 We're going to pick up the exercise that we started last time. I decided to turn this into one of these engaged exercises because I actually want some information from you. I want to know your thought process as we're going through the light bulb game. And so I handed out to everybody a worksheet that they can use to record calculations that they do as they go through the class and then measurements that I'll take up here from the setup as we go through. So after the quiz, what I'm going to have you do is partner up with somebody near you. So we have 2, 4, 6, 8, 10, 12, or we have an odd number. Okay. So we can have a group of three. So I don't know if Arianna, Tina and Lynn, if you guys want to group up in the back or something like that, or maybe Christian is going to be here in a minute. But just to remind you what we were doing last time, we have this nice setup here with a couple of chief and easy to obtain resistors. They're just light bulbs. One of them is 100 watts. This is the one we've been calling bulb one. One of them is 40 watts. And we've been calling that one bulb two. And the manufacturer says that if you plug one of these into a 120 volt outlet, it's actually written right on the top of the bulb here. This one's easier to see. I will bring this up. There we go. All right. So that's terrible. There we go. 120 volts. Let's see if I can not shake here. Okay. There we go. You get 40 watts out. Okay. So the manufacturer is nice enough to tell us if we plug it into that voltage, that potential difference, what power we can expect to get from the bulb. And so we basically did that last time. We plugged the 40 watt bulb into 125 volts. So this thing supplies 120 and close enough. This we plug this one in by itself. And of course, the 100 watt bulb glows a whole lot brighter than the 40 watt bulb. And one of the things that we found out was that the resistance, if you calculate using Ohm's law and taking the information basically from our experiment or from the manufacturer, you find out that the resistance of the 100 watt bulb is a whole lot less than the resistance of the 40 watt bulb. And so it might already be a little bit weird as to why is it that the worst resistor that is the better conductor makes a brighter bulb. And we'll come back to that later. I'm going to bust out the microscopic model of the material in a bit and we'll play around with that. And I have a little meter on here that I can essentially use to measure the energy output of that system when we drop built, what are they called, ball bearings through it. And we can take a look at what happens if I turn that into a better conductor and then see what the power output by the system actually is. Okay. So with that in mind, we're going to pick that game up again in a moment and I will put some of the data we measured last time and calculated last time up on the board. But let me get the quiz handed out. You guys can get started on that and that will keep you entertained while I. Okay. So to review what we've managed to do so far, we have managed to skip past this here. We've managed so far to just look at this system of two light bulbs in parallel. So I have placed one side of each of the resistor at one into basically one prong of the outlet and the other side of each resistor into the other prong of the outlet. And incidentally, the sides of the resistor in a light bulb are determined by what part is plugged into this metal screw base here and what part is plugged into this little metal contact on the bottom. See here. All right. So the metal on the bottom is one side of the resistor and the metal on the side there is the other side of the resistor. So it's a bit funny looking, but the socket is meant to make sure that those two things don't touch each other. Okay. And here we go. What we did was we plugged in each bulb individually and we found the, we'll hear the voltage that was across each bulb individually was 125 and we plugged that in to the power that we were given. We're given by the manufacturer that one of them is 100 watt bulb and one of them is a 40 watt bulb and we were able to use this equation to solve for R. Okay. So this is just P equals IV with Ohm's law substituted into it. You can get a relationship between voltage, power and resistance and we solve for resistance and that's how we got these. So these are calculated based on basically our wall voltage and the manufacturer's information. One of them has a resistance, the 100 watt bulb has a resistance that's about half or so a little less than half the resistance of the 40 watt bulb. So 156 Ohms and 391 Ohms. And so what you can do on your worksheet right away is since we did this last time you can fill in first of all the measured voltage at the top here was 125 and I wrote that here almost spot on. There's some uncertainty in these numbers from the voltmeter and the ammeter we'll assume that it's sort of at the level of a few percent like four or five percent later. That's a probably a bit of an overestimate for the error on this but not too bad. Okay. Then the bulbs alone in the circuit. Well, we already have the calculated resistance of bulb one alone and the calculated resistance of both two alone R1 and R2. So that's these numbers here. You can write those down on the worksheet right away. No work involved there. Okay, then we put them in parallel and when you put them in parallel. Okay, they glow just as brightly as they would if they were each hooked up individually. So you get a nice bright 100 watt and a softer light from the 40 watt here. And one of the things we did last time was we do this again if back here. Okay, so I've got the meter here set up to measure current. This number is an amps. And so this is 1.18 amps that's going into the circuit from the wall. So I've put the, where is it here, the ammeter in the circuit itself. If I take it out, the circuit's broken. This will spark if I do that. Okay. And I've got it in the circuit