Hi, my name is Augustinus Christiadi, and I'm going to present about our paper titled Learnable Insurmountability under Laplace Approximations. This is a joint work between me, Mattias Sain and Filip Henig, and we're from the University of Tubingen, Germany. So we know that Bayesian neural networks promise calibrated predictive uncertainty. And Laplace approximations provide a cheap way to construct Bayesian neural networks. But perhaps due to their approximate natures, in practice, their uncertainty is not necessarily calibrated. And in this talk, we want to answer whether we can improve the uncertainty of Laplace approximations in Bosch hoc and thus cheap manner. So the standard way we train deep neural network is by finding just a single point estimate in the parameter space. And this point estimate will induce predictions while they might be accurate, they are overconfident because they don't have notion of uncertainty. Laplace approximation can be used to mitigate this issue. Laplace approximations basically just surrounding the map estimate with a Gaussian that depends on the lost landscape geometry at that point. And by marginalization, we can arrive at a prediction that is as accurate as the map estimate prediction but it has accompanying error bar or uncertainty. But not this far from the data, we can still have narrow error bar, meaning that Laplace approximations can still be overconfident and thus not calibrated. And our work fixes this issue. We do so by augmenting the lost landscape such that we can find another point which under a Laplace approximation will induce the same prediction but with better uncertainty characteristic. And to do so, we propose a method called Lula. And the premise of Lula is very, very simple. So suppose we have a pre-trained network with its map estimated parameters, then we simply add additional hidden units at each layer of this network. And because we have changed architecture of the network, then we have to adjust the parameters. And we do so as follows. We form block matrices and block vectors where its block element is either the map estimated values of the weights the zero constant weights or free parameters. And all those free parameters are Lula parameters. That is the parameters associated with the additional hidden units that we can set freely. This construction might be very simple but it has interesting properties. The first properties is that Lula construction will not change the output of the network and that's the prediction. Secondly, under a particular patient treatment over the network, we can show that the output variance will change. And lastly, the gradient of the loss with respect to the additional weight in the last layer of the augmented network is nonlinear in Lula's parameters. The first and the second properties imply that Lula units are indeed unsung to units because they only affect the variance and not the prediction itself. And the construction of Lula, along with the third property here, imply that Lula units are learnable because we have free parameters and non-triphily affect the loss landscape. And remember that Laplace approximations exploit the loss landscape geometry. And therefore, Lula is particularly suitable for manipulating Laplace approximations. And we can train Lula as follows. First, we form a Laplace approximation over the augmented network. Then, of course, this approximate posterior will induce a predictive distribution. And we want to find a value of Lula parameters such that they induce good predictive distribution in the following sense. We use uncertainty over the loss, which basically simply says that we want to be more uncertain on odd layers while remain confident on the true data. And all of this can be done post hoc, meaning that this amounts to a post-processing step after the map training. And thus, this procedure is cheap. Now let's see some results. The first one is toy classification data set. Here, we can see that a Laplace approximation can be used to give notion of uncertainty in the input space. But notice that far from the data, we can still have high confident predictions and Lula fixes this. In more challenging scenarios, in data sets shift robustness tasks with rotated Amnist data set and corrupted Cypher 10 data sets, we can see that Lula in green improves the base Laplace approximation in yellow in all metrics considered. The same observation can also be seen in odd of distribution data detection. Here, we use two different kinds of Laplace approximations to show that Lula consistently improves Laplace approximations of various kinds. And indeed, Lula makes both Laplace approximations better in all tasks considered in general. And so to conclude, we have introduced Lula which is a very simple construction where we add hidden units to a pre-trail network and we show that this unit can be used as answer the unit that does not affect the prediction. But they can be trained in a positive manner such that they induce good predictive uncertainty under a Laplace approximation. And empirically, we have seen that Lula is effective in a various kinds of quantification tasks. Thank you very much for your attention.


All right, let's go ahead and get rolling today. We're also subdued. So let me grunt a little you further. So here's some class announcements. You've got some reading to do for Thursday. You've got a video to watch. You've got a homework due Thursday. Grand Challenge Problems. I will be getting an example solution right up that I'll circulate to the teams. Actually, I'll probably start doing this at the team meetings. And speaking of team meetings, the first ones begin this week. In fact, they begin today. This week is a little truncated for me, however, because I will be away at the CERN laboratory from Thursday afternoon until fall break. So I have nine days of research that I can get done. It's about all I'm going to get this semester. So I'm going to take advantage of it. So classes next week will be led by the Honors Physics Teaching Assistant, Eric Godot. So those of you who take Honors Physics know who he is. There will, of course, still be reading quizzes. Everything is the same. All right, so he's getting all the material that I would have done in class with you guys. And the key thing is I'll work with NUSH to add extra office hours. Eric has volunteered to do office hours, if needed, as well. So between the two of them, you'll get plenty of support. And of course, I'm still available by this amazing thing called email. And if you have any really urgent issues, we can chat via Skype or whatever you guys use these days to send messages in real time to one another, okay? All right, so any questions? I'll try to bring you guys back a present from CERN. So let's have a quiz. What's that? Where is that? CERN? Yeah. Switch or what? You're going to switch? Oh, this is like Big Bang Theory stuff. Yes. Okay, that's really cool. It's exactly like Big Bang Theory stuff, yes. Sometimes, not a huge difference like them.  