 Well, welcome to this special lecture on simple circuits, complex circuits, and the use of Kirchhoff's rules. This will supplement the material for Physics 1308, and will help to fill the void in your hearts that is left by having fall break between us and the next lecture. So to begin, we've been talking about applying the basic principles of Ohm's law and energy conservation and charge conservation in order to analyze situations where you have a source of electric potential difference, an EMF device like a battery, an ideal battery, that can create an electric potential difference and drive charge through a system. We looked in class at the situation of two resistors in either series or in parallel. We used light bulbs as resistors, and we did experiments in class and calculations in class to see how Kirchhoff's, not how Kirchhoff's law is, to see how Ohm's law can be used and what situations it does and doesn't apply to. We're going to go back now to thinking about ideal situations or mostly ideal situations involving very simple circuits, and then we'll move on to the more complex circuit examples, and I'll motivate why it's important to learn this material in the context of something I've mentioned before, which is electrical transmission down the length of a neuron. So let's consider the following circuit. This is not terribly different from what we did in class with two light bulbs in series. What we have over here is we have an ideal EMF device, and we have a resistor. And you'll notice that those are enclosed in a green box. I'll come back to those in a moment. And over here, of course, we have another resistor whose resistance is labeled as 4 ohms, and this little R resistor that's grouped with the ideal battery has a resistance of 2 ohms. What's being depicted here is something that's much more akin to a real battery situation. This is how one would represent a more realistic battery situation in a simple circuit. The source of the electric potential is ideal, but the battery itself does contain an internal resistance. Batteries are not perfect conductors of electric charge, and so there is always going to be some resistance when moving charge between the terminals that is the minus end and the plus end of a battery. And a way to represent that schematically using the bits and pieces of circuit symbols that we've built up so far is to put an ideal battery here represented with the EMF symbol of this calligraphy E next to a small resistor. Now, the typical internal resistance of a battery that you might get off the shelf, like a 9-volt battery or a 12-volt battery or something like that, is about an ohm, 2 ohms, something like that. It's not a whole lot, but it is not zero. And so the green box is meant to indicate the boundary in the diagram of the real battery. So a real battery consists schematically of an ideal source of electromotive force, an EMF device, and a resistance. And that resistance may be very small, but it is not zero. The positive terminal is again labeled with the plus, the negative terminal is again labeled with the minus, and just as always in a simple circuit like this, you can very quickly figure out what the direction of current flow will be. Current flows in the direction positive charge is moving and positive charge is emitted from the positive side of the battery. And so the arrows that are drawn here are correct for this situation. We would expect the current I to flow clockwise through the system. Now, the circuit consists of an additional resistor, R equals 4 ohms, which is much larger than the internal resistance. And one could then do things like, for instance, solve the current that's going through the battery. Now, how would one do that? Well, you want to use the principle of Ohm's law that for every resistor in the circuit, there is a relationship between the current moving through the resistor and the voltage across the resistor that's given by this representation of the consequences of Ohm's law that R is not dependent on V or I, and so it's just a constant that relates the two. One can use the rules of adding resistors in series. We have here that the total resistance in this circuit will be equal to the battery resistance plus the external resistance of four ohms. And so you can see very quickly that this is equal to six ohms. And so with that in mind, then one could go and divide the voltage, 12 volts by the resistance, six ohms. So let's go ahead and do that. And we see that we expect to be flowing through this circuit a current of two amps. With V in volts and R in ohms, we automatically get amps out the other side when we take the ratio of V and R. All right, well, there's nothing exciting about this particular diagram, but it does help us to now begin to think about the following question. What happens if one short circuits a battery? Short circuiting means that you, for instance, take a nine-volt battery. It's got those two little plugs on the top of it that you could connect into a device like a smoke alarm or a carbon monoxide monitor or something like that. Imagine that you take a paperclip and you drop the paperclip on the terminals. This creates something known as a short circuit. The paperclip is made of a highly conductive metal that offers very little resistance to the flow of current between the plus side and the minus side of the battery. And this results in a situation known as short circuit. You're essentially letting current flow from plus to minus on the battery with almost no external resistance at all. That's the situation depicted here. We have now removed the four-ohm resistor from the picture. We only have the internal resistance of the battery present, and that means that this system will now drive current through the loop from the positive terminal to the negative terminal. What you can do then, of course, is assess this using Ohm's law. You can solve for the current moving through the circuit. It's very simple now. It's simply the voltage of the battery divided by the internal resistance present in the real battery. That's nine volts divided by two Ohms, and this gives you a current of four and a half amps. That is a lot of current that would suddenly be coursing through a nine-volt battery and, most importantly, through the circuit that connects the plus terminal of the battery to the minus terminal of the battery. How much power is being dumped through the battery's resistance? So if you imagine again that the battery is a device that looks something like this with two terminals, a plus terminal and a minus terminal, and you suddenly short-circuit them like this, the resistance is somewhere in here. That's where the internal resistance is located. That's gonna begin to dissipate power, and that is going to cause heat to build up in the battery, which will then be transmitted. So this will begin to emanate heat, radiating out through the battery, and eventually into whatever environment contains the battery. That could be a sensitive piece of electronics. That could be your pocket. If you put a nine-volt battery in your pocket and you have coins in your pocket, you could accidentally short the battery. Well, how much power is being