so that the current must flow through the meter before it gets to any of the bulbs. So it's now in series with the bulb system over here. Okay. Do not like touching that. And so it's measuring a current of about 1.18 amps total going through this entire circuit. Now what we could do is we can compare that to what we would have calculated the current to be. So we're going to make what's called a post-diction. That is you're going to calculate the expected number after you already know the measured answer. Those don't count in science. A theory of nature is not successful if all you ever do is predict numbers you already know because you can tune your theory to get the answer you want. A successful theory is one that predicts numbers before you measure them. And so everything we do after this today is going to be the act of prediction, making calculations before we get the data from the experiment and then comparing what we expect using Ohm's law. Okay. And Ohm's law is based on the assumption that the voltage and current are related linearly to one another and that the resistance of the circuit doesn't depend on, for instance, voltage. All right. And we will test that assumption as we go forward today. But the beautiful thing about Ohm's law, the power equation, they're very simple equations. And once you master them at a basic level, you can actually start to predict all kinds of things about the world around you involving electricity. So for instance, I keep mentioning the Chevy Volt that my spouse and I have, it can be charged at either an 8 amp current draw or a 12 amp current draw. And I mentioned why they limited to 12 amp last time. If you want to calculate the change in time required to charge the batteries from zero to full, you can use P equals I squared R to calculate the ratio of the power delivered to the car batteries and based on that get an estimate of the time that it takes to charge the car compared at 12 amp versus 8 amp. And actually it's almost, it's pretty accurate considering you're basically treating an entire car like a resistor, which is a pretty bold assumption for a complicated vehicle like that. All right. So you can start making all kinds of really neat predictions about the world around you and getting right answers. And so I hope that this exercise today will illustrate both that and the other side of prediction, which is that sometimes you might encounter situations where things just don't quite work out the way you expected and now is your opportunity to learn something new. All right. So what I want you to do first is we have to get the total current predicted in this circuit. So what I want you to do is first we have to get the total resistance in the circuit. So we have the voltage source. We have the two resistors in parallel. So here's our electromotive force, our E. Here is our one. Here is our two. And we can combine these using the rules we derived last time into a single resistance, our total. Okay. So work with your partner and spend a couple of minutes and calculate our total. All right. And talk. I want you to talk to each other. And I'll wait till it sounds like people are kind of wrapping their things up and then we'll move on to the next step. Okay. And again, sit near each other so you can actually talk to one another without yelling across a gap. And if you're uncomfortable with your partner, then pick a different partner. That's all I can suggest. Okay. And if you want, yeah, here's some people talking about the current. If you want to now go and calculate the total current that's passing through the circuit, that is through the resistor network and, you know, driven by this electromotive force from the wall, go ahead and get iTotal as well. Go ahead and fill that blank in on your worksheet. Okay. Does anybody need another 10 seconds or so? Raise your hand if you urgently need time. All right. Yeah. And don't change your answers after this. I want to see the process that people go through. All right. So make sure your names of each of your partners is down on both your worksheets. Circle your name so I know who's who. But put both of your names down and then circle your name. Okay. All right. So let's kind of go through a few of the teams here and figure out what we've got. So let's do our total first. Okay. Don't change your answers. I'm not grading you on correct answers when I look at these engaged exercises. I'm looking to see if people kind of tried things. I want to see what you tried and what you did and then how you thought. I don't really care so much about your answer, I care about your process. Okay. So yes, you'll be graded on these, but, you know, if you've seen the previous engaged exercises, I created it pretty easy because I'm looking at your process and not your answers. Okay. So what, let's see, what did Team Ashley Morgan get at the back for the rTotal? One 12. One 12. Okay. And we'll write homes. All right. So can anyone confirm or refute that? Does anyone disagree with that? Catherine? Okay. We got one eight. All right. So what was your process? How did you, how did you get your answer? We did one of our rTotal equal. Great. Okay. So it sounds like it might just have been a math thing, maybe? Yeah. Something like that or 3.91. Okay. So it might just have been a math thing. So something that a bunch of you who came to Office Hours saw from me with capacitors. Capacitors in series, you have to do that same kind of algebra. You have to do one over total capacitance is one over C1 plus one over C2. But if you put the fractions in terms of common denominators and add them, and you can do the same thing for resistors, and this is in the book too, you can get a fraction like this that just directly gives you the total resistance. And it's the product divided by the sum. That's how I remember it. Product divided by sum. So for resistors in parallel and for capacitors in series, that formula will apply. For capacitors, replace the r's with c's. For resistors in series and for capacitors in parallel, you just add them straight away. Just r1 plus r2 or c1 plus c2. Okay. So this formula is very helpful and its analog for capacitors is very helpful. And again, I like the symmetry that you have between resistors and capacitors if you remember that there's a relationship that's reversed for them in series and in parallel, it will save you memory space. You can use that memory for other things. Okay. Okay. And what current did we get? So let's see, Ethan, Lindsay, what did you guys get for the current? Did you get that far? The current total running through the whole circuit. So once you've gotten the total resistance, you can use the potential across the whole thing. 