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0:00
Welcome to the 16th lecture for Physics 4B. In this and the next lecture, we cover a little bit of circuit analysis we can cover with the amount of physics we have learned so far. Depending on your approach, you may be looking at this lecture immediately after Lecture 15, which introduced electric current and Ohm’s law, or you may be returning to this material after having studied magnetism first. Either approach should be fine for the sequencing of topics.
0:36
We start this lecture with Kirchhoff’s rules, which are not telling you any law of physics that you don’t already know if you really think about it, but what Kirchhoff’s rules tell you is what to focus on when you are analyzing circuits.
0:55
We show examples of applications of Kirchhoff’s rules in the remainder of the lecture. We use Kirchhoff’s rules to derive the resistor-addition formulas that you may already know. As we get into more general applications of Kirchhoff’s rules—where we encounter situations involving resistors that cannot be added as series or parallel—we make our definitions of series and parallel precise so that we don’t accidentally add resistors that we shouldn’t.
1:30
We wrap up with an example of how we can model batteries in circuits more realistically.
1:38
That’s it for this lecture. We will cover more examples, both easier and harder, of applications of Kirchhoff’s rules in the next lecture as part of our coverage of DC circuit analysis. This is Ohm’s law from last time, and this should be in the back of your mind for everything you do, because when you’re dealing with circuits this is what circuits really come down to. Let me
2:15
save a little time and, instead of sketching a circuit, use one of the
2:22
posted problems as an example of a circuit that you will soon be able to analyze. We won’t work it out right now; I’ll just use it as an example.
2:42
Let’s say this is a circuit you are given. By the way, is everyone familiar with the circuit symbols? I don’t think we actually talked about them in class, so let me write them down so that I can use the shorthand that’s in the diagram and it will make sense to everyone. These are the circuit symbols. Right now on the diagram you see three different kinds of symbols. Here’s the first one: the symbol that looks like straight lines. What do you think that symbol represents?
3:25
No, not the arrow—just the straight lines. These lines are part of a standard circuit symbol. When you see those lines, all that means is wires. We didn’t explicitly cover this in class, but
3:53
you know what wires are: conducting material. What properties should you attribute to these wires in a way that’s different from the other symbols you see? Any guesses? Constant voltage—constant voltage throughout the whole wire, right? Why do you say that? Because it’s a conductor, between two resistors. When you compare the resistance of the wire to the resistance of a resistor, whatever small
4:42
resistance in the wire there might be is so small that we are going to approximate it as being zero. You saw last time that it wasn’t exactly zero; when I had a long piece of wire, it had a resistance of like one or two ohms. But for the purpose of circuit analysis, we are going to assume that wires have zero resistance, and as a result of assuming that wires have zero resistance, they have the same voltage throughout. Whatever parts are
5:17
connected by a wire have the same voltage. That’s the most common and important symbol. The second symbol is the squiggly line; it represents a resistor. I don’t have any resistors out here—let me bring out some resistors so that you can see what they look like in case you haven’t seen them before. What I am passing out are resistors. This is a 1-kilohm resistor. When you think about a resistor, this is the typical resistor I want you to think about:
6:05
it has a value of 1,000 ohm (1 kΩ). One ohm is actually too small—like a wire has about one ohm—and sometimes you will see bigger resistors or smaller ones. This is a low-power resistor; it is rated for a quarter of a watt, and when it dissipates more than a quarter of a watt it will burn out. Resistors that are rated for higher power will be bigger so that they can dissipate more energy without burning out.
6:43
Whenever you see this squiggly line in your circuit diagram, it represents resistors characterized by some resistance R. R can range a lot; it can be as small as a few ohms and it can be as large as gigaohms, but I want you to be in the mindset that a typical value is around 1,000 ohm (1 kΩ). Sometimes in homework problems you will see an author talk about a resistor having 1 ohm or 2 ohm—just chuckle a little bit and then move on. One ohm and two ohm are not reasonable
7:26
values of resistance. A resistor is an element that has this resistance and obeys Ohm’s law. The last symbol consists of two lines of unequal length. It is important that the lengths are unequal: one is longer and one is shorter. Any ideas what this symbol represents? Power supply. I like to call it a battery, because this particular symbol is meant for voltage sources—in fact, a DC constant-voltage source. If we can change the voltage, there will be some different symbol.
8:14
So this is what we would call a battery, and the longer end—that’s supposed to be the positive end. It’s the higher-voltage end, and the shorter end is the negative end, or the lower-voltage end. The property of a battery is that the ideal battery would provide some constant voltage. So the voltage difference from the low end to the high end—this voltage difference—is supposed to be whatever the voltage of the battery is. It provides a voltage difference
8:55
of whatever the voltage of the battery is. That’s the property of a battery. So that’s it. Those are the circuit elements. What’s represented here is a circuit that you can imagine building, now that you know what each of these elements represents. You have two batteries here, somehow separated, and you have three resistors this way. And you can imagine connecting this battery to this end with a wire, take another wire connected to a resistor here,
9:33
and make all these other connections with however many wires you need. Yeah. So, when we have a circuit like this, then I want first to describe what we mean when we say “solve a circuit,” because it’s like—anybody here into game theory or computer science, artificial intelligence—what does it mean when people say tic-tac-toe is a solved game, like actually solved the game? What does that mean? "Finding the best possible answer for the manner of winning it."
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Yeah. So, in other words, given the rules of the game, tic-tac-toe is a very simple game: you know exactly what the best moves are in every situation. On the other hand, chess and Go—they are not a solved game, as in computers can be programmed to play it, but it’s just… yeah. So here I want to describe what it means to solve a circuit. When you are given a circuit, some of the parameters will be given, like in a board game like tic-tac-toe or chess—some of the things will
10:51
be given to you; the rules of the game are given to you. Here what I want to say is: when you have that circuit, you might be given these. Usually, the typical values that will be given to you will be the voltages. This will be given, because when you’re building a circuit you know what battery you are using, so some voltages will be given. And typically the resistances will be given, because
11:20
this marking on the resistor—there are supposed to be rules on reading it so that you can figure out what resistance this is, or I usually measure it; I don’t bother reading it. So these resistances will usually be given, and a description of the setup that gives all these values will have completely determined this setup. And what it means to solve this circuit is for us to figure out what other things are there in the circuit for us to know about, might be interested in.
11:58
For example, we might be interested in how long these batteries are going to last, and if we want to know that, we’ll have to figure out how quickly the energy is draining from these batteries. Or we might want to know how much these are heating up. There’s going to be one particular parameter that we are going to want to know to answer any other questions we might have about this circuit. What is that one physical observable,
12:30
or something that we can measure, that we would want to be able to figure out from this setup? There’s one very particular type of quantity that we would like to know. Yeah. Current. So, having determined all the voltages—actually, maybe not all the voltages; I know the voltage difference here and here, but I don’t know the voltage here, right? Okay, all right. But having determined the resistances, it turns out current is the one information that,
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if we have it about this circuit, then any other questions we have about the circuit we can answer.
13:16
So here in this circuit there can be three different currents. One is already labeled there—and actually, I think if you read the problem, this is the given, but whatever. So here’s one current; let me label this I₁. And there’s going to be another set of current that’s flowing through this branch; let me label that I₂. And there might be another set of current that’s flowing through this branch here; let me label that I₃.
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So this is how I want to describe solving a circuit. When you have solved a circuit, you will know all these parameters, so that now when someone asks any question about power dissipation or what is the voltage here at that point, it’s now a single-step process to just answer that based on your knowledge of the voltages and currents of this circuit. It’s a very general problem-solving approach. That’s why I’m trying to talk in general terms.