Awesome. Except all the time. That's so cool. Does that sound? Okay, let's go over the quiz question. So what's true about electric current? Is it one, unlike charge, it's never conserved? Two, it's conserved, but only in resistors. Okay, three, it is positive in the direction that positive charges are flowing. Okay, and four, it's positive in the direction that negative charges are flowing. Okay, the answer is it's positive in the direction that positive charges are flowing. Again, thanks to Ben Franklin, we're sort of stuck with the definition that the positive things happen in the direction that positive charges move, even though we know from the atomic theory of matter that it's really the electrons that get ripped off their atoms and then sent along on a journey. The atoms usually stay in place. That's not always true. And there are some materials where positive charges do move, but that's a sort of special class of materials we won't worry too much about in this course. It is absolutely conserved because current is just charge moving past a point in a material per unit time. And if charge is conserved, current is conserved. And as far as we know, charge is always conserved. Okay, all right, so current density, a new concept introduced in this section of the course, is it one, current per unit length? Two, current per unit volume. Three, increases in value when a current I goes from a smaller area conductor into a larger area conductor. Okay, and four, increases in value when a current I goes from a larger area conductor into a smaller area conductor. Okay, yeah, yeah, you can think of, we're gonna dig into this a little bit more as the course goes on, and we're gonna really play around with something called Ohm's Law Next Class. But in general, you can think of electrons, current moving through a conductor, like water moving through a pipe. And so if you have a steady flow of water in a pipe, if the pipe is big or the pipe is small, the water's moving at the same rate no matter what, but the density of water in the small pipe goes way up. And so it's number four. It increases in value when the current I goes from a large area conductor into a small area conductor. So you might still have the same current I, but because current density is current per unit area, if the area decreases, the density goes up. And then finally in the video, what was true of the experiment using two light bulbs in sequence in a circuit? Was it one, the 100 watt bulb glowed brightly while the 40 watt bulb was faint? Okay, was it two, the 40 watt bulb glowed brightly while the 100 watt bulb was faint? Okay, was it three, neither of the bulbs lit despite being plugged into wall voltage? And four, both bulbs glowed with exactly the same brightness. Okay, and the answer is that the 40 watt bulb glows brightly while the 100 watt bulb doesn't. And we're gonna explore that in great detail next class where you guys are really gonna dig into an exercise. Like the whole, it's gonna be a whole class exercise and I know it's gonna be silly, it's called the light bulb game. But so it sounds kind of innocuous, but actually it really will help you to try to build up some understanding of current and how you can split current and rejoin current and how much work current does and things like that. And what it means to be a 40 watt bulb versus a 100 watt bulb. So we're really gonna dig into that next time. All right, so let's go over some basic principles at this point in the course. We've made it through exam one. We're working on the material for exam two. So the leftover stuff that lingers with us throughout the entire course is that electric charge exists, comes into kinds positive and negative. Charges have associated electric potentials or just potentials for short. That's work that they can do per unit charge on another charge. And those potentials lead to conservative fields and forces that is you can store energy in them that can be released in the form of, for instance, kinetic energy. A device that stores charge and thus energy via an electric field is a capacitor. So this is a newer idea that we've been playing around with. We'll play around with it more today. Capacitors are very useful for saving energy for later use in a circuit. And a circuit is nothing more than a, that combines an electric potential difference with other components and uses the movement of current to do work. Okay, so it's in the same sense that you can pump water and use the motion of water against like a paddle wheel to get the wheel to spin and then get it to do some work. You can do the same thing with current, okay? But the forces that current exerts has to do with the collisions of the free charges moving in a circuit with other materials like the atoms in the circuit. Okay, so you can get heat, for instance, by having charge smash into atoms, vibrate them and those atomic vibrations propagate through the material like the metal and then that can hit, the vibrations of those atoms can hit the air molecules just outside the conductor. And then of course those air molecules now have more kinetic energy. That's what we call heat. Okay, we don't study them in this course. Actually, we don't really study heat in any meaningful way in any of the introductory courses, although I'm angling to get that included in this class next semester. I think I've found enough places that I can have savings that I can get heat and thermodynamics in. But heat is just the kinetic, it's just increased kinetic energy for atoms and you feel it, that wave of heat that you feel when something is hot and you put your hand near it or you touch it, that's atoms crashing into your atoms. And your body has a limit to how much energy can absorb before it gets damaged. And that's, pain is a response to too much energy input, either electrical or thermal or pressure. Now, the neat thing about capacitors is that you don't have to go out and build exactly the capacitor that you need for a circuit, custom built. Your body does this and you can do this as an engineer. You can design your own capacitor from other capacitors. So biology does this, chemistry does this and engineers do this. There are rules. Those rules are based on conservation of energy or conservation of charge. Pick, take your pick, both of them apply in different situations. We'll explore those rules today. And finally, some new ideas here are really just combining old ideas. If you have an electric potential difference, that is a voltage. And I'm just gonna shorthand that as voltage going forward. That causes charge to accelerate. We've already explored that in sort of phase one of the course. That means by accelerating electric charge, you can establish electric currents. That is a flow of charge per unit time. Now, currents are not free to move without resistance in whatever medium they're traveling in, like a metal, like copper or aluminum or tin or something like that. That's because they collide with atoms, as I mentioned before. And a material that obeys Ohm's law has a proportional relationship between the voltage applied to the material and the amount of current that's driven in the material. That's all Ohm's law is. If I apply a certain V, this material's properties give me a certain I. And if I detect a certain I in that material, I can infer the voltage that's driving