Leslie Kaelbling: 6.01 Lecture 07

PROFESSOR KAELBLING: Ok, so welcome back from spring break. I'm so sure you're all sort of sorry to not still be on the beach or wherever it was that was fun. Let's see, if you remember, before we all took a week off, we have been looking at circuits.

And what was kind of odd about circuits, compared to the previous things that we've looked at-- that is to say, computer programs and the linear systems, is that they didn't have an easy characterization that really helped us think about how you could characterize what was going on with this part of the circuit in such a way that, if you attached it to something else, it wouldn't change your characterization. So what we saw in all the exercises that we've done about, for instance, attaching a load to resistor divider was that, when you put the load on there, it kind of changes what the divider's doing. It isn't really acting like a divider anymore.

So what we did in the lab in the week before you came, the week before you went away to break, rather, was to try to get back a certain kind of modularity by thinking about putting buffers in our circuits. So remember that a buffer is an op-amp wired up in the negative feedback configuration, then just with something attached to the input, something to attached to the output. And what that does is, it ensures that the voltages are the same on the two sides, but it doesn't let any current flow through. And so we used that idea to good effect to make those motor controllers last time.

And the op-amps, the buffers that we made with the op-amps, made it so that what was going on with one side of the motor didn't pull our voltage down, right? We could hold the voltage on one side of the motor, hold another voltage on the other side of the motor, and by having a voltage difference across the motor, we could be sure that it was going to turn in the way that we wanted it to go. OK.

So we've seen that op-amps are a way to get modularity in circuits. What we're going to do today is study a different way of characterizing circuits and see that there is a kind of modularity. It's just a kind of different way of thinking about things, again. So that's the plan for today. So right.

So last time we looked at putting buffers in circuits as a way to kind of get modularity. But we're going to think about something else. And in particular, we're going to think about building abstractions of circuits. Now we actually, on the first circuits lecture, we already talked about building abstractions. We talked about how, if you had two resistors in series, you could abstract that as a single resistor. So we're going to take that idea and apply it to bigger pieces of circuitry.

So a question you might ask is, as we change this resistor, how does that change some piece of behavior of this circuit? And so, actually, I'm going to skip this example. I don't love it. So let me just go on here. So what's interesting is, if you look at this piece of circuitry over here, not the part in the red box, but the piece of circuitry here, as you change the value of this resistor, and you look at the relationship between the current flowing through here and the voltage across here, well, you already know this, right?

There is a linear relationship. This resistor exerts a linear relationship between the current flowing through it, and the voltage across it, right? That's the V equals IR. That's Ohm's law. We've been doing this stuff with it all the time. And that means you can think of this relationship between IO and RO in this very simple case as being a linear relationship. So if you plot on the IV curve, you get something like that.

OK, so that works great for a resistor. And the question is, well, will it work for any kind of circuitry we want? And what we're going to see is that, if we take any piece of circuitry that just has linear elements in it, we're going to be able to describe how it works using a linear relationship.

So we're going to introduce, now, an abstraction. Remember when we did system functions, we had sort of systems as an abstraction. So now we're going to introduce an abstraction of a circuit. And this abstraction of a circuit is called a one-port. I know, it's got two wires. But two wires constitute a port. So it's a one-port. A voltage-controlled voltage source, right, so an op-amp, that's a two-port This is a one-port.

So our circuit elements-- resistors and current sources and voltage sources, we can think of those all as one-ports. And what we're going to be able to do, what we're going to study today is, how it is that we can take any combination of those guys and think of it also as a one-port.

So a one-port is some sort of a circuit element. It has two connections. There's a current that flows through it, and there's a voltage difference across it. And we're going to be able to characterize all one-ports using two parameters. There's a parameter V0, which is the voltage difference across it, and a parameter minus I0, the minus I0 is, again, it's a kind of a hereditary thing about how people describe these things. So there's this current, I0, and so we're going to be able to think about, if we knew those two numbers, this I0 and V0, and we'll see how that works out, if we knew these two numbers, they completely characterize the way that one of these one-ports works.

