 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 we 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 going to, for simplicity, for shortcut, try it this way. Mean of what? Of squares of the currents. So I'm going to write it 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 one squared plus i two 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 squared, then take the mean value and then finally take the square root of it. And then I'll get my rms. All right. So let's do it stepwise. 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 not square times sine squared omega t. Yay. First step done. Okay. Next, I think that's the most difficult step. Let's calculate the mean of the current. I think this is the most difficult step because the third step is just to calculate the square root. That's also something we can do. So 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 not square sine square omega t1 plus at some time t2, i not square sine square omega t2 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 not square is a common everywhere i not, there will be an i not square, i not 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 not 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 i1, i2, i3, i4 conceptually that's what you're doing. But if you're doing it continuously, you have to do an integral. But luckily, we're dealing with sinusoid, there's a logical way of figuring this out without any integral. Only works for specific functions, it'll work for science. And I'm going to show you 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. 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 not square by two. And what does that happen? What is that equals? That equals i not divided by root two. And there we go. We have proved that for sinusoids, the RMS value equals the peak value divided by root two. And okay, you can also write this as i not times one over root two is about point seven zero seven. 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 a 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 I add up all the values of i's over here and over here, they're going to 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 one here, I have some minus i one here for some i two here, I have some minus i two here. So they will all cancel out and I'll get zero. 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 sinusoid, 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 we're all we have to do to calculate the mean values 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 square? It's not all that easy, right? Well, we can use a trigonometric relation. You might recall the trigonometric relation cost two theta equals one minus two sine square theta. And why I like this is because from this, now I can say, hey, sine square theta has to be, if I just rearrange this, has to be one minus cost two theta divided by two. 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 cost, not cost square costs. And the graph of costs is going to be same. It's just shifted a little up, you know, shifted a little bit left. So when I draw costs, I get a sinusoid. If I add something to a cost, I will get a sinusoid. If I divide some number by cost, 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 want 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 one minus cost two omega t divided by two. Okay, now let's graph this. So first we'll graph what cost to omega t looks like. And then we'll do minus cost to omega t, then we'll add one to it, then we'll divide this whole thing by two. Let's do it together. So here's our axis. I want you to first think about what cost to omega t would look like just graph that cost to omega t, it's going to be a sinusoid. And you might remember when you when you cost zero is one, 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 to 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 two 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 around 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 cost. Now let's draw minus cost. What would minus cost look like? Well, just take every point and multiply by minus one. That means this point when I multiply minus one would come down. This point when I multiply minus one 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 cost to omega t. It's still a sinusoid. Okay, now let's add one to this. What will happen? Again, I want you to pause and think about it. Add one and divided by two. Imagine what's going to happen. Okay, let's see. When I add at every point, the value of every point increases by one. So this point, which is at minus one, when you add one comes to zero. This point comes to one. This point comes to two. So the whole graph shifts up and this will come to zero and this will come to two. So it'll look like this. This is one minus cost to omega t. We're coming close. This is still a sinusoid. Finally, let's divide the whole thing by two. What will happen? Well, every value gets divided by two. So this will stay zero. This divided by two makes it one. 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 one minus cost to omega t. We have grafted. 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 zero and one. And that is half booya. 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 prove 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 two, whether it's a current or a voltage or some other quantity.