12. LCR Circuits—DC Voltage
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Shankar, Ramamurti
Like capacitors, inductors act as energy storage devices in circuits. The relationship between voltage, inductance and current in a variety of circuits with DC voltages is described.
 I left you guys thinking about inductors last time, so I should start there. So an inductor is very different from the resistor when you start doing circuit theory, because when you have a resistor and you connect some voltage to it, the current is determined by this equation. This is what I call an algebraic equation. It's an equation algebra, one unknown. You solve for it by dividing and you get I at time t, as V at time t divided by r. And you can make the network more complicated, put few more resistors, put few more there, few more here resistors within resistors. It doesn't matter, because we can always, in the end, combine the ones in series, then lump them with the stuff in parallel, till the whole thing to the right of this is one single effective resistor. When you find the current coming out of the source as V divided by that guy, then every time it comes to a branch, we sort of know how to divide it. So resistor circuits are very easy. Now when you bring inductors, things are different, so I'm just going to tell you a little more about them. So here is an inductor. This is a symbol for an inductor. It's got some inductance L measured in Henry's. And the implication of that is the following. If you have a current going through this inductor, that is going to be necessarily a voltage with this voltage, then the polarity of the current is increasing. And the voltage you need is LdIdt. This is the bottom line. Even if you skip all the other stuff about conservative forces, this, that, this is something you need to know how to do problems. The difference you notice is that the relation between voltage and current is not an algebraic equation, but a differential equation. This is called a differential equation. Now you don't have to worry because I will tell you how to solve it, because I don't know if it's prerequisite or not for this course. But I once had a kid in summer school who, when I started using partial derivatives, got very upset and said, he's not a prerequisite. Why are you using partial derivatives? So I told him that when you come to class, you always have to learn something new. You just have to learn to live with that. So I'm not going to leave you in the cold. If I do something you've not done, I will tell you how to do it, but I cannot avoid using certain notions. So this one, even though it's called a differential equation and so on, if v is a constant, we can easily find the integral. I will be v over L times t. Okay, but now we are going to be interested in cases where v is a constant, a constant. It could be varying with time in an arbitrary way. So you take this guy, the first thing you notice is that the relation between current and voltage is given by derivatives. The second thing is, when a current flows through a resistor, whatever energy you provide is gone in the form of heat. It is dissipated, the light bulb glows and that's the end. With an inductor, when you drive a current, you're building a magnetic field inside the inductor and that's an energy associated with the magnetic field and that stored energy will be given back to you later on. So it's like a capacitor. It takes a lot of work to charge a capacitor because you've got to take charges from this plate and keep on piling them in that plate. It's a lot of work. But then if you connect it to a bulb and you squeeze the trigger in your camera, this charge is giving you the energy back. So the two things that you see in the inductor and capacitor are different from the resistor in the sense in two ways. One is the voltage across the capacitor is the charge on the capacitor divided by C, which if you like is the integral of the current up to that time divided by C. Again, calculus comes in. So these are the two differences. The relation between current and voltage involved is the same and these are energy storing circuit elements compared to this one where energy is dissipated. So let's start with a simple problem where I have a voltage source. I'm going to take a fixed voltage V0. There is this which, there is a resistor R, and there is this inductor L. And I'm going to assume current is flowing this way, in which case I'm sorry. This picture is wrong. What's wrong with this picture? What is wrong? Switch is open. Okay? That's what's wrong. So nothing is flowing yet. But when I close the switch, then it's fine. So, well, now it's flowing. And the question is what does it do? How does it flow? So, overall, the inductor is a resistance-free wire. So you may think current equal to V0 over R will start flowing immediately, but that's wrong. That is wrong because if such a current were flowing, there'll be an energy one-half Li squared with that current in it, and that energy cannot come from 0 to all of that in no time. There's no way to do that. Or if you like, if the current really flows rapidly or instantaneously from 0 to some value, that's a current, DID is infinite here. DID is infinite, LDID is infinite. There's an infinite voltage needed somewhere in the circuit, and we don't have it. So that will not happen. So current will have to be continuous, start from 0, and it'll have to start climbing. The minute the current starts climbing up, it starts flowing down. There is a resistance times current voltage drop across this. Therefore, the amount of voltage available to drive current through this one, or to increase the current through this one, is reduced from V0 to V0 minus Ri. That's equal to LDID. So you can see as the current increases, the motivation to increase the current decreases because this resistor starts swallowing up more and more of this voltage. So the growth in current will be very, very slow, and we expect that after a very long time, it'll settle down at some value. What is that value? We can find that value by saying, wait till the current has stopped growing. That means DID is 0. That current will be V0 over R, and I'm going to give it the name I sub infinity, because it will turn out that you will reach that value only after infinite time. But you can get very, very close to that by waiting some reasonable amount of time, and part of what we want to do is, how long should I wait? Suppose I'm settling for 90% or maximum. When does that happen? You say that for 1 second or 10 second or 10 hours? What's going to decide that? Well, once you have the right equations, you can figure everything out. So we're going to do that. Basically, we are going to solve the equation LdI dt. So let me show you one more time how it is done, this equation. You should know how to write it. Start anywhere you like. Go around this guy. You gain voltage V0. Then go around this guy. You drop by no more minus Ri. You drop by no more Ri. And here you jump from this end to this end. You drop by no more minus Ri. That's a bad zone. That's a zone where the potential is not even defined. But I've told you multiple times that even though there are nonconservative things going on inside this, once you come outside, you just find a regular electrostatic potential difference equal to LdI dt. So you drop again, and you come back to where you start, the whole thing should add up to 0, which means the equation is LdI dt plus Ri equal to V0. That's what we want to solve. So when you have this thing, as I told you, solving differential equations a matter of guesswork. There is no algorithm. You rely on what you've seen before, and you guess the answer, you put some free parameters, you fiddle with them until everything works. And in this kind of problem, we can make everything work. Some problems we cannot solve. We can write the equations, but we cannot solve them. This is something you're not used to when you take an elementary course like this one. But there are situations. For example, inside the proton there are these quarks, and the quarks interact with each other with entities called gluons. They're called gluons because they glue the quarks and form the protons and neutrons. We know the equations governing the behavior of the gluons. We cannot solve them. It's really strange. You have an equation, but you cannot solve it. It's completely possible because certain equations, every equation has a solution, but it does not mean you can write it down analytically. So quite often, one may be on top of the right answer, but one may not know how to solve it, so one doesn't even know if the theory is right. So suppose this was the equation you got, and you could not solve it. It happens with the right equation. You don't know what the answer looks like. You can never be sure this is the right equation. Similarly, when Mr. Schrodinger invented his quantum mechanics, he wrote an equation that will tell you the energy levels of hydrogen, but luckily he was able to solve it. By solving it, he was able to show that his equation gives the energy levels in the spectra you expect from the hydrogen atom. So you have to remember that not every interesting equation can be solved, and I gave you the equation for these gluons which we can write down but cannot solve. There is no function whose derivative we cannot take. There are lots of functions whose integrals we cannot do. That's simply asymmetry between those two. Here's a very simple function. e to the minus x squared dx from 0 to 19, or some fixed number, 0 to some x max. It has a definite value. It's a formula under the graph, but no one knows how to write a formula for this in terms of x max, let us say in closed form. Okay, but anyway, this guy we can trample to depth as follows. First, we say at very, very long times the current I told you is V0 over R. That's when DIDT is 0. So we are going to write the actual current in our problem as the constant at infinity plus some difference I twiddle, the standard thing. You take out the part you know and solve for the rest of it. So let's take that assumed form and put it into the differential equation. So I get when I do LdI dt, this is a number V over R. It has no d by dt. So dI dt becomes dI twiddle dt plus R times I twiddle dt plus R times I infinity equals V0. Now you can see why we did that. We did that because now these two guys cancel and you get LdI dt plus Ri equal to 0. That means dI twiddle dt is minus R over L times I. So they're saying get me a function whose time derivative looks like the function itself up to a constant. That's something we know from high school. It's an exponential function. So I twiddle looks like some constant. I twiddle 0 e to the minus RT over L. You can take the derivative of this guy and out will come minus R over L times the function itself. And I0 twiddle is completely arbitrary. It's not determined by this equation. There's another useful property to know. This is a linear equation. It means if you multiply both sides by 19, you get another solution which is 19 times the old solution. So the solution's overall scale is not determined by the equation. If I give you one solution, you can multiply it by any number you like. It's also a solution because if you multiply by a number, say 9, the 9 goes into the derivative, 9 comes here. You can see 9 times I twiddle satisfies the same equation. We don't know this number just from the equation, but we go back to what we know, which is I of t is I infinity, which is V0 over R, plus I twiddle 0 e to the minus RT over L. But I know that at t equal to 0, this guy should vanish. T equal to 0, this guy should vanish because there was no current when I just threw the switch. So that tells me that is equal to 0. That tells me I twiddle 0 is equal to minus V0 over R, and we get our final result, which is I