dumped through the resistance of the battery? A lot, 40.5 watts. Again, you just take P equals IV, take 4.5 amps multiplied by nine volts, and you get 40.5 watts. This is a lot of power. That's as much energy as is emitted by a 40-watt light bulb, right? Imagine grabbing a 40-watt light bulb, plugging it in, and shoving it in your pocket. That's going to get very hot very fast. This is extremely bad news for your skin if the heat is transmitted very quickly to your skin. It can also cause the battery to explode. It can mechanically and chemically fail building up pressure inside the battery, which then lets go in an explosion, and that can do serious physical damage to your skin, your clothes, and other things, especially if there are dangerous chemicals inside the battery that will not be splattered everywhere. So this is why it's extremely important never to short a battery. You should always put a significant resistive load on the battery, never operate it outside of the conditions under which the manufacturer has specified for it. That's why it's bad. Now, let's think about another situation that's very common in the world around us, especially now in the era of cheaper, more affordable, and more abundant rechargeable batteries. A battery charger is a very simple device. You usually have a cord going into the wall outlet that then runs into some circuitry inside the battery charger, and then at the end of the battery charger, there's a pair of metal plates. You stick your AAAA batteries between the plates, wait a few hours, come back, and the charger will have them fully charged for you if they're rechargeable. How does recharging work? Well, basically what you're doing is you're taking one source of electric potential difference that is much larger than another source of electric potential difference. So a typical AAAA or AAA battery, for instance, has a voltage of about one and a half volts. Rechargeables have a little bit less, something closer to 1.2 or 1.3 volts across their terminals, but eventually one and a half. The voltage into which they're connected, however, will be much larger than the voltage that they can deliver. The circuit for the charger will represent some kind of resistance between the source of voltage for the charging, so this would be the source of voltage that does the charging, and this is what gets charged over here. And what you're essentially doing is you're overwhelming the EMF of the battery by driving a big current through the circuit that goes backward in the battery. So of course the one and a half volt battery wants to push current out of its positive terminal this way, but it's being overwhelmed by the current that's being driven by the 12 volt potential in this picture here. So the net current will flow not up out of the one and a half volt batteries, positive terminal, but rather back through it. And this puts energy back into the battery. It undoes the chemical reaction inside the battery. It allows this charge to separate again, for instance, and then when you unplug it, the chemical reaction that provides the electric potential difference in the first place can run forward. So this is how you recharge a battery. You basically have an overwhelming potential difference of some kind, a resistive load represented by the circuit that handles the charging, regulating the charging process, and then finally you have the battery you actually want to charge at the other end of this process and everything's connected together in a single circuit. So you can do things like ask questions like what's the net voltage across the resistor? What direction is the net current in the circuit flowing? And which EMF device is supplying energy and which is absorbing energy? These are the questions that we would like to answer about this kind of circuit. In order to get started on answering these questions, we have to do our analysis of this a little bit more complex circuit with two EMF devices in a few steps. And these steps are going to come in handy because they are the basis of much more complicated circuit analysis. So the first thing we want to do in order to establish the voltage changes in a circuit like this so that we can figure out where the current is flowing is to just follow two simple steps, okay? In order to find the voltage changes in the circuit, we basically have to walk a path through the circuit and we want to figure out what that path is using some conventions. So first of all, we're going to choose a direction in this above loop that we believe current is flowing. It doesn't have to be correct at all, but once you make a choice, you have to stick with it, okay? So that's the first step. Simply draw arrows indicating where you believe current could be flowing in the circuit. Now use conservation of charge. That is an important rule that you need to keep in mind. Charge is conserved in a closed system like this. So you don't get to draw current flowing one way in the circuit and then in the next leg flowing back this way. That doesn't make any sense. At that corner, current should continue to flow continuously into the next segment of the circuit. There are no branches here. Branches are something we'll learn to deal with in a bit. So the right way to draw the guess at which way current is flowing is a consistent, in this case, clockwise direction. You can draw it counterclockwise. Whatever direction you choose, choose it and stick with it for the rest of the problem, all right? So let me just say that these are my chosen eyes. I believe current is flowing in this direction. Now if I'm wrong, we'll learn that at the end of solving the problem. It's okay to be wrong, it's just a guess and your guess will be corrected by following the steps of using energy and charge conservation to understand this circuit. The second thing you need to do is once you've established where you believe current may be flowing in this loop, you then have to pick a direction that you are going to walk through the loop. It doesn't have to be the direction in which current is flowing. If you really like going counterclockwise through a loop, you're welcome to do that. But once you pick the direction to walk the loop, you have to walk in that direction and stick with it. So let's demonstrate this. So this is a much prettier version of the hand sketched drawing I just did. This shows the arrows that I have selected based on my guess as to where I think current i is flowing in the circuit. Now we'll find out later that I'm wrong, I have the direction wrong, but that's okay. You do not have to be right when you make this guess, you must only be consistent. The second thing you have to do is choose a direction to walk the loop. So let's go ahead and do that. So I'm going to choose to walk the loop in the same direction that current is flowing. You don't have to do that, but once you make a choice, you have to do it and stick with it. So everything that happens after this will be based on these two steps. I chose the current, the direction I think current is flowing and I chose the direction that I would like to walk through the loop to set up and solve what will turn out to be the energy conservation