that current. That's what Ohm's law lets you do. That's really all it tells you, all right? There are non-Ohmic materials that is materials where there is not a proportional constant relationship between V and I. We're mostly gonna stick to Ohmic materials in this course. All right, so let's start today by looking at capacitors. And in particular, how one handles a complex system of capacitors. All right, so here's a scary picture. Nothing like this to start off a Tuesday morning. We have a circuit diagram, all right? And to sort of set this up a little bit, a simple circuit. In fact, the simplest circuit you can write down where there's actually current flowing through the system consists of a battery or some other source of voltage. Source of voltage. And again, voltage is electric potential difference. That's all I mean by voltage. There's a delta V in this thing, okay? So this thing has some delta V that it generates. And we're gonna explore non-ideal batteries a little bit, but for now we're gonna assume that this is an ideal battery and that means that it always delivers the same voltage no matter what conditions it's operating under. It will always give you, if it's a nine-volt battery, it will always give you exactly nine volts no matter what you plug into it, no matter what you do to it. Heat it up, heat it down, always nine bits, okay? That's an ideal battery. Of course, the real world is full of non-ideal things and that's what makes life interesting. And we'll explore some of the consequences of non-ideality in these sorts of situations. So you have a source of electric potential difference and quite literally to establish a current, all you need to do is connect one end of the battery to the other. So these lines are conductors and we're going to assume that they are perfect. That is no resistance to electric current flow. Nothing going on in those lines creates any additional source of electric potential difference. Resistance, as you'll see from using Ohm's law, resistance can induce a voltage of its own, okay? But we're assuming this is a resistance-free medium. Now, in the real world, even with a non-ideal battery, you never want to do this. You never want to take a nine-volt battery and a screwdriver and tap the connectors of the battery together with the screwdriver. Screwdriver is made of aluminum or something like that, a steel. Okay, it's a pretty good conductor. And basically what you're doing is you are, this right here is what's known as a short circuit. That's specifically what I've just drawn here is something called a short circuit. There is no resistance in the circuit anywhere. And so this thing, as you'll learn from Ohm's law, is capable of driving a very large current. In Ohm's law, it would be an infinite current, although there are no infinities in the resistance. Some resistance in this material. And that's really bad. So at the very minimum, that can cause batteries to explode. And so this is why, for instance, now when you fly on airlines, they ask you if you've stored any lithium-ion batteries in your checked baggage. Because there have been incidents where lithium-ion batteries have shorted during flights, just because something happens in the electronics and they explode in the cargo hold and that can actually start a fire. And that's really bad. It's much harder to put out a fire in the cargo hold than it would be if this happened in your backpack while you're sitting on the plane. You could do something. You could pick up the backpack, you could get it to where there's a fire extinguisher on the plane. It's a little harder to do that in the baggage hold. I mean, I'm assuming they don't have an automatic fire suppression system in there. Somebody would have to climb down and put out the fire. So not fun. And short circuits are often what result in batteries exploding. Of course, you can also have a bad battery design, but the principle is the same. You get a runaway current and boom, the thing explodes. Okay. This is also the simplest case of something known as a simple circuit. That is a circuit with one loop that current can travel in. So if you imagine current traveling in here, we have the plus side of the battery, oops, emits positive charge. That's what that plus means on the battery symbol. Positive charge comes out of the top. Okay, and the direction that positive charge is flowing is the direction of positive current flow. So we can write a little arrow in here that tells us the direction that current is traveling. Okay. Now the direction opposite that is the direction electrons are flowing. So if you think of this as the source of positive charge, that must mean this one's emitting negative charge in the opposite direction. So negative charge flowing counterclockwise is the same in terms of physics as positive charge flowing clockwise. All the changes is the sign of the flow. That's it. Okay. So this is just to kind of get you to get this picture in your mind of what a circuit is and in particular, a simple circuit. Is that a simple circuit? Why not? The definition of a simple circuit is there's one loop for current to flow. Why is that not a simple circuit? It's not necessarily. Is there three loops? Yeah, there's multiple loops. Yeah. There's the big outer loop. Okay. There's the left inner loop and there's even this right loop over here. So there's three loops in this picture. Now the neat thing is that we can use conservation of energy and conservation of charge and we can turn this from a complex circuit into a simple circuit. And that ultimately is what we're going to do. We're going to learn to do that today. And we're going to learn to do that by trying to answer a few questions. So in this picture, a complex circuit, I have a battery with a potential difference V and I tell you it's nine volts. This circuit consists of multiple capacitors. C1, C2 and C3. All right, and I gave you the capacitances here. So C1 is four and a half times C2 and that's also equal to seven and a half times C3, okay? And then I gave C1 as well, all right? So you can figure out based on those relationships what are the individual capacitances of each of these capacitors. All right, so you can sit right down all those little numbers or you could just leave them as symbols for now and hold off on plugging in numbers until you've finished up any algebra on the problem. I recommend that path, but I'll leave it up to you. You know your math skills better than I do, okay? The first question we wanna answer is, given this picture, given the fact that there's a battery hooked up to a bunch of capacitors, we know that that electric potential difference is piling up charge in different locations in the capacitor system. So for instance, if I modify the picture over here on the left and just put a single capacitor into this picture, so we'll just call this C and this potential difference is V, okay? For every capacitor in a system, there is a capacitor equation, Q equals CV. So if I know C and I know V, I know immediately what Q is stored on this capacitor. That's what the capacitance equation does for you. It's a relationship of proportionality between the voltage applied