So what we're going to spend our time most of the time today on you see how we can take any complicated piece of circuitry, think of the relationship it exerts between the voltage across those two wires that are coming out and the current that goes through, and see how we can compute that line, given whatever's inside the box. OK, so one port, two terminals. Current coming, voltage across. So always remember that we can conventionally think of the current coming into the box, that's I, the voltage across in the same direction-- that's V.

OK, so now we might want to ask the question, well, when is this relationship a straight line? And so let's think about that. So first of all, we know that for each of our individual elements-- so if we look at just a plain old resistor, if we took the resistor, and we drew a box around and said, oh this resistor, I'm going to think of it as a one-port, right? It's got a current going through, and a voltage going across, and we can describe the relationship it exerts.

So this resistor constrains the relationship between the I and the V. It doesn't tell us what I has to be. It doesn't tell us what V has to be. But it says, if you tell me I, I could tell you V. If you tell me V, I can tell you I. So it's exerting a constraint. And the constraint is this straight line that goes through the origin.

For a voltage source, the constraint is that the voltage has to be V0. It's says, I don't care what the current it. Current, give me anything you want to, but the voltage has to be V0. And for a current source, I don't care about the voltage, that can be anything you want. But the current has to be minus I0.

And this is a funny convention. And we'll try to keep it straight from here on out. We think of the current as flowing into the one-port, but for whatever reason, when people draw the current source, they make it go pointing up with current I0. So I is equal to minus I0. I can't take responsibility for that. I refuse to take responsibility for that, but it's a convention, and we're going to go with it. OK, so current always points up.

OK, does this make sense? Right, so for each of our individual things, the relationship between the current and the voltage is a straight line. Do you buy that? OK. So now, what I asserted to you was that, for and crazy combination of these things, now, no matter what it is, and how we wire it up, the relationship between the current and the voltage is going to be a straight line.

And we're not going to do, like, a whole formal proof of that, but I'm going to kind of try to convince you buy a couple of examples. Oh, so first, let's think about this, though. This is a good example. So what's the relationship-- so here's a little piece of stuff. We've got a resistor in series with a voltage source. And if we drew a box around it like this, and we said, OK. We're going to think of this as a one-port, what is the relationship between I and V? So look at that and see if you can figure out the answer to that one.

OK, so put up one of the number of fingers corresponding to A, B, C, or D, which one do you think is the right answer? I see one, two-- I see three. I see one, two, three. I see one, two, three. I don't see any fours.

OK, so you we've got a little bit of dissent on this one. So let's think about it. So what we have is, a resistor, and a voltage source, like that. And of voltage source is 5 volts. And the resistors is two ohms. OK, so let's just temporarily call this 0. Right, so if that's 0 volts down here, what's the voltage at this node? 5, good.

And if we have a current, I, flowing into this thing, what's the voltage drop across this resistor? 2I. So what's the voltage here? What's the voltage here?

STUDENT: 5 plus 2I.

PROFESSOR: 5 plus 2I. So the voltage difference from the top to the bottom is 5 plus 2I. Do you buy that? So we get V is Equal to 5 plus 2I. And so which of these guys is 5 plus 2I?

Well, let's see. So another kind of useful thing to think about is, when the current is 0, the voltage is 5. Right, so on one of these things, when the current is 0, the voltage is 5. So it looks like we've got that point there, and a positive slope. So it looks to me like A. Yeah?

Does that makes sense? Ask me questions. That's why we're here. No, OK? So there you go. That make sense to everybody.

OK, so that was an example of two components in series. And what we saw was that we got out a line. And in fact, actually, let's just go to the series case because that's sort of an interesting case. So let's draw the pictures for these two guys, right?

So for this to ohm resistor, we have, if this is V and this is I, then when V is 0, I is 0, I is 0. That's pretty clear. When V is 1, what's I? Oops, excuse me. When V is 1, what's I? For a two ohm resistor. So for this guy. When V is 1, I is a 1/2. So we're going to get a line that looks like that, roughly. Right? Good with that? OK, good. Now, 5 volt voltage, 5 volt voltage source, V, I, that says V is 5, I don't care what I is, right? So a 5 volt voltage source has just got-- a line over there.

OK, so now, how can I think about what it would mean to take those two guys and put them in series? So if you think about it, what happens when you put things in series? When you put them in series, so here's a kind of a schematic picture putting two, now two generic one-ports in series. We're not really here making a commitment about what particular one-ports they are, then what do we have?