equation. So I've used conservation of charge already to indicate the direction that current is flowing. Charge can't get to a corner in reverse direction. It has to keep going consistently and smoothly through the loop. That's the expression of charge conservation here. And I've also now written a random and arbitrary guess direction about what direction I would like to go through the circuit as I traverse the components in the circuit and that is in input to the energy conservation equation application. Okay, now a couple of additional rules to keep in mind. What we're going to do now is we're going to set up an equation that looks something like this. The total change in the electric potential differences in the circuit will be zero. If I walk a closed path through the circuit going through one EMF device, then a resistor, then another EMF device, and then back to where I started, the net sum of all changes in electric potential difference must be zero. That is, in a closed system with no external source of energy coming in and no energy leaving the system, all the changes in energy must sum to zero. So this is the equivalent in a gravitational problem of moving an object through a closed path in a gravitational field when you return back to the original place where you started, the net work done in making the journey is zero. So in our walk, which is going to be a walk through the circuit that takes us through an EMF device, a resistor, another EMF device, and then back to where we started, you have to keep in mind a few rules for the changes in potential difference through each of the devices. If your walk takes you through an EMF device, for instance, an ideal battery, from negative to positive terminal, then the change in potential is plus E. That is, the symbol for the potential change in an EMF device is this calligraphy E. And if you go from the negative terminal to the positive terminal, the change is plus E. That is, you should be gaining potential energy. If you go through an EMF device, for instance, again, an ideal battery, but you go from the positive terminal to the negative terminal, that is, you walk backward through the device, the change in potential is negative E. If you go through a resistor in the direction in which you chose current to be flowing, that is, if you follow the flow of current through a resistor, the resistor is pushing back against you, and so the change in potential is negative IR. That's just using Ohm's law to write the potential change V through the resistor, and that is going to be negative IR, because again, the resistor's pushing back against the flow of current, and you have elected to draw the flow of current in a particular direction. So if you follow the current through the resistor, you're being resisted, and the change in current and change in voltage is negative IR. Finally, if you go through a resistor against the direction that you chose current to flow, the change in the potential is IR. You're moving in the direction now that the resistor is pushing against the current, and so you get a bump in potential. So let's apply these rules. My path is indicated in blue. I am first, I'm gonna start here, and I am going to walk through the circuit in a clockwise motion according to this blue line. That also happens to be the direction that I chose current to flow in order to set up this problem. So the very first device that I will encounter in my walk is this one and a half volt potential difference, E1. And the positive terminal's up here, and the negative terminal is down here, that's the convention for the EMF device or battery symbol. And so I'm going from the negative to the positive terminal, and so my change in potential here is plus little curly E1. I then continue my walk through the circuit, and the next device that I encounter is the resistor. I have gone through the resistor in the direction that I have chosen current to flow, and so the potential change through here is negative IR1. That's the potential change that I experience traveling through the resistor. I now continue my walk through the circuit and I encounter another EMF device and I walk through it, but this time I'm going from the positive terminal to the negative terminal. So the potential change that I experience here is curly E2 with a minus sign in front of it. And finally my walk takes me through the conductor, there are no further changes in potential until I get back to my starting point and I'm done. And according to the conservation of energy, the sum of all of these changes in potential must equal zero. Nice simple equation to have to set up and solve. So that's what I've written down here. I have plus E1 minus I1R1 minus E2 equals zero. Now I'm given the potential difference in the first battery. I'm given the potential difference in the second battery or second source of electric potential difference. I'm given the resistance, the circuit has an overall resistance of 100 ohm. And so from this I can solve for the current that is going through the circuit. And I get this number. Negative 0.105 amps. What does a minus sign mean here? Let's go to the frequently asked questions page. What does it mean if I solve for current I and that number is negative? Well all it means is that when you chose the direction that you believed current was flowing, you chose the wrong direction. It's okay, you just have to make that guess and stick with it. At the end, if you do all your math correctly and you stick with your original guesses, you will get an answer that is correct for the circuit. And what the circuit is telling you is that no, no, no, no, you drew the arrow going clockwise, but actually the net current in the circuit is going counterclockwise. And that's kind of what you would have expected. The potential on the right hand side of this circuit diagram is far larger than the potential on the left hand side. And as a result you would expect this potential to overwhelm this potential and drive a current counterclockwise through the circuit. But now you can see it mathematically that that's true. If you apply Ohm's law, conservation of charge and conservation of energy, the circuit tells you that the current actually is flowing in the counterclockwise direction opposite the direction you originally drew and that's where you get the minus sign from. Now if you had been very far-sighted at the beginning of the problem and you said no, I know just by looking at this that the current is gonna go counterclockwise and you drew your arrows to be counterclockwise at the end of the day, your current would have been positive 0.105 amps, meaning you just happened to choose correctly at the beginning of the problem. Now let's take a look at a more complex situation. And we've already basically set up what are the things which are known as Kirchhoff's rules merely by writing down and using current and energy conservation or charge and energy conservation, but we will formalize the process now to solve much more complicated circuit problems. Why should you care about this? Well, the reality is, is that the world is full of complex circuits. To be honest, the ones that humans have come up with for the