and the charge stored. Now, let's think about this from the current perspective. We have a battery, all right? And let's say we start off with this circuit not connected at all. That is, there's a break in the conductor, current cannot flow, that battery can't drive any current through the circuit. So nothing has happened. Often what you'll do is in a diagram like this, you'll put in a picture that looks like this. This is known as a switch and it's identical to this. That's a switch, okay? Yeah, sorry, sorry, sorry Lucy, all right? So a switch interrupts a circuit and of course the results are dramatic. I stopped current to the lights and I switched them off, okay? The current can no longer do work in the lights because there's no path for the current to get to the lights anymore. And that's very similar to this picture here. Switches, however, can be closed. If I close this switch, if I flick it down and I connect up that little ravine there in the circuit, if I bridge it, now current can be driven. And what happens? Well, there is no electric potential difference over here to begin with and so my battery fires one positive charge, like a little cannon out and it gets over here to the capacitor. That one little positive charge, this is all electrically neutral to begin with. How much work does it take to move that one charge? There's no net charge anywhere else in the circuit to begin with. How much work does it take for the battery to move that one charge? And think about summing up charges, bringing them in from infinity. How much work does it take to move the first charge if there are no charges present to begin with? Zero, yeah. So the battery, easy, that first charge is free. Now it wants to move at second charge. So, shoots another charge over here. There's already an electric potential due to this first charge, so work has to be done by the battery to place the second charge near this one. And of course, these charges don't want to be near each other at all. So they're probably gonna move as far apart as they possibly can get on these conductor plates in the capacitor. All right, and so forth. You keep shooting positive charge. The third one costs more energy. The fourth one costs even more. The fifth one costs a lot more. Six, seven, eight, nine, okay, up to like Avogadro's number of charges. Or however many can be stored on here before this electric field inside the capacitor opposes the battery completely and no charge moves anymore. We're gonna explore this in a little bit more detail next week. Ultimately, this is something known as a resistor capacitor circuit, and there's actually a time behavior to this. It takes time for a capacitor to charge up. But because you're piling up positive charges over here, these are emitting electric fields, okay? And any positive charges here are repelled, okay? So the positive charges all move away from this other side of the plate and you're left with negative charges. The positive charges scoot through the battery and they keep getting dumped over here on the other side of the capacitor until this electric field completely opposes the battery and a whole thing stops. And now you've stored up all the energy you're gonna store. And I kind of demonstrated that with that 12-volt power supply. I hooked up that big cylindrical capacitor to it. Very quickly, the capacitor had the same voltage as the power supply. That means no more charge is flowing. Energy's stopped being stored. We're good to go, okay? All right, so in this complex picture, positive charge is piling up on the left plate of C1, negative charge must be piling up on the right plate of C1, okay? And that must be pushing positive charges down to the top plates of C2 and C3, which causes negative charges to pile up over here and positive charges to be shoved through the battery. And that whole thing's gonna keep going until finally all the voltages and the rest of the circuit oppose the battery and everything comes to a stop, okay? This is a mess, but the good news is that we can combine capacitors from this complex circuit into a single equivalent capacitor in a simple circuit. Our goal is to get to that picture. One battery, one capacitor, that's it. But we have that mess, all right? But we can pick it apart. And we can pick it apart using the bits and pieces written over here on the left. All right, so I'll leave that there, because that's our goal. That's what we want. Capacitors that are side-by-side, where you can draw a line between one plate and another plate on the capacitor with no interruption of conductor by another device, okay? No other capacitor, no resistance, of course there's no resistance in this circuit at all from the materials, these are all perfect conductors, but you do have capacitors. If I can draw a line along a conductor unbroken to one plate of the capacitor, and similarly from the other plate of the capacitor drawn unbroken line to the other plate of the neighboring capacitor, those two are said to be in parallel with one another. So two and three are in parallel with one another. In parallel, the electric potential difference across the two capacitors is the same. So by looking at this picture, we already know something. Whatever the voltage is across capacitor two, at the end of this, when all the charges have stopped moving, it's equal to the voltage across capacitor three because they are in parallel. When you have capacitors in parallel, you can combine their capacitances into a single equivalent capacitance. So I can take, here's my battery, here's C one, I can find some combined capacitance, C two three, which means I've added two and three together somehow, that leaves me with a simpler circuit. Almost perfectly one loop, but two capacitors, C one, which I haven't dealt with yet, and this thing, C two three, which is the combination of C one, or sorry, C two and C three. Now, when I have parallel capacitors, the way you do that is you simply sum the capacitances. So C two three, because two and three are in parallel is just C two plus C three, not so bad, okay? And the reason you can do this mentally, you can think of this as, if I were to take these parallel capacitors and bring them closer and closer and closer together, okay? I'm not changing anything about the circuit. I'm not adding any resistance. I've got perfect conductor and I bring them so close together so they start off looking like this over on the right hand side, here's two, here's three, okay? And I bring them closer and closer together until finally their plates touch, okay? Because this is just conductor with no resistance here, I can just kinda swoosh this conductor down until there's just one line sticking out of both sides. So what I've done is I've taken the area of the two plates and I've added the areas together. And if you remember the parallel plate capacitor equation, capacitance is epsilon naught A over D. If I double the area, I double the capacitance. If I add the areas of A one and A two, I increase the capacitance simply by adding A one and A two in that numerator, assuming that the separation of the capacitors is the same, okay? So parallel capacitors can just be pushed