of t is equal to V0 over R times 1 minus e to the minus RT over L. So at this level, you are really doing physics the way we like to do it. So we have to study things, write down some equations and solve the equations. Then we get a very precise prediction on what will happen. Because now we don't have to guess when the current will come to 90% of its maximum value. We can pick any number you like because we can now plot this graph. Let's plot this graph, I of t versus t. At t equal to 0, e to the minus 0 is 1. You get 1 minus 1, you get nothing. At t equal to infinity, e to the minus infinity is 1. Yes? Student is asked, over on the right where it says I twiddle some twiddle, e to the minus e not over R, is that true? This one? Yes. Yes, right, thank you. Okay, so now this exponential which has a full strength of 1 at time equal to 0 goes to 0 at time equal to infinity. Therefore, it takes away exponentially small stuff at very large times. And if you want to know roughly how long should I wait, the answer is the time you should wait is roughly of order L over R, because you can write this as e to the minus r, e to the minus t over tau, where tau is L over R. Whenever you have an exponential, it'll always be e to the minus a physical quantity divided by some number that sets a scale for the physical quantity. Exponentials fall exponentially, we know that. But how long should I wait? If you wait t equal to tau seconds, then it'll be 1 over e, which is roughly 1 over 3rd of the starting value, and 1 minus that will be roughly 2 thirds of its value. If you wait t equal to 3 tau, e to the minus 3 is 1 over 20, then you will be what? 95%. So this is, you can calculate everything you need in the simple problem. Now let's come to the same problem. The switch has been closed for a long time and the current is stabilized. So here is the switch. It has been closed. That's my V0, that's r, and the current you remember is V0 over r. This is L. So now I want to do something else. So how do I want to open that switch? So what do you think will happen if you open that switch? Yep? Student is asking, when will this change in time? But how can it resist? Student is asking, by reducing the magnetic field. It can do what it wants, but how does it drive a current through the open circuit? Student is asking, from energy storage to magnetic storage, in what manner will the current flow, given that I've broken the circuit? No, no, but how will it manage to go around? Because it's like removing the bridge, right? How will it go around? I agree with everything you said, but how will it make the circuit? Yes? Student is asking, if you have a higher concentration of charges on one side of the way? Yes, but how will you really get rid of the current in a circuit that seems open? Have you ever done this? Yes, that's just fine. But we know it has to get rid of that current, because otherwise what happened to the energy? It has to go through. So you know what? You guys don't know what will happen if I pull that switch. Yes? Student is asking, is there a current in an electric field? Because the change in the magnetic field will create an electric field which doesn't need the closer... Right. No, I agree. It'll have an electric field. How will the electrons flow when the wire has been interrupted? That's all I'm asking you. Yep? Student is asking, will the current oscillate? No? Yes? Student is asking, do you have a static discharge? You will have a discharge. You will have an arc. You will have a flash, and that's when you will hear a zip, and then it'll jump this gas. So whenever you remove the switch, you have to be careful that it's not stored energy somewhere. You might think, hey, I'm pulling the plug on. Well, I won't say who. You're pulling the plug, okay? And what can it do to me? Well, it can zap you because this is a very dangerous place to pull the plug. So you know what people do? They have another resistor. Let's call this guy R1. This is R2, and R2 is a huge number, 10,000 ohms, let's say. Most of the time, R2 doesn't do anything, because when the current comes here, in any situation, it looks at the inductor and looks at 10,000 ohms, and it says, I'm going this way. So primarily, people all go that way. Okay? In the end, when the switch has been closed for a very long time in our old experiments, when the current is stabilized, you remember LdI dt is zero here, so there's zero volts between these two, that's the same zero volt across R2. No current will flow in R2. Everything will flow through L. That current, in fact, will be B0 over R, because this guy is being cut out of the loop. This is how it's going. But now, if you throw this switch open, you have given it a path to discharge. So the inductor will discharge through the resistor. The current will continue to flow in this direction. Of course, it cannot flow forever, and the resistor will burn up the energy. But you're giving it a path. You're avoiding. In a way, if you like, this is also a path with very, very high resistance. The air is a path with very high resistance. That means by and large, there are no free carriers in the air. But if the two tips get really charged, just like you said, there'll be an EMF, there'll pile up charges, they're jumping, they're waiting to jump the gap, and eventually they'll polarize the air, which is electrically neutral into plus and minus parts. Then the minus will go one way, the plus will go the other way, and eventually you'll recognize that as a discharge. But you don't have to worry about that now, because R2 will take up your current. So now the equation we write is LdI dt plus Ri is equal to zero, because if you start anywhere and you go around a loop, you have no change in anything. There's no voltage. And the answer to this one, R2. So I is equal to I0 e to the minus R2 over L times T. So now the current will decay exponentially. Again, with the time constant R2 over L, which is the one over the time constant, or the time constant, is L over R2. So you put L in Henry's and R in Ohm's, you'll get some time. And I'll give you an idea of how many times that time you have to wait before the inductor is completely discharged. Well, it's never going to be fully discharged. It's going to take forever. But if you say, look, one thousandth of the original current is safe enough, how long they have to wait, you put.001 here and see what time you get. So you want e to the minus T over Tau to be.001. So this is how the inductor... So every big inductor you have in a circuit you will find as God had resisted it in parallel with it to take up the energy it has. So let's do a little energy calculation. Energy calculation will be I had energy in the inductor in the beginning, which is 1 half L I0 square. What happened to it? Well, we know what happened. There's a current in the circuit, and when the current is discharged, it is dissipating heat in the resistor. Therefore, the power in the resistor will be I square R2. I square is I0 square e to the minus twice R2T over L times another R2. Now we integrate this power from 0 to infinity. You will get I0 square R2. So the integral of e to the minus something is e to the minus that divided by that infinity to 0. I can promise you that what you will get will be L over 2 R2. That'll give you then 1 over half, 1 over 2 L I0 square. So this is not a surprise that the stored energy in the inductor is driving a current. If you want one way to store energy, it's a capacitor, you charge it, and you discharge it when you want the current to flow. Or you can have an inductor in which a current has been set up. Then when the power shuts off in your house, somebody throws a switch, that inductor can keep the light bulb burning. In fact, forever. It'll be infinite time before the bulb shuts down, but you won't be able to see anything after a while, and this graph is exponentially falling. Okay? This is a very simple problem where we can do the math and we can understand what it's doing. Yep? Student's question This one here? This is called tau. You mean that symbol? Student's question Tau is a unit of time whose numerical values are 2 over L, and we can write I as I0 e to the minus t over tau. We always like to write it in this form so that if tau is equal to 100 seconds, you will have to make many multiples of 100 seconds before the current is negligible. Oh, in this graph here? This is time, and this is the current in the loop. It starts out at some initial value. Okay, now I'm going to do the next slightly more complicated circuit, and that is an LC circuit. If you have a circuit like that, it won't do anything. You can take these guys, hook them up, and wait all day. Nothing will happen because it's got no energy. Why should it do anything? It's just going to sit there. But if you are charged up your capacitor like that, and there's an open switch that you then closed, then these plus charges are going to leave and find their way around the other side and utilize them, and the capacitor will discharge. But you know that once it discharges, it's not the end of the story. So why is that? Why doesn't it stop when the capacitor is discharged? Yep? Student's question is kind of like the inductor has momentum and it keeps pulling. The inductor would have had a current by then. So the inductor is not going to suddenly stop having its current. So it's going to keep driving more current for a while until the capacitor is charged up the opposite way to an extent that this fights that one, that voltage, and then it'll go the other way. So it'll go back and forth. So let's write down the equation that tells you how it goes back and forth. So in other words, you should draw pictures. Now textbooks have nice pictures and I won't even try it. In the beginning it may look like, I'm sorry, in my example, in the beginning there's an electric field here, nothing in the capacitor. There's a quarter cycle, there's nothing here, there's a magnetic field in the inductor. Another quarter cycle, it's back here, but with the opposite polarity and it goes back and forth. So we can get all that by writing the equation LdI dt plus a q over c equal to 0. So let's write everything in terms of q. Then q dq dt is I, because as the current flows this way it builds up charge. That's my convention. So I can write it as Ld2q over dt squared over cq equal to 0. Now we have seen exactly this equation before. We saw the equation for a mass coupled to a spring, the equation md2x over dt squared plus kx equal to 0. There we know what it does. Now it oscillates back and forth. And L, this is an SAT question. L is to m, what 1 over c is to k. It's very useful to bear in your mind this analogy, because mathematically these equations, these equations have exactly the same solution. It's no more difficult to solve this one than that one. You may involve electric charges, that may involve masses. You don't care. So mathematically, here's an equation. Cow times d2 dog over dt squared plus, let's say, elephant times dog equal to 0, where dog is a function of time, has exactly the same solution. Because what does it matter what you call the unknown variables, right? But you've got to tell me that cow and elephant are time dependent, and dog is the only thing that depends on time, because otherwise the equation is not the same. So here m and k don't change, L and c don't change. So what's the answer to this one? Once again, it's a differential equation. You write dq dt squared is minus 1 over lcq, and you're asking yourself, give me a function which, when I differentiate twice, looks like itself. There's always the exponential. There's also sines in the cosines, so you can take any combination you like, and the one acceptable solution is some constant A times cosine square root of 1 over lc t. Right? Yes. If you take two derivatives of q, you'll pull 1 over root