most part are relatively simple. They don't require a tremendously difficult set of analysis in order to figure out what's going on inside of them. But nature is an excellent engineer, of very complex circuit situations. And so if we want to model nature using the laws of physics, using charge conservation and energy conservation, we have to do things like represent natural phenomena using circuit diagrams. And one example of that is the neuron. The neuron is the beating heart of the transmission and storage of information in something like a brain. It is a complex network of neurons and the configuration of the neurons is set by learning procedures. And those learning procedures involve the transmission of electrical signals across the length of the neuron itself. So you have some impulse that's transmitted from a stimulus that comes into the neuron through the dendrites. That stimulus is transmitted from the cell body of the neuron down the axon of the neuron to the terminal bundle. And the terminal bundle itself involves chemical transmission of information to other neurons through their dendrites. So if we think about this as a cable or some kind of complicated circuit body that's carrying impulses, electrical impulses along the length, we can actually make a great deal of progress in understanding just how information is transmitted through this system, which is in and of itself very complex. And in fact, this is one way that you can represent in an oversimplified way what the circuit looks like for the axon. So this is actually a circuit diagram which gets the basic transmission of electrical information through the axon about right. And it involves not just resistors but resistors and capacitors. The capacitors can be charged up and discharged by changes in electric potential across them. When an electric potential is built up across a capacitor, it will store charge. If the electric potential from the external source is released, then the capacitor can release its charge back into the system. And you can imagine that there's a very complicated transmission of information down the length of this repeating system of resistor, capacitor, and resistor, and resistor, capacitor, resistor, resistor, resistor, capacitor, resistor, resistor, resistor, all the way down the axon, down to the terminal nodes of the axon with a signal originating in the dendrites. The circuit, the capacitor and resistor segments you see here are physical representation of what are known as the nodes of Ranvier. The nodes of Ranvier are capable of storing and discharging charge. And so they are very well represented by capacitors. But they contain, of course, resistance to the flow of charge. And so one also has to insert a resistance into that circuit in order to model it more accurately. In between the nodes of Ranvier, you essentially can represent the interstitial fluid and the intracellular fluid as resistive media. They don't have to have the same resistances. And so electrical signals will propagate in a very complex way through this system with energy being stored in the capacitors released and that's pulsed to the next capacitor resistor system and that's pulsed to the next capacitor resistor system and so forth. So a signal can be transmitted down the length of this by applying a voltage on one end. That's the stimulus. The stimulus goes away. But meanwhile, the capacitor has been charged. When the stimulus goes away, this thing then discharges its charge to the next capacitor. That charges up when its charge goes away. This then transmits to the next capacitor and so forth. And so by doing this, you can do signal propagation down the length of an axon and model it using just a circuit diagram that looks like this. This leads to a very complicated set of outcomes that are actually fairly difficult to model with a very simple circuit I just showed you. Nonetheless, if you add more complexity to the circuit, you can essentially reproduce the gross features of the behavior of the neuron. A stimulus enters at some time. This causes a voltage to swing in the neuron from negative 70 millivolts, which is the resting potential of the neuron, all the way up to positive 30 millivolts. Basically what you're getting here is you're getting a closure of gates to the flow of sodium. Then sodium is allowed into the system. You get a net positive charge buildup which creates a positive potential difference. And then at some point you get potassium gates that open. This causes a large downswing of the charge and thus a great decrease of the potential that overshoots the original resting potential goes as low as negative 90 millivolts. And then over time, this system will slowly go back to its resting potential by the potassium gates closing and potassium being trapped in the system again. All right, so you have an active relationship between sodium and potassium pumps that move sodium in and potassium out. And this allows you to alter the electric potential difference in the system which creates a voltage change. And this voltage change can be transmitted down the length of the axon through the resistor capacitor network to the terminal nodes and then potentially onto other neurons in the system. And all of this can be represented by a complex circuit system. So that's why it's important to master some of the basic skills of circuit analysis. There are many complex structures in nature. Biology is home to many of these. If one wants to come to a basic understanding of how signals are propagating through the system, a little circuit analysis under your belt will go a long way. As an example of a more complex circuit than the one that we looked at earlier with two batteries in it, consider the following one which has two batteries in it, but also branches and multiple resistors. So let's imagine that we were to start out taking a little journey around this circuit here from just before the left battery. So I'm going to just put a little dot here to remember. If we were a positive electric charge, you could imagine that maybe we are moved through battery one and then we turn the corner at point A and we continue through resistor one and then we turn the corner down here and we get to point D. And at this point we could go along one of two branches, one that takes us through resistor number three or one that takes us through resistor number two and then possibly backward through this second battery over here on the right until eventually we might return to point B and then the point where we actually started our journey. This is a multi-loop circuit. You see it has a left loop, it has a right loop that's about the same size in this drawing and it has a big outer loop as well. And in order to analyze these circuits, we just have to keep applying the principles of charge conservation and energy conservation to understand what's going on in the circuit. Now in a problem involving complex circuits, you'll often be given some amount of information. So for instance, in this particular case, we are told the electric potential delivered by battery number one. We're told the electric potential that's delivered by battery number two. We're also told that