together so that their plates essentially touch and get one giant capacitor out of it. And to get the total capacitance, you just add them, that's it. So if I look back here, I see that, let's see, so how do I wanna do this? C2, C3, well I can plug in numbers at this point, so why don't I just go ahead and do that? So this winds up being 2.67 times 10 to the negative five and the units of capacitance are farads. They're named after Michael Faraday, okay? So farads and that's just farads are just coulombs per volt. So just take C equals Q over V coulombs per volt, that's all a farad is, it gets its own name though, okay? So now I have this picture, all right? And I need to add these two capacitors together because I wanna get that, I want one capacitor in the circuit. Now it's not possible for me to go from each plate of these capacitors over to the other without encountering some other device. So up here, no problem, I can connect the top plate to the right plate of C1 with an unbroken line. But in order to connect the other plate to the remaining plate on C1, so the bottom plate here to the left plate here, I've gotta go through the battery, all right? So there is an unbroken path on one side, but not on the other. And when you have that situation, if you have capacitors where you can do one unbroken path on one side, but not on the other, then these are said to be in series with one another. So C1 and C23 are in series, that is they're sequential in the circuit, but there's no way I can get from the other side of C23 back to C1 without going through some other device. In this case, the battery, okay? Series capacitors store the same charge, okay? And you can kind of see that here. If you imagine that I put one positive charge over on the left plate of C1, then because this is a capacitor, I get one negative charge over on the right plate of C1. In order for that to happen, I must have pushed one positive charge down to the top plate of C23. That causes an equivalent negative charge on the other side, okay? That's the only way charge can be conserved in this circuit is for that to happen. So charge conservation forces the Qs, Q1 and Q23 to be the same. Whatever they are, they're the same. All right? And that's because they're in series. Now the voltages are not necessarily the same, and that's because if one of these capacitors is bigger than the other, it's capable of storing more electric field. And that means it can store a larger electric potential difference. Potentially, okay? So we'll keep that rule in mind. To add series capacitors, I take one over each capacitance, add those together, and I get one over the total. So one over C total is one over C1 plus one over C23. That's the rule, okay? It's a consequence of charge conservation, but that's the rule. Now, when you have two capacitors, just two, you can rewrite this equation and solve for C total. It's a bit of algebra, but it's not too bad. C total will be equal to C1 times C23 over C1 plus C23, okay? And you can figure that out by just doing the common denominator trick on the right-hand side, and then invert the whole equation when you're done, simply getting the right-hand side combined, okay? That's only when you have two. If you have three, it doesn't work this way. You can write a formula, but it's much more complicated than this, okay? But thankfully we have two capacitors, and so the total capacitance will turn out to be that. And let's see, so in that case, that is equal to 1.97 times 10 to the minus five, I'm assuming five ferrets. So 1.97 times 10 to the minus five ferrets. Okay, wow, we just walked a mile here, all right? Who cares? Why did I do all of this? I wanted to find the charge on C3. Here's why you do this, okay? Once you get to this stage, you can apply this equation for the total circuit. Q total equals C total, V total. Okay, let's stop for a moment. Do we have the total charge stored in the circuit? No, we don't know that, that was never given to us. Do we know the total capacitance of the circuit? Yeah, we just figured that out. Do we know the total voltage applied to the circuit? No nodding heads, faith you're making a face. Do we know the total voltage applied to the circuit? We know it's a nine volt battery, and that is the sole source of external voltage for everything else in the circuit. So in fact, we know V total. So V total was given, that's nine volts. C total we figured out, so we can calculate the total charge in the circuit. Let's see if I did that in my notes. So we just take C total times V total, and we get 1.77 times 10 to the negative four coulombs. Okay, awesome. Now, let's pull our capacitor network back apart. We got it down to this picture, V total, C total, Q total. Let's yank C1 and C23 back apart, and get back to this picture again. What was the same between the two capacitors, C1 and C23, because they're in series? What's the same for both of them? Q, charge. And whatever the charge is on this circuit, on this capacitor, must be the same as the charge on this capacitor, okay? And so what we find out is that Q total equals Q1 equals Q23. The total charge stored on the combination is equal to the charge is stored on either of the two. And the way you can see that is, remember what I said here, you put one charge here, you get a negative charge here. That pushes a positive charge here, which induces a negative charge here. If I bring these capacitors together so that I merge their interior plates, what's the net charge of the interior plate? So I bring all those plates really close together, smush them together into a single conductor. What's the charge in that interior plate that I've smushed together from the two capacitors? Could you hint? You got a minus here and an equal-sized plus here. So when you add those charges together, you get neutral, zero, right? So when I add these capacitors together, the charge on that interior plate is zero, leaving me with just plus one and minus one here. So I have plus one, minus one, plus one, minus one. When I smush them together, I still have plus one and minus one. It's not that the charges add up. It's that that net charge in the interior plate sums to zero and doesn't add anything to the whole capacitor. And that's how you can get away with this, okay? So we already know q1 and q23 because we got qtotal, okay? So we know a lot. Now what we're doing is applying conservation of charge and conservation of energy. The total energy available in the circuit can't exceed the energy provided by the battery and the total charge can't exceed the total charge stored on the equivalent capacitor, okay? We want the charge on C3. So what do I have to do to C23 to figure out the charge on C3?  