 You might know that the voltages that we get from our socket in our houses is an alternating voltage But I was very curious when I saw that in India the voltage that we get is 230 volts. I was thinking hey, the voltage is fluctuating right? It's it's continuously changing So what does this to a 30 volt represent? At first I thought obviously it must represent the peak value That's what it means, but then I was surprised to see that. That's wrong. That's not what it represents In fact, it turns out that it represents something called the RMS value And the goal of this video is to figure out what exactly is this RMS value and why should we care about it? So let's start by looking at a situation imagine I pass an alternating current through a bulb So let's say we are passing a current that is fluctuating between some peak value i0 and minus i0 Then we know that current passing through any resistor which is inside the bulb over here dissipates power Generates heat and because the current is fluctuating the heat generated or the power dissipated also Fluctuates when the current is zero the power dissipated over here would be zero and when the current increases and goes all the way to Maximum the power dissipated over here, which I'll show as the brightness of the bulb will also increase and go all the way to Maximum so as the current fluctuates the power dissipated also fluctuates But here's the thing in India this fluctuation is happening about 50 times per second And as a result the fluctuations are happening so fast our eyes will not be able to make that out Instead, you know what we will see we will see not the maximum power not zero not the fluctuations But we will see some kind of an average Power or average brightness somewhere between maximum and zero So what our eyes will end up seeing and what I'm denoting over here is the average power that is being Desipated in this resistor due to this alternating current All right. Now, let's say I take that same bulb and pass a direct current through it DC constant current not alternating Okay, then that's also going to dissipate some power over here But the power dissipated over here, which I'm going to call the DC power the power dissipated over here DC power This power will be a constant So we don't have to talk about average values over here because the current is a constant the power dissipated over here is A constant and now here's my question Okay, ready for this my question is what should be the value of this DC current so that the power dissipated over here is Exactly equal To the average power that's dissipated over here I repeat What current should you send over here so that the constant power that you find over here Equals the average power that you are getting over here due to the alternating current Did you get that question? Let's do a quick sense check because it's not a very straightforward question. I used to get confused a lot So let's take an example Imagine the current that is alternating over here is alternating between say 10 amperes and minus 10 amperes That's the peak value. Let's say, okay And as a result of that the power will also fluctuate between some maximum and zero Now my question to you is the DC current that I need to send over here So as to get the same power as the average power here Do you think that current should be equal to 10 amperes? more than 10 amperes or Less than 10 amperes. Can you pause the video and think a little bit about that? Okay, let's see what would happen if I were to send 10 amps of current over here Well 10 amps represents the peak value of the current over here When when the current is 10 amps the power dissipated over here would be maximum And what you're seeing now is not the maximum power what you're seeing now is the average power So if you were to send 10 amps of current in this example The bulb would be going with maximum power and so that would not be equal to the average power Therefore the current that you need to send over here should be less than 10 amps. Does that make sense? But how much exactly? Would it be 5 amps? Well, that's it. We'll not guess but that's what we want to find out And you may be wondering why we need to find this out. We'll talk about that in a few minutes But the question now is how do I figure out what that current should be? What this current should be? Well, the way I like to do this is we already know how to calculate power in any circuit We know that the power power dissipated at least through resistors. We can always write that to be equal to Because current is given I square R. So we can say it is the I DC square times R Where R is the resistance of this bulb All right, and we want this To equal the average power Decimated over here and so the question now is how do I calculate the average power that's dissipated over here? Well, we can try calculating the average just like we always do So we're here since the current is changing at every point There is some value for current and so there will be some value of power at each point So we can add up all those powers and divide by n and that was that will give us the average power Don't worry. We don't have to actually do it. It's just the concept that it's what matters So here's I'm thinking so let's say here are some points that I've marked So at this point the power would be I one squared R at this point will be I two squared R And so on and so forth you take all those power values add them up divide by n that usually the average power And I want you to pause the video and see if you can equate that to this and see what expression you end up for IDC Then you might be able to rediscover why this is called the RMS value yourself So hopefully you're excited and pumped to give this a try. All right so this should equal the average power and Average is just adding up individual powers at