Well, we know that the current flowing through the whole box is the same as the current flowing through each of individual elements because we have them in series. And we know that the voltage across the whole box is the sum of the voltages across the two individual boxes. So that means that if we want to try to construct a description of how the voltage relates to the current, in fact, it's kind of like we're going to add these two things up. But it says that for the same place on the current plot, we're going to add V1 and V2. So it's as if we were going to take these two figures and add them together in that direction. We're going to add in the voltage direction. Make sense?

So in this case, it's like taking this line and this line and adding them up in this direction, which means, basically, taking this guy and displacing it over by five. So we're going to end up with something that looks like-- right? We just took that guy and pushed him over by five.

So that's kind of an informal argument that says, if you give me two boxes that are describable-- who's relationship between voltage and current is linear, and you put them in series, then, for the whole box, you can describe the relationship between the voltage and the current as being linear. Do you buy that? Yeah, maybe, OK?

So now let's do parallel. It's a similar idea. Oh, and I just did these two slides backwards. But it will be OK. OK, so in parallel, it's a similar idea.

Imagine that I take two generic one-port and I wire them up in parallel. Now what happens is that the voltage across the whole box, this VP, is the same as the voltage across the first component and the same as the voltage across the second component, right? So we have VP is equal to V1, which is equal to V2. All those voltage differences are the same.

But now, the currents add. So that the current [INAUDIBLE] goes through the whole box is the sum of the current through the first box and the current through the second box. So that means that if we can use the line to describe what's going on in box 1, and use a line to describe what's going on in a box 2, then, for any given point in the voltage curve, the current is going to be the sum of the two currents in the previous cases. And so these guys are going to add up. We could take this line, and this line, and add them up sort of vertically, and get the line that describes the behavior of the whole box. You OK with that? Does that make sense?

OK, so that was just kind of an informal proof that says that, now, so if, at least for a series in parallel combinations, if you start with any of our simple little components and you wire them up in series or in parallel, you're going to keep getting out things that are describable with lines. And in fact, if you wire them up in any kind of a network, ultimately, you're going to get out something that's describable with a line.

So this is kind of a good thing, right? Because it means that we could take a whole complicated circuit-- in fact, no matter how complicated you can make it-- as long as it's made of resistors, and voltage sources, and current sources, we can study it. We can figure out how it works. And we can say, OK. If we wanted to take that piece and make it a separate circuit, sell it on eBay, whatever, we can characterize what good it is, what it does for you, by understanding the relationship between the current going in and the voltage across these two points.

Now, of course, if I drew the box in a different place, if I picked a different place in the circuit to put my two probes, I would get a different VI relationship. So this is a way to characterize how this thing works from the perspective of those two probes that we're sticking into the circuit. OK, and what we know is that, wherever we stick our two probes into the circuit, we won't always know what the V is because that depends on how it's connected up. We won't always know what the I is. That depends on how it's connected up.

But we will this VI relationship, no matter what it's connected to. So that's the thing that stays the same. Not the V, not the I, but the VI relationship. It's a linear equation.

All right, so now here's the cool part. And I'm going to do it, let's see, in a slightly different way. Let's actually go back to this circuit. And let's do it on the board. So the first step for something like that is to figure out-- so I said that there's a linear relationship that describes was going on in that box. And what we want to do is figure out what that linear relationship is.

And that might seem like it's sort of a hard job, but, in fact, the idea is maybe not so hard, is that two points define a line, so to find the line, we only have to find two points. And then two easiest points to think about what are the intercepts. So if we can figure out the value of I when V is 0, and the value of V when I is 0, then we know the line. Do you buy that? OK, good. Elementary whatever it is.

So let's do this circuit. So let's take that one that's in the picture. It looks like this. And we're interested in the relationship between V and I. This is 3 ohms. This is 6 ohms.

So what's the linear relationship here? So we're going to ask ourselves two questions. So the first question is, when I is equal to 0, what's V? So that's the first question we're going to ask. So we're I is equal to 0, what's V? All right, so how can we ask the question when I is equal to 0.

