Sine wave RMS value - derivation (without calculus)

when we pass alternating current through any resistor or say a bulb then the power dissipated in that resistor which i'm going to show as the brightness of the bulb will keep fluctuating because the current is alternating and it's for that reason most of the times you are interested in the average power that gets dissipated and we saw in a previous video to calculate average power all you need is an rms value of the current if you know what the rms value of the current is we can just say average power equals i square times r or if you know the rms voltage we can say it's v square over r now the goal of this video is to figure out the rms values for sinusoidal currents or voltages and we're going to derive we're going to prove that the rms value for sinusoids turns out to be the peak value divided by root 2. before we begin if this looks new to you or you feel like what's going on what is this rms business what are we talking about don't worry we've talked a lot about this in our previous video on rms values and power dissipated so feel free to go back and check that out okay let's begin so our goal is to figure out what the rms value of the current is and i'm coloring them because there are three operations what does rms mean r stands for root so you have to take the square root square root of the mean and i'm just going to write mean this way this brackets means calculating mean meaning you calculate various add up all the different values and divide by n i'm just gonna for simplicity for short shortcut write it this way mean of what of squares of the currents so i'm going to write this way i squared where i is given over here it's a sinusoid okay and what i mean by this let me just write that somewhere over here what i mean by this is what i mean by this is you just take different values of i so i 1 squared plus i 2 squared and so on and then divide that by n i just didn't want to write the whole thing over here so that's what i'm writing okay so there are three steps involved first take my current square it then take the mean value and then finally take the square root of it and then i'll get my irms all right so let's do it step wise so let's first do the first step i think that's one of the easiest steps square my current so that's easy i just have to square this number this value so what will i get if i square that i get i naught square times sine squared omega t yay first step done okay next i think that's the most difficult test step let's calculate the mean of the current i think this is the most difficult step because the third step is just to carry the calculate the square root that's also something we can do so the calculating the mean is the harder part okay so we'll have to calculate the mean or the average value of this number that's going to be i not square sine square omega t now this calculation you can imagine it to be something like at some time t1 i'm calculating i naught square sine square omega t1 plus at some time t2 i naught square sine square omega t 2 and so on and so forth and then dividing the whole thing by n now one thing you can see when i do that i naught square is a common everywhere i not there will be an i naught square i naught square and i can take that common out and therefore i can always write this i can say that this is i not square times the mean value of sine square omega t does that make sense because i know square is going to be the same everywhere and the question now is what is this and this is the hardest part actually of the whole calculation this requires you know this requires mathematics because it's a continuous function and so you can't just say i 1 i 2 y 3 i 4 conceptually that's what you're doing but if you're doing it continuously you have to do an integral but luckily we are dealing with sinusoids there is a logical way of figuring this out without any integral only works for specific functions it will work for science and i'm going to show you in in a minute or two that if you calculate the average value it turns out to be half you have to wait for it i'll show you that in a second in fact both you and i can do it together we can actually together without integration uh see this happening so wait for it so so this will be the average value that's the step two and now step three we can take the square root square root of the average or the mean of the square of the current so that's going to be the square root of this number i naught square by 2 and what does that happen what is that equals that equals i naught divided by root 2. and there we go we have proved that for sinusoids the rms value equals the peak value divided by root 2. and okay you can also write this as i naught times 1 over root 2 is about 0.707 so you can also remember this you can also think of it this way all right now the last thing and i think is the most interesting thing is to actually convince ourselves that the average value or the mean value of sine square is half so let's now just focus on that let's convince ourselves of just that so i'm going to dim everything else so we need to prove this let me keep this up let me move this thing also up okay so here's my first question to you forget about sine square what do you think is the mean value of just sine if i take any sine function say sine of omega t what do you think is the average value of just that what's that going to be well because sine is a symmetric function i can say if i take the mean value over say one complete cycle then when i add up all the values of i's over here and over here they're gonna give me zero because for every positive value over here i will find a negative value for i naught i have minus i naught for some i 1 here i have some minus i 1 here for some i 2 here i have