lc each time, then you will get a minus sign and the cosine becomes minus sign, and one more derivative will bring it back to minus cosine. So this is called the frequency, which is 1 over root lc. That's the analog of square root of k over m. Anyway, this will oscillate. It's a mechanical analog, this electrical analog of the mechanical problem. So I don't want to do too much of this, because I think you should be quite familiar with this, at least the oscillatory behavior of this. The analog is perfect. For example, to say that I pulled the mass by 1 meter is to say that x was given a non-zero value and dx dt was 0. Analogous thing here will be q was given a non-zero value and dq dt, which is i, is 0. That means the capacitor was charged, the inductor was not carrying any current, then I let it go. What happens? The problems are identical. Yep? Student is asked about the differential equation of a white element v of sin. Yes, so I should explain to you. That's an important point. So q of t can be one number cosine omega 0 t, omega 0 is this guy, plus another number sin omega 0 t, and both are acceptable. And the property of this differential equation is, if you have one answer cos omega t and another possible answer sin omega t, then any constant times cos plus any constant times sin is also a solution. You can verify that if you have one solution q1, a second solution q2, you can add the two equations and you can show q1 plus q2 obeys the same equation. Not only that, suppose there's one problem with q1, dq1 over dt square is minus omega nr square q1. That is some function of time. And here's another function of time obeying the same equation. Now you can multiply this by a number a, and you can multiply this by a number b. Take the a inside the derivative, b inside the derivative, and add the two sides. On the left-hand side, you will find the second side, the aq1 plus bq2. Right-hand side, you'll find omega nr square times aq1 plus bq2. That means aq1 plus bq2 obeys this same differential equation. This is called superposing two solutions. You've got one solution, another one. You can multiply one by a constant. It's very important it's a constant because that's what lets you take it inside the derivative. Therefore, the correct answer here is really a cos omega d plus b sin omega t. And you can ask, why did you write a cos omega nrt? First of all, that's not a good solution. I can put a phi there, which is arbitrary. And it's not hard to show. Let me call that c, if you like. C. This can be written as c cos omega nrt minus phi. In other words, it's possible to take either a cos omega nrt and a cos omega nrt. The solution is with a cos and a sin, with no extra phases inside, to another one with one amplitude and one phase. That phase is the delay in the cos sign. Anybody here has trouble who does not know the details? I'll be happy to explain why that's correct. You don't know, you should tell me now. You can check that if you take this thing, c, b and c, and an angle phi here, then a can be written as c cos phi and b can be written as c sin phi. And that just happens to be cos omega t minus phi, just for some trigonometry. So the differential equation, the second order in time, will always have two unknown parameters in it. So you can choose them to be these numbers a and b or this number c and phi. And the equation won't tell you what they are, because once you tell me there's a mass coupled to a spring, and I say, okay, this is spring constant, this is the mass, and I ask you, where is the mass right now? I don't know, because it depends on when you started it and how you released it. So I need to know what's called initial conditions, which is the value of q or the value of x at the initial time, and the value of the velocity or the current at an initial time. With those two pieces of information, I can solve for a and b. This I think you should know, this kind of stuff you should know, so I won't say too much about that. All right, so now we do another problem. We put an alternating voltage on these guys. This is V0 cosine omega t, and this is c and this is l. Now we ask, what's the current? This omega is not the natural frequency of oscillation. It is some externally given omega, like 60 Hz, from your power supply or from your socket. So that's driving the circuit. And you can ask, what happens now? The answer will be, you write the same equation. You can write l d2q over dt squared, plus 1 over cq, is equal to V0 cosine omega t. Now what are you going to do? You have to again guess the solution. So I want a function q of t, so that when I take two derivatives and add it to some multiple of itself, I get something, something times a cosine. So what should that function look like? Yes? No, no, what function q of t do you think will satisfy this equation? I'm asked for a function of time, which when I put into this has some chance of obeying the equation. So I want to get a function of time, which is a functional form. Yes? Student's question Right. In other words, I can take it to be some constant c times cosine omega t, period, not even times phi, because all right, yeah, take this one. See what happens now? When you take one derivative, you get minus omega l c cosine omega t. When you take another derivative, you get minus omega square l squared, I'm sorry, this becomes sin times cosine omega t, and this one is just, oh my God, this is my nightmare. You can see the nightmare now. How many people see the nightmare? Yes? Yes? So if I pick that c, it will be the same as this c, so we'll put a hat on this guy so we can tell them apart. So c twiddle over c times c twiddle is v0 cosine omega t. And what this tells you is, if you take two derivatives, I'm sorry, this is so sloppy, if you take two derivatives, you will get plus omega square cos omega t, also with a minus sign. Q over c. Okay, you guys buy that? There is no l squared, they're just an l. So c twiddle cos omega t is equal to v0 cos omega t. Right? Every derivative brings an omega and there's a net minus sign. So you find out that your guess c twiddle cos omega t is going to work, provided this equation is satisfied, and the cosine omega t can be cancelled. That's the whole point of doing this thing. So when you get c twiddle, it's equal to v0 minus v0 divided by omega square l minus 1 over c. Or if you like, minus v0 over l divided by omega square minus omega 0 square. Don't worry about the numerator, look at the denominator. The denominator says that if your driving frequency is equal to the resonant frequency, c twiddle blows up. That means when a system has got a resonant frequency, yeah? I took two derivatives. Oh, here. Oh, I'm sorry. Yes, thank you. You know, it's good to do these things slowly and not rely on what you heard somewhere. So let me do it for you again, I'm sorry about that. This time I take two derivatives. Each one brings an omega, there's an l, and the minus sign comes because the cosine becomes minus sign, differential, one more you get minus the cosine. This guy is nothing, he just says divide me by c. That's equal to that. And then I wrote 1 over lc is omega 0 square. Thank you very much. Okay? I was really hung up on the final result, which tells you you better not drive this at the resonant frequency, because then current amplitude will build up indefinitely. That's also true in a swing. If you've got a swing and the kid's coming back and forth, and you're reading a newspaper just pushing the kid, you've got to push at the right time to get the best result, and that kid's not going to fly off because there's one more term in the equation for the kid, which is friction. But if you have frictionless swing and you're doing this, you've got to watch out, because soon there'll be nobody who's going to be able to do this, because the amplitude will keep growing. Well, these are unrealistic problems, but I want you to notice one thing. Here's what I want you to notice. The voltage you are given look like v0 cosine omega t, and the current that you got, I can obtain by taking the derivative of this charge, which is like dq dt. This is the final sine omega t, and c twiddle, we can write as v0 over L divided by omega square minus omega 0 square sine omega t. This is I. Here's what I want you to notice. It looks a lot like Ohm's law, because the current looks like voltage divided by some number. That's a very big difference. The voltage is a cosine, and the current is a sine. That's something I want you to think about. The current is not in step with the voltage, whereas in a resistive circuit, the current follows the voltage. It's the same profile as the voltage, except you divide by r. Here, one is a cosine, one is a sine. That means when one guy is a maximum, other is a minimum. It's called outer phase. In fact, outer phase by 90 degrees. Yes? Student's question is, what's the proof by the sine that you got where the omega is? Here? Student's question is, so on the other side, together, we will find that proof by the sine where the omega is. Oh, omega did not go. It's back. You're right. That's the omega that came from. Okay, these problems, the only thing I'm focusing on right now, I would say, of all the things I wrote there, to be the most important thing, is number one. This is what I want you to bear in mind. I want you to understand why sometimes not paying attention to some constants. What is important to notice is that this is an equation you can solve by inspection. Why was it easier to solve it by inspection? Because you are trying to get in the end a cosine to balance the right-hand side. You're trying to find a function which is itself a cosine, or whose second derivative is a cosine. And we know the answer to that as a cosine, so we can guess the answer. Once you guess the answer, you put it in and you analyze it. You notice also that the current and the voltage are not in step. One is a cosine. You know what the cosine does? It does that. Other is a sine, so they are out of step. When one is a maximum, the other is a minimum, and so on. That means a current as a function of time is not equal to the voltage as a function of time divided by anything. That is nothing you can divide a cosine by to turn it into a sine. Whereas with resistors, you just divide by r, and you get a current. Now for the more realistic problem, the realistic problem has got a capacitor, an inductor, and a resistor, and no other power supply. The equation obeyed by this one will be L d t squared over Q over dt square, that stands for L dI dt, plus r dQ dt, plus Q over C equal to 0. This is analogous to m d2x over dt squared, plus I don't know how you guys wrote this in the last semester. It doesn't matter, gamma dx dt plus Kx equal to 0. Notice that it's the same form, and I'm going to give this to you as a homework problem to analyze the answer to this one. So this describes a problem where you have a mass and a spring and some friction on the table. That means if you pull it and let it go, the oscillations will eventually get dapped, and it'll die down. And the general solution for the case when it's oscillating will look like sum number A e to the minus sum number alpha t times cosine times some frequency omega prime t. Let me see. I had a phase phi, but I'm not going to do that. If you want, you can put an extra phi, but I will choose my origin of time so that I don't have that. This is what the answer is going to look like, where omega prime and alpha are going to be controlled by L, R and C. That's the thing I don't want to do in class. Have you seen this before? Professor Harris tells me you have seen this last time, so the hints in the homework will guide you on how to do this. You just assume the solution except t looks like sum A e to the minus, I don't want to call it alpha now, beta times t. You put it into the equation and solve for beta, and