1.12. Great. 1.12 amps. Anyone want to confirm or refute that? James, Rachel, Yannick, we've seen some nodding there. Okay. Yeah, that's what I got to. All right. So if you take a look at this, we've done a post-diction. After making the measurement, we predicted, well, we've post-dicted, that we should have seen about 1.12 amps going through the circuit. We measured 1.18. You know, within uncertainties from the instrument probably, that's not so bad. I mean, you're only off by something at the level of about 5%. That's not the worst thing in the world. So let's call that a success, okay? But now we want to exercise the machinery as prediction, then measurement, because we're interested in testing prediction, not trying to fudge the answer to get the right numbers after the fact. All right. So what I want you to do now is do what Ethan, I think, just suggested. I want you to calculate the individual currents in each of the bulbs. So spend a couple of minutes doing that. All right, so get the individual currents, I1, I2, and these will be calculated. And if you're using formulas, you know, jot them down next year blanks. Again, that'll help you to know your process. You know, more notes is better for this exercise. I know I didn't leave a lot of room, but just write small. I'll know by looking at it what you were up to. Okay, does anybody need more time? Like a lot more time? Okay, so let's go through some of the teams here. Okay, so Jasmine and Avery, what'd you guys get for I1? This is the current, this would be the current through the 100 watt bulb. We got 0.80 amps. 0.80 amps, excellent. Good use of units, good, excellent. All right, anyone want to confirm or refute that? Confirmed? Anyone want to refute? Yeah, okay, confirmed. All right, I2, let's, Jen and Emily, let's have it for me, guys. 0.32. 0.32. Now, let's do a test. Conservation of charge dictates that those numbers have to sum up to what? The total, do they? Yeah, 0.8 and 0.32 gives you 1.12, which is what you, this is the calculated here. Okay, so that's good, that's a cross check. If those numbers don't add up, you've made a math mistake. Okay, because all of this was based anyway on charge conservation and energy conservation. All right, so that's just a built-in little cross check if you do. Those things don't add up to 1.12, there's a problem. Could be a calculator problem, could be a rounding problem, or it could be a calculation error where you just used the formula incorrectly, all right? Great, all right, so let's compare that to measurement. So again, I have the meter set up to measure current. What I've done is I have now placed it in the circuit, but after the 40 watt bulb and before the current flows back out into the wall. All right, so now we're measuring the stream of current coming out of resistor two, all right? So this is gonna be resistor two. So this is the 40 watt bulb. And if I plug it in, we get a current of 0.33 amp. Okay, so you can write that down. Let me disconnect this dangerous toy. And now I will simply repeat that, but this time placing the current meter in the path of the circuit after the 100 watt bulb. And now, let's see what we get. 0.83, we are like gods. We can master the universe around us, okay? A little knowledge in physics can go a long way, but it can also be dangerous, all right? So let's be careful as we go forward here. So 0.8, 0.32, we got 0.83 and 0.33 for the measured values. That's fantastic, okay? And we expected that we probably would come close, given that we did get the total very close to the expected total, all right? But we can actually measure it. And then we can see what we got compared to our math, all right? The math is based on assumptions. Those assumptions in this case are energy conservation, charge conservation, and Ohm's law. Okay, so those were the assumptions that were built into this. All right, let's go one step further and I want you to tell me the voltages across these bulbs. And I want to now rewire this to do voltage measurements. So go ahead and work on that. This is a bit easier for me. One meter, I know. And again, you're trying to get the calculated voltage across bulb one, and if you flip the page over, the calculated voltage across bulb two. Again, using your calculated numbers. Don't use any of the measured numbers because that's cheating. You don't wanna mix calculations and measurements. You wanna make pure predictions here based on the voltage that we have there and the resistances that we've calculated. Okay, does anybody need more time to calculate the voltages? I'll take that as a resounding no. All right, so Arian and Laura, what did you guys do to get the voltage across bulb one? And then what voltage did you get? So how did you calculate the voltage across bulb one? Okay, so you used what for I? Point eight, okay, great, and R? Great, okay, and what did you get for the voltage? 