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And what we are going to do now is introduce tools for doing exactly that—introduce tools for, if you are given all the voltages and resistances, figuring out the current. Or it can be the other way around; in this question, it’s actually voltage and this current that’s given, and you’re supposed to figure out these other things. And I want you to be reminded of the Standard Strategy in Physics 4A. In Physics 4A,
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it didn’t almost matter what kind of problem you’re dealing with; if we dealt with the forces, then you started out with these set of steps. It’s a very similar approach we are taking to circuits. It doesn’t matter what kind of circuit it is—if it’s a circuit problem, these exact same set of steps that we are going through will help you come up with a set of equations which you can then solve to completely determine everything to know about this.
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Okay. So, the tool that we are going to use—it’s a set of two rules. It’s called—so, I just found out
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during the summer break that I’ve been misspelling this name forever. There are two H’s. And I’ve been also mispronouncing the name probably forever. I’ve been saying “Kirchov,” but I think it’s supposed to be—it’s a German name—I think it’s supposed to be "Kirch-hoff"; is that right? Whatever. I’ll try to say it right. However you pronounce this name—Kirchhoff’s rules. These rules are the beginning point for solving any circuit, and it’s almost like Newton’s laws of circuit. Let
16:19
me write down this, but I want you to realize that when I have given you Kirchhoff’s rules, you will feel like I haven’t told you anything new—that’s the feeling I want you to have. So let me state Kirchhoff’s rules—two H’s. There are two rules, and I think usually people state the loop rule first, but I like to state the junction rule first, for a reason that I’ll explain. So this is what the junction rule says. It’s a rule regarding junctions. What do we mean by
17:11
“junction”? Well, it’s kind of what it sounds like. This is a junction—like an interstate junction. This is another junction. And to tell you the truth, I can treat this as a junction too, but this becomes a trivial junction. You’ll see when you see the rule. So this rule involves junctions, and it involves a current involved with this junction. What the junction rule says is: let’s say you have—let’s say you have—I’m just going to draw
17:48
a part of a circuit. Let’s say you have a part of a circuit where several wires meet at this point, and maybe there are some currents coming in and there are several different currents going out. What the junction rule says is: add up all these currents. Add up all the currents that you label as coming in. Add up all the currents you label as going out. Then, when you add up all those currents, current coming in is equal to current going out.
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Do you agree that this should be true? You would have guessed this, right? If I just asked you, what would be the relationship between the currents coming in and going out? This doesn’t surprise anyone, that current coming in equals current going out. Why is this true? Yeah—conservation of charge. And we talked about how it’s really hard to accumulate charges anywhere. So yeah, this is true.
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The value of Kirchhoff’s rules is that they highlight the very basic facts about circuits that you will be using. They’re not telling you anything you don’t already know, but what they’re telling you is that these are important facts for you to utilize in solving a circuit. Let me give you the loop rule. So the loop rule tells you—this involves a loop, as the name says.
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I guess let me ask you this question, just to help you start thinking about what loops are. How many loops do you see here in this one circuit? How many loops do you see?
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Okay, I hear “two.” Okay, let me go: how many people see only one loop? Okay, how many people see only two loops? Okay, how many people see up to three loops? Okay, how many people see four loops or more? Anyone see four loops or more? No one says four loops or more? All right, let me draw the three loops. Then some of you are thinking, yeah, there are at least three loops here—three sort of independent loops. So there’s one loop here. Imagine this is
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like a road, and you start at one point, you take a little journey, go around, and come back to the same point that you started. That’s one loop. And you can do the same thing here: start here, go around, and come back to the same point. That’s another loop. And the third loop that some of you are realizing is: I can start here and go around the entire circuit and then come back to the same point. That’s another loop that’s not like the other two loops.
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To tell you the truth, there should be an unlimited number of loops—not that it’s useful. But let me draw one other loop that, if you’re trying to be difficult, you could call a loop. Let’s see—I could start from here, go around this path, go down this way, come down here, refuse to end here, keep going, and then go around here, and then come back here. That’s a loop. I’m not saying that you should define loops that way.
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Here the maximum number of useful loops is three. But with loops, what I want to start you thinking in terms of is: really the only thing that’s required in a loop is that you’re starting and ending at the same point. Other than that—and you can jump around in the circuit—there’s nothing really restricting how you define your loops. So
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when we talk about the loop rule in Kirchhoff’s rules, that’s the kind of loop it’s referring to. You know what, let’s just be difficult, and let’s say we are talking about the orange loop that I drew, not any of the other loops that are here. I want to be able to make some kind of statement about these loops that I have drawn. I’m just going to erase these current marks because that’s easier. So let’s say we are going with the craziest loop I’ve drawn—the orange one—and I want to
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say something about it. One thing I can say, as I imagine going around this crazy loop, is I can think about voltage change. When you’re dealing with a circuit, there are always going to be only three things you’re interested in: voltage, resistance, and current. In fact—actually—two things: voltage and current, and resistance happens to be one of the ways voltage and current can be related. As we start introducing other circuit elements like a capacitor—actually,
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you have seen a capacitor—you will realize there are really two things that you are always interested in in the circuit: voltage and current. It’s a question of how you relate those two. So as I think about this loop—current sounds complicated, so I’m going to focus on the voltage.
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As I start at this point—this is my starting point on the loop—what I can look at is: what are my changes in voltage? How much does the voltage change as I go across this element? Everyone agrees voltage should change as I go across the battery. Okay. How much does voltage change as I go across this resistor? How much does voltage change as I go across here? How much does voltage change as I go across here? And so on—skip over here—and then how much
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does voltage change as I go across here, and then back across here, and then across here.
24:30
I’m going to imagine I add up all the changes of voltage as I go around the loop. What value do you think this should be? Zero. Why zero? It’s the correct answer. It’s the correct intuition, but I want you to be able to justify it. Here is the same—sorry, I didn’t catch the first half. “Voltage…” So “voltage out from the…” There’s no “voltage out.” I guess this is the intuition I want you to have. Voltage, as we talked about,
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relates to potential energy, right? What’s the one parameter that potential energy depends on? There’s one dynamic quantity that all potential energies depend on—gravitational potential energy, electrical potential energy—position. Yeah, potential energy is energy of position. So, when I had this thing going through this loop, the gravitational analog of that would be this: I take this mass and I’m going to take it on loops. Sometimes it’ll move sideways, sometimes it’ll
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move up and then down, and then some complicated path, completes a loop, comes back to the same point where it was. And I ask you: what is the total change in the potential energy of this mass? And you say “zero,” like Crystal says. Why is it zero? Position. It’s at the same position. So when you add up all those complicated changes, as long as you come back to the same point, all those changes are guaranteed to add up to zero. Sometimes people say
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potential energy is path-independent, and all that stuff. All of that comes down to: when you add up all those changes over a loop, it must add up to zero. So this is something else that you already kind of knew. I’m going through this long discussion to make sure that it makes intuitive sense, but the short version is that it should make intuitive sense. I’m not telling you anything new here.
27:07
These are the two rules that we use for solving any circuit. The rest of the class today will be that demonstration. I’m going to use these two rules—the fact that these are true—and Ohm’s law from time to time to derive some of the formulas that you might already know. How many people here know how to add resistors in series? How many people here know how to add them in parallel? Some of you. I haven’t given you those formulas yet;
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I’m actually just going to derive it in class today using these two rules.
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I’m going to use these two rules [... same line repeated ...] So let’s do that. Let me draw the two circuits for doing exactly that. So the first circuit that we’ll look at, using Kirchhoff’s rules, is resistors in series. This is actually the easier one. It kind of looks like this. I can have a battery here—I’m trying to draw it small so that I can use the rest of the space—battery here, resistor here,
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another resistor right after that, and then it comes back to the battery. So this is a circuit; there’s a path for current to flow, like this. Let’s say that this battery has some voltage of V₀, and the resistances of these resistors are some values R₁ and R₂.
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Here’s a question that someone could ask—the question that
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is a term you will see in your textbook also. There’s a phrase “equivalent resistance,” or, if you are looking at a really fancy textbook (I don’t think that one’s ours), they might even say the Thévenin equivalent resistance, which is—there’s some theorem proven by a guy named Thévenin. But it comes down to: what does the phrase “equivalent resistance” of these two resistors refer to? Any ideas what that might refer to?