Well, when this circuit's just sitting here, as it is, drawn on the board, what's I? 0. OK, on this wire, there's no current flowing. It's not connected to anything. It's just hanging out there. I is equal to 0. So cool? So we don't have to do anything, right? Here on the board is a picture of I equals 0.

OK, I equals 0, what's V? So when we say, what's V, we want to know what's the voltage difference between here and here. So are you going to tell me what V is? 90? I hear 90. Can I here another answer?

STUDENT: 60.

PROFESSOR: 60. 60's another answer. I think I like 60 better. Let's see why. What this look like to you, in terms of an idiom we know now in circuits? It's a voltage divider. We've got a voltage divider. It loses one third here and two thirds there. And we put 90 volts into it. We lose one third of the 90 volts here, and two thirds there. So that's going to be 60 volts. Do you buy that?

So for example over there, it's 60 volts. So we're going to call this answer V0, 60 volts. And there's a name for V when I equals 0. It's called the Open-Circuit Voltage. I'm sorry I'm going off road with respect to the slides. We made the slides, and then, last night, I thought the lecture through in a different way. So here we are.

OK, good, so 60 volts. Good? OK, so now what's the other easy question to ask? Well, the other easy question to ask is, when V is equal to 0, what's I? All right. When V is equal to 0, what's I? So V equals 0. What can I do to my circuit to make V equal to 0? There's a simple thing.

STUDENT: Attach a wire?

PROFESSOR: Attach a wire. The quickest way to make the voltage difference between two points in your circuit equal to 0 is to connect them together. So we're going to do that. We're just going to, temporarily, in our heads, we're going to ask the question, now what if I just wired my two ports together? I shouldn't call them ports-- my two probes together. What if I wired my two probes together. That would certainly make the voltage equal to 0.

OK, so let's think about that. If I put a wire here, and now we want to ask the question, what's the current? So if I wire these two things together, now there is definitely current flowing on this wire. So I is not 0 anymore. So we can ask the question, what's I?

OK, how do we think about that? So we have now, we have a circuit. The easiest way to think about it is, basically, once you put a wire in parallel with a resistor, that resistor's not playing the game anymore, right? Because all the current, it's got a choice between going this way and going this way. It's going to go this way.

So basically, we just have a voltage source and the 3 ohm resistor. And so what's the current? 30 amps. We've got 30 amps. We've got 30 amps going this way. So I, I's going to be minus 30. And because we do, we're going to traffic in a slightly different parameter called I0. And this parameter I0 is equal to minus I. And that's going to be equal to 30 amps.

So this quantity, the I when V is equal to 0, that's called the short-circuit current. So we've got open-circuit voltage, short circuit current. So if we know those two things, we can now draw our plot. We can draw the plot, and the plot is going to just have those two points on it.

So our two points are, let's see. When I is equal to 0, V is 60. So we have a point-- we're interested in VI. So we have 60, 0, and 0, minus 30. right? Those are the two points that we just found? Right? So when V is 60, I is 0. 0 minus 30, we get a relationship like that.

STUDENT: [? I'm not. ?]

PROFESSOR: Right. This is a confusion, and, in fact, I think that some of the slides, they're not wrong, but they might be slightly inconsistent. So here's the story. We were making these curves. We are always thinking about the one-port having I go this way, and V.

OK, so we've got I and V. Those are two, like, generic variables. And we're drawing these pictures, we're drawing them in terms of that I and that V. So that's the relationship that we're talking about constraining-- the relationship between this I and that V.

Now, when we we're describing what's going on in a particular box, right, now I'm going to go buy this particular box from Radio Shack. And I'm going to buy that other particular box from Radio Shack. They're going to come with a spec sheet. And they're going to say, oh look, I'm a particular box. And I have a V0 and an I0. That's for this particular box.

So this particular box, given it's V0 and I0, that constrains-- so V0, more generally, these two points, are V0, 0, and 0 minus I0. Those are particular points, V0 and I0 are constants on a graph whose variables are V and I. Is that OK?

So these are particular values for this particular box. And this is a generic graph that shows how I and V relate in general. So right, so when I go to Radio Shack and I buy this thing, so when I go to Radio Shack, and I buy this particular thing, it's going to say in its spec sheet, oh yeah, my V0 is 60, and my I0 is 30.