you'll find beta will have a real part and an imaginary part, and you have to combine the two to get this answer. That brings me to another thing, so I don't know how prepared you guys are for what's about to happen next, which is the use of complex numbers. Everybody familiar with complex numbers? Who doesn't know complex numbers? How do you do your taxes? You don't know imaginary numbers? I will give you a lightning review. I'm assuming everybody has seen them in high school. I will only tell you the part you need, but I'm going to assume that I can use them fluently. I don't want to stop every time and worry about you guys. So, I really need a show of hands. Anybody never seen x plus Iy and x minus Iy, and how about e to the i theta? You know that guy? Okay, that's all you need, so I'm going to tell you what the deal is. Everyone knows what a complex number is. So, if I is square root of minus 1, then we know that we can write a complex number as x plus Iy, and visualize it in the complex plane where you measure x this way and Iy is measured that way. That's usually called a generic complex number z. But now, let us find the length of that. The complex number is a single point in the complex plane. The length of that is square root of x squared plus y squared. So, I can also write z as x divided by square root of x squared plus y squared, plus I times y divided by square root of x squared plus y squared times square root of x squared plus y squared. Just nothing, just rearrange stuff. This is going to be called the modulus of z. What is this? This angle is θ here. This is really cosine θ plus I sin θ times modulus of z. And that, thanks to this great identity by Euler, is e to the i θ. Now, I don't know how much you know about this great formula that relates e to the i θ to cosine plus sin. How many people know where it comes from? What does it mean to raise e to a complex power? You know where it really comes from? Yep? Come from power series. Turns out you can write a power series for cosine. You can write a power series for sin. Then, for example, cosine of θ is equal to sin minus θ square over 2 factorial, θ over 4 factorial over 6 factorial, etc. And you can write a power series for e to the θ, if you like, which is 1 plus θ plus θ square over 2. These, this defines cosine θ in the sense that if you put θ equal to π over 2 in this infinite series, you will get 0. If you put θ equal to π, you will get π. So, in spite of this funny looking form, it really is the same guy. It will oscillate, it will have 0, it will be periodic. None of it is obvious. But this power series is numerically equal to this one if you keep the infinite number of terms. Likewise, e to the θ is defined by this series. And similarly, there's a formula for sin θ. Now, once a power series is defined, you can put e raised to anything you want there. So, you know what I'm going to put there. E raised to dog is 1 plus dog plus dog squared. This is not a joke. This is really true. This is the definition. Someone says, how do I raise e to the power dog? You do this. This will have all the properties of the exponential. It doesn't matter what's in the exponent, that's the key. People originally put real numbers, then they put complex numbers, now they put matrices, operators, anything you want. E raised to anything, you formally define to be this infinite series. So, you have the infinite series converges and gives you a meaningful answer. In that sense, if you put e to the i θ here and put, compare the result to cosine θ plus i sin θ, it matches. That's all. I'm just going to keep using that result. Then, you should also know that for every complex number, there's a complex conjugate which is x minus i y. That means i goes to minus i. That means z, this angle θ, will change to minus θ. And given any complex number, the real part of z, which is what, which is x, is z plus z star over 2, and the imaginary part of z, which is y, is z minus z star over 2i. So, this is what you need to know. Every complex number has a real part and an imaginary part. In fact, z star looks like this. And if you add z plus z star, the vertical parts cancel and you get double the horizontal part, that's why you divide by 2. If you subtract and divide by 2i, you get y. Basically, that's all I want you to know, I want you to be able to manipulate them rapidly enough. Here's something very useful about complex numbers. Let's take a complex number z1, which is mod z1 e to the i θ1, and another complex number z2, which is mod z2 e to the i θ2. So, what do they look like? Z1 looks like this, at some angle θ1. Z2 may look like that. It's got some angle θ2. So, every complex number has a length and an angle. You can think of it two ways. It's got an x part and a y part, namely real and imaginary, or it's got a modulus and a phase. That's the Cartesian version of the number and the polar version of the number. And the connection between them comes from Euler's formula. Now, look what happens when I take the product z1 z2. I get mod z1 mod z2 e to the i θ1 plus θ2, because exponentials add when you multiply them. Well, this allows you to immediately guess what the product is going to look like. The product is going to have a length equal to the length of z1 times the length of z2. It's going to have an angle. I've not drawn a good picture here, but I'm going to draw a good picture here. So, this is the sum of the two angles, because it's θ1 plus θ2. So, listen to this statement very carefully. When you take a complex number, say number 1, and you multiply it by a second complex number, you do two things to the first guy. You rescale it and you rotate it. So, two operations are done in one shot when you use complex numbers. You get a real number that's equal to the length of z1 plus θ2. So, you get a real number that's equal to the length of z1 plus θ2 plus θ2. It adds to θ1. Likewise, if you divide, if you take z1 over z2, it is mod z1 over mod z2 times e to the i θ1 minus θ2, because exponential θ2 is downstairs. So, multiplication by real numbers is very easy. You take a real number 4. You take a real number 5 here. With complex numbers, you take a number, you multiply another number, you rescale and you rotate. That's what I want you to know. That's going to be very important for what I'm saying. Okay, now I come to the problem I really want to solve, which is an LCR circuit driven by an alternating source, B0 cosine ωt. It's R. This is C. And this is L. The equation we have to solve is LdI dt plus Ri plus 1 over C integral I dt is equal to B0 cosine ωt, because Q is the integral up to time t of the current. That's just Q over C. It's the same equation. Now, unlike in this problem, where I had no driving voltage, I have a driving voltage, and I've also written it in terms of current rather than in terms of charge, because in electrical engineering, you don't really watch the charge in the capacitor. You look at the current flowing through the circuit. So, you have to solve this equation. The question is, can you guess the answer? That's the only way we know. You've got to guess. You've got to try to find a function I of t so that when you differentiate it, you get a cosine ωt. When you integrate it, you get a cosine ωt. So far, we can do it. Sin ωt will do it. But when you leave it alone, then also you get a cosine ωt. You cannot do that, because sin will become a cosine and cosine will become a sin. In the LC circuit, when you didn't have this guy, you were okay, because if you pick a sin, this became a cosine, and that also became a cosine, and you can combine them to get a cosine. But you cannot have dI dt, I, and integral of I all in one equation, and all satisfied by trigonometric function. The only time we can guess the answer is if it looks like v 0 e to the alpha t. Suppose that was the voltage. Then, can you guess the answer? What form the answer will have? It will have to be e to the alpha t times some number, because then this will look like e to the alpha t. That will look like e to the alpha t. That will look like e to the alpha t. The e to the alpha t has a great property that whether you multiply it, whether you integrate it, differentiate it, or leave it alone, it looks the same. Unfortunately, no one's interested in this voltage, because it's growing exponentially fast, or if you put a minus sign, it's dying exponentially. What we really want is this. So the question is, how do we solve this problem with cos omega t? So we're going to use a certain trick. The trick I'm going to use is the following. Let me take a general case where this is some function v of t. Then, but I'm not even assuming v is real. I'm not assuming i is real, but l and r and c are real. So something else can be complex numbers. Now, once you have an equation and you've found a solution, if you take the complex conjugate of both sides, they will still match. Do you understand that? If two things are equal, the complex conjugates are equal, because this thing is the real plus imaginary. That balances the real plus imaginary on the other side. Complex conjugate just reverses the imaginary terms, so they will still match when you flip both sides. So this means if you take l dI star of dt plus r times i star plus 1 over c integral i star of t prime dt prime of t equal to v star of t. So star is the rule by which if v had a real and imaginary part, you flip the imaginary part. So that solution for v of t implies a second equation. In other words, if the voltage v drives a current i, the conjugate of the voltage drives the conjugate of the current. That simply follows from the equation. It comes from the fact that when I conjugated it, l, r and c are real. Do you understand that? Suppose l were imaginary, or had an imaginary part. Then I must also put a conjugate on l, then i star obeys a different equation, because this had l in it, this had l star in it. So if everything is real, i star obeys the same equation when you have a v start driving it. I'm almost done now. I want you to add the left hand side, the left hand side, and the right hand side, the right hand side. And what do I get? Well, if I write it here, you may not be able to see it. Let me put it here then. If you add these two, you get l d by dt of v. So this is the equation. I plus i star plus r times i plus i star plus 1 over c integral of i plus i star dt is equal to v plus v star. This again comes because the equation is a linear equation. You can always add the left hand side and the right hand side to get another problem where the driving voltage is v plus i plus i star, etc. But i plus i star is 2 times the real part of i, and v plus v star is 2 times the real part of v. Therefore, this implies that l d by dt of the real part of i is r times the real part of i plus 1 over c times integral of the real part of i will be equal to the real part of v. So let me say in words what the equation means. If you, by luck, solve an equation with a complex potential v, and you get a current, then the answer to the problem where the potential is only the real part of the actual potential applied, the current will be the real part of the answer you got. That's all I want you to know. If you solve a complex problem, the driving voltage is complex, which is just a mathematical fiction, real life you don't have that, then the current that you get will also be complex, but its real part will be due to the real part of your voltage. And you can also show the imaginary part of the current will be due to the imaginary part of the voltage. This is just the principle of superposition. So here is the trick we are going to use. We are going to go back to the original equation, LdI dt plus Ri plus 1 over c integral I dt equal to v0 cos omega t. And we're going to solve a new problem. We're going to solve a problem where this is v0 e to the i omega t. And