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“Equality,” equivalent—like equivalent in what sense? Because if I have two resistors, how much resistance is there in the circuit, in the way they are arranged in the circuit? This is a black-box view that I want you to think about. Imagine that these two resistors are put inside the box—literally inside a black box—and you have no way of looking directly inside the box. All you can do is: you might be able to apply voltage, and you might be able to measure
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the current that’s going in. If you have this black-box imagined, then this is the “equivalent circuit” you can imagine building: you have a battery here, and inside this black box, instead of imagining some complicated arrangement with the two resistors, you imagine that maybe it’s just a single resistor that’s in there—that it’s only one resistor in there. And what you see outside, with the same voltage that somehow produces the same amount of current, will be consistent with
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the resistance of this resistor here being what we are going to call the equivalent resistance. So that’s the sense in which this is the equivalent resistance. Does that make sense? Yes? No? I feel like I’m going a long way to explain something that is actually simple. In fact, if I asked you, what is the equivalent resistance, what would you have said? Any guesses for the formula? R₁ + R₂. You just added them,
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right? And that turns out to be the correct answer. So when it’s in series, it’s pretty simple, but we’ll go through the rule to prove that that’s the case. Where it gets more interesting is when they are in parallel. So let me draw the picture here first, and then we’ll go through the mathematics for both of them together. When resistors are in parallel, this is what it looks like. I will draw them “parallel,” but
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we’ll talk about that a little bit later. I have two resistors in parallel lines hooked up to the same battery, but being in parallel lines actually is not what “parallel” means. We’ll go through the actual definition of parallel in a circuit. We are applying some voltage V₀, which results in some amount of current out of this battery, and the resistances have values R₁ and R₂. Someone could ask the same question again: what is the equivalent resistance?
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What do you think? Just taking a guess—what do you think the equivalent resistance of these two resistors together should be? Should the equivalent resistance also be R₁ + R₂,
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like before? No, it’s not. You have seen in your circuit-introduction lab that when you build the circuit this way versus this way with your two light bulbs, they actually behave differently, and this circuit is different from this. Let’s see if we can apply the definition we used here in a consistent way. I’m going to define equivalent resistance in this sense. These two resistors—I’m going to once again imagine that they are in a black box. I put a box around them,
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and the only access I have to these two resistors is one set of two leads to apply a voltage difference with. That’s all the access I have. So I imagine looking at the circuit with this black box here. I’m applying the same procedure I applied to that other circuit. What I can say is: this black box has two leads going into it, and I can apply some voltage to those two leads using a battery. When I do that,
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I measure some current going in. Inside the box, this is what I’m going to imagine: instead of there being some complicated arrangement of two resistors—either this or that—I’m going to say maybe there is only one resistor in there, and based on what I see I try to deduce what that resistance is. Whatever this resistance is, that is the equivalent resistance. Okay, so the formulas that you might know about adding resistors
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are formulas for trying to figure out the equivalent resistance in these two limited circumstances. Some of you might know the formulas already, but let’s derive it. One more preliminary item to take care of before we actually do that. I told you before that to solve a circuit means to figure out all the currents in the different branches of the circuit, because once you know that, you can answer any other question that people ask you. I
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didn’t say “to solve a circuit” means to figure out equivalent resistances, because sometimes there is no such thing as equivalent resistance. So here, if what I said was true, then I should be able to figure out the current, and from knowing the current I should be able to figure out the equivalent resistance. What do you think that final connection is? Once I figure out the current, how would I use that to figure out equivalent resistance?
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Okay, so I know the current and voltage. How does that relate to the equivalent resistance? Yeah, you relate it through Ohm’s law. So whatever single resistor there is, it must be consistent with this expression: the equivalent resistance is equal to whatever voltage I applied divided by the current that’s passing through. Good. In fact, it will be the same formula here: whatever that equivalent resistance is, it must be consistent with (voltage applied) over (the
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current). If that’s not the case, then this model of the black box is completely wrong. And the theorem that Thévenin proved is that, as long as your circuit consists only of resistors, no matter how complicated this is you can reduce it down to this. There’s one more part to that theorem, but we’ll go into that later.
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So we are going to try to find the current, and once we do that we’ll put it in here, and we’ll try to express this equivalent resistance in terms of these resistances alone. So let’s do the simple one first. Do you see any junctions here? No junctions, right? Okay, so that means as I try to apply Kirchhoff’s rules, I can forget about applying the junction rule because there’s no junction to use it on, and I’ll just use the loop rule. Right?
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So what I’m going to say is: when I go through this circuit in a loop, all the changes of voltage must add up to zero. All right, that seems simple enough. I need to define the loop first—I mean, there’s only one loop, but let me be careful in saying what that loop looks like. Let me start from this point here—it looks like a good starting point—and starting from here I’m going to go across the battery, keep going,
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go across the resistor, across the second resistor, and then return to the same point.
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Let me write that down—let me write down the loop rule with all those changes of voltage. Loop rule says that change of voltage, added up all around the loop, is zero. As I go across the battery, what’s my change in voltage—plus V or minus V? Does the voltage go up or go down as I go across this battery? Up, right? I’m going from the negative to the positive terminal. So it’s a +V₀.
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I keep going. Now, as I go across this resistor—it’s going to be given by Ohm’s law—but is it up or down? Down. The voltage will drop as you go over a resistor, because the charges lose energy. So: − I R₁. And for the second resistor it would be the same thing except with the resistance of the second resistor: − I R₂. Then I come back to the same point, so the loop rule says this must add up to zero. This is your system of one equation that you’re trying to solve for. How
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many unknowns do you have? Two? Wait—what are you counting as unknowns? Is V₀ unknown? No. Is I unknown? Current—yes, it is unknown; I don’t know the current. It’s not given. Oh, I see—you’re treating the resistances as unknown. Resistances have to be given, right? Otherwise… yeah. So you have to assume the resistances are given. If both resistances were not given, there would be no way to solve this. You can go from here to here, but there’s no way
41:36
to go backward. So here the only unknown is the current. Usually the situation is that voltage and the resistances are specified, because in a real physical situation that’s usually the case. So: one unknown, one equation—I can solve it. Let me solve it for current. I’ll
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skip the algebra steps. Does this look reasonable?
42:15
Okay—yeah. What you have to do is move this to the other side, factor out I, so you get (R₁ + R₂); divide it out, then you get this: I = V₀ / (R₁ + R₂). Okay. Now that I have the current, I can use this formula here. According to the formula that we guessed before we did all of this, the equivalent resistance is equal to the voltage divided by the current: R_eq = V / I = V₀ / I. Well, the V₀ cancels with I = V₀ / (R₁ + R₂), and you end up with—let me flip it back around:
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R_eq = R₁ + R₂. This is what you would have guessed, and that’s what applying Kirchhoff’s rules says. Good. Maybe this is not all that impressive—like we just took the long way to get something that you already knew. So let’s do this one. Unless you’ve been reading on it in the textbook,
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you might not know what the formula is. There is actually a way to kind of guess at it. One way to guess at the formula would be to use this relationship that we covered in Monday’s class: the resistance can be related to the resistivity by R = ρ L / A of the conducting thing. You could use this to try to guess at it, but I want to spend most of our time in the circuits, so let me use the circuit problem-solving tools—Kirchhoff’s
44:02
rules—to figure out what the equivalent resistance here is.
44:11
Do you see any junctions? How many junctions do you see? Two. You see two junctions. All right, let me label them. And, by the way, all I’m saying now is summarized in the handout that you see, but I want to give you some intuition for the advice I’m giving you in that handout. So I have two junctions. All right. I think I’m going to be using the junction rule then, right? I need to add up all the currents coming into a junction,
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add up all the currents going out of a junction, set them equal to each other. That would be one equation I get. And the junction rule applies to each junction. So there’s a junction rule that applies to this, and there’s a junction rule that applies to this.