124.8 for the voltage across it? Great, okay, and let's see. Who haven't I picked on yet? I've picked on everybody. Okay, we'll just keep going down the line then. Catherine, Alexa, what'd you guys get? 124.8, four, V2? Okay, great, yep, so they should be the same because this is parallel, okay? So let's just check that real fast and plug in. All right, so now what I'm gonna do is I'm just going to bridge across the part of the circuit that contains the resistors. And what do I? They should both be the same thing. Are you got different numbers? Are you got different numbers? These are in parallel, so both ends of the resistor are hooked up to the same sides of the battery, right? So this end and this end are on this side, this end and this on that side, so. Just when we did the math and the resistors. But they're not exact. Okay, yeah, that's fine, it could just be a rounding thing. Yeah. What did you think of the calculator? What's that? The V2 calculator. Yeah, you're using the calculated numbers to do all this. All right, so 124.3, 124.8, good enough for government work, and that was the 100 watt bulb, so yeah, that was one. So, 23 volts and 124.3 for this one as well. Great, okay, doing great so far. Very good, very good. Okay, now let's go to the situation that perplexed us all the first time, all right? But before we do that, here's what I want you to do. Use a little comment box while I get this thing set up in series. So what I want you guys to do, and you can do this as a team, chat about it and then write your answers down and then just note maybe who made, if there's a specific contribution that came from one person in the team, note their name next to it on both of your sheets, all right? But in the box, just jot down a few sentences and I want you to write your assessment of the predictions. How do they compare to the measurements that were actually made of the current and voltages? Assume that there's some small uncertainties on the measured values at the level of it, sort of a few percent, so five percent seems like a reasonable thing here. And I want you to assess the agreement under those conditions, yeah, actually. Well, can I actually, okay, so though we... Okay, so let's move on to series. So the first exercise is not gonna come as a huge surprise. What you wanna do first is you wanna calculate the total resistance of this system, because what we wanna do is find the current moving through this circuit. So from conservation of charge, if the current is moving this way, for instance, driven by the electromotive force, okay? So there's the I in the circuit. Then whatever goes into R1 comes out of R1, goes into R2, comes out of R2 and goes back to the electromotive force, the source of the electric potential difference. So the current going in has to be the current going through here, has to be the current that comes out and goes into R2, has to be the current going through R2, has to be the current that comes back this way because charge isn't leaking anywhere. There are no branches in the circuit. There's no place for current to split off. So already we can use the fact that in series, from charge conservation, we expect the current to be the same everywhere in this circuit, all right? So go ahead and calculate R total. You can work with your partner to do that. And then I want you to make a prediction of what I total is going to be for this system. And again, just to remind you what this looks like, this is what they look like when you hook them up in series. The 100 watt bulb is sort of barely glowing if one stares at the filament, it's faint orange. The 40 watt bulb is a nice soft color now, but it's not as bright as it was before, but it's certainly brighter than the 100 watt bulb. Okay, so let's see here, James and Rachel, what did you guys get for R total and how did you get it? You got 537 nodes. 547 and how did you get that? Just adding, yeah, they're in series. This is the easy case for resistors. You just add them together. Anyone want to dispute that number at all? An argument is part of science. Don't be afraid to do it, okay? In fact, that's what makes it better. Okay, let's look at some numbers for the total current now. Jenna and Emily, what did you guys get? 0.23. 0.23, 0.23, okay? Anyone want to confirm or refute that? Laura, Ariana, did you guys get that number? Are you just agreeing with me because I asked her? No, okay, all right. You got it, okay, good. That's what I wanted to hear, defense. All right, so let's measure it. So what I've done is I have again inserted the meter, now in current measuring mode, into the circuit. So in order for current to flow through the circuit, it has to go in the red wire from the wall, flow through the meter and then out and then it goes into this bulb, goes through this bulb, goes through the next bulb and comes back. So, okay, so that's certainly the case. That's barely glowing orange. There it is, it just warmed up a little bit and there's the current that I get, all right? 0.3 oh amps, interesting. Let's press this forward, okay? So what I want you to do now is I want you to move on to the next blanks, all right? So we got predicted 0.23, got 0.3. I want you to now calculate the current going through just bulb one and I want you to calculate the current just going through bulb two and similarly the voltages. So let's just do that in one big block. Give me I1, I2, V1 and V2, calculate it. What do you predict those to be? Ever assume the premise. That's anti-scientific. Make the measurement and then assess the results. Okay, does anybody need more time to do calculations? I hear non-physics related chatter. Okay, cool. So let's go ahead and get these banged out here. All right, so the currents. I'll give this, I'll throw this softball to Linitin. The currents. What are the currents predicted to be as these two resistors? Yeah, and so okay, so you use P equals I squared R. Right, but the P you use, what P did you use? Yeah, now is that thing giving up 100 watts of power anymore? No. The lot last is the filament's barely glowing. So what you did was you took the manufacturer's power output for the bulb when it's across a voltage of about 120 volts and shoved that into an equation and got a wrong answer as a result. The situation has very clearly