the current for that I'm going to call some i twiddle. It's a complex current because the driving voltage is a complex voltage. So think of v e to the i omega t as the v in the previous problem. The actual problem I was given had v0 cos omega t, but you know that v0 e to the i omega t is v0 cos omega t plus i times v0 sin omega t. Consequently, this potential has a real part, a cos sign, an imaginary part which is a sign, and the current will have a real and imaginary part. And the answer to our original question is the real part of the answer to this question. This is what you should think about this point, you should understand. So I've manufactured a problem with an unphysical, complex voltage, and all I know is that at the end of the day, if I take the real part of the current, I'll get the answer to the real part of the potential, which happens to be the actual potential. This is not cos omega t. So why would I do all this? Why would I take a problem that's bad enough with a cos sign and turn it into a complex exponential? I think you can sort of guess what the reason is. The reason is that it may be complex, but it is still an exponential. That means its derivative is going to look just the same. That means I can make a guess that the current i twiddle is proportional to e to the i omega t. So, you make the assumption that I twiddle of T is some constant I twiddle times e to the iωT. Make that assumption and put it in this equation and see what you get. You will find you'll get iωL times I twiddle plus R times I twiddle plus 1 over iωC times I twiddle times e to the iωT times V0 e to the iωT. This is because if you assume the current is an exponential, every derivative is equal to an iω and every integral is equal to 1 over iω and multiplying is just multiplying. So, d by dt is got replaced by iω and you get this equation for I twiddle. That means you will satisfy the equation provided I twiddle times all of this is equal to V twiddle. And I'm going to write this as Z times I twiddle, equal to V, where Z is the name for all the guys multiplying I twiddle. So, let's start with that here. I wrote IωL plus R plus 1 over iωC, I twiddle, e to the iωT is V0 e to the iωT. Everybody get that? That's because differentiating on an exponential is just multiplying by the exponent. So, this is a complex number, R plus i times ωL minus 1 over ωC. I want you to know that i in the bottom is like minus i in the top. And this is called the impedance, Z. It's a complex number. You can visualize the complex number as follows. It's got a real part which is R, got imaginary parts, I twiddle minus 1 over ωC, and the impedance Z is a complex number with some modulus and some phase. If you know the real and imaginary parts, you can construct the modulus in the phase like that. You want this, the modulus of Z. Therefore, canceling this, I find I twiddle is equal to V0 over Z. So, this becomes V0 over the modulus of Z times e to the iφ, where φ is its angle defined by tan φ equal to ωL minus 1 over ωC divided by R. So, don't worry too much about writing all of this down, because this you're going to find in every book. There's nothing novel or different. So, this is the formula for I twiddle. The formula for I itself, you remember, is equal to the I twiddle of 0. I'm sorry, this I twiddle of 0. Did I define it that way? Yeah, I should write it, I'm sorry, I should write it as I twiddle of 0, with a subscript 0 there, because the full I twiddle has got e to the iωt in it. So, this looks like V0 over mod Z times e to the iωt minus φ. But this is not the physical current, because it's the result of a complex voltage. The physical current is a real part of this, without the twiddle, that we write as real part of the I twiddle. Where do I get the real part? V0 and absolute value of Z are both real. The real part of e to the i something is cosine of something. And that's our answer, that's the final answer you get. This tells you that the current has a magnitude, which is V0 over the absolute value of Z, and it's got a phase φ, by which it is behind the driving voltage. The magnitude of Z, for as in any complex number, if someone says what's the magnitude of Z, it is simply square root of real part square, plus imaginary part square. So, I want you to think about the magic of complex numbers. So, the final answer has a current, which is a cos ωt minus φ. Your driving voltage is a cos ωt. There's no way you can divide the voltage by anything to get that current. But in the complex language, the complex current is the complex voltage divided by the complex impedance. How does that happen? That happens because the impedance Z has got a magnitude and a phase, therefore, if you had a voltage V0 e to the iωt, and you divide by mod Z e to the i φ, that becomes e to the minus i φ upstairs, you're able to change the magnitude and you're able to change the phase by dividing. So, if you want to take the voltage and you want to rescale it and shift its phase all in one shot, you can do it if you divide by a complex number, because a complex number will rescale this V and also the i φ in the same phase. And then you take the real part. In other words, cos ωt divided by nothing will give you cos ωt minus φ. But e to the iωt, when divided by e to the i φ, will in fact give you e to the ωt minus i φ. So that phase shift you can produce only by dividing the complex number. So, I'll come back and explain to you a little bit more on how you do circuit problems using the complex impedance. But I think it'll be helpful for those of you who don't know complex numbers to get used to this part of it.
 I left you guys thinking about inductors last time,  so I should start there.  So an inductor is very different from the resistor when  you start doing circuit theory, because when you have a  resistor and you connect some voltage to it,  the current is determined by this equation.  This is what I call an algebraic equation.  It's an equation algebra, one unknown.  You solve for it by dividing and you get I at time t,  as V at time t divided by r.  And you can make the network more complicated,  put few more resistors, put few more there,  few more here resistors within resistors.  It doesn't matter, because we can always,  in the end, combine the ones in series,  then lump them with the stuff in parallel,  till the whole thing to the right of this is one single  effective resistor.  When you find the current coming out of the source as V  divided by that guy, then every time it comes to a  branch, we sort of know how to divide it.  So resistor circuits are very easy.  Now when you bring inductors, things are different,  so I'm just going to tell you a little more about them.  So here is an inductor.  This is a symbol for an inductor.  It's got some inductance L measured in Henry's.  And the implication of that is the following.  If you have a current going through this inductor,  that is going to be necessarily a voltage with this  voltage, then the polarity of the current is increasing.  And the voltage you need is LdIdt.  This is the bottom line.  Even if you skip all the other stuff about conservative  forces, this, that, this is something you need to  know how to do problems.  The difference you notice is that the relation between  voltage and current is not an algebraic equation,  but a differential equation.  This is called a differential equation.  Now you don't have to worry because I will tell you how to  solve it, because I don't know if it's  prerequisite or not for this course.  But I once had a kid in summer school who,  when I started using partial derivatives,  got very upset and said, he's not a prerequisite.  Why are you using partial derivatives?  So I told him that when you come to class,  you always have to learn something new.  You just have to learn to live with that.  So I'm not going to leave you in the cold.  If I do something you've not done, I will tell you how to do  it, but I cannot avoid using certain notions.  So this one, even though it's called a differential  equation and so on, if v is a constant,  we can easily find the integral.  I will be v over L times t.  Okay, but now we are going to be interested in cases where v is  a constant, a constant.  It could be varying with time in an arbitrary way.  So you take this guy, the first thing you notice is that  the relation between current and voltage is given by  derivatives.  The second thing is, when a current flows through  a resistor, whatever energy you provide is gone in the form of  heat.  It is dissipated, the light bulb glows and that's the end.  With an inductor, when you drive a current,  you're building a magnetic field inside the inductor and  that's an energy associated with the magnetic field and that  stored energy will be given back to you later on.  So it's like a capacitor.  It takes a lot of work to charge a capacitor because you've  got to take charges from this plate and keep on piling them in  that plate.  It's a lot of work.  But then if you connect it to a bulb and you squeeze the  trigger in your camera, this charge is giving you the  energy back.  So the two things that you see in the inductor and capacitor  are different from the resistor in the sense in two ways.  One is the voltage across the capacitor is the charge on the  capacitor divided by C, which if you like is the  integral of the current up to that time divided by C.  Again, calculus comes in.  So these are the two differences.  The relation between current and voltage involved is the  same and these are energy storing circuit elements  compared to this one where energy is dissipated.  So let's start with a simple problem where I have a voltage  source.  I'm going to take a fixed voltage V0.  There is this which, there is a resistor R,  and there is this inductor L.  And I'm going to assume current is flowing this way,  in which case I'm sorry.  This picture is wrong.  What's wrong with this picture?  What is wrong?  Switch is open.  Okay?  That's what's wrong.  So nothing is flowing yet.  But when I close the switch, then it's fine.  So, well, now it's flowing.  And the question is what does it do?  How does it flow?  So, overall, the inductor is a resistance-free wire.  So you may think current equal to V0 over R will start  flowing immediately, but that's wrong.  That is wrong because if such a current were flowing,  there'll be an energy one-half Li squared with that current in  it, and that energy cannot come from 0 to all of that in no  time.  There's no way to do that.  Or if you like, if the current really  flows rapidly or instantaneously from 0 to some value,  that's a current, DID is infinite here.  DID is infinite, LDID is infinite.  There's an infinite voltage needed somewhere in the circuit,  and we don't have it.  So that will not happen.  So current will have to be continuous, start from 0,  and it'll have to start climbing.  The minute the current starts climbing up, it starts flowing  down. There is a resistance times current  voltage drop across this.  Therefore, the amount of voltage available to drive current  through this one, or to increase the current  through this one, is reduced from V0 to V0  minus Ri.  That's equal to LDID.  So you can see as the current increases,  the motivation to increase the current decreases because this  resistor starts swallowing up more and more of this voltage.  So the growth in current will be very, very slow,  and we expect that after a very long time,  it'll settle down at some value.  What is that value?  We can find that value by saying, wait till the current  has stopped growing.  That means DID is 0.  That current will be V0 over R, and I'm going to give it the  name I sub infinity, because it will turn out that  you will reach that value only after infinite time.  But you can get very, very close to that by waiting some  reasonable amount of time, and part of what we want to do  is, how long should I wait?  