45:02
Before I can write down my junction rules, anything I should do first? Yeah, I’m kind of hinting at it. I need to label my current. Here, this current was the one same current throughout the entire circuit—yes? Right? Here, can you say this current is the one same current throughout the entire circuit? No. Once you come across this junction, you face a choice: the current could go up or to the right,
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which means it splits off. There’s going to be a current that goes here and a current that goes here. Let me label them I₁ and I₂. And I’ll keep using the same symbol I. Good. Once you enter each branch, within the branch it’s the same current until it meets another junction. Within this branch it’s the same current until it meets another junction. All right. So that’s the first thing we had to do: label the currents in each of the branches so that we
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can write down this equation. Let me write down the junction rule. Since I see two junctions, let me use both of them. Let me label the junctions as junction 1 and junction 2. Let me write down the junction-rule equation for 1. I add up all the currents coming in—what are the currents coming in? Just I, at least the way I have it labeled. So I = I₁ + I₂.
46:50
All right, that’s one equation. Let me get myself another equation here—equation 2. By the way, how many unknowns do you see right now? Three. What are those three unknowns? Yeah, they’re all unknowns. I have unknowns in terms of other unknowns. So I’m looking for at least three equations, which is why I’m looking for as many equations as possible. In fact, this is the area where Portable TA and I differ. Portable TA will tell you:
47:28
using these two rules, come up with as many equations as possible. I don’t tell you that. I tell you the exactly correct number of equations—not one more, not one less. And this is really why. So let’s say I’m going for as many equations as possible, so I’m going to use all my junctions. I used junction 1; let me use junction 2. What are my currents coming in? I₁ + I₂. What are my currents going out? Yeah, the I is the
48:02
current that’s coming out here. Haven’t I seen this equation before? Yeah. So if you include both of these equations into your system, you have created a dependent system of equations—meaning, solving a dependent system gets way more complicated than solving an independent system.
48:22
So the advice I’m giving you in circuit solving (that’s also in the handout) is geared toward avoiding dependency. I’m going to do what I need to do to avoid introducing the second equation which doesn’t actually add anything. This is the advice in the handout: when you see the number of junctions in your circuit, the number of times you should use the junction rule is the total number minus one. Leave one
48:50
junction unused. That’s what you see here. When you have two junctions, the very last junction you use is guaranteed to give you an equation that’s already dependent on your system of equations. Here it’s easy to see they are the exact same equation. When you have more junctions, sometimes it won’t look like any of the other equations, but it will be dependent in some other way. So that’s why I want you to use the junction rule first—it’s
49:16
the easiest one to say, all right, if I have two junctions I can use only one of those; if there are three junctions I can use only two of those.
49:26
"Does it depend on which junction you use?" It doesn’t really matter. I could have used this junction and kept this, and got rid of this [... inaudible ...] Yeah. As long as you leave one unused, you’ll be guaranteed to have an independent system of equations. Independent, but not enough. I only have one equation. I need two more—at least two more. Where do you think I can get those two other equations?
49:54
Yeah, the loop rule. This is the rule that I haven’t used. Here’s the benefit of having used the junction rule first: how many times do you think I should use the loop rule? Twice. I need to use it exactly twice. I want three equations; I don’t want four; I don’t want two. So I need three total equations. Now that I’ve used the junction rule, I have the benefit of knowing I need to use the loop rule exactly this number of times.
50:24
So I’m going to try to find two loops here. The general advice I would give is: pick the loops in a way they overlap with each other as little as possible—as in, you would not do this. You wouldn’t say, “Okay, I pick this loop—call that loop 1—and for my second loop I’m going to pick the exact same loop.” Would anyone here do that? No, right—you know not to do that.
51:01
When the situation gets more complicated, the equivalent version is that you want to avoid overlaps as much as possible. The loops will always overlap a little bit. Let me pick this as loop 1, and for loop 2 let me pick this one—starting from this point, going around this way. These two loops overlap a little bit; they share the same branch here. Some overlap is fine—they’re going to have to overlap.
51:30
This is an illustration I could give: if, in addition to having picked loops 1 and 2, I pick a third loop that goes this way, it’s going to give you another dependent equation. Can you see why the third loop would give me a dependent equation? Look at how the third loop overlaps with the first two. Does it completely overlap with the first two? Right. If I imagine going over this loop—say this is my third loop—for every part of the branch I’m going through,
52:13
has it already been covered by the other loop? Yeah. So that’s what I mean by overlap. When you have a situation where the branch you’re covering with your loop has already been covered by another loop, then it’s going to give you a dependent equation. So the advice I give you is: define your loops in a way that minimizes overlap.
52:40
With these three loops I could have used loops 1 and 2, or 1 and 3, or 2 and 3—just not all three, because if I do that, that’s when there’s complete overlap. Okay, so let me not use loop 3; I’ll use loops 1 and 2. That begins to sound promising, right? I wanted exactly two equations, and there will be exactly two equations that will probably give me exactly what I was looking for. Let me relabel these loops so that I don’t confuse my numbering system. Let me label this 2,
53:16
because it’ll be giving me equation 2, and let me label this loop 3, since it’ll be giving me equation 3. And let me erase this other junction that I’m no longer using. All right—equation 2 comes from this loop. I’m going to go through the same thing I did before. I imagine starting from here, and as I go across each element I figure out the change in the voltage. For loop 2, I first go over the battery—that gives me a voltage change of +V₀—and
53:55
then I go across the resistor and my voltage drops—everyone agrees?—so +V₀ − I₂ R₂ = 0. All right, that seems fine. It didn’t introduce any new unknowns. Hopefully the last equation also looks fine. Loop 3. I start at this point. As I go across this resistor from right to left, does my voltage increase or decrease? How many people say “increase”? How many say “decrease”?
54:59
At some point we will summarize this in a sign convention, but here I want you to gain the intuition that as you go from here to here, voltage is increasing. In fact, that’s the only way you can be consistent with what you said here, because here you said going from here to here voltage was decreasing. So if you are going the other way, it must be increasing. In the sign convention that we are going to come up with, it’ll have to
55:24
do with the direction of the path you are going relative to the direction of current. When you are going with the current, then your voltage is decreasing; when you’re going against the current, then your voltage is increasing—for the resistor. It’s a resistor-specific rule. So as I go across this resistor for loop 3, it’s actually + I₂ R₂, and then as you go across here, this time with the current going across R₁, it’ll be − I₁ R₁ = 0. Good. I have
56:04
three equations, three unknowns. I should be able to solve it. Good. All right, let’s just go through it and do it. Now that it’s three equations, I’m not going to do the algebra in my head; I’ll go through it step by step. Whenever you do algebra, you should take some time to look at it—try to sketch out a path that’s going to be the quickest. I stare at this for a while. I see that this has a lot of unknowns; I see these two both deal with only one unknown at a time.
56:38
So if I solve these each for I₁ and I₂ and then plug them in here, then I’ll get an expression for I that’s going to be nice and clean—an expression for I that’s not in terms of any other unknown. Okay, so let me solve these each for their unknowns. That’s the easy one—I’m going to do that in my head. I₂ is—oh, wait, wait, wait. Sorry, I missed this. It’s okay. Now that I realize it, let me change the step a little bit. I’m going to solve 2 for I₂, plug that in here, and then
57:20
solve the whole thing for I₁. Then I’ll have I₁ to plug in here, and I₂ to plug in here. Yeah. So I₂ is simple: I₂ = V₀ / R₂.
57:37
I plug this in here to get the intermediate expression: (V₀ / R₂) × R₂—oh, wait, that’s just V₀—minus I₁ R₁ = 0.
58:01
So the R₂’s cancel out. I get a nice simple expression for I₁: I₁ = V₀ / R₁. All right, they all look simple.
