 In this video, we're going to explore power dissipated in an LCR series circuit and we'll introduce the idea behind power factor. But before we look at an LCR circuit, let's go back to our familiar pure resistive circuit. So let's say we have a resistor of 100 ohms connected across a source, an alternating voltage source, whose RMS voltage is 100 volts. Immediately, the first thing I want to talk about is what is the current in the circuit? The RMS value of the current is going to be the RMS voltage divided by the total opposition that is 100, 100 by 100 is one ampere. Now, the beauty about RMS current is that it can help you calculate the average power very easily. And we've seen before, the average power consumed by the resistor, or you can say it's the power dissipated at the resistor, that's going to be just the RMS current squared I RMS squared times R. So in our example, the average power dissipated would be one square times 100, that's going to be 100 watt. All right, now here's my question to you. Suppose this time I have an LCR circuit whose impedance is also 100 ohms and it is also connected across the same voltage source as before, 100 volt RMS voltage. Then the current in this circuit, I RMS, the RMS value of the current here, will also be just like before, RMS voltage divided by total opposition this time, V divided by impedance this time, it'll be 100 by 100 is going to be one ampere. So my question to you is, what will be the power consumed in this circuit? Would be the same as before? Do you think it'll be more than before? Or do you think it'll be less than before? So can you pause and think a little bit about it? Okay, now it might be reasonable to think just like earlier, here average power might be I RMS squared times total opposition, which is Z. Let me use red for that, Z. And we might think, okay, maybe we end up with the same answer, which kind of makes sense, right? Because I have the same voltage, same current, same opposition as before, everything is the same. So maybe the power dissipation also remains the same. But this is not true. This is wrong, why? This is wrong, why? Because the entire opposition does not consume power. Here's what I mean. Among LC and R, resistor consumes power just like before I square R. But what about the inductors and the capacitors? They do not consume any power. For example, if you take a capacitor, then what you'll find is that when the capacitor is being charged, it definitely consumes power, it stores energy in its electric field. But then when it gets discharged, it transfers that power back to the source. Therefore, on an average, it does not consume any power. And same is the case with an inductor as well. On average, it consumes zero power, which means the entire opposition, the entire impedance does not consume power. Only the resistive part of it consumes power, which means even in this circuit, even in this circuit, the average power consumed would only be the power consumed by the resistor. So it'll still be I RMS squared times R. And so would it be the same as before? No, because before, the entire resistance was 100 ohms. This time, the total impedance is 100 ohms, and therefore, resistance would definitely be less than that. Now, how much, that depends upon the details, we don't know, but let's say, for the sake of some example, let's say the resistance in this case is just 10 ohms. That's totally possible. Maybe the reactances are very high, and as a result, I'm getting total impedance to be very high. It's totally possible in this circuit that I can have resistance to be only just 10 ohms. Then in this particular example, the average power that gets consumed by this circuit is going to be one squared times 10, which is going to be just 10 watt. At first, this did not seem like a big deal to me. I mean, smaller resistance, less power consumed, larger resistance, more power consumed. So what's the big deal? This is where the electric company calls me and says, oh, Mahesh, you are consuming the same voltage as before, same current as before, yet you're consuming way less power than before. And I say, yeah, so what's the big deal? The big deal is, and this is where the electric company is trying to explain to me the problem, see, whenever electric current is passing through a transmission lines, it gets heated up. So there's some energy loss happening, right? Who's gonna pay for those energy losses? Let's say that energy loss over here was, I don't know, maybe 20 watt. Now, when I'm consuming 100 watt, energy loss of 20 watt, the electric company might say, fine, not a big deal. Don't worry about the 20 watt energy loss happening. You're consuming 100, you're paying for that, it's compensated, I have enough profit margin. But over here, the energy loss stays the same because the current is the same and the voltage is the same. So the energy loss here is 20 watt. I'm only consuming 10 watt. So the electric company says, you, this time it's not profitable for me. There's so much energy that loss that is happening just for your 20 watt of consumption. And so the person says, you have to pay more over here. You have to pay for my losses also. Does that make sense? Well, I'm sitting here, I'm thinking