So now we know that if we have a circuit, any circuit in the whole world made up of resistors and voltage sources and current sources, we can ask these two questions. What's the open circuit voltage? What's the short circuit current? We can ask these two questions. We can find these two parameters that describe this particular circuit, the V0 and the I0, and then we know how to make that plot. So we now know the linear relationship between V and I. Does that make sense to everybody?

So knowing that, we can take another step. Because if we wanted to do is summarize the goings on in this box, we know that this is the way to summarize it. But we're not totally sure what to do with that. If we're going to go now and solve a more complicated circuit that has this box connected up to something else, knowing that it has this linear relationship, we're not sure how to put that into the NVCC method, or something.

So instead of, or in addition to, thinking about what's going on in this box as having this linear relationship between V and I, we can also think about what's going on in this box as just being a simpler circuit. Just as when we looked into resistors in series, and we said, oh, we can think of that as one restistor, what we're going to be able to do is, look at what the stuff inside this box and see it as either a combination of a resistor and a voltage source, or a combination of a resistors and a current source. So let's see how that goes.

So there's two equivalent idea. No, they're not equivalent ideas. Excuse me, there are two ideas about circuit equivalents, which are not the same. One is called a Thevenin equivalent, and the next we'll see, is called a Norton equivalent. So the Thevenin equivalent says that, basically, once you know the linear relationship between V and I, you can come up with a circuit that's equivalent to the one that you started with that is its equivalent in the sense that it has the same VI relationship.

And it has a resistor in series with a voltage source. OK so now, we can think about this as going in two steps. What we just saw was how to go from any circuit to this linear relationship. And then the next question is, if you know the linear relationship, how can you come up with a circuit? How can you come up with a circuit, in particular, that just has this simple form.

All right, so what we can find is that what we need is d source with voltage V0. And a resistor with resistance-- and in fact there's one more parameter, which we don't really need to write down because it's derivable from the other ones. But we'll often use that that R0 is equal to I0 over V0. Is that right? Yeah, it's the other way around. What am I doing?

V equals IR, that's one of those things, you know? I probably wrote this wrong in the mini-lecture, too. I'll have to look at that. OK, good. So if you could give me the V and the I, can find the R. So if you find those two, the open circuit voltage and the short circuit current, then what you can do is, you can make a new circuit that's equivalent to this one that's a lot simpler.

And it's just going to be a voltage source. So let's actually draw the equivalent circuit to that one over there. It's going to be voltage source. How many volts? [INAUDIBLE] How many volts are we going to have in the voltage source in the equivalent circuit? 60. So 60 volt voltage source and a resistor, which is V over I, which is two ohms. And this thing, from the perspective of those two lines coming in, is going to be indistinguishable from this one.

So that is the idea of a Thevenin equivalent. And so, if you have a big, complicated piece of circuitry, you can find those parameters and then make Thevenin equivalent circuit, and put that in instead, and use it to do some analysis. So we'll see an example of why that's a useful idea in a little bit. For right now, does it make sense? Yeah? Maybe, sorta?

OK, so then, you know, Thevenin had to do this. And then Norton came along. I actually don't know much about Thevenin and Norton, in particular, as people. But you could probably guess that anything you could do with a voltage source, you could do with a current source. You just have to do it in a slightly different way. So you could do a similar thing, called a Norton equivalent.

So again, for any VI relationship that's linear, you can make a very simple arrangement of a current source and a resistor. This time, the current source is in parallel with the resistor. And so the current source-- and what we're going to find is that, if they're in this arrangement, and the current source has current I0 going this way, and the resistor here is that same resistance that we had there-- it's the same resistor, which always surprises. It seems like maybe it should be something different. But it's the same resistor-- then it will also have the relationship, the same VI relationship as the circuit that we started with.

OK, so Thevenin equivalent is a resistor in series with a voltage source. The Norton equivalent is resistor in parallel with a current source. And you can figure out what the values of the sources are directly from knowing those two intercept parameters that describe the line that relates V and I. So now we'll go through some examples of this.

OK, now in the slides, this is where we do the thing that I did first. I just decided it was nicer to do it first. So let's look at this circuit. This gives us another example. OK, so here's a circuit. We have of a voltage source in parallel with a two ohm resistor, in series with a two ohm resistor up here. And so we want to find these points on the plot.