the resistances all have numerical values and they're all equal to two ohms. So they're all identical resistors, although your mileage may vary and a problem that you get, you may not be granted identical resistances for all of the resistors and you may have to combine resistors, for instance, if you can. You'll notice that what makes this problem more difficult than ones we've looked at before is that there's no simple way to combine neighboring resistors. For instance, this resistor is definitely in parallel with this resistor here, so two is in parallel with three. But because you have two batteries in the system, it's not clear how you combine these resistors and suck the battery in at the same time in order to get one battery hooked up to one resistor and then back to itself. And so that's what makes these multi-loop more complex circuits at first an apparently intellectual challenge to understanding what's going on in the circuit. But we are simply going to apply the rules of charge conservation and energy conservation. We're gonna use Ohm's law and we are going to tear this thing apart and figure out what's going on. And the things that we want to figure out are we want to find current number one, we want to find current number three, the current that moves down through the middle vertical branch of the circuit and we want to find current number two, which here is the current going through resistor two in the right branch of the circuit. So we have three currents, we don't know what they are. And so at this point, we are tasked with attempting to figure out exactly what those currents are. So what we want to do first is the same trick we played with the simpler circuits before. We first want to figure out what direction that we believe current is moving in the circuit. Now, this picture has sort of handed us a treat. The picture indicates some suggestions for how current might be flowing in the circuit. So for instance, if we were to imagine the current that's flowing through the top left battery then down through resistor one, the drawing that we've been given already conveniently gives us the current direction as pointing downward through R1. Now that may not turn out to be the actual direction that current is flowing in this particular circuit in that part of the circuit. It's merely an assumed direction of current flow. But quite often, you will be given a situation where in fact you do not have this information handed to you. So for instance, I might white out this arrow here. And then we can repeat this exercise for the other arrows that are here. We can clear that one out as well. So we'll just erase that current arrow, get rid of the helpful suggestion placed here by the original drawing from Wiley, and then finally we can erase this arrow as well. This is a much more common situation that you might be handed. In one of your homework problems, you're given suggestions for the direction that the current might be flowing already through the circuit. But that isn't necessarily what will happen on an arbitrary circuit analysis problem. You may simply be given a picture that looks something like this and be left to completely figure it out on your own. So for instance, the way that you would figure it out on your own is you would make some guesses about which way you think current is flowing. And remember, when you get to a branch in the circuit, that is a location like this, a node in the circuit where current can flow one way or another, you have to use conservation of charge to determine how you're gonna break up the currents. So let's start with the left branch of the circuit. We might imagine that current could be flowing like this, just the way that that original arrow had been drawn. And if we follow that current I1 through resistor one, eventually we get over to this first node, node D. Let me erase this here so it's clear that that's a D. So we get to node D. And now we could have a few situations that are occurring. It could be that this current from I1 splits into two currents, I3 and I2. And we see those written over here. I2 going through resistor two, I3 going through resistor three. But that's not the only option. We could instead guess that, well maybe what's happening is that current is coming down through resistor three and it meets the current from resistor one and it then merges like two streams merging together to become a single stream or river. And then that current then flows through I2 and up through I2 like this. All right, so you see you have a tremendous amount of freedom here to write a picture of where you think current is flowing and your only constraint is that at a node, the current going in, I in, must be equal to the current going out. So if you have one current coming in like I1 to this node and another current coming in to the node like I3, the resulting current must add up to I2. So in the picture I've drawn, this is what is called the node equation for node D. There's a similar node equation for node B. Let's take a look and see if we can figure out what it is. So at node B, we know that over here on the left branch of the circuit, the current is traveling leftward and then downward through the resistor. So the current I1 is coming out of node B. We also have current going down through resistor three, which means that that current is coming out of node B as well. And then finally, we have the current coming in from resistor two backward through battery two and then entering in to node B. So see if you can before I write it down here, see if you can figure out what the equation is that represents conservation of charge at this node and I'll give you a few seconds to think about it. Well, if we think of the current two as coming into the node, then I2 must be splitting and becoming I1 plus I3. That's the only way that this could possibly work. And what's great about this is if you've done a good job with your circuits and you've actually consistently used the direction of current flow in your nodes, you'll notice that these two equations are exactly the same. They must both be true. They can't conflict about what's going on in the circuit if you've consistently used your assumed direction of current flow. And you'll notice that here we have I1 plus I3 equals I2 at node D and then at node B, bravo, we have I2 equals I1 plus I3, which is exactly the same formula. So this is a good built-in cross check that you have consistently used the assumed current directions that you write down. If you've inconsistently used them, you will get conflicting equations that give you different answers for how I1, I2, and I3 relate to one another and you'll be unable to solve the problem as a result. So I always recommend that the current conservation at the nodes in the circuit, places where you have current splitting or merging, okay? Those are the places where you wanna try writing down after you've drawn your arrows, try writing down your current conservation equations and make sure that they don't give you conflicting pictures of how the currents relate to one another. If they do, you've made a mistake someplace and you should go back and double check your work. So let's take a look