58:20
I probably should have expected it, because this is actually a simple circuit—I’m not supposed to get a super-complicated result. So let’s plug this into equation 1 and get an expression for I. The current I is equal to I₁ + I₂, or I can factor out V₀ and get I = V₀(1/R₁ + 1/R₂). Good. Yeah. To find the equivalent resistance, I can plug this in here and solve—well, plug this in here. But let me re-express it in a way that’s
59:15
more convenient, because it keeps the expression simpler. Instead of trying to solve for the equivalent resistance R_eq, let me solve for its reciprocal, 1/R_eq.
59:32
Okay, so if I’m doing that, then what I’m writing down here is I/V₀. This makes it simpler because I’m going to avoid nested fractions this way. So I is just this: V₀(1/R₁ + 1/R₂),
59:56
divided by V₀. So the V₀’s cancel, and you get this result: 1/R_eq = 1/R₁ + 1/R₂.
1:00:15
So that’s the formula for adding resistors in parallel. In shorthand, sometimes we say: in parallel, resistors add in reciprocals. There are useful things we can talk about here, but this is the formula, and we just derived it. The most important thing to note is this—and you can numerically verify this—the equivalent resistance is guaranteed to be smaller than either of these two resistors. It’s always smaller than R₁ or R₂. Yeah.
1:00:58
So you saw the application of Kirchhoff’s rules in very simple cases, and we even derived these formulas. Truth is, there are a lot of problems you can solve with this, and I hope it’s intuitive for everyone how you would extend this in case you’re adding more than two resistors. If you’re adding three resistors in series, what do you think the modified version of this formula will be? Adding just one more term, right? What about the modified version
1:01:29
of this one if you’re adding three resistors in parallel? Add it here, in reciprocal again. Yeah. So these are pretty useful formulas. There are some homework questions that you can do using only this. What I want to point out is that there will be some questions where you cannot solve it using these alone. In fact, the circuit that you see here is one such example.
1:02:00
Let me draw a version of this—one that looks similar—on that board over there, and I will try to explain why it’s impossible for you to answer this question only by trying to add resistors in series and parallel and simplifying the circuit that way. This is what that circuit looks like. I have a resistor up here—I’m going to give them generic labels: R₁—and another resistor here, label this R₂—and another resistor here, label this R₃. And there are two batteries: let me call
1:02:52
this V₁, and I’m going to say there’s another battery here, V₂, and not give you the current. In this circuit, to solve this circuit would mean to find this current. There’s a current that’s flowing through this battery—for my own reasons, I’m going to call that current I₃. And there’s going to be a current that’s flowing through this battery—again, I’m going to call that I₂. And there’s going to be a current maybe flowing this way here—let me call that I₁.
1:03:37
So to solve this circuit would mean to find what each of these currents are in terms of the voltages and the resistances that are given to you. Good.
1:03:52
I keep telling you that you cannot do this using these formulas alone. Why not? I mean, couldn’t I add these resistors R₁ and R₂, simplify this upper half, and then simplify that with this R₃ again and do it that way? Let’s imagine that’s what you are trying to do—even though none of you seem to be willing to do that. Then, if that was R₁ and R₂, are they in series or parallel?
1:04:27
They might appear to be parallel. So I should go over the definition of what it means for two resistors to be in series, and what it means for them to be in parallel, because
1:04:40
the language we are using is rooted in what it looks like geometrically. Let me give you this picture and have you tell me if this represents resistors in series or parallel. Let’s say I have two resistors—they are arranged along parallel lines—but let’s say I add it to a battery this way. I have a battery here, and I add it to the battery using these lines. These wires—wire here connects here; this wire here connects here;
1:05:13
and this wire here brings it back down to here. That’s like what you’re doing in lab, right?
1:05:23
Why do you say it’s in series? "It’s series because there is no—if there is no consistent voltage for each resistor being…" I don’t want to define series that way, because you are looking at it—there will be situations where two resistors do not have the same voltage across, and it still won’t be in series. So “series” does not mean that. Yeah. I want to define series based on what it is, not what it’s not. How would you define this as a series circuit based on what it is?
1:05:58
One path for current to follow—you know, they are in series in the sense one comes first and then the other comes. In fact, that’s how we are always going to define “series”: for two circuit elements to be in series, when a current flows through one element, that same current must flow through the other element. Whenever that happens, we call that series. So in that sense, are R₁ and R₂ in series? Look carefully. Current that’s going through
1:06:31
R₂—must that flow through R₁? As in, can you say that I₂ must be equal to I₁? Look at it carefully. If you’re saying I₂ is equal to I₁, what are you saying about I₃?
1:06:54
Not equal? If you’re saying I₂ is equal to I₁, what is I₃? No, no, no—use the junction rule. I₂, I₃, and I₁ all meet together at this junction. This is I₂; this is I₁; this is I₃. Can you say that I₂ is equal to I₁ without knowing anything more? Why not? "I₃." Yeah—because of this junction being here, you cannot say I₂ is equal to I₁. I mean, there is a path for you to go from here to there, but that’s not the question being asked when we’re
1:07:36
saying “they are in series.” When we are saying they are in series, we are not saying that there is a path; we are saying that path is the only path. That’s what you see here: this path is the only path—that’s what we mean. Current that flows through here must—not “can”—must flow through the other element. Here, the current through here doesn’t “must” flow through R₁; it can flow through the other elements also. Or current that’s flowing through R₁ could
1:08:09
have come from elsewhere beyond here. So in that sense, R₁ and R₂ are not in series. Okay. So we’ve satisfied ourselves that they are not in series, and the same thing holds for any other pairs of resistors here—none of them are in series. Let me go over the definition of “parallel,” because I really want to emphasize this: “parallel” does not mean “not series,” and vice versa. “Series” does not mean “not parallel.”
1:08:44
They are mutually exclusive, but they don’t cover all possibilities. Once again, the same arrangement here. If I’m imagining connecting them in parallel, this is how I would do it—this is what you did in lab. You connected the battery to one set of resistors, and then you used another set of wires to connect this here. Now you would call this in parallel, right? So give me some definition of why you would—or would not—call this parallel.
1:09:25
While you’re thinking about that, let me draw an alternate version of this circuit. If I have a circuit diagram that’s drawn like this, would you say that these two resistors are in parallel?
1:09:39
In fact, let me ask you a different question: is this circuit equivalent to this circuit? Right? As in, you can imagine taking this, moving it around—without disconnecting anything. Maybe wires might stretch, but not disconnecting anything. You can imagine moving it around to make them look exactly the same. So, it’s kind of a question of topology. There’s a topology joke saying that if you’re a topologist,
1:10:09
coffee mugs and donuts look the same to you, because a donut has a hole in it, and a coffee mug—its handle—has a hole in it, and the other parts of the shape can be deformed without changing anything. It’s the same thing with circuits. When you deform some part of the circuit, as long as you are not messing up any of the junctions—as long as you are maintaining the same connections—you can deform the circuit to your heart’s content without changing anything.
1:10:38
So these two circuits are the same. If this is “parallel,” this must also be “parallel.” I want to have some definition of these things being parallel that applies to both of them.
1:10:52
How would you define resistors as parallel—or circuit elements as parallel? "[... inaudible ... ] Current?" Well, in this circuit you see current dividing into two, right? One more and I’m going to end up saying this is not in parallel with this. With parallel, I want you to pay attention to the voltage, because it comes down to this: when we look at current, there are just way too many possibilities. With series, remember, we nailed down one thing—that when you look at the current,
1:11:39
one same current passes through all the elements. I want to have something similar here, and if I try to do it in terms of current I have no chance of doing that ever. So what other thing remains? "Voltage stays the same throughout its travel through the circuit?" Okay—with voltage, I want to be careful in stating things about voltage. Voltage does not move. Current does describe flow—so I talk about current through a resistor. Voltage does not move;
1:12:09
voltage is measured at a single point. You might do this: you might say the voltage at this point is 0 V, and you might measure all the other voltages relative to this. So when I say two circuit elements are in parallel, I’m saying something about voltages at particular points—without describing anything as moving, because voltage has nothing to do with movement; it has to do with the values of voltage at particular points. No path is for you to go through. You
1:12:49
are talking about voltage here—so what would you say about the voltage of these elements? You can look at change in voltage, but I want to be more specific than that.