So if the current is zero-- this is basically the example, if the current is zero, then that just means letting these terminals hang out loose in the breeze. And we have to figure out what the voltage difference is. So the voltage difference between here and here, if that's a voltage source, the voltage difference between here and here has got to be one volt because that's what the voltage source does, right?

So the voltage difference V0 is going to be 1 volt in this circuit. Actually, it's interestingly different than the one we had. Yeah, OK. Everybody good with that? OK, so that gives us this point. It says V0 is a 1 volt. That's the intercept on this axis.

OK, now we ask what's the short circuit current. So we sneakily wire together these two probes. And we ask what's the current flowing through in this direction. OK, so how do we figure out the current? We can see that we have a loop here with a the two ohm resistor and a 1 volt voltage source. And so the current has to be half an amp. Now the half an amp going in this direction, because the voltage, this is positive terminal to the negative terminal, so I is going to be minus a half an amp. I0 is going to be half an amp. So that gives us this other point.

So now we know the line. We can compute R0. That's the voltage over the current. That's 2 ohms. So now we get our two equivalent circuits. Once we know those three parameters, we can make these equivalent circuits very easily. This is a 1 volt voltage source in series with a 2 ohm resistor. This is a half an amp current source in parallel with a 2 ohm resistor. These guys are equivalent to each other. So that means by putting probes in from the outside, you can't tell the difference, and you can't tell the difference between these guys and the original circuit. Good? OK.

Do we want to do one more? We already did this example, essentially. I don't think we want to do that one. Nah, we don't want to do it. OK, good. I'm going to skip this for now, too, OK.

Let's do this one in detail, though, because I like this example. OK, so why do we care about this Norton and Thevenin business, right? So one reason to care is just that it's kind of an interesting thing. It's kind of a mathematically and scientifically interesting thing that you could take this contraption, and no matter how complicated it is, see that the relationship it exerts is something linear. As why would it be useful, let's consider this example.

So imagine that we're trying to kind of design a big, complicated piece of circuitry, or it's already designed, and we're trying to analyze it. And here, maybe we want to ask the question, well, what happens? We're interested in the current that's flowing along here for some reason, and we want to know what happens if the switch is open versus closed? How does that change the current that's flowing here?

And we know well enough now that we could use our mad algebra skills or, even better, our Python program to just write down the circuit description in both cases and solve it. But that feels like kind of a lot of duplicated work. And early on, when we were doing computer programming, I said that if you're writing code twice, then you're doing something wrong because programmers are lazy. Well, circuit analysts are lazy, too. So if we're analyzing the same big chunk of the same wiring twice, that seems like kind of waste.

So if we're looking at this problem, what we can do is, we can say well, really this piece of the circuit, if we're trying to understand the current here, this piece of the circuit here is going to be the same in both cases. And so maybe I can just take it and abstract it away. Maybe I can simplify it down to something much simpler. Maybe I could put the Thevenin equivalent in there.

And then I can do that work once. And then I can explore the question of how changing the stuff on this side changes as I go along. And so that's a good reason for wanting to do things, by reducing something to an equivalent circuit.

So let's actually do this one out as an example. I have it in my notes. We'll do it over here. OK, that's let's do this circuit. So we've got 20 volts. You can watch me do it, or you could do it secretly all by yourselves. Find my mistakes. 4 ohms, 4 ohms.

OK, so let's just think about this much of the circuit for right now. That's the stuff on the left half. And imagine that we want to draw a box around it, kind of an abstraction box like that. And so now the question is, well, how can we summarize that? So what's the Thevenin equivalent, or the Norton equivalent, or just the once I find the V0 and the I0, I'll know.

So what's the equivalent of this circuit? How can we think about the relationship that it exerts on the current flowing in here and the voltage difference across the terminals? And so that's what we want to think about.

OK so let's see. What's the voltage? So remember, that we've got two questions to ask. We have to figure out the open circuit voltage and the closed circuit current.

So let's think about the open circuit voltage. So if we just leave this thing alone, we don't make any changes to it, and we put our probes here, what's the voltage difference going to be across those two terminals? So how do you want to think about that? I'm going to let you guys actually work this and then tell me.