at the next step. I'm gonna put the arrows back that come from the original problem in the textbook. On the left branch through resistor one, current one is moving downward. Through the middle branch through resistor three, current is also moving downward. And then as a result of current conservation, it has to be moving upward through resistor two because at this node, you have two currents meeting and you can't have a third current meeting there or all that would ever be happening is that charge would be building up at node D and never escaping. If it comes in from two branches, it has to go out the other. That's the only way for this to work out and actually for charge to be conserved so that it's not building up someplace in your circuit secretly and quietly out of your view. So I too has to go up once you've chosen one and three to go down through the left and middle branches. Now the next step is to apply energy conservation in the various loops inside of your circuit. Now as I said earlier, there are three loops in this circuit. There's a large outer loop, there's a left small loop and a right small loop. And I've labeled the left small loop, loop one and the right small loop, loop two. And then to be conscientious, I should, oh, actually I do have that over here. I've labeled the large outer loop, loop three. Now what do you have to do at this point? Well remember earlier when we defined loops, we were doing it because we were going to take a walk through our circuit. And you're going to do the same thing here. You're going to take a walk through each of the loops in your circuit and you're going to write down equations that represent the conservation of energy along that walk. The sum of the changes in potential along your walk through your circuit must equal zero. That is the guiding principle of energy conservation here. The net change in energy in any closed system is zero. Conservation of energy. We've already used conservation of charge at the nodes. Now we're going to use conservation of energy. So what I have done here on this blank page is to merely reproduce the salient features of the formal diagram from the textbook for this particular problem. And if I go back just one slide here, I'm going to draw the loops just so that we have them here. So the loop one is the loop around the little left side of the circuit. And I've decided to walk that loop in a clockwise manner. Similarly, I've decided to walk loop two, the right hand little loop in a clockwise manner. And just to be extremely consistent, I've decided to walk the whole outer loop of the circuit, loop three, in a clockwise manner. So we're going to be taking little journeys through the circuit now and using energy conservation to write down another set of equations. So we know from the nodes in this equation here. Let me write those down first. Actually, to be nice and consistent with the colors I've shown here, let me do blue. So the node equations, which we can get from applying the exercise we went through a few moments ago with current conservation, are going to be that at node B, we have the equation that I2 flows in. And I1 and I3 flow out. And so the sum of I1 and I3 must be equal to I2. And at node D, we have I1 and I3 flowing in. And those are equal to the only other thing left here, which is I2, which is flowing out. And you see we have here consistent equations with one another based on the picture that we drew of which way we assume current to be traveling. May not actually be traveling that way, but we'll figure that out when we solve for these currents. OK, but we're not quite there yet. We've got to do some walking. So we're going to take a counterclockwise journey through loop one. So now we have to write down what are known as the loop equations. And this exercises conservation of energy. So that's the operating principle for this stage of the setup of the problem. And remember, our goal is to figure out numerically what are I1, what are I2, and what are I3. And to do that, we have to apply these principles, conservation of charge, conservation of energy, and then we can get answers. OK, great. So loop one, that's the first loop that I've drawn here. It's this little loop on the left. And I'm going to start walking it from just below this resistor. I could pick anywhere in the loop as long as I walk in the direction I just indicated, clockwise. So I'm going to start from R1, and I'm going to follow my loop. And the first thing I'm going to do is I'm going to go through R1. And I'm going to do that in the direction opposite to which current is flowing. The resistor is pushing against current, so this gives me a boost in potential of I1, R1. I then go through the battery, E1, on the next leg of my journey. And I do so from the positive terminal to the negative terminal. And so that's going to cost me energy. And I lose by a potential change of negative E1. Finally, I'm going to get to point B, and I'm going to come down through resistor 3. And when I travel through resistor 3, I'm going to be doing so in the direction that current is flowing. That's going to cost me. So I have to deduct I3, R3. I get down to point D. I coast back to my starting point, and I'm right back where I started. And so that's the end of my journey. And the sum of all of those terms must be equal to 0. Let's repeat. We're going to do loop 2 next. And again, I'm going to walk this in a clockwise fashion. And I'm going to similarly start my journey just before a resistor. So when I go through R2, I'm going against the current. And I'm going to gain in potential when I do that. So that's I2, R2. I then come down the bottom of the loop, head over to node D, head up through resistor 3. When I go through resistor 3, I'm opposing the flow of current. So I'm going to gain again by I3, R3. That's my gain in potential. And then finally, I'm going to coast through node B. And then through battery 2, going from the negative terminal to the positive terminal, the direction that it would like to push current. And in that case, I'm going to get a boost in potential as well. Now you might be wondering, wait a second. I have a bunch of sums of terms equal to 0. How can that be? Well, remember, it's not necessarily true that we have drawn the current arrows in the correct direction. And so it's possible that one or more of the currents, I2 and I3, could be, in fact, negative numbers. And they will actually subtract in this equation. We just have to faithfully write the equation down. We traveled backward through resistor 2 against the flow of current. We traveled backward through resistor 3 against the through of current. We went through the battery, the direction it wants to push us. Our arrows for current could be backward. We'll find out at the end of this exercise. And finally, we have the big outer loop, the beltway of this particular problem, going all the way around resistor 3 and avoiding it. So for that, what I'm going to do is I'm going to start my journey from node D. And I'm going to walk the big outer loop, again, in a clockwise fashion like this along the outside. It's good to make pictures for this. It helps you to remember what you're intended to