changed, right? That thing is no longer the 100 watt bulb, no longer putting out 100 watts of power. It's barely glowing at all. So you should have just gone with your first instinct which is that they're all just the same because they're in series, the prediction is they all have to be the same. So they should be the same as the total. The total was predicted to be 0.23 amps, okay? Now let's actually check that, since I have this thing still set up to measure current, let me redo this. All right, so I'll just do one of these as an exercise. Okay? I'll use 0.23, yeah, I get that. Okay, so I'm gonna plug this back in and let's see what we get. So the current that comes out of the 40 watt bulb which would be measured I2 is about 0.3. And that's what we measure the total current going into the 40 watt bulb to be. So hooray, charge is conserved, but not quite in the way perhaps that we had predicted based on whatever assumptions we made to get 0.23. Okay, and if I repeat this, if I did this again for the second bulb, why don't I do that just to demonstrate this? I mean, why not? This is science, I might as well act like it. So let's make the measurement. Don't assume the conclusion. Let the data speak for itself, okay? So here's the current coming out of the 100 watt bulb and it's also 0.3, okay, and there we go. So indeed, the current coming out of the 40 watt bulb and the 100 watt bulb is the same as the current going into it. Good, that's a good place to be. Now let's look at voltages, all right? And while I'm getting this set up, let's see. Jasmine and Avery, what did you guys get for the voltage across the 100 watt bulb? So 35.7, so let's call that 36 volts, rounded. Anyone wanna agree or disagree with that? Let me see if even I agree or disagree. Haven't looked at these numbers in a bit. No, no one wants to disagree? Okay, so everyone agrees 36 is what they got as well, right? That's what I got, too. All right, so let's check it. Let us check. So this is going back into voltage measuring mode. There we go, let's see if I got this all hooked up again. So current comes in, goes through, goes out, goes through, comes back, and we're good. All right, so the voltage across the, back red, red voltage across the 100 watt bulb is a mystery. There we go. All right, what about V2? So let's see. Patrick and Alexa, what did you guys get? 89.7, okay? So yeah, all right, 89.7 or 90-ish. Anyone wanna agree, disagree, including me? What did I get? No one wants to disagree? Okay, all right. Oh, you're all geniuses. Excellent, all right, excellent. So let's see what we measure. This is an easy change. Wait for it. 104, 103, something. Fine, you wanna be difficult? 105, which is average at 105. So how's that looking? Are we doing, are we batting 1,000 here on estimating? No, okay, so what I want you to do, here's the bonus. I want you to calculate, and this shouldn't, I hope take too long, if I can find a shock. I want you to calculate now what, based on your calculations of the current, okay, and the voltage, your calculations, not the measured numbers. What powers do you think are now being dissipated by the bulbs? So by the 100 watt bulb and by the 40 watt bulb, okay? And so this now is at a voltage, well this is the, these are the powers that you now calculate using your expected values for I and V, okay, what are the powers? You can jot this down on your, you can put this in the box that's provided, okay, on the last side. You can put it below the box, wherever you'd like to put that, and you can leave the box for your comments. All right, so calculate the powers, and we'll go 30 more seconds. Okay, so what do we get for P1? Morgan and Ashley, what'd you guys get for P1? 8.3 watts, okay? And let's see, Ethan and Lizzie, what'd you guys get for P2? 20.6, okay. Now, great, does anyone wanna agree or disagree with those numbers? Okay, great. Now what I want you to do is spend the next couple of minutes talking to your partner and filling in the text box. I want you again to think about, I want you to assess, and there's two avenues of assessment that I want you to sort of frame this in as you think about this. One, just comparing the numbers. How do they compare? Did we do a good job of making predictions compared to what the data actually tells us? And then the other way of looking at these numbers is to ask, well, whether the numbers were accurate or not compared one by one, is the pattern at least reproduced? For instance, if we predicted that V1 was supposed to be lower than V2, did we find V1 to be lower than V2? Even if the numbers themselves are not right. If we expected by looking at this that the emitted power from bulb one is supposed to be less than the emitted power from bulb two, did we predict that at all? Okay, that is a bit of a post-diction, all right, because we already can look at it and know, well, that's a lot less than that because the light from that is a lot less than that. So the 100 watts giving off a whole lot less power than the 40 watt, and they're both giving off less than they were when they were in parallel. Did we get that right? Okay, so go ahead and talk to your partner if you need to, but jot your assessments down, put your thoughts down, and assess these results, and then we'll talk a little bit about what's going on here before the next class comes in. Okay, so go ahead and hand in the quizzes, I'll start with the exercise sheets, and then let's talk a little bit about what's going on. There's some very interesting quizzes here that we have to think through for a bit to understand a little bit. All right, so the last part of this sort of, I'll give you guys another few seconds to finish writing. Any more here? Yeah, great, thank you. Anybody else? Thank you. Okay, oh, sorry for walking around the ample of all memories. Thank you. Okay, so there's a couple of mysteries here that we want to discuss. The, one of these mysteries has to do with, that's gonna work fine. This