every instant and dividing by n there are infinite values actually So the P1 plus P2 plus P3 plus P4 Up to Pn and then divide all of that by n and like I said They're actually infinite values and you can't really do it this way You have to do an infinite summation which is called an integral, but don't worry about all of that We're not actually going to calculate what the average power is going to be over here We just want to understand the concept and therefore we just leave it at that Okay, so this would equal what? Well P1 would equal the power dissipated at this point. That'll be I1 squared So that'll be I1 squared times R That's the power dissipated at this point in time Plus P2 would be power dissipated at this point in time. So that would be I2 squared I2 squared Times R and it keeps going on Divide this whole thing by n Okay, our goal is to figure out what IDC is right? So from this two equations you can see R can be pulled out and you have R is common and you can cancel R on both sides I'm not gonna do that. I'm sure we can do that in our head and therefore what do we end up with? We end up with if you cancel R on both sides IDC squared ends up being I1 squared Plus I2 squared and so on Whole thing divided by n But I want IDC. So I'm gonna take square root on both sides So if I take square root on both sides, I will end up with the Square root of I1 squared plus I2 squared and so on divided by n And that's how you calculate the current needed to give the same power as the average power over here Now the question is what should we give a name to this? You can call it as IDC but it doesn't tell you much It just tells you it's a constant current But what should we name this? And this is where we name it based on what we are doing over here So what we are doing is we are taking a square root So we put an R, we are taking a root Root of what? Root of what is this operation called when you add things up and divide by n Hey, that's average or mean So we are taking the root of mean of Mean of what? What are you adding and then dividing by n? Oh currents, not currents, square of the currents Oh, why are we adding the square of the currents? Well that's because when you calculate power you get a current square So we are doing root of mean of square of the currents That's what this is and that's why this is called IRMS Let me just write it as small That's why this is called the RMS value And so the power dissipated over here I can just say it's IRMS squared times R And since this power is the same as the average power over here I can now say that the average power dissipated Due to any alternating current through a resistor Equals the IRMS squared times R And this is why we love RMS values Because if you give me the complete alternating current function With the peak value and the frequency and all of that Calculating average power is a nightmare You have to actually go ahead and do this calculation And this will require integral as we spoke about earlier But if you give me directly the RMS value I say thank you, thank you Because all I have to do now I don't care about alternating currents I don't care about the fluctuations I don't care about anything I just take I squared RMS times R And I get my average power And of course we could have done the same thing In terms of voltage And so if you want to calculate the average power In terms of voltage We can just do V squared by R But the voltage we take is the RMS value of the voltage So I can also say this equals I'm running out of space over here But we can also call this as V RMS squared divided by R Everything is similar And now we can come back to that original question The reason why the voltage rating is Given as the RMS value for us Is because now if we attack some Say resistor across it And we want to calculate the average power There's also an alternating current Power is fluctuating But if I want to calculate the average power I say I have the RMS value So I'm just going to do V squared over R Now finally you may have a question What if RMS values are not given to us Not all questions are nice to us, right? What if they give you say a current function With the peak value And they give you the frequency and all of that But they don't give you the RMS value Then how do you calculate the average power? Well then you have to calculate The RMS value first And it turns out If you do this for sinusoidal currents And sinusoidal voltages You will see that RMS value Happens to be the peak value Divided by root 2 The same thing will be true for Voltages as well V RMS happens to be The peak value V naught divided by root 2 And if you're wondering where is this coming from We have to actually go ahead and solve this And we will do that We will derive this expression in a future video And in fact there is a logical way of doing it as well We'll do that We'll get to that in a future video But remember this is only true If you take sinusoidal currents and voltages Because there can be other kinds of alternating currents And voltages like they can be You have triangular waves Or you can have sawtooth or square waves So many different kinds are available And for each one the RMS value would be different So in our case we can use this formula Because we are dealing with sinusoid So in our case if we calculate what the RMS value is This is the RMS value It happens to be I naught divided by root 2 10 divided by root 2 And that turns out to be close to 7 amperes So long story short RMS value represents that constant current Which gives you the same power As the average power due to the alternating current You calculate it as the root of the mean of the squares Of all the currents at every instance of time And why we love it so much is because To calculate average power If you have the RMS values You can just do I square R or V square over R