do and then check that you actually did what you intended. So for loop 3, we're going to start at point D. We're going to move around here. And the first thing we're going to do is we're going to go through resistor 1. And we're going to do it against the flow of current. So we're going to gain in potential by I1R1. We're going to go through battery number 1 from positive to negative. We're going to lose potential there, negative E1. Now we're going to coast through node B and head through battery 2 in the direction it wants to push us. So we're going to gain by adding E2. And we're going to come down through resistor 2 against the flow of current. We're again going to gain by I2R2. And then we're going to finish our journey back at node D. And so the sum of all that must be equal to 0. What we have just done is apply Kirchhoff's rules. Kirchhoff is essentially famous for having applied charge conservation and energy conservation to the analysis of circuits. And so the rules that we just wrote down bear his name. The node equations and the loop equations, the node rules and the loop rules. Current must be conserved at the nodes, charge conservation. Energy must be conserved and sum to 0 in terms of changes in energy if you walk a closed path. That's what we did in the circuit for the loop equations. And we set those equations equal to 0, applying energy conservation. Now here comes the beautiful part of the application of Kirchhoff's rules, the loop equations and the node equations to solving the problems. What do we want? That's the question we have to ask ourselves at the end of any exercise and just mindlessly writing down equations. Just writing down the equations is not going to get you to the answer. You have to know what you want. And what we want are to find I1, I2, and I3. So we have three unknowns. And you'll notice that many of these equations all contain at least one or two of those unknowns. The node equations contain all three of them. And the loop equations tend to contain the currents in pairs because of the way this circuit happens to be laid out. And so here's the beautiful part. The rule of thumb is in the application of Kirchhoff's rules, you just need to use all but one of your loop equations and all but one of your node equations to solve the problem. In this case, you'll see that that gives you exactly, let's say we were to pick loop one and loop two as our two loop equations. It doesn't matter which two you pick. You just have to pick two of them, all but one of them. So two in our case. I could pick the loop one equation and the loop two equation. That gives me two equations. And then if I pick, for instance, the node B equation, because it doesn't matter, the two node equations are identical to one another. I have three equations. I have three unknowns. And algebraically, I can solve that situation. So this is your strategy. In applying Kirchhoff's rules to solve a problem, you should first choose the direction that you think current is flowing, conserving current at the nodes. That gives you the node equations and a picture. You should then choose loops in the circuit and what direction you're going to walk through them to set up the loop equations. That will give you a number of loop equations and how you move through resistors with regards to the direction of the current flow you wrote down determines changes in potential, how you move through the batteries forward or backward. We'll determine the changes in potential through the batteries. You sum all those changes up. You set them equal to zero. You do that for each loop. And you get loop equations from that. And then in this case, you have three unknowns. So you're going to need three equations. And you can, in this case, use all but one of your loop equations and all but one of your node equations and you can solve the problem. Algebraically, you should be able to do this at this point. So let's walk through this. What we'll do is we will pick the loop one and loop two equations and we will pick the node b equation because it doesn't matter whether we pick the node b or node d one. And we'll see if we can solve the problem for this particular situation. So on this page, I've rewritten the node equation that I selected and the two loop equations that I selected from the previous slide. And now what I want to do is I want to start solving these equations for, for instance, one of the currents and substitute those solutions from one equation into another until finally I isolate only a single one of the currents. I can solve for it and then I can put that answer back into my equations and solve for the others. Eventually, for instance, if I can figure out i1 and i3, I get i2 for free from the first equation. Or if I figure out i2 and i1, I get i3 for free from the first equation. So your goal should be in this particular situation to try to isolate two of the currents, solve for them exactly, and then plug those back into the node equation at the top and solve for the third one. All right, so let's go ahead and do that. One of the things that's kind of convenient is that by combining a couple of these equations, we can actually eliminate already a few things. So for instance, you'll notice that if you were to add together the two loop equations, one of them contains a term that's negative i3, r3, and the other one contains a term that's positive i3, r3. And you could eliminate that term right away. The other thing, of course, that you could do is you could, in fact, subtract these two equations. And that would give you a term that involves two i3, r3, and another term that involves, for instance, the differences in currents, i1 and i2, multiplied by their resistances, respectively. So we have a bunch of things that we can do. There's no one rule of thumb I would give you at this point for how to go through this. It's algebra. You're going to have to grind through it. There are easier ways, and there are hard ways. And it's sort of up to your skill level at this point. I'm going to just try to mindlessly plow through this just to demonstrate what we do. So, for instance, one of the things that I could do would be to rewrite this equation, the first loop equation here, in solve for i1. So I think I will go ahead and do that. So that will give me i1 in terms of i3, and these resistances and voltages from the batteries, which we already know. So let's take the loop 1 equation, and I'm just going to reproduce it over here. i1r1 minus e1 minus i3r3 equals 0. And I'm going to rewrite it and solve for, for instance, i1. So I will isolate the term with i1 on the left-hand side, and this will be equal to e1 plus i3r3. And now I can divide both sides by r1, and I've accomplished my goal of having an equation that has i1 on one side and the other terms on the other side. So that now gives me an equation for i1. Let me box that up there. And I'm now going to insert that into one of my other equations. Well, the only other equation that contains i1 is the loop equation or the node equation up on the top. So let's take the node equation, and that's i2 equals i1 plus i3. OK, great. So now I can go ahead and plug in i1. And this will give me now a