain number of outlets which have two wires one of them is zero and other is a phase and in some cases if you have something like an electric stove or refrigerator well summer refrigerator different refrigerators on the market then you will have an outlet two wires of which are actually two different phases and in one particular case it was really curious thing I needed 220 in one particular place I didn't have it I didn't have any outlet I had for 220 but I have one outlet from phase one and zero which was 120 and then another outlet from two to zero so what I did is I took one phase from one outlet and another from one from another outlet and the difference between them was 220 don't do it at home so in any case as you see my purpose was to basically convey to you that the difference in voltage between two phases is square root of three greater in peak and in effective voltage then the difference between any phase and zero now if you will open some textbooks more or less technical textbooks they will tell you that the voltage between two phases is 1.73 greater than the voltage between phase and and and the zero but usually you don't have any explanation explanation is very simple it's plain trigonometry that's where the square root of three is and square root of three is 1.73 and etc etc okay so that's the end of this lecture I do suggest you to read the notes for this lecture on Unisor.com the calculations are all there obviously do it again yourself it's a you know good exercise in trigonometry and again let me just emphasize it again you do need masks to study physics seriously okay that's it for today thank you very much and good luck