that kind of makes sense. It's kind of like delivery charges. When you're ordering something big and you're paying a lot for it, then the petrol charges, delivery charges are covered because they're small compared to how much you're paying. But if you're ordering, say, a five rupee expensive, just exaggerating, then you have to end up paying for the petrol charges and everything, right? So from a power consumption point of view, this circuit has less efficiency compared to this circuit, right? Because for every watt that you're consuming here, the energy that is lost that is happening is much higher compared to over here. So immediately you might be wondering what do we do about it? Well, the first step is to actually put this in numbers. When I say the power consumption efficiency is smaller, how do I sort of represent it officially? Well, to represent it, what we can do is we can look at this expression for the power consume and we can try to write this in terms of Z, okay? And I want you to again give it a shot. How can you write the power dissipated in this circuit? Not in terms of R, but in terms of Z. How do you bring Z into the picture? Well, I can help you a little bit. We can bring back the phase diagram that we have seen before. Just to quickly remind you, I naught represents the current. Then the voltage across the resistor, which is I naught times R is in phase. The voltages across the inductors and capacitors are 90 degrees out of phase. And if we are assuming that the capacitor reactance is larger, then the voltage becomes 90 degrees behind the inductor. And then if we add all the voltages together, we get the total voltage. And we can now say that total voltage, we can represent it to be I naught times Z, right? We can represent it this way. So what we can do now is we can look at this triangle and try to find the relationship between R, Z, and phi, and then substitute over here to get power in terms of Z. So feel free to pause and try, all right? So since I want to connect the adjacent side and the hypotenuse, I'm gonna use cos. Cos phi is this divided by this. I naught cancels out, and I end up with R divided by Z. And since I want to substitute for R, I can say R equals Z times cos phi. And as a result, now I can say, hey, average power, therefore, equals I RMS squared times, instead of R, I'll say Z impedance times cos phi. So this is the same expression as this. You'll get 10 watts itself, but now we are representing this in terms of the total opposition. And this number, cos phi, tells me how poor or how good my circuit is when it comes to power consumption. And so this number is called the power factor. Okay, let's dig a little deeper. For a pure resistive circuit, can you think about what the power factor would be? Well, here, impedance is equal to resistance. So Z is the same as R, R cancels, and you get power factor to be one. So here, power factor is one. Another way to think about this is we could say, hey, current and voltage are in phase with each other, so phi value is zero, no phase difference. So cos zero is one, you get the same answer. And so this means you get the maximum power consumption. Your entire impedance is consuming power because your entire impedance is just resistance. Okay, what about circuits that only have inductors or capacitors? No resistors at all. What would be the power factor there? Well, then R is zero, power factor becomes zero. And that kind of makes sense because we just said that inductors or capacitors, on an average, do not consume any power. So here, the entire impedance is not consuming any power. And so since there is a current running over here, but there is no power consumption due to that current, we call these currents watt less current, watt less current. And from an efficiency point of view, these are the worst because there will be a heating effect in the transmission lines, there'll be energy losses, but for what? You're not consuming any power. So this is the worst circuit you can think of. Now, of course, real circuits will always have all components, and so their power factor will always be somewhere between zero and one. The closer it is to one, the better it is, better efficient it is when it comes to power consumption. And you can now also appreciate why we like resonance here. We have seen before that at resonance, the impedance is pure resistive, and therefore at resonance, the power factor is one. And so if you're consuming power at resonant frequency, then you will be consuming the maximum power. Okay, I want to end this video with a question. What if we have an LCR circuit whose LCR and the frequency values are all fixed and it has a very poor power factor? What can you do to improve the power factor of that circuit? I'll give you a clue. If you have poor power factor, it means the five value is very large. It means the phase difference between the voltage and the current is very large. So the question is, how can you reduce the phase difference without changing this? But you can always attach something, something more to that circuit, maybe in series or maybe in parallel, okay? So that's the clue. Ponder upon this, think about it, and maybe discuss with your friends or teachers.