I gotta yank C2 and C3 back apart in the picture and go back to that picture. Now, what was the same between C2 and C3? They're in parallel. So what's the same? Voltage, okay? And I know I wrote it there, but I really, you know, repetition is the essence of learning, okay? So C2 and C3 are at the same electric potential difference. V2 equals V3. So I can write down a capacitor equation for each of them. Q3 equals C3, V3, and Q2 equals C2, V2. For every capacitor equivalent or otherwise in a circuit, there is a capacitor equation. That's the helpful tip in all of this, okay? So that's true, that's true, that's true. There's one for Q1, I didn't write down because we don't need it right now, but there's a Q1 equals C1 times V1 as well, okay? Now, whatever V3 and V2 are, they're the same, okay? So at this point, I know, well, let's see, how do I wanna do this? And there are many ways you can tackle this at this point. Let's see. Well, I can go back, actually, let's go ahead and do that. Say Q1 equals C1, V1, okay? So I can figure out the voltage across V1, and I can figure out the voltage across, our voltage across C1, and I can figure out the voltage across C2,3, okay? So the, so going back to, well, there's another one here, Q2,3 equals C2,3, V2,3, okay? So I know, C2,3, Q2,3 I've got because it's just Q total. So I can figure out V2,3. So V2,3 is, where are you, 6.64 volts, okay? And when I split C2,3 into two components, capacitors, the voltage across each of them, V2 equals V3 equals V2,3. I don't add any new electric potential differences by splitting these back up again. They're at the same potential difference when they're together, they're at the same potential difference when they're apart. So by energy conservation, V2 equals V3 equals V2,3. So I now have V2,3, which means I know V3, I know C3, I can figure out Q3. And Q3 is, yoink, 6.6 times 10 to the minus five coulombs. All right, and that is what I wanted at the beginning. Let me pause here, okay? So what did I do procedureally? For every capacitor in a circuit, there's a capacitor equation. For a complex circuit like this, the first thing you want to do is combine all the capacitors together to get a simple circuit out of the complex one. So you want to go from that picture to this picture. Once you've done this, you can unpack the capacitor network and figure out all the little things about each of the individual capacitors. What's the voltage across them? What's the charge stored on them? And the way you do that is by remembering that capacitors in parallel have the same voltage. Capacitors in series have the same charge, okay? So by working backward, once you've simplified the circuit, you can complexify it again. And as you go, you can figure out numbers you didn't know to begin with. Okay? And so at the end of it, we get to Q3 is 6.6 times 10 to the minus five coulombs. And then the second question I wanted here, we actually get the answer for this one for free by having gone through this whole thing. What's the voltage on C2? Well, the voltage on C2 is V2, which is equal to V3, which is equal to V23. It gets 6.64 volts, okay? So this stuff just sort of cascades out of the problem. Once you start unpacking the network, you'll very rapidly start to answer questions like what's the voltage on this capacitor? What's the charge on that capacitor? What's the voltage on the equivalent capacitance? Things like that, okay? So with those things in mind, you do a problem, okay? Well, I give you a biology-ish problem. Biology, like, again, going back to my picture of, you know, when I, a physicist, I look at a bird, I see lots of bird, I see bird, bear, bird, bear, bird, bear, bear, bear, there. So when I see a cell wall, as you guys explored the last time we had problem solving in class, and you calculated the thickness of the cell membrane, given some information and assuming it's a capacitor, which in fact is, it is. Now we're gonna kind of think again about the capacitance of the cell membrane, all right?  So that's kind of what we're gonna do here to just this part of a cell membrane, all right? So you've got this cell membrane, you've got a separation of charge, you've got sodium ions on the left, anions on the right. There should be, well, there aren't, but there should be roughly equal numbers of anions over here as there are for sodium ions over here. I didn't crop this picture very nicely, but there should be plus Q and there should be minus Q over here, all right? So you can model the cell membrane as a large number of very small parallel plate capacitors, that's a hint, parallel plate capacitors, all in parallel, all actually in parallel with one another, all right? So you could slice here at the cell membrane and that would be one parallel plate capacitor and then you could slice here. That's a second, that's adjacent to it and a third and a fourth and so forth, okay? So consider breaking this cartoon into five identical capacitors all in parallel with one another. If the thickness of this membrane is 10 nanometers and it's dielectric constant kappa is three and the area of the wall shown is one times 10 to the minus 14 meters squared. So the surface area of either side of this thing is that number, one times 10 to the minus 14 meters squared. What's the capacitance of each of those five capacitors? What's their total capacitance? And if the voltage across the entire cell membrane is 70 millivolts, how much charge is stored on each of those five capacitors? So go ahead and partner up like you usually do and start working through this, okay? So the first thing you probably wanna do is figure out how you're gonna use the parallel plate capacitor equation, which let me rewrite it so you can all see it. Can you all see over here? Okay, this is okay. Can anyone not see this corner of the board? This thing's kind of in the way. So for a parallel plate capacitor, okay? You have kappa epsilon naught A over D. Kappa's the dielectric constant. Epsilon naught is a constant of nature. It's how much the empty space of the universe allows electric fields to permeate. Area of the capacitor plate on either side, separation of the two plates, D, okay? So there's your parallel plate capacitance equation. First thing you probably wanna do is start with the total capacitance of this system if you can figure it out, okay? And then start thinking about what it means to dice that into five equivalent parallel capacitors, okay? So start talking, do whatever it is you guys do on a Tuesday morning at 10.15. We'll do this for about 10 or 15 minutes, and then we're gonna go on to a current problem, okay? So make haste. You can measure quantum effects. You'll measure the quantization of energy, so. Okay, so let's go ahead and move on to solving problems involving current and resistance since you've had reading and a lecture video on this. So this will be kind of a little easy breezy for the most part, but I wanna relate stuff that is around you all the time to these concepts.  