Suppose I'm settling for 90% or maximum.  When does that happen?  You say that for 1 second or 10 second or 10 hours?  What's going to decide that?  Well, once you have the right equations,  you can figure everything out.  So we're going to do that.  Basically, we are going to solve the equation LdI dt.  So let me show you one more time how it is done,  this equation.  You should know how to write it.  Start anywhere you like.  Go around this guy.  You gain voltage V0.  Then go around this guy.  You drop by no more minus Ri.  You drop by no more Ri.  And here you jump from this end to this end.  You drop by no more minus Ri.  That's a bad zone.  That's a zone where the potential is not even defined.  But I've told you multiple times that even though there are  nonconservative things going on inside this,  once you come outside, you just find a regular  electrostatic potential difference equal to LdI dt.  So you drop again, and you come back to where  you start, the whole thing should add up to 0,  which means the equation is LdI dt plus Ri equal to V0.  That's what we want to solve.  So when you have this thing, as I told you,  solving differential equations a matter of guesswork.  There is no algorithm.  You rely on what you've seen before, and you guess the answer,  you put some free parameters, you fiddle with them until  everything works.  And in this kind of problem, we can make everything work.  Some problems we cannot solve.  We can write the equations, but we cannot solve them.  This is something you're not used to when you take an  elementary course like this one.  But there are situations.  For example, inside the proton there are these quarks,  and the quarks interact with each other with entities called  gluons.  They're called gluons because they glue the quarks and form  the protons and neutrons.  We know the equations governing the behavior of the gluons.  We cannot solve them.  It's really strange.  You have an equation, but you cannot solve it.  It's completely possible because certain equations,  every equation has a solution, but it does not mean you can  write it down analytically.  So quite often, one may be on top of the right  answer, but one may not know how to solve it,  so one doesn't even know if the theory is right.  So suppose this was the equation you got,  and you could not solve it.  It happens with the right equation.  You don't know what the answer looks like.  You can never be sure this is the right equation.  Similarly, when Mr. Schrodinger invented his  quantum mechanics, he wrote an equation that will  tell you the energy levels of hydrogen,  but luckily he was able to solve it.  By solving it, he was able to show that his  equation gives the energy levels in the spectra you expect from  the hydrogen atom.  So you have to remember that not every interesting equation  can be solved, and I gave you the  equation for these gluons which we can write down but cannot  solve.  There is no function whose derivative we cannot take.  There are lots of functions whose integrals we cannot do.  That's simply asymmetry between those two.  Here's a very simple function.  e to the minus x squared dx from 0 to 19,  or some fixed number, 0 to some x max.  It has a definite value.  It's a formula under the graph, but no one knows how to write a  formula for this in terms of x max, let us say in closed form.  Okay, but anyway, this guy we can trample to  depth as follows.  First, we say at very, very long times the current I told  you is V0 over R.  That's when DIDT is 0.  So we are going to write the actual current in our problem as  the constant at infinity plus some difference I twiddle,  the standard thing.  You take out the part you know and solve for the rest of it.  So let's take that assumed form and put it into the  differential equation.  So I get when I do LdI dt, this is a number V over R.  It has no d by dt.  So dI dt becomes dI twiddle dt plus R times I twiddle  dt plus R times I infinity equals V0.  Now you can see why we did that.  We did that because now these two guys cancel and you get  LdI dt plus Ri equal to 0.  That means dI twiddle dt is minus R over L times I.  So they're saying get me a function whose time derivative  looks like the function itself up to a constant.  That's something we know from high school.  It's an exponential function.  So I twiddle looks like some constant.  I twiddle 0 e to the minus RT over L.  You can take the derivative of this guy and out will come  minus R over L times the function itself.  And I0 twiddle is completely arbitrary.  It's not determined by this equation.  There's another useful property to know.  This is a linear equation.  It means if you multiply both sides by 19,  you get another solution which is 19 times the old solution.  So the solution's overall scale is not determined by the  equation.  If I give you one solution, you can multiply it by any  number you like.  It's also a solution because if you multiply by a number,  say 9, the 9 goes into the derivative,  9 comes here.  You can see 9 times I twiddle satisfies the same equation.  We don't know this number just from the equation,  but we go back to what we know, which is I of t is I  infinity, which is V0 over R,  plus I twiddle 0 e to the minus RT over L.  But I know that at t equal to 0, this guy should vanish.  T equal to 0, this guy should vanish because there was no  current when I just threw the switch.  So that tells me that is equal to 0.  That tells me I twiddle 0 is equal to minus V0 over R,  and we get our final result, which is I of t is equal to  V0 over R times 1 minus e to the minus RT over L.  So at this level, you are really doing physics  the way we like to do it.  So we have to study things, write down some equations and  solve the equations.  Then we get a very precise prediction on what will happen.  Because now we don't have to guess when the current will come  to 90% of its maximum value.  We can pick any number you like because we can now plot this  graph.  Let's plot this graph, I of t versus t.  At t equal to 0, e to the minus 0 is 1.  You get 1 minus 1, you get nothing.  At t equal to infinity, e to the minus infinity is  1.  Yes?  Student is asked, over on the right where it says I  twiddle some twiddle, e to the minus e not over R,  is that true?  This one?  Yes.  Yes, right, thank you.  Okay, so now this exponential which has a full strength of 1  at time equal to 0 goes to 0 at time equal to infinity.  Therefore, it takes away exponentially small stuff at very  large times.  And if you want to know roughly how long should I wait,  the answer is the time you should wait is roughly of order L  over R, because you can write this as e to the minus r,  e to the minus t over tau, where tau is L over R.  Whenever you have an exponential, it'll always be e to the  minus a physical quantity divided by some number that sets  a scale for the physical quantity.  Exponentials fall exponentially, we know that.  But how long should I wait?  If you wait t equal to tau seconds, then it'll be 1 over  e, which is roughly 1 over 3rd of the starting value,  and 1 minus that will be roughly 2 thirds of its value.  If you wait t equal to 3 tau, e to the minus 3 is 1 over 20,  then you will be what?  95%.  So this is, you can calculate everything you need in the  simple problem.  Now let's come to the same problem.  The switch has been closed for a long time and the current is  stabilized.  So here is the switch.  It has been closed.  That's my V0, that's r, and the current you remember is  V0 over r.  This is L.  So now I want to do something else.  So how do I want to open that switch?  So what do you think will happen if you open that switch?  Yep?  Student is asking, when will this change in  time?  But how can it resist?  Student is asking, by reducing the magnetic field.  It can do what it wants, but how does it drive a current  through the open circuit?  Student is asking, from energy storage to magnetic  storage, in what manner will the current flow,  given that I've broken the circuit?  No, no, but how will it manage to go around?  Because it's like removing the bridge, right?  How will it go around?  I agree with everything you said, but how will it make the  circuit?  Yes?  Student is asking, if you have a higher concentration of  charges on one side of the way?  Yes, but how will you really get rid of the current in a  circuit that seems open?  Have you ever done this?  Yes, that's just fine.  But we know it has to get rid of that current,  because otherwise what happened to the energy?  It has to go through.  So you know what?  You guys don't know what will happen if I pull that switch.  Yes?  Student is asking, is there a current in an electric  field?  Because the change in the magnetic field will create an  electric field which doesn't need the closer...  Right.  No, I agree.  It'll have an electric field.  How will the electrons flow when the wire has been  interrupted?  That's all I'm asking you.  Yep?  Student is asking, will the current oscillate?  No?  Yes?  Student is asking, do you have a static discharge?  You will have a discharge.  You will have an arc.  You will have a flash, and that's when you will hear a  zip, and then it'll jump this gas.  So whenever you remove the switch, you have to be careful  that it's not stored energy somewhere.  You might think, hey, I'm pulling the plug on.  Well, I won't say who.  You're pulling the plug, okay?  And what can it do to me?  Well, it can zap you because this is a very dangerous place to  pull the plug.  So you know what people do?  They have another resistor.  Let's call this guy R1.  This is R2, and R2 is a huge number,  10,000 ohms, let's say.  Most of the time, R2 doesn't do anything,  because when the current comes here, in any situation,  it looks at the inductor and looks at 10,000 ohms,  and it says, I'm going this way.  So primarily, people all go that way.  Okay?  In the end, when the switch has been closed for a very long time  in our old experiments, when the current is stabilized,  you remember LdI dt is zero here, so there's zero volts between  these two, that's the same zero volt across R2.  No current will flow in R2.  Everything will flow through L.  That current, in fact, will be B0 over R,  because this guy is being cut out of the loop.  This is how it's going.  But now, if you throw this switch open,  you have given it a path to discharge.  So the inductor will discharge through the resistor.  The current will continue to flow in this direction.  Of course, it cannot flow forever,  and the resistor will burn up the energy.  But you're giving it a path.  You're avoiding.  In a way, if you like, this is also a path with very,  very high resistance.  The air is a path with very high resistance.  That means by and large, there are no free carriers in  the air.  But if the two tips get really charged,  just like you said, there'll be an EMF,  there'll pile up charges, they're jumping,  they're waiting to jump the gap,  and eventually they'll polarize the air,  which is electrically neutral into plus and minus parts.  Then the minus will go one way, the plus will go the other way,  and eventually you'll recognize that as a discharge.  But you don't have to worry about that now,  because R2 will take up your current.  So now the equation we write is LdI dt plus Ri is equal to  zero, because if you start anywhere and you go around a  loop, you have no change in anything.  There's no voltage.  And the answer to this one, R2.  So I is equal to I0 e to the minus R2 over L times T.  So now the current will decay exponentially.  Again, with the time constant R2 over L,  which is the one over the time constant,  or the time constant, is L over R2.  So you put L in Henry's and R in Ohm's,  you'll get some time.  