business of the results here in series. So we ran into a situation where everything was going great in parallel. We basically had the bulbs exactly at the potential differences that the manufacturer recommended for getting their rated output, 100 watts and 40 watts. Okay, and if you look at the parallel system, they're both at the same voltage, they're both at the same electric potential difference, everything looks hunky-dory. You do the calculations, you get exactly the currents that you expect in each of the bulbs, everything's fine. But, when we put them in series, we altered the voltages that were now across each of the bulbs. And it seems that something changed as a result. When we changed the voltages over which the bulbs are placed, okay, something was different. And so, something fundamental has changed and that then makes us unable to reliably predict any longer, for instance, the total current through the bulbs in series, their respective voltage sharing. We're getting the patterns right. We know that the 100 watt bulb was glowing less than the 40 watt bulb. And the physics behind that is because the 40 watt bulb has the bigger resistance, and we'll come back to that in a moment. It gets the lion's share of the voltage in that series circuit. And so, it just has more work per unit charge to drive charge through it. And as a result of that, you get a brighter light from the 40 watt bulb, even though it's the one that you think ought to be giving you less light because the 100 watt bulb is rated for a higher power. The reason it's rated for a higher power is at the same voltage. One of them will give you more power than the other. But in series, they're not at the same voltage anymore. One of them is at about 100, and one of them is at about 20. And the 100 watt bulb is getting a lot less work per unit charge out of that circuit. And so it's just not going to glow as bright. It's going to give you less power, and that's exactly what we see from the calculations and in reality. OK, so the physics of what's going on here in terms of who's glowing brighter and who's glowing fainter is entirely driven by the way that resistances split voltage in series. And that, because the currents are the same through them, we measured that, and that was true. We calculated it. It was true. The voltages were expected to be different, and we saw that they were. The 40 watt bulb gets the higher voltage, and the 100 watt bulb gets less voltage, and that's entirely driven by the fact that the 40 watt bulb appears to have a higher resistance than the 100 watt bulb. And so it gets the lion's share of the voltage, and so it gets more power output as a result. So that's one thing that's going on. But what the heck is changing that we're getting these different numbers? And what's going on here is that an assumption turns out to not hold for light bulbs, and that is Ohm's law. When you change the voltage across a light bulb filament, it alters the resistive property of the material. OK, so some people asked about that earlier. Light bulb filaments are an excellent example of a non-Ohmic material. So everything's hunky-dory when you're operating at exactly the rated voltage that the manufacturer tells you to operate at to get the desired power. But once you change the voltage, you can no longer assume that the resistance is the same anymore. That resistance only holds at about 120 volts. So the numbers we calculated here only hold at about 120 volts. Actually, oh yeah, absolutely. Can you, Avery? Or Avery, sorry. Whoa, yeah. Can you repeat what you said about the non-Ohmic response? That was a little confusing. So an Ohmic material is one where the resistance is not a function of voltage. And as you increase voltage, resistance remains a constant. And that's why you get to use V equals IR. But in this material, the filaments probably tungsten that this is made from. It's a finely-wrapped coil of tungsten. That material alters its resistance with voltage. More voltage, more resistance. Less voltage, less resistance. And in the end, this is driven entirely by the fact that as you drive more work per unit charge through the bulb, it heats up. You can even see that because the filament glows. And as materials heat, their resistances can often change. So heat makes the atoms. Basically, if you think about this is a cold crystal of copper. If you were to put copper at absolute zero, where all atomic motion stops, this is what it might look like. But at room temperature, this thing is jittering all over the place. Because the atoms are jostling each other because they're being jostled by air molecules that are all around the material. If you heat up this material, the atoms will jostle even more. And the collisions will become more fierce. And as you get more collisions, you slow current down even more. And so that alters the resistance of the material. So if you want to change the resistance of a material, the first rule of thumb is heat it up. That's a great way to change the resistance of material. If you want to make it into a better conductor, cool it down. And actually, this is how we get to superconducting states. Superconducting states occur at very low temperatures, approaching absolute zero, where, essentially, if you want to think about it in this picture, all the atomic motion stops, and the electrons can pass through the system without basically scattering off of anything. There is a special quantum mechanics reason why you get zero resistance. But you can imagine that as you lower the temperature, and this whole thing quiets down, the collisions become fewer and fewer and fewer, and resistance should drop. Superconductivity was a bit of a surprise when it was discovered. And it took a very careful work with quantum mechanics to understand it. So if we go and look at Wikipedia, and believe it or not, look at the article on incandescent light bulbs. I freaking love the web, OK? It says, the actual