1:13:06
This is how I can describe it in a way that’s as simple as possible and true in every single instance you will see. Let me define points A, B, C, D—these are points at the ends of each of these elements. When I want to say these two are in parallel, this is what I mean:
1:13:32
V(A) = V(C) and V(B) = V(D). Good. In this particular case, what Kevin was saying also turns out to be true: if this is true, then the difference would be the same also. But I want to state it this way because when you get into semiconductor circuits—deal with op-amps—there’s a very strange way of making things that are parallel that are not actually connected. So when we say some things are parallel, it really means the voltages at the end points
1:14:13
are the same. That’s the definition of something being parallel. Here, the voltage at this end point is the same because it’s the same end point, and voltages at these two end points are the same because they’re connected by wire alone. Okay. Using that definition, R₁ and R₂—are they in parallel? As in, voltages at these two end points—is the voltage at this end point the same as this? Oh, I guess it is;
1:14:48
it’s connected by wire. Is the voltage at this end point the same as the voltage at that end point? Why not? Yeah—because of this battery here. In fact, that’s the whole reason I inserted a battery there. If you didn’t have this battery, you could have simplified this circuit using these rules: add these two as parallel, then the remaining is a series; add them as series, and you’re done. But because of this battery, all of that is ruined. These are not in series
1:15:19
because of this junction, and they’re also not in parallel because of this battery. So this is a good example of a general circuit that you might have to solve where your circuit elements are not in series or parallel with any other element. If all your problem-solving tool was this—something you learned in high school physics—then you’re kind of stuck here. There’s no tool you can use that’s based
1:15:50
on these two. But if you know the very basic circuit-solving tools, we can go back here, come up with a system of equations, and then solve the system of equations to find I₁, I₂, and I₃.
1:16:13
What I want to point out is that there will be some questions where you cannot solve it using these alone. In fact, the circuit that you see here is one such example. So to solve this circuit would mean to find what each of these currents are in terms of the voltages and the resistances that are given to you. This is a good example of a general circuit that you might have to solve where your circuit elements are not in series or parallel with any other element.
1:16:56
If all your problem-solving tool was this—something you might have learned in high school physics—then you’re kind of stuck here. There’s no tool that you can use that’s based on these two. But if you know the very basic circuit-solving tools, we can go back here, come up with a system of equations, and then solve the system of equations to find I₁, I₂, and I₃.
1:17:24
Now that I’ve convinced you that the methods you might have used in high school physics don’t work, let’s use the one that actually works—that we cover in college physics. Technically this is university physics, but I can imagine this being covered in college physics since it’s algebra.
1:17:45
All right, so we are going to use Kirchhoff’s rules. Which of the two rules do I usually use first—junction rule or loop rule? Junction rule, for the reasons I explained earlier. How many times can I use my junction rule here? I see two junctions. Can I use it twice? How many times? I’m going to be able to use it only once. So let me pick just the one to use. I’ll use this one. I’m going to use this junction to write down my junction rule.
1:18:19
Before I do that, how many unknowns do you think I’ll have that I need to solve for? Three. I mean, that’s labeled here—the currents that I labeled. That’s, by the way, one of the reasons to actually label the currents in each of the branches: that’ll tell you the number of unknowns before you write down any equation. So I’m going to need three equations total. All right, I got one taken care of. I need two more. Where am I going to get those two?
1:18:52
Yeah—I’m going to use the loop rule twice to come up with my total of three equations.
1:19:04
I can pick the same loops I did last time. Let me do that; I think there are some interesting possibilities here. The loops I’m going to pick are: loop involving this—starting from this point here, I’ll go around and come back here. This is going to be my loop number one. And my loop number two: let me start from this point here, just behind the battery, and then go around and come back to
1:19:38
this point—my loop number two. Let me label them by the equation number I’m going to be using: 2 and 3. You might want to verify that they avoid overlap as much as possible. They do overlap in one branch, but that’s it; each loop has parts that don’t overlap with the other. So this is my plan for coming up with one, two, three equations, which will hopefully allow me to solve for one, two, three unknowns.
1:20:17
Now that I have the plan, let me write down those equations and see if they satisfy my criteria. Equation 1 is based on the junction. What are my currents coming in? By the way, before I do that, let me ask you this question. Current I₁—I’m pretty sure that has to be the direction, because based on the batteries I’m pretty sure the voltage here is higher than the voltage here, right? So I₁ has to flow that way. With the currents I₂ and I₃,
1:21:00
is it guaranteed to be this direction? There’s a possibility that it might be different. You can imagine V₂ being very, very small; in that case, I₂ would actually flow this way.
1:21:17
So, should I worry about that? Should I try to figure out the actual direction of current before deciding on one, or can I label one direction and stick with it? Maybe you can label it and use the signs later? Yeah. With the direction of currents, that’s going to be true in every single case. I could have labeled I₁ going the other way, and as long as I follow a consistent set of rules, the result I get will be correct. When my final answer gives me
1:21:53
negative current, that will be my indication that I should flip the actual direction of the current. So here, even though I don’t really know the actual direction of I₂ and I₃, I labeled it some way, and I’m going to assume this is the correct direction and do the rest of the problem that way. So, for junction 1, what’s my current coming in? Just the I₁—it comes into the junction. So I₁ = I₂ + I₃.
1:22:29
All right, that seems like a good start.
1:22:33
Equation 2 comes from my first loop. As I go across my battery, I’m going to gain a voltage of +V₁. As I go across this resistor, does my voltage rise or drop? Drop—I’m going with the current, so it drops. So it’s − I₃R₃. I still have another half of the loop. From here I’m going to go across this resistor. Am I gaining or losing voltage? It looks like I’m losing voltage, but
1:23:31
that’s based on the direction of the current that I labeled, right? What if it’s the other way? Is that something for me to worry about? The answer is no. Let me write it down and I’ll tell you why. Here I would say: if the current is actually going this way, I say I gain the voltage of + I₂R₂. Now suppose I was wrong about the direction of the current—that it actually would have been a voltage
1:24:00
drop going from here to here—then what that would mean is my answer for I₂ would be negative, so this expression would still represent the voltage. When you’re trying to figure out the rise or drop of voltage, you worry about what direction of current you labeled, not what is the actual direction of current. The direction of current that you labeled overrules everything else, because you’re trying to be consistent with the direction that you picked.
1:24:34
My last element before I come back to complete my loop: as I go across this battery, am I gaining or losing voltage? Losing—I’m going from positive terminal to negative terminal. Now, what if my direction of current were to flip? Then should I say I’m actually gaining voltage? Yeah, still losing. So with the batteries, direction of current doesn’t matter—it’s the direction of your path that matters. That’s actually the sign convention.
1:25:05
The sign convention for resistors says: you lose voltage when you’re going with current; you gain voltage when you are going against current. The second sign convention is the one for a battery: you lose voltage when you are going from positive to negative terminal; you gain voltage when you go from negative to positive terminal—without regard to current direction. Those are the sign rules. They’re fairly intuitive once you understand them.
1:25:31
So this should be −V₁—sorry, −V₂—that completes my loop. So this is equal to zero. All right, let me finish loop 3. I’ll go through this more quickly: across the battery, +V₂ (from negative to positive); across the resistor with the current, − I₂R₂; across the resistor with the current again, − I₁R₁; come back to the same point: = 0.
1:26:03
All right, these are my three equations. Three unknowns—I should be able to solve it, right? Yeah.