Wall potential, copper wiring in the walls of your home. You've got materials that make up the appliances in your house. This is a heating element from an oven, okay? It's just a metal that has resistance, and the metal is chosen in such a way that you get a particular kind of resistance, and so you can get lots of heat off the element for cooking, you know, food, cake, spread, lasagna, things like that, okay? Whatever, whatever your heart desires. So let's think about home electricity for a second. So home electrical wiring has to be done very carefully, and that's because some of your appliances are quite current-hungry. That is, even for the same electric potential difference, 110 volts, let's say, coming out of the wall, they're designed to drive a lot of current through the device, in some cases, to generate a lot of heat. So your clothes dryer, okay? If you've ever used one of those, and I hope to God, as college students, you've at least seen one, okay? Those get really hot, okay? And the way they get really hot is they use what's called a heating element. It's just a conductor with some resistance, and because it has resistance, all the collisions of charge cause heat to be emanated from the conductor when current is driven through it, okay? Now the cost here is that there is not an infinite amount of current you can drive through a conductor of a certain size. There's a limit. If you dump too much heat or too strong an electric field into a conductor, you can chemically alter it. You can rip all the electrons off the conductor, their chemical bonds will break, and the material will change composition. It can crack, it can fatigue, it can melt, okay? That's bad. That can even vaporize. You can take a conductor and turn it into a cloud of ions with no resistance, which can ignite clothes on fire, okay? That's really bad. And in fact, in the grand challenge problem, that is possibly a consequence that you're exploring in the grand challenge problem, all right? So current hungry appliances need thick wires to carry a lot of current, all right? And this is known as gauge. So if you have a bigger area, the gauge of the wire is said to be smaller. So it's a little weird, but that's the way this is defined. Don't ask me who makes these standards up. I think they're insane, okay? So a bigger area equals a smaller gauge of wire, and here's a table at the left. So there's gauge four, gauge six, gauge eight, 10, 12, 14, et cetera. This is known as the American wire gauge, or AWG, okay? So it's a standard that electrical codes use in the United States. And it tells you here, for each of these gauges, what's the corresponding diameter of the wire in inches? So gauge four has a two tenths of an inch diameter, okay? Gauge 14 has a six one hundredths of an inch, sorry, yeah, six one hundredths of an inch diameter, okay? So it's a much smaller diameter wire than gauge four. So as gauge goes up, diameter goes down, okay? Now, what you'll notice in the third column is the maximum Coulombs per second, or amps. Coulombs per second equals amperes, or just amps, A, okay? Named after Ampere, who was instrumental in the history of electricity and magnetism, okay? Amps, there's a maximum that can be driven through these wires given their size. So for instance, gauge four wire can drive a maximum of 135 amps. Gauge 14 wire can drive a maximum of 32 amps, all right? And then over here, for power transmission, you have the maximum over here. So anyway, this is just a lot of information, but basically the thinner the wire, the higher the gauge, the lower the current it can maximally carry before catastrophic failure, gets it, okay? Most homes have 14 American wire gauge or 14 gauge wire in the walls, okay? 14 is the standard, and it's placed at 110 volts. So there's 14 gauge wire in this box hooked up to the little slots here that provides 110 volts. Now imagine you plug a hairdryer into this, okay? So a hairdryer is capable of delivering 1200 watts, that's 1200 joules per second of power when plugged into this potential difference. Why so much power? Because it requires four joules of energy roughly to raise one gram of water, one degree Celsius.  So if you wanna take water in your hair at roughly body temperature and get it to evaporate, you've gotta crank its temperature up to get it to go from the liquid phase to the gas phase. You've gotta get it over 212 degrees Fahrenheit. So that's why you have to dump 1200 watts of power out of a device like this. You wanna, your hair to dry fast, you want a lot of power, because water takes a lot of energy to go up one degree Celsius, okay? So when you plug in this hairdryer, it can generate 1200 watts of power, and it's connected via the 14 American wire gauge wire, okay? So the first thing we wanna figure out is what's the resistance of the hairdryer, all right? So we know the power, I'm gonna get rid of all this stuff because we don't need it, we know the power that the hairdryer is capable of delivering through its heating element. So we can draw a simple diagram to represent this. I have a voltage source, which is the wall. I have a plug, which is a perfect conductor. And then I have the hairdryer element, which is a resistor. It resists the flow of electric current, and then I come back to wall voltage, okay? That's it, that's a hairdryer. That's how you can represent a hairdryer with a circuit. It's a nice, simple circuit, okay? There's a lot of circuitry inside the thing that handles the regulation of current and voltage, but that's all details. Basically, if you wanted to make a hairdryer for cheap, just get a fan, okay? Take a wire, plug it into the wall, let the wire heat up and blow the air over your head. That's what a hairdryer does, okay? Now, I don't recommend you do this because you're very likely to set yourself on fire, set the wire on fire, because you didn't choose the resistance properly, or any number of other horrible things. The nicest thing that will likely happen if you do that is you'll trip a breaker in your apartment or home. That's a safety mechanism that prevents your house from catching on fire, okay? Because it detects the too much currents going through the wall, okay? So that's a hairdryer. The power output of this is 1200 watts. And we know the voltage. The voltage is given. It's 110 volts of wall voltage that this thing's plugged into. Now, what's the resistance of the hairdryer? Well, Ohm's law is the first thing we can write down. Ohm's law relates the voltage applied to a resistor and the current driven through the resistor by this equation, V equals IR, okay?  But the good news is, is that this is Latin for strong, vir, okay? So if any of you like no church Latin or like ancient Latin or any of that stuff, you can remember this by going Ohm's law is strong, vir. And then you can remember this. No, nothing, nothing, okay? This is like the one of the three things that Latin did for me over five years. It was a huge waste of time. I can't look at French and figure it out from Latin. So then I speak French way more than I speak Latin. So anyway, getting back to Ohm's law, we know the voltage. We don't know the current going through this thing. We don't know the resistance, okay? But the good news is, is that a resistor that is dissipating energy in the form of joules per second power, okay? There are other equations that we can use. So power is current times voltage. If you wanna know the power that's being dissipated by a circuit, take the current and multiply by the voltage, P equals IV. All right, well, we don't know the current in the hairdryer. Okay, but we do know the power. We do know the voltage. We know the voltage. We don't know the resistance, so we don't know I and we don't know R. But the good news is, is we have two equations and two unknowns. Okay, we don't know I, we don't