resistance of the filament is temperature dependent. The cold resistance of tungsten filament lamps, those things, is about 1 15th the hot filament resistance when the lamp is operating. For example, a 100 watt, 120 volt lamp has a resistance of about 144 ohms when it's lit. But the cold resistance is much lower. It's more like 9 and 1 half ohms. And actually, before the class, I took the voltmeter, which is now buried in my pile of garbage down there. I set it into resistance measuring mode. And I measured the cold resistances of the two bulbs. And what I got when I did that was 11 ohms for the 100 watt bulb, so about 9 and 1 half, close enough, and 30 ohms for the 40 watt bulb. So indeed, even at cold operating temperatures where there's no voltage on the bulb, the 100 watt bulb has the lower resistance. So that pattern is also repeated even if you remove the voltage from the system. The 40 watt bulb has a higher resistance than the 100 watt bulb, even cold. It's just that it changes with temperature, which is changing with applied voltage. And that's basically what's going on. Now, the last mystery here is why the heck, and by the way, I should say that we had this adjunct professor who taught computational physics in the department. He came to us from Texas Instruments. He has a home shop. And he actually did home experiments measuring the voltage, the current through bulbs, and their resistance at different voltages. And what he found was that the resistance, as you increase the voltage, you drive the current up, but the relationship between current and voltage keeps changing because the resistance keeps changing. And you find that when you go from basically 0.6 volts, it has a resistance of 454 ohms for this particular light bulb he was looking at. When you double the voltage, you increase the resistance by about 25%. When you double the voltage again, you increase the resistance by 32%. In an ohmic material, this should be 0, 0, 0, 0. That's an ohmic material where there is no change in resistance with applied voltage. So filaments and light bulbs are explicitly non-ohmic. And we've learned something here about bulbs by playing around with Ohm's law and getting some wrong answers. You could do a whole thesis on non-ohmic materials. You see, there's a whole bunch of science you can do if you're really bored and have a home shop. We all do this in science. It's cool. Don't worry about it. I know the biologists are messing around with deadly viruses at home, so I don't even want to hear about that. This is harmless by comparison. So there you go. A little failure is a doorway into a whole lot of understanding about the natural world. So could you theoretically produce a light bulb with ohmic material, and then it would require almost no power and be bright? Well, no, it's not that it wouldn't require any power. It's that as you alter the voltage, its resistance will remain constant. So maybe what you're asking is, what if you made a light bulb filament out of superconducting material, then it would offer almost no resistance. And if the resistance goes exactly to zero, with these rules, you'd expect v squared over r, power would go to infinity as resistance goes to zero. In fact, superconducting materials are used to make the very high field strength magnets for magnetic resonance imaging machines. And there's a reason for that. In order to get the big magnetic fields, which we're about to get into in a couple of weeks, that you need in order to get all the spins in the protons in your body to line up in one direction along a magnetic field. That's what an MRI does to you. It takes every nucleus in your body, and it lines them up along a magnetic field, like little bar magnets. And then you blast the body with radio waves, and that causes the magnets to flip, and then they flip back in the field. And the frequency with which they do this tells you the material that you are imaging at that time. So you can tell the difference between hydrogen, oxygen, carbon, and other atoms in the body and build a huge image of where materials are distributed in the body, identify foreign materials, and things like that. Those high field swings are only possible because of superconductivity. It would be extremely dangerous to operate an MRI machine with normal conducting material, because first of all, if you're giving off so much heat, you'd probably cook the patient. And even if you didn't, if that thing were to fail in any way, the entire hospital would catch on fire. Because if so much current being driven through normal wire to make a field that big, no one would want to go near that thing. It would be dangerous, at best, deadly, at its worst. So OK. So since I have just about five or six minutes left, let me show you the reason why the 40 watt bulb, which has more resistance, is not so great at delivering power. And then just a comment. So before people leave for the cancer bio exam, so there will be homework assigned today. It's already up on Wiley. But I'd like to make it do the Friday after break by, like, I don't know, 11 PM or something like that. Is that OK? Because we have the exam the week after fall break. So not the Thursday after fall break, but a week after that. And I'd like to get that material back in and grade it and back to you so that you can use it to study for the exam. So Friday at 11 PM after the fall break, and I'll add some special office hours in on Thursday to compensate for that. So yeah, OK? Are you turning our work to your office? Yeah, that's fine. All right, so only the people who need to get to the cancer bio exam and get a seat leave, I'm going to show you the microscopic model one more time. And so all the people that are leaving will have no idea now why a 40 watt bulb gives off less light even though it has more resistance than 100 watt bulb. Well, at the same voltage, if you put them both at 125 volts. Oh, you're saying why they're OK? Yeah, so now let's put them