1:26:15
Almost all the circuit problems you deal with will be a system of three equations (or more). That’s why I say the algebra is going to be more complicated than what you remember from Physics 4A. You should take time to come up with a plan when you see something like this. I have three equations, three unknowns. It looks complicated enough that I should have an actual plan before I do anything.
1:26:45
I look for one of two things. First: is there any equation that’s in terms of only one unknown? That’s the one I would handle first—and that’s what we did earlier. There was an equation in terms of only one unknown; I solved for it, and that was useful. Here, there’s nothing like that. Then I look for the second thing: is there an unknown that occurs a lot? I₂ occurs in every single one of the equations. So I want to get rid of I₂ as
1:27:25
soon as possible. I’m going to solve one of the three equations in terms of I₂ to eliminate it, then I’ll reduce my system down to two equations in two unknowns.
1:27:40
So let me do that. I’m going to solve the first equation in terms of I₂. Solve this for I₂: I₂ = I₁ − I₃.
1:27:56
And I’m going to plug this into both of these to get a new system of equations that has only two unknowns.
1:28:08
Plug it in here—oops, here. This is my new equation 2: It’ll be V₁ − I₃R₃, and I’m going to plug it in. Do I want to distribute it? Yeah, let me distribute as I do it. I’m going to plug it in here, so I get:
1:28:29
I₁R₂ − I₃R₂ − V₂ = 0.
1:28:40
Let me do some simplification in place. I have two like terms with the same I₃, right? So I am going to combine these two. That would be: − I₃(R₃ + R₂).
1:28:56
Everyone agree? Yes. So let me do that. Let me move this around: V₁ − I₃(R₂ + R₃) − I₁R₂ − V₂ = 0. [sign error fixed later]
1:29:19
Good. All right. So this is one of my equations—the new system of equations. I need to do the same thing for equation three. So my new equation 3 is going to be: V₂ − … all right, I have this I₂, I’m going to plug this in and do the substitution on the fly. In fact, let me plan this out ahead of time: when I plug in I₁ that’s going to give me a like term with this, right? So I’ll have − I₁R₂ − I₁R₁,
1:29:56
so I’m going to write it down from the beginning as − I₁(R₂ + R₁). Good. So I handled this. Let me handle I₃ there—so it’s “minus minus I₃,” so it’ll be
1:30:11
I₃R₂ = 0.
1:30:19
Good. All right. I have two equations, two unknowns, I₁ and I₃. So I solve for it. From here the choices are equivalent; you can do one or the other and they’d be equally complex. Let me just do one: I will solve this for I₃ to plug it in here. Question? "For your second equation—should it be + I₁R₂?" Sorry, are you looking at this— "Your second number two?" This one? Yeah. Should it be + I₁R₂? Oh, it was + I₂, but this I₁ that’s coming from… Yes, I think you’re right. Thank you.
1:31:12
Yeah. That’s why—thank you. It should be plus. All right, good—good that we caught that mistake now.
1:31:24
Okay. So I’m going to solve this for I₃. That means I can imagine moving this to the other side and divide out by R₂ + R₃. When I do that, this is what I’m going to get: I₃ = [ V₁ − V₂ + I₁R₂ ] / (R₂ + R₃). Yeah. Plug this in here. Then you end up with this equation—I’m just going to alternate between purple and black. So this equation will turn into: V₂ − I₁(R₂ + R₁) + ( V₁ − V₂ + I₁R₂ ) / (R₂ + R₃) × R₂ = 0.
1:32:49
All right, it’s just going to look complicated. Let me collect all the like terms. I think I can collect the like terms in terms of I₁ on the other side. When I do that, this is what I get:
1:33:18
V₂ + (V₁ − V₂) · [ R₂ / (R₂ + R₃) ] + … and the other terms I’m going to move to the other side, so it’ll be
1:33:42
I₁(R₁ + R₂) − I₁ · [ R₂² / (R₂ + R₃) ].
1:34:00
Good. I can actually simplify this a little bit, I think—yes. If I combine the denominators—make them share the same denominator R₂ + R₃—then I have to multiply this by R₂ + R₃. Believe it or not, that will actually simplify things a little bit. Let me write it out so that you can see what simplifies. When I write it out, this is going to be: I₁ / (R₂ + R₃) … I’m factoring out I₁ and (R₂ + R₃) to the front. And this is the rest of the remaining terms:
1:34:45
R₁·R₂, R₁·R₃, … R₂², plus R₂·R₃, … so what I’m saying simplifies is these two things cancel. Okay. Yes. So we can say all of this left-hand side is equal to I₁ / (R₂ + R₃) × [R₁R₂ + R₁R₃ + R₂R₃]. It’s actually, in some sense, simple—1, 2, 3—these three terms are all three possible pairs of R₁, R₂, and R₃. So it’s easier to remember that way: R₁R₂ + R₁R₃ + R₂R₃.
1:35:50
And to actually solve for I₁, I would have to divide both sides by this, and then that will be I₁. I’m kind of out of space, so I’m going to leave it there. With circuit problems, you already know how much I hate plugging in numbers, right? But at some point with circuit problems like this I do have to give you numbers, because the algebra gets horrendous with all these different parameters.
1:36:20
With this, you can see that I can solve for I₁ in terms of everything that’s known, right? In this equation the only unknown is I₁; if I have numerical values for everything else, I can solve for I₁, which means I can plug that I₁ into here to get I₃, and then I can plug I₁ and I₃ into here to get I₂. That’s how you solve for all the currents in this circuit.
1:36:52
So this is an example of a problem that can only be done using the general problem-solving technique—using Kirchhoff’s rules.
1:37:06
This is all that you need; the rest is something you figure out in the course of working out a particular problem. Let me pose you this hypothetical—which is the reason for this simulation being here. I’m not going to do this with a real battery; as I keep telling you, it’s dangerous. And when I do this with a power supply, it doesn’t quite work as a demonstration because of all the safety features built into the power supply that make them behave not like a battery.
1:37:44
What is the property of a battery in a circuit—what is the one thing that batteries do? “Provide power.” It does provide power, but you will find that’s not very useful in problem solving. The fact that a battery is like a power supply is useful when you’re describing things conceptually—like in Physics 10—but this is Physics 4B; I expect you to analyze a circuit. So what is the property of a battery in that sense? One thing a battery does is it provides a voltage;
1:38:22
it provides a voltage difference. Imagine you have a battery—what this battery does is provide a voltage difference between one end and the other end. Really, this is the one thing you use about the battery when you are solving a circuit problem. This fact comes into play when you are applying Kirchhoff’s rules. Good.
1:38:51
Let’s say I’m doing the thing I keep telling you not to do during lab, and I do this: I take a
1:39:02
battery and connect the two terminals—negative and positive—with a piece of wire. In other words, I have shorted the battery. If I short a battery, the question is: what happens? When you look at this circuit, it’s a kind of self-contradictory circuit. We said
1:39:42
the defining property of a battery is it provides a voltage difference—as in, if you look at points A and B, one thing you can say about these two points is that their voltages differ by this much, and that’s because this is a battery. What’s one property of wire that you also use when you’re solving circuits? It’s a conductor—yes—but when you analyze a circuit using Kirchhoff’s rules, the one property about the wire that you need to use over and over is:
1:40:28
voltage is constant throughout the whole wire. In other words, if you take, say, these two points C and D, the voltage difference between these two is equal to zero because they’re connected by wire. So that’s where this contradiction comes up. What is the voltage difference between A and B? One part of the circuit says that this should be their voltage difference. The other part of the circuit I’ve drawn says what? Zero—because they
1:41:10
are connected just by wire. Well, something’s got to be wrong here.
1:41:15
I say this is a contradiction, but it’s not as though you cannot build that circuit—you can actually build it. You can short a battery. I keep telling you not to do it, but nothing actually stops you from doing it. What do you think will happen when you actually build this circuit? Will the voltage difference between A and B be closer to nonzero—closer to this—or closer to zero? "Closer to zero." Closer to zero?—yeah.