know R, but we've only got two unknowns and we've got two equations and algebraically we can solve this. And what you find out if you kind of do some substitution here, so get rid of I is that P is equal to V squared over R. So when you have a resistor that's dissipating power in a circuit, if you don't know the current, but you know the voltage and you know the power, you can substitute and you can use Ohm's law to solve for the unknown. And that's all we're gonna do here, all right? So now of course you can rewrite this equation and you can say R is equal to V squared over P. We know V, 110, we know P, 1200 watts. So we take 110 and square it and we divide by 1200 and we find out that the resistance of this is just 10, yeah, 10 Ohms, okay? So this is Ohms, it's named after George Simon Ohm, German scientist who played around with batteries a lot, and this unit of resistance is named after him and it is resistance is equal to V over I, so this is just volts per amp. That's all an Ohm is, Ohm say with me, right? Very relaxing, okay? So Ohms, that's the unit of resistance, macroscopic resistance, okay? So this is a very low resistance thing, all right? We notice right away this, so 10 Ohms, that's not a big resistance. Your skin has a resistance of maybe thousands of Ohms, tens of thousands of Ohms, millions of Ohms, okay? But we know your skin can conduct electricity, right? You can certainly electrocute yourself with a total of car battery or if you plug your fingers into that wall, you'll get a shock, okay? So your skin is very capable of conducting electricity and it hurts, all right? This is a lot lower resistance. So for that same wall voltage, there's a whole hell of a lot of current going through this device. How much current? Well, we can figure it out, all right? We got R, 10 Ohms, so we can go back to this equation and plug in R and solve for current. I equals V over R. See what we're doing here, we're good, okay? And what we find is that this is 11 amps, 11 Coulombs per second. Now, does anybody know what the breaker limit in a typical home is on a given circuit? So if you have a bunch of plugs in a wall and they're all part of one circuit in your home, does anybody know how much current maximal can be driven through home wiring before the breaker will trip and save your home? Typically, it's actually a fairly standard number in the U.S., it turns out. It's 15 amps. So 15 amps of current is enough for your home to go, nope, and trip a breaker. A breaker is just a pair of metals. They normally touch each other, but when enough current is driven through them, they heat up and the metals are dissimilar, maybe like zinc and copper or something like that. So they heat at different rates and they will bend away from each other. So a very simple breaker is just a pair of metals that normally touch when they're cool. Even when they're heated a little bit, they'll touch. But if they get too hot, they'll yank apart from one another. Just the metals will contract or expand at different rates and they'll move away from each other. And that will cause the current to stop in the circuit. Now, they tend to be much more dramatic than that. A real breaker unit will literally throw a switch, okay? It's all done with heat and current and mechanical force.  That's protecting you, okay? So if you were to plug two of these hair dryers into one 15 amp limited circuit in your house, you trip the breaker and they're a problem. Cause they're both gonna try to draw 11 amps out of that circuit and that's way above what the electrical safety codes allow in 14 American wire gauge wire. Now, wires won't fail at 15 amps but this is a safety margin, right? Wires will, 14 gauge wires will fail at 32 amps but nobody wants to get close to that and test it, right? So they cut off at 15, about half of the maximum, okay? All right, then the last thing we can do I'll leave this for the notes though cause I'd like you guys to play around with the problem real fast. The last thing we can do is calculate the current density in the wire. You just take the current and divide by the area and you assume that this is a circular wire so the area is pi r squared and you can use the diameter to figure out r and then get the area and then take i over a. So current density, j, its magnitude is current divided by area, okay? So the area is just getting pi r squared assuming that the wire is a circular cross section wire which is a fair assumption, okay?  So imagine you have a heating element in an oven, okay? Which is pretty typical for ovens. It's really just the length of wire. I showed you one of these already, right? So it's just, that's a heating element from an oven. Looks very similar to that. All right, it's really just a length of wire. It's got some resistance. It can be hooked into a 110 volt potential difference. Consider an element with a thickness of 0.75 centimeters. That's about the thickness of this, okay? It's capable of delivering 3000 watts of power for cooking, right, big, more than a hairdryer which is what you'd hope for for cooking food. Treat the wire as a long cylinder. It's a pretty good assumption. That's how it looks like a long cylinder to me, all right? All right, so what's the original resistance of the heating element given that information? Now over time, the metal in the element wears down. It fatigues. When it came from the factory, it was bad metal, okay? So maybe you use it for a couple of years, but then something catastrophic happens. And catastrophically in a short segment of the wire, its area is cut in half for some reason, okay? And this drives the power output to 10,000 watts. What's the current density at that point in the wire, okay? So the first thing to figure out what's the original resistance of the heating element? Catastrophically it fails, okay? So now the resistance is not necessarily the same anymore. Its area winds up getting cut in half and the power output is driven to 10,000 watts. Can anyone guess what 10,000 watts will do to this heating element? Give you a clue. This used to be one heating element, okay? So this is a heating element that failed in my house one day when we were cooking and it exploded and scattered soot and metal all over the inside of the oven and ruined what was going on inside the oven at the time. And it was very bad, this could have caught fire. Luckily the failure was so catastrophic that these two pieces of metal which were once fused, you know, I'll pass this around, they're melted now. Be careful they're a little sharp, okay? Literally melted the metal and thankfully they stopped making contact with one another or it could have run away and really started a fire inside the oven. I mean a real fire, like a dangerous one, not a controlled one you use for cooking bread, okay? So play around with this, you've got about four minutes left and again the class notes will be posted so you guys can pass this around. So essentially I'm just finding the resistance so it's just gonna be far equal to square root hour and then it's just gonna be, I got like four million homes. Let's see, over, that seems excessive. Give me a square. Give me a check, give me a check.