So what's the voltage-- really, what we need to know, right-- it always confuses people, when we're doing this stuff, we often end up with resistors kind of hanging out in space, not really in part of a circuit. And the answer is that there's no voltage. There's not going to be any current flowing along here. Our premise is that I is equal to 0. So there's no current along here. So you could just ignore this guy. So what's the voltage?

Not too hard, right? What's the answer? 10? Why? Because it's a voltage divider again, right? 4 ohms, 4 ohms, 20 volts, 10 volts. So V0 is 10 volts. That good?

All right, so now, closed-circuit currents. Tiny bit trickier. So we come in and we temporarily weld these wires together, like that, temporarily welded together. And now, what we need to know is, the current, right, you can think of this, really, as the current going through this component. All right? That's the I that we're interested in. And I don't know, I approach this problem by finding the voltage at this node. And then the current's easy.

So I can we find the voltage at that node? We'll write the equation out. Let's see. We'll put that guy that way, this guy this way. And then we can write a KCL equation for that node. So let's see. And we know, we'll call this 20 volts. And we'll call that ground. We'll call that ground. So we know that 20, I'll call this V question mark.

OK, so 20 minus V question mark over 4, that's going in. And our guy, let's see, plus V question mark, no, the other way around, plus 0, oh, this is not the way I did it before. OK, that's OK. 0 minus V question mark over 2 Is equal to V question mark minus 0 over 4. Do you buy that?

So just again, just, this is practice. I took each of these guys and followed the currents in the direction that they're drawn, and said 20 minus V question mark over four. V question mark minus a zero over 4, zero minus V question mark over 2-- those are the relevant currents. And then they have to add up to 0, respecting the directions.

So we solve this, we get 20 minus V question mark minus 2 V question marks is equal to V question mark. And so 20 is equal to 4 V question mark. This is what I got, yeah. And so V question mark is equal to 5. So now we know that this is V bolts here, in this case. So what is the current? That's a two ohm resistor, so what's the closed-circuit current?

STUDENT: 2.5?

PROFESSOR: 2.5, good. So which direction? It's actually going to be minus 2.5. OK, so I is minus 2.5. I0 is 2.5.

So we know everything-- everything-- there is to know about the circuit from the perspective of what it will do when we connected it up to something else. Because again, these two numbers characterize that whole linear relationship between V and I. So they don't specify the V, the don't specify the I. But they say how they're going to relate to one another.

OK, and then the good news is that we can now make an equivalent circuit, right? So we can say, abstract totally away from that. And we can say, oh yeah. that big hairy thing over there, we can just pretend it's a 10 volt voltage source in series with a-- what size resistor? Let's see. It's going to be 4. All right? And so that means that we can do the analysis of the left hand half of this big, complicated circuit all by itself once. And then we don't have to mess with that stuff anymore. We have a very simple version.

And then we can think about what happens when we connect it to the stuff on the right hand side. And actually, we can use the same ideas with the stuff on the right hand side. When the switch is open, we can summarize this whole circuit on the right hand side as being a two ohm resistor. Because when the switch is open, this guy's not in our game.

When the switch is closed, we remember what we remember about resistors in parallel, right, that's 2 times 2 over 2 plus 2-- that gives us one. So when the switch is close, this is a 1 ohm resistor. And so now we have two very simple circuit analysis problems. So closing the circuit makes this resistance go down, which increases the current. So now we can answer the question that we started with. Any questions about this?

STUDENT: Do we know that's it's better to use Thevenin or Norton?

PROFESSOR: When is it better to use Thevenin or Norton? It's really a question of what you're more comfortable with. You can look at the circuit that you're going to plug it into. And if you're happier with a current source in there-- maybe if you maybe few more things in parallel, the current sources are easier to deal with. If you have more things in series, maybe voltage sources are easier to deal with. It's completely a matter of your convenience. Other questions?

OK, so what we have got there now is, two ideas. We can summarize a complicated circuit with this linear relationship. And in terms of analysis, you can go and find a chunk of stuff and simplify. You could simplify because you're going to use it twice, or, maybe, sometimes just simplifying it helps you see what to do with the rest of the circuit. So that's another reason that you might want to do this.