some minus i 2 here so they will all cancel out and i'll get 0. so in general we could say whenever you have a sinusoidal function sinusoidal function the average value of a sinusoid in general we can say is at the center now of course here the center was zero because the sinusoid is centered around zero and that's why this one's mean value turns out to be zero but in general you can have sinusoids however you want and we can say that its mean value or average value has to be at the center and the beauty is we will see that when you square a sinusoidal function we will end up with another sinusoidal function and what i mean by sinusoid is you're going to get a similar function just maybe stretched a little bit or squeezed or maybe moved up or down but it's going to stay a sinusoid which means where all we have to do to calculate the mean value is look at the center okay so the first step is i want to convince you that the sine square is in fact a sinusoid and draw a graph for that but how do i do that how do i draw just by looking at this how do i draw a graph for sine squared it's not all that easy right well we can use a trigonometric relation you might recall the trigonometric relation cos 2 theta equals 1 minus 2 sine square theta and why i like this is because from this now i can say hey sine squared theta has to be if i just rearrange this has to be my 1 minus cos 2 theta divided by 2. when i look at this and i say ah so if you take a sine square it's the same thing as writing this function and notice this function has a cos not cos square cos and the graph of cos is going to be same it's just shifted a little up shifted a little bit left so when i draw cos i get a sinusoid if i add something to a cause i will get a sinusoid if i divide some number by cos i get a sinusoid okay so you can kind of see this whole thing once we graph it will still be a sinusoid and as a result we can just look at its center once we graph it and that will give us the mean value so let's go ahead and together graph this so the way i'm going to graph this wait wait i'm jumping ahead i'm so excited so from this let's first write this means sine square omega t can be written as 1 minus cos 2 omega t divided by 2. okay now let's graph this so first we'll graph what cos 2 omega t looks like and then we'll do minus cos omega t then we'll add 1 to it then we'll divide this whole thing by 2. let's do it together so here's our axis i want you to first think about what cos 2 omega t would look like just graph that cos 2 omega t it's going to be a sinusoid and you might remember when you when you cos 0 is 1 so it starts over here so can you just imagine what that graph is going to look like pause and think about it all right here we go this is what cost omega t would look like so it swings between plus one and minus one it starts at plus one and then swings like a sinusoid and you can see because it is 2 omega t the frequency is double and therefore it's oscillating much faster and that's why it's kind of like shrunk in the x axis or along the time axis does that make sense compared to this one and if you're logically thinking about why it is shrinking well for now i can say it's a trigonometric relationship but maybe later on we might be able to make sense of why it is the frequency has become twice but anyways this is cos now let's draw minus cos what would minus cos look like well just take every point and multiply by minus 1 that means this point when i multiply minus 1 would come down this point when i multiply minus 1 will come up and so on so the whole thing will flip can you see can you imagine here goes it's going to flip like this this is minus cos 2 omega t it's still a sinusoid okay now let's add 1 to this what will happen again i want you to pause and think about it add 1 and divided by 2 imagine what's going to happen okay let's see when i add at every point uh the the value of every point increases by 1. so this point which is at minus 1 when you add 1 comes to 0. this point comes to one this point comes to two so the whole graph shifts off shifts up and this will come to zero and this will come to two so it'll look like this this is one minus cos omega 2 we are coming close this is still a sinusoid finally let's divide the whole thing by 2. what will happen well every value gets divided by 2 so this will stay 0. this divided by 2 makes it 1. so it's kind of like gets compressed along the vertical and so this will come to one and so you can kind of imagine this is what it would look like and again compressing a sinusoid will still give us a sinusoid so this is still a symmetric sinusoidal graph and we have gotten 1 minus cos 2 omega t you have graphed it so that means this is the graph of sine square omega t and you're seeing right in front of your eyes it's a sinusoid and therefore i can say the mean value of this should be right in between and where is the center of this it's between 0 and 1 and that is half booyah therefore mean value of sine square is half when i first learned about this my mind was blown because i thought integration was the only way in general you have to integrate but because sinusoids are such nice functions we can do it this way beautiful right and finally can you sort of see why the frequency of this sine square is twice the frequency of this one well that's because when you square this you see this lobe which is down over here also comes back on top and that's why you end up getting a sign that looks like this and that's why the frequency becomes double but the main point was we proved that sine square mean function is half and so if i go back we've conclusively proved that the rms values of any sinusoid is going to be the peak value divided by root 2. whether it's a current or a voltage or some other quantity