And I'll give you an idea of how many times that time you have  to wait before the inductor is completely discharged.  Well, it's never going to be fully discharged.  It's going to take forever.  But if you say, look, one thousandth of the original  current is safe enough, how long they have to wait,  you put.001 here and see what time you get.  So you want e to the minus T over Tau to be.001.  So this is how the inductor...  So every big inductor you have in a circuit you will find as  God had resisted it in parallel with it to take up the energy  it has.  So let's do a little energy calculation.  Energy calculation will be I had energy in the inductor in  the beginning, which is 1 half L I0  square.  What happened to it?  Well, we know what happened.  There's a current in the circuit, and when the current is  discharged, it is dissipating heat in the resistor.  Therefore, the power in the resistor will be I square R2.  I square is I0 square e to the minus twice R2T over L times  another R2.  Now we integrate this power from 0 to infinity.  You will get I0 square R2.  So the integral of e to the minus something is e to the  minus that divided by that infinity to 0.  I can promise you that what you will get will be L over 2 R2.  That'll give you then 1 over half, 1 over 2 L I0 square.  So this is not a surprise that the stored energy in the  inductor is driving a current.  If you want one way to store energy, it's a capacitor,  you charge it, and you discharge it when you  want the current to flow.  Or you can have an inductor in which a current has been set  up.  Then when the power shuts off in your house,  somebody throws a switch, that inductor can keep the light  bulb burning.  In fact, forever.  It'll be infinite time before the bulb shuts down,  but you won't be able to see anything after a while,  and this graph is exponentially falling.  Okay?  This is a very simple problem where we can do the math and we  can understand what it's doing.  Yep?  Student's question  This one here?  This is called tau.  You mean that symbol?  Student's question  Tau is a unit of time whose numerical values are 2 over L,  and we can write I as I0 e to the minus t over tau.  We always like to write it in this form so that if tau is  equal to 100 seconds, you will have to make many  multiples of 100 seconds before the current is negligible.  Oh, in this graph here?  This is time, and this is the current in the loop.  It starts out at some initial value.  Okay, now I'm going to do the next slightly more complicated  circuit, and that is an LC circuit.  If you have a circuit like that, it won't do anything.  You can take these guys, hook them up,  and wait all day.  Nothing will happen because it's got no energy.  Why should it do anything?  It's just going to sit there.  But if you are charged up your capacitor like that,  and there's an open switch that you then closed,  then these plus charges are going to leave and find their way  around the other side and utilize them,  and the capacitor will discharge.  But you know that once it discharges,  it's not the end of the story.  So why is that?  Why doesn't it stop when the capacitor is discharged?  Yep?  Student's question is kind of like the inductor has  momentum and it keeps pulling.  The inductor would have had a current by then.  So the inductor is not going to suddenly stop having its  current.  So it's going to keep driving more current for a while until  the capacitor is charged up the opposite way to an extent that  this fights that one, that voltage,  and then it'll go the other way.  So it'll go back and forth.  So let's write down the equation that tells you how it goes  back and forth.  So in other words, you should draw pictures.  Now textbooks have nice pictures and I won't even try it.  In the beginning it may look like, I'm sorry,  in my example, in the beginning there's an  electric field here, nothing in the capacitor.  There's a quarter cycle, there's nothing here,  there's a magnetic field in the inductor.  Another quarter cycle, it's back here,  but with the opposite polarity and it goes back and forth.  So we can get all that by writing the equation LdI dt  plus a q over c equal to 0.  So let's write everything in terms of q.  Then q dq dt is I, because as the current flows  this way it builds up charge.  That's my convention.  So I can write it as Ld2q over dt squared  over cq equal to 0.  Now we have seen exactly this equation before.  We saw the equation for a mass coupled to a spring,  the equation md2x over dt squared  plus kx equal to 0.  There we know what it does.  Now it oscillates back and forth.  And L, this is an SAT question.  L is to m, what 1 over c is to k.  It's very useful to bear in your mind this analogy,  because mathematically these equations,  these equations have exactly the same solution.  It's no more difficult to solve this one than that one.  You may involve electric charges, that may involve masses.  You don't care.  So mathematically, here's an equation.  Cow times d2 dog over dt squared plus,  let's say, elephant times dog equal to 0,  where dog is a function of time, has exactly the same solution.  Because what does it matter what you call the unknown  variables, right?  But you've got to tell me that cow and elephant are time  dependent, and dog is the only thing that  depends on time, because otherwise the  equation is not the same.  So here m and k don't change, L and c don't change.  So what's the answer to this one?  Once again, it's a differential equation.  You write dq dt squared is minus 1 over lcq,  and you're asking yourself, give me a function which,  when I differentiate twice, looks like itself.  There's always the exponential.  There's also sines in the cosines,  so you can take any combination you like,  and the one acceptable solution is some constant A times  cosine square root of 1 over lc t.  Right?  Yes.  If you take two derivatives of q, you'll pull 1 over root lc  each time, then you will get a minus sign  and the cosine becomes minus sign, and one more derivative  will bring it back to minus cosine.  So this is called the frequency, which is 1 over root lc.  That's the analog of square root of k over m.  Anyway, this will oscillate.  It's a mechanical analog, this electrical analog of the  mechanical problem.  So I don't want to do too much of this, because I think you  should be quite familiar with this, at least the oscillatory  behavior of this.  The analog is perfect.  For example, to say that I pulled the mass by 1 meter is to  say that x was given a non-zero value and dx dt was 0.  Analogous thing here will be q was given a non-zero value and  dq dt, which is i, is 0.  That means the capacitor was charged, the inductor was not  carrying any current, then I let it go.  What happens?  The problems are identical.  Yep?  Student is asked about the differential equation of a  white element v of sin.  Yes, so I should explain to you.  That's an important point.  So q of t can be one number cosine omega 0 t,  omega 0 is this guy, plus another number sin  omega 0 t, and both are acceptable.  And the property of this differential equation is,  if you have one answer cos omega t and another possible answer  sin omega t, then any constant times cos  plus any constant times sin is also a solution.  You can verify that if you have one solution q1,  a second solution q2, you can add the two equations  and you can show q1 plus q2 obeys the same equation.  Not only that, suppose there's one problem  with q1, dq1 over dt square is minus  omega nr square q1.  That is some function of time.  And here's another function of time obeying the same  equation.  Now you can multiply this by a number a,  and you can multiply this by a number b.  Take the a inside the derivative, b inside the derivative,  and add the two sides.  On the left-hand side, you will find the second  side, the aq1 plus bq2.  Right-hand side, you'll find omega nr  square times aq1 plus bq2.  That means aq1 plus bq2 obeys this same differential  equation.  This is called superposing two solutions.  You've got one solution, another one.  You can multiply one by a constant.  It's very important it's a constant because that's what  lets you take it inside the derivative.  Therefore, the correct answer here is really a cos  omega d plus b sin omega t.  And you can ask, why did you write a cos  omega nrt?  First of all, that's not a good solution.  I can put a phi there, which is arbitrary.  And it's not hard to show.  Let me call that c, if you like.  C.  This can be written as c cos omega nrt minus  phi.  In other words, it's possible to take either a  cos omega nrt and a cos omega nrt.  The solution is with a cos and a sin, with no extra phases  inside, to another one with one amplitude and one phase.  That phase is the delay in the cos sign.  Anybody here has trouble who does not know the details?  I'll be happy to explain why that's correct.  You don't know, you should tell me now.  You can check that if you take this thing,  c, b and c, and an angle phi here,  then a can be written as c cos phi and b can be written as c  sin phi.  And that just happens to be cos omega t minus phi,  just for some trigonometry.  So the differential equation, the second order in time,  will always have two unknown parameters in it.  So you can choose them to be these numbers a and b or this  number c and phi.  And the equation won't tell you what they are,  because once you tell me there's a mass coupled to a spring,  and I say, okay, this is spring constant,  this is the mass, and I ask you,  where is the mass right now?  I don't know, because it depends on when you  started it and how you released it.  So I need to know what's called initial conditions,  which is the value of q or the value of x at the initial time,  and the value of the velocity or the current at an initial  time.  With those two pieces of information, I can solve for a  and b.  This I think you should know, this kind of stuff you should  know, so I won't say too much about that.  All right, so now we do another problem.  We put an alternating voltage on these guys.  This is V0 cosine omega t, and this is c and this is l.  Now we ask, what's the current?  This omega is not the natural frequency of oscillation.  It is some externally given omega, like 60 Hz,  from your power supply or from your socket.  So that's driving the circuit.  And you can ask, what happens now?  The answer will be, you write the same equation.  You can write l d2q over dt squared,  plus 1 over cq, is equal to V0 cosine omega t.  Now what are you going to do?  You have to again guess the solution.  So I want a function q of t, so that when I take two  derivatives and add it to some multiple of itself,  I get something, something times a cosine.  So what should that function look like?  Yes?  No, no, what function q of t do you think will satisfy this  equation?  I'm asked for a function of time, which when I put into this  has some chance of obeying the equation.  So I want to get a function of time, which is a functional  form.  Yes?  Student's question  Right.  In other words, I can take it to be some constant c  times cosine omega t, period, not even times  phi, because all right, yeah, take this one.  See what happens now?  When you take one derivative, you get minus omega l c  cosine omega t.  When you take another derivative, you get minus  omega square l squared, I'm sorry, this becomes sin  times cosine omega t, and this one is just,  oh my God, this is my nightmare.  You can see the nightmare now.  How many people see the nightmare?  Yes?  Yes?  So if I pick that c, it will be the same as this  c, so we'll put a hat on this guy so we can tell them apart.  So c twiddle over c times c twiddle is v0 cosine omega t.  And what this tells you is, if you take two derivatives,  I'm sorry, this is so sloppy, if you take two  derivatives, you will get plus omega square cos omega t,  also with a minus sign.  