1:41:43
Let me try that in the simulation. This is a reasonable simulation; I can build reasonable circuits here—as in, I can hook up a light bulb and it’ll light up. (It’s possible that… I think I was using resistors—oh, yeah, sorry. That’s why I couldn’t extend it—which just looks like wires.) So I could hook this up to a light bulb, but it’s probably not going to be quite bright. Let me do it with two batteries so that it’ll be brighter. Hook it up to
1:42:30
this—then it lights up the light bulb. What I’m asking is—what I’m going to try to do now is—so you can take it for given that if this light bulb lights up, it would be because… wait, can I—oh, I do have a voltmeter. All right, so I can actually use a voltmeter here. So
1:42:58
this is the positive end, negative end—okay, sorry, the other way around. So that’s the voltage. When I connect it to the light bulb, then I still get the same voltage. So this is what would happen in a real circuit. I’m doing it in simulation, not with a real circuit, for safety issues. In a real circuit, if I connect these two wires—oops,
1:43:30
why can’t I… all right—if I connected these two wires together, then this is what you would get. In fact, I think this is making it… sorry, I should have tried it before class. One thing that’s realistic here is that the batteries are “burning,” because that’s what you would see in a real circuit. And as the battery burns, you wouldn’t expect to measure 14.4 V here; it should actually be something closer to 0 V,
1:44:01
because as the current flows… So what I want to do now is—how we are modeling things here—the model is wrong. I have to modify something here to make this model more correct.
1:44:19
Here’s one thing I can do (that I won’t do this time, because that’s not what I’m trying to demonstrate). When we introduced circuits, the resistance of a wire—right? So instead of drawing a wire as zero resistance, I could modify my model so that wherever I’m drawing a wire, it’s actually a wire with some amount of resistance. I could express things that way—that would be one way to
1:44:51
correct this model. As in, here with this circuit, I could have done the thing that I did mistakenly; I can imagine that the wire I’m connecting to this light bulb with actually has some resistance. So it’s in series with some resistor. It lights up the light bulb, and these resistors actually slow things down a little. I can model things that way. With that model, everything would be correct—there
1:45:24
wouldn’t be any contradiction—because when I “short” this battery, then this resistance still makes things slow down; all the voltage drop happens over these resistors, and you wouldn’t
1:45:39
have this statement here, and everything will be consistent with how you would analyze a circuit. So that’s one way to look at it, and that’s how you saw it when we introduced the circuit with the power supply: it was able to provide a large amount of current—like 1 ampere—with a small amount of voltage. I want to look at things the other way. When you build the circuit in such a way that the resistances of the wires are negligible,
1:46:10
then how would we analyze that? Instead of fixing my model of this physical situation by acknowledging that wires have resistance, I want to fix it the other way. I want to acknowledge that my battery may not be the perfect battery. So this is what we call an ideal battery. An ideal battery will enforce this voltage difference no matter what happens. And this is the contradiction that you come across if
1:46:44
you are too rigid about that “no matter what happens,” because any physical battery you have will have imperfections—it’ll burn when connected that way. So how would you model the imperfections that any physical battery has to have? That’s the situation where the idea of internal resistance comes up. You will see that as you work through some of the homework questions. So, instead of—let’s say these points A and B represent actual terminals of a real battery:
1:47:23
B is the positive terminal and A is the negative terminal—when we try to model this realistically, instead of saying that inside a battery it’s an ideal battery, we acknowledge, all right, it’s a real thing; it’s not going to be ideal. Instead, we model it as an ideal battery in series with something that’s non-ideal. The part of the battery that’s non-ideal is that it does not have zero resistance; it has some quantity of internal resistance.
1:48:04
This is where the black-box picture is useful, because this internal resistance is not something that you can take out of the battery and measure. This is just how we model the characteristics of the battery. So when you look at it here—let’s say we are trying to get the maximum current possible out of this battery—how would you… if you have this battery (or this set of two batteries) and you’re trying to get maximum current out of it, how would you connect this battery so
1:48:39
that you get the maximum possible current? Let me show you an example of measuring some current. I could connect this battery to a light bulb here to get some current out of it, like the way you did in lab. Right? And this measures 1.8 ampere of current. How would I modify this circuit so that I draw more than 1.8 amperes out of this battery?
1:49:14
Oh—the length? So here the wires are actually ideal wires. Actual length of wire doesn’t matter, because it’s a matter of how things are being modeled. If I somehow make it so that I can connect it with a shorter wire, that won’t actually change anything, because the length of the wire doesn’t mean anything in this model. This still reads 1.8. So not “shorter wire,” but how would you connect this so that the amount of current is greater?
1:49:44
You have actually seen one in the lab—yeah, put more light bulbs in parallel. That’s another way I can do it. So let me do it this way. I’m going to be sure to measure current coming straight out of the battery, so let me put it this way so that I can read it more easily. I’m going to hook up this wire, and just one more light bulb—put them in parallel, meaning I have a connection coming
1:50:17
from here to here. So that’s one; that will give me 1.8 ampere again. And I have another connection coming from here to here, and when that’s done, it’ll draw double the current—3.6 ampere. I want you to imagine extending this to the most extreme possible. What is the arrangement of the circuit that would draw the maximum possible current?
1:50:46
So, when you connect more light bulbs in parallel, what’s the effect of doing that in terms of, let’s say, resistance of the circuit? When you add more light bulbs in parallel, what does that do to the resistance? Lower. You could say that’s why there’s more current here, according to Ohm’s law. Okay. What’s the lowest the resistance can be? Zero, right? How would you build a circuit with zero resistance?
1:51:23
Just with the wires, right? That circuit. So, let me do that. I would never do this in an actual physical circuit, but with this simulation, I can imagine: all right, the minimum amount of resistance possible—I would just connect it with the wires. I can do it this way. In this simulation, I think it has an unreasonable expectation for the battery. But when you actually connect something this way, you can measure the current, and you can measure essentially
1:51:52
what is the maximum current output of a particular battery. This would be another way of stating the imperfection of this battery, because ideally it would have been infinite—there would be no maximum limit to how much current it can provide, the way we say with the normal force there’s no such thing as a maximum to the normal force. But in reality there is: if I push down with enough force on this table, at some point I’ll break the table instead of there being enough normal force.
1:52:21
So with a battery, we can describe the imperfection of this battery either with the internal resistance or with some idea of the maximum current that the battery can provide. How do you think these two are related—the maximum current that the battery can provide and the internal resistance that we use as a model for expressing the imperfection of the battery? They are related in a pretty simple way—it’s not a trick question. They’re related through
1:53:02
Ohm’s law. So you could say that maximum current there—this is the actual circuit where maximum possible current out of this battery would flow (before it burns or explodes or whatever)—and this maximum current is given by Ohm’s law by V₀. This is what’s sometimes called open-circuit voltage—that’s the voltage you would measure if you had an open circuit (the circuit is broken somewhere, so no current flowing). So this maximum current—let me write it down that way:
1:53:38
I_max = (open-circuit voltage of the battery) / (internal resistance) = V₀ / r_int. So this is the idea behind internal resistance: it’s some resistance that’s internal, inherent to whatever device you are looking at, and it’s a way of modeling this actual physical device. And when you have modeled the battery this way—the physical battery this way—then even for this circuit you will see that it no longer has any self-contradiction.
1:54:14
The thing that we say about a battery—it no longer says that between the two physical terminals of the battery this must be the voltage. Instead, it says: between one
1:54:28
terminal and the other point (which is internal to the battery, and we can never access it), the rule about the battery says that voltage difference is V₀; but before you actually get the measurable voltage here, there can be additional voltage drop if there’s current flowing. Yeah. So that’s internal resistance. We spent more time on it than I intended, but you will see it in some of the homeworks. Once you recognize this idea, then in terms of applying it,
1:55:00
it’s actually pretty simple. Once you write in the internal resistance, you just analyze the circuit as you would analyze any other circuit.
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