relationship purely between i2 and i3. And this will come in handy, because you'll notice I have i2 and i3 in this equation down here. And if I can get i2 written in terms of i3 in the same way that I got i1 written in terms of i3, I can substitute it back into the bottom equation, and I can solve for i3. And that's going to be my goal right now. I'm going to try to solve for i3 first. All right, so writing here with my substitution, I'm going to have 1 over r1 e1 plus i3r3 plus i3. That's going to be equal to i2. And I think at this point what I will do is I will go ahead and substitute. I will try to group the terms in terms of i3 here. So for instance, I have, because I'm eventually going to want to isolate i3 anyway. So why don't I go ahead and rewrite this as e1 over r1 plus r3 over r1 i3 plus i3. And now you can see very easily how to group the i3 terms. e1 over r1 plus this whole thing, r3 over r1 plus 1 times i3. And that, again, that whole thing is equal to i2. So now I can put that into the loop 2 equation. And I know this looks like a mess, but hey, it's algebra. Welcome to the real world. All right, so I'm going to put that in here. So I've got i2r2 plus i3r3 plus e2 equals 0. And I'm going to go ahead and substitute my i2 that I just got into this. So that won't look pretty, but that's OK. So we're going to have e1 over r1 plus r3 over r1 plus 1 all times i3. And that's all times r2 plus i3r3 plus e2 equals 0. This looks like a nightmare, but it's really not that bad. Now our goal just has to be to isolate the terms involving i3, move everything else to the other side of the equation, and solve for i3. So let's go ahead and isolate the terms involving i3 first. So what I'm going to do to get this done is I'm going to multiply through here by r2. All right, so I'm going to wind up with e1 times r2 over r1 plus, might as well distribute this in here, r3r2 over r1 plus r2 times i3 plus i3r3 plus e2 equals 0. I haven't exactly picked an easy road here, but that's OK because now I can group my terms. So I'm going to have the following, e1r2 over r1. Let me move e2 over here and group these together. Plus, all right, well, I've got r3r2 over r1 plus r2 plus r3, and that's all times i3 equals 0. And now I can move stuff to the other side here. So I'm going to wind up with the following equation. i3 is equal to the negative of e1r2 over r1 plus e2 all divided by this mess, r3r2 over r1 plus r2 plus r3. And now I could plug my numbers in. I've got numbers for all this stuff. One thing that I could do to make this look a little prettier would be to actually take r1 out of the denominator here. So if I pull an r1 out of the denominator down here or conversely pull the r1 out of the denominator up here, I'll get a slightly simpler equation out of this. It will wind up just be something like this. e1r2 plus e2r1 all divided by r3r2 plus r2r1 plus r3r1. And that's not so bad. And so at this point, we could go ahead and plug that into, for instance, a calculator and get a number out the other end. And the numerical answer that we get is negative 3.5 amps. Again, think back to the frequently asked questions. What does that minus sign mean? Well, what that minus sign means is if we go back to our picture and we look at i3, we originally drew i3 to be pointing down. But in fact, the math tells us that we chose incorrectly, but that's OK. What's really going on is that there are 3.5 amps flowing upward through resistor 3. So if one wanted to go ahead and represent this, this means that we have upward current in resistor 3. So it turns out at this point, we're nearly there because we've already got an equation that relates i1 and i3 to one another. And we just got a numerical answer for i3. We know r1, e1, and r3. We can go plug in and solve for i1. So that's what we're going to do. So here is our answer for i3. We have our equation for i1. And all you have to do now is plug negative 3.5 amps in here, put all your other numbers in, and calculate. And you'll very quickly get a numerical answer for this. And the numerical answer that we get if we grind through the math, put it into our calculator is we find out that this is one amp. So the good news is we've got a plus sign on this one, which means if we go back to our picture one more time and look at where we drew the current i1, we said that, oh, I'm going to guess that that's pointing down through r1. Total guess. Turned out to be right. In fact, the current is in this circuit pointing downward. And it has a numerical value of one amp in that left branch of the circuit. And so we're nearly there now. We've got now finally our answer for i1. And remember, we have the other node law to help us to relate i1 and i3 and i2. So we have that i2 is the sum of i1 and i3. And this one is probably the easiest situation we have. We've already gotten i3 and i1. We have one amp. We have negative 3.5 amp. And so the answer is negative 2.5 amp. So let's make sure that everything makes sense. We find that this current is in fact moving down in r1. We found that current i3 was moving upward through resistor 3, so upward instead of downward where we originally drew it. And we find that the current i2 is moving downward, because we got a minus sign on that one, downward in r2. So let's just quickly sketch our drawing one more time and make sure that this all makes sense. So here's resistor 1, here's resistor 3, and here's resistor 2. And what the answers we got say are that the current should be moving down in r1. It should be moving down in r2. And finally, if we look back at 3, it's supposed to be moving upward in r3. Let's make sure that this works. If we have current coming down here and flowing into this node, and we also have current coming down here and flowing into this node from this side, the other current has to go out of the node in order for current to be conserved. And in fact, it is. So everything is self-consistent in the picture. Current is flowing into nodes, but not exclusively into nodes. It's flowing in, and some of it's flowing out. And so current can be conserved in that situation. These all look like sound answers, based on the physics principles of current conservation and energy conservation that have been applied here. So in summary, if you exercise Ohm's law, current conservation, charge conservation, and energy conservation, and apply that set of principles to circuits, you can solve in a fairly neat manner seemingly complex situations involving multiple batteries, multiple resistors, and even capacitors in the future. Kirchhoff's rules are merely a restatement of charge conservation and energy conservation, and applying those rules by making some assumptions about current flow, and making little loops, little walks around loops in the circuit, and applying the principles of charge conservation and energy conservation will let you solve even seemingly complex situations involving multiple voltages and current flows in a complex circuit. And I hope that this will help you with homework problems and get you started on training how to set up and solve these sorts of difficult problems by applying core physics principles and basic rules of algebra to tackle them.