OK, there's one more cool property-- so the reason that we can do this is because of, right, what we really exploited was linearity here. Because each of the individual elements had the property of exerting a linear relationship between voltage and current, we're able to see that, when we put them together in combinations, we still got something that exerts of this linear relationship.

And we can take advantage of that again using another idea called superposition. Now this isn't really an abstraction method. It's not a way to think about a piece of the circuit by itself. But it's still a way to kind of decompose this circuit into a simpler circuits. It's just a very different notion of a simpler circuit.

And the story goes like this. It says, again, if your circuit only have linear parts, so resistors, or current, or voltage sources, then any quantity that you're interested in, so any question you might want to ask about the circuit, like what's the current here? What's the voltage difference somewhere? Any quantity that you want to know, you can compute in a step-wise manner, considering only one of the sources at a time.

So I know from the questions on Piazza that circuits with multiple sources in them freak people out. There's just all this, oh my gosh, I don't know what to do because there's a current this way, and a current that way. And how can there be two currents, and oh, I don't know. And it is. It's sort of-- all your intuition that you kind of built up about circuits, at least for most of us, who are not super duper circuit gurus-- for most of us, when there's one source, you can kind of some intuition about what's going on there. When there's multiple sources, it's really hard to know, it's hard to guess which way stuff is going.

And although that shouldn't matter if you can just put your head down and do the algebra, your intuition is usually gone at that point. So the good news about superposition is that it lets us analyze a whole bunch of circuits, each of which has only one source in it. And if you take your circuit and consider one source at a time, by deleting the other two sources, and I'm going to say, very carefully, what it means to delete a source because that's a little bit of a tricky thing, then you can ask the question, what's the current here with only the voltage source, with the current source deleted? And then, what's the current here with only the current source and this voltage source deleted? And if you can answer that question in both of those cases, and add them together, that's the answer.

OK, so that's kind of very powerful, sort of, maybe, counter intuitive, but very powerful. So again, obliterate one source and ask the question. Obliterate the other source and ask a question. And then add the answers together. So that's the idea of superposition.

OK, so here's really an important thing. And then we'll do a couple of examples that have numbers in them. Example that's first here has just Rs in it, and that adds an extra kind of fun. OK, but it's so, an important thing to remember is, what does it mean to delete an element? Because it's different between current sources and voltage sources.

Deleting a current source means leaving that circuit open, leaving a gap. Because if you think about it, what can I do to my circuit to make there be no current source? Or to be no current here? Well, that is to leave a gap in the wire. So deleting a current source means just taking out and leaving an empty hole.

Deleting a voltage source, over here, deleting a voltage source, that's putting a wire in. Because think about it, to delete a voltage source is to say, oh, you know what, there's not really a voltage difference between these two places. I lied. So to say there's not really a voltage difference is to put in a wire.

So getting rid of a voltage source means putting in a wire. Getting rid of a current source means putting in a gap. So you have to be sure that, when you're taking these guys out, you take them out in the appropriate way. [COUGH] Excuse me.

OK, so let's do this Check Yourself problem. And then I have a slightly more complicated one to do for fun. And then I think we'll probably have this idea. OK, and we'll just go through this together. So actually, this is a familiar problem. We've looked at it a couple times already. So now we're going to look at it in terms of superposition. And so the question is, what's the voltage difference here?

And so now, instead of writing out the node equations and solving them, we're going to solve two circuit problems that are probably going to be easier. And then we're going to add the answers together. Oh, OK. It's not worked out in here in the kind of detail we wanted. All right, I'll just do it on the board.

All right, so we have like that. And it's 1 amp, and 1 volt. OK, so 1 volt, 1 amp, and 1 ohm. And we're interested in knowing the voltage difference here. OK, so let's start considering just the voltage source. So we're going to start considering just the voltage source. We're going to delete this guy.

What does it mean to delete this guy? Tell me what do when I get there. OK, I'm keeping my voltage source. I'm keeping my resistor. And what do I do about this part of the circuit? I leave it empty, right? So I'm just draw the gap there, just to kind of help us see that. But I just leave it empty. And so I might as well not have those guys there. And I ask the question, what's the voltage difference between here and here in this circuit. What is it? 1 volt, OK.