back the way the manufacturer intended them, so that each bulb is at 120 volts. So you're setting them back in parallel? Yeah, exactly. Put them back in parallel where they're both operating at exactly the same voltage, so the same work per unit charge is available to drive current through the bulb. The 100 watt bulb at the same voltage as the 40 watt bulb will give you more light. And the question is why? And you can figure that out from the atomic model of the material. So the model that we have here is that, again, gravity is our electric field. It's the potential difference that's been set up by some electromotive force, battery or wall socket or something like that. The pins are the atoms in the material. And the bulb bearings, which I apparently can't. This is why I don't play sports. All right, there we go. The bulb bearings act like the little electrons that are being accelerated by the electric field. They suffer collisions as they go through the pin board. And as a result of that, they slow down. They don't go as fast as they would if I were to drop it from the same height but with no resistance. Now, for normal materials at room temperature, there are better conductors and worse conductors. So for the purposes of this analogy, this is my worst conductor. And air, where the collision happens much later with the table, that's going to be my better conductor. So it takes a lot longer for a collision to occur when I drop it outside of the pin board than it does inside the pin board. The collisions happen about every centimeter or so. And here, out here, it's more like a foot and a half before a collision occurs. So with that in mind, what I'm going to do is I'm going to drop the same amount of charge through this pin board and then drop that charge also through the air. And the way we're going to measure the power output by the collisions is we're going to use the sound meter. Now, we all have to be real quiet, no shuffling of bags to bias the results. I'm already annoyed that the air handler's on. And I will stop talking in my stentorian voice. What I want you to see is you see right here these numbers. So right now, I'm creating a noise level, a power output in sound waves of about 80 decibels, all right? You go louder than that. You can shatter eardrums and things like that. So this is essentially a measure of the power output by my voice right now. What we're going to do is we're going to measure the power output from the sound waves of these charges, these ball bearings, striking the pin board. So let me see if I can get this close. One of my reservoirs over here. All right. All quiet, please. Let's do this. OK, here we go. So what was the sound power that we roughly saw from that? Yeah, about 75, 76 decibels. So now let's simulate having a more conductive material with fewer collisions per unit distance so that when the collisions occur, the kinetic energy of the collision is far higher because the charges have been accelerating longer in the potential. So let's see. Closer to 80, 79, 80 was about the highest I saw that get. We dropped a whole lot more, which would make a tremendous mess. And I've already made a tremendous mess. I could get this number to really be radically different. But what's going on in the 100 watt bulb is that the time between collisions is a lot longer because it's a better conductor. And so the electrons are having more running room to accelerate in the electric field created by the wall potential, the wall socket, than the electrons in the 40 watt bulb, which are suffering far more collisions per unit distance. So while it's true that the 100 watt bulb is giving you more light, the reason it's giving you more light is that the kinetic energy of the collisions of the electrons with the atoms in the filament in the 100 watt bulb are much higher because the electrons have far more running room and time to accelerate before they collide. All right, so it's the difference between bumper cars, where if you're in a bumper car arena and people are kind of running into each other, they're running into each other like every few seconds. So the cars never really have a chance to get up to really high speeds. But take the same cars with the same accelerating power, put them at opposite ends of a long rink, and then have them accelerate for a long time until they collide head-on, and you'll break necks because the kinetic energy is much higher when the collision occurs. And you get a lot more energy out as a result. So in that case, you damage the bumper cars, you break bones, you snap necks, it's bad stuff. This is why bumper cars are so densely packed in a bumper car arena is to prevent people from accelerating too high in velocity. And really, I mean, you know how kids are, anyone ever done bumper cars before? Yeah, I mean, you want to tear the heads off those other kids, but there's a reason they pack them so densely so that can't really happen. So you don't get a lot of V, and thus not a whole lot of one half MV squared as a result because the masses are all about the same. Kids all have about the same mass in those things compared to the mass of the bumper cars. So the reason a hundred level gives you more light even though it's a better conductor is because the collisions that do occur inevitably are much higher energy. They release a tremendous amount more light from the atom as a result of really exciting the electrons in the atom and then having them fall down a long distance and emit photons. And so you get a much more intense light out of this as a result, okay? So that's it for the stuff before break. So we didn't quite get to Kirkhoff's laws. What I'm gonna do is do a little video and show you some Kirkhoff's laws tricks that you can use to start the homework over fall break because I know that is what everybody is going to be doing over fall break. I'm the only one who's gonna be doing that, isn't it? Damn it, all right, so have a good break everybody.