Q over c.  Okay, you guys buy that?  There is no l squared, they're just an l.  So c twiddle cos omega t is equal to v0 cos omega t.  Right?  Every derivative brings an omega and there's a net minus sign.  So you find out that your guess c twiddle cos omega t is going  to work, provided this equation is satisfied,  and the cosine omega t can be cancelled.  That's the whole point of doing this thing.  So when you get c twiddle, it's equal to v0 minus v0  divided by omega square l minus 1 over c.  Or if you like, minus v0 over l divided by  omega square minus omega 0 square.  Don't worry about the numerator, look at the denominator.  The denominator says that if your driving frequency is equal  to the resonant frequency, c twiddle blows up.  That means when a system has got a resonant frequency,  yeah?  I took two derivatives.  Oh, here.  Oh, I'm sorry.  Yes, thank you.  You know, it's good to do these things slowly and not rely on  what you heard somewhere.  So let me do it for you again, I'm sorry about that.  This time I take two derivatives.  Each one brings an omega, there's an l,  and the minus sign comes because the cosine becomes minus  sign, differential, one more you get minus the  cosine.  This guy is nothing, he just says divide me by c.  That's equal to that.  And then I wrote 1 over lc is omega 0 square.  Thank you very much.  Okay?  I was really hung up on the final result,  which tells you you better not drive this at the resonant  frequency, because then current amplitude will build up  indefinitely.  That's also true in a swing.  If you've got a swing and the kid's coming back and forth,  and you're reading a newspaper just pushing the kid,  you've got to push at the right time to get the best result,  and that kid's not going to fly off because there's one more  term in the equation for the kid, which is friction.  But if you have frictionless swing and you're doing this,  you've got to watch out, because soon there'll be nobody  who's going to be able to do this, because the amplitude  will keep growing.  Well, these are unrealistic problems, but I want you to  notice one thing.  Here's what I want you to notice.  The voltage you are given look like v0 cosine omega t,  and the current that you got, I can obtain by taking the  derivative of this charge, which is like dq dt.  This is the final sine omega t, and c twiddle,  we can write as v0 over L divided by omega square  minus omega 0 square sine omega t.  This is I.  Here's what I want you to notice.  It looks a lot like Ohm's law, because the current looks like  voltage divided by some number.  That's a very big difference.  The voltage is a cosine, and the current is a sine.  That's something I want you to think about.  The current is not in step with the voltage,  whereas in a resistive circuit, the current follows the  voltage.  It's the same profile as the voltage, except you divide by  r.  Here, one is a cosine, one is a sine.  That means when one guy is a maximum, other is a minimum.  It's called outer phase.  In fact, outer phase by 90 degrees.  Yes?  Student's question is, what's the proof by the  sine that you got where the omega is?  Here?  Student's question is, so on the other side,  together, we will find that proof by the sine  where the omega is.  Oh, omega did not go.  It's back.  You're right.  That's the omega that came from.  Okay, these problems, the only thing I'm focusing on  right now, I would say, of all the things I wrote  there, to be the most important thing,  is number one.  This is what I want you to bear in mind.  I want you to understand why sometimes not paying attention  to some constants.  What is important to notice is that this is an equation you  can solve by inspection.  Why was it easier to solve it by inspection?  Because you are trying to get in the end a cosine to balance the  right-hand side.  You're trying to find a function which is itself a cosine,  or whose second derivative is a cosine.  And we know the answer to that as a cosine,  so we can guess the answer.  Once you guess the answer, you put it in and you analyze  it.  You notice also that the current and the voltage are not in  step.  One is a cosine.  You know what the cosine does?  It does that.  Other is a sine, so they are out of step.  When one is a maximum, the other is a minimum,  and so on.  That means a current as a function of time is not equal to  the voltage as a function of time divided by anything.  That is nothing you can divide a cosine by to turn it into a  sine.  Whereas with resistors, you just divide by r,  and you get a current.  Now for the more realistic problem, the realistic problem  has got a capacitor, an inductor,  and a resistor, and no other power supply.  The equation obeyed by this one will be L d  t squared over Q over dt square, that stands for L  dI dt, plus r dQ dt,  plus Q over C equal to 0.  This is analogous to m d2x over dt squared,  plus I don't know how you guys wrote this in the last  semester.  It doesn't matter, gamma dx dt  plus Kx equal to 0.  Notice that it's the same form, and I'm going to give this to  you as a homework problem to analyze the answer to this one.  So this describes a problem where you have a mass and a  spring and some friction on the table.  That means if you pull it and let it go, the oscillations will  eventually get dapped, and it'll die down.  And the general solution for the case when it's oscillating  will look like sum number A e to the minus sum number alpha t  times cosine times some frequency omega prime t.  Let me see.  I had a phase phi, but I'm not going to do that.  If you want, you can put an extra phi,  but I will choose my origin of time so that I don't have that.  This is what the answer is going to look like,  where omega prime and alpha are going to be controlled by L,  R and C.  That's the thing I don't want to do in class.  Have you seen this before?  Professor Harris tells me you have seen this last time,  so the hints in the homework will guide you on how to do this.  You just assume the solution except t looks like sum A e to  the minus, I don't want to call it alpha now,  beta times t.  You put it into the equation and solve for beta,  and you'll find beta will have a real part and an imaginary  part, and you have to combine the two to get this answer.  That brings me to another thing,  so I don't know how prepared you guys are for what's about to  happen next, which is the use of complex numbers.  Everybody familiar with complex numbers?  Who doesn't know complex numbers?  How do you do your taxes?  You don't know imaginary numbers?  I will give you a lightning review.  I'm assuming everybody has seen them in high school.  I will only tell you the part you need,  but I'm going to assume that I can use them fluently.  I don't want to stop every time and worry about you guys.  So, I really need a show of hands.  Anybody never seen x plus Iy and x minus Iy,  and how about e to the i theta?  You know that guy?  Okay, that's all you need, so I'm going to tell you what the  deal is.  Everyone knows what a complex number is.  So, if I is square root of minus 1,  then we know that we can write a complex number as x  plus Iy, and visualize it in the complex plane where you  measure x this way and Iy is measured that way.  That's usually called a generic complex number z.  But now, let us find the length of that.  The complex number is a single point in the complex plane.  The length of that is square root of x squared  plus y squared.  So, I can also write z as x divided by square root of x  squared plus y squared, plus I times y divided by  square root of x squared plus y squared times square root of x  squared plus y squared.  Just nothing, just rearrange stuff.  This is going to be called the modulus of z.  What is this?  This angle is θ here.  This is really cosine θ plus I sin θ times  modulus of z.  And that, thanks to this great identity by Euler,  is e to the i θ.  Now, I don't know how much you know about this great formula  that relates e to the i θ to cosine plus sin.  How many people know where it comes from?  What does it mean to raise e to a complex power?  You know where it really comes from?  Yep?  Come from power series.  Turns out you can write a power series for cosine.  You can write a power series for sin.  Then, for example, cosine of θ is equal to  sin minus θ square over 2 factorial,  θ over 4 factorial over 6 factorial, etc.  And you can write a power series for e to the θ,  if you like, which is 1 plus θ plus  θ square over 2.  These, this defines cosine θ in the sense that if you put  θ equal to π over 2 in this infinite series,  you will get 0.  If you put θ equal to π, you will get π.  So, in spite of this funny looking form,  it really is the same guy.  It will oscillate, it will have 0,  it will be periodic.  None of it is obvious.  But this power series is numerically equal to this one if  you keep the infinite number of terms.  Likewise, e to the θ is defined by this series.  And similarly, there's a formula for sin θ.  Now, once a power series is defined, you can put e raised  to anything you want there.  So, you know what I'm going to put there.  E raised to dog is 1 plus dog plus dog squared.  This is not a joke.  This is really true.  This is the definition.  Someone says, how do I raise e to the power dog?  You do this.  This will have all the properties of the exponential.  It doesn't matter what's in the exponent, that's the key.  People originally put real numbers, then they put complex  numbers, now they put matrices,  operators, anything you want.  E raised to anything, you formally define to be this  infinite series.  So, you have the infinite series converges and gives you  a meaningful answer.  In that sense, if you put e to the i θ here and  put, compare the result to cosine θ plus  i sin θ, it matches.  That's all.  I'm just going to keep using that result.  Then, you should also know that for every complex number,  there's a complex conjugate which is x minus i y.  That means i goes to minus i.  That means z, this angle θ, will change to minus  θ.  And given any complex number, the real part of z,  which is what, which is x, is z plus z star over 2,  and the imaginary part of z, which is y,  is z minus z star over 2i.  So, this is what you need to know.  Every complex number has a real part and an imaginary part.  In fact, z star looks like this.  And if you add z plus z star, the vertical parts cancel and  you get double the horizontal part, that's why you divide by  2.  If you subtract and divide by 2i, you get y.  Basically, that's all I want you to know,  I want you to be able to manipulate them rapidly enough.  Here's something very useful about complex numbers.  Let's take a complex number z1, which is mod z1 e to the i  θ1, and another complex number z2,  which is mod z2 e to the i θ2.  So, what do they look like?  Z1 looks like this, at some angle θ1.  Z2 may look like that.  It's got some angle θ2.  So, every complex number has a length and an angle.  You can think of it two ways.  It's got an x part and a y part, namely real and imaginary,  or it's got a modulus and a phase.  That's the Cartesian version of the number and the polar version  of the number.  And the connection between them comes from Euler's formula.  Now, look what happens when I take the product z1 z2.  I get mod z1 mod z2 e to the i θ1  plus θ2, because exponentials add when you  multiply them.  Well, this allows you to immediately guess what the  product is going to look like.  The product is going to have a length equal to the length of  z1 times the length of z2.  It's going to have an angle.  I've not drawn a good picture here,  but I'm going to draw a good picture here.  So, this is the sum of the two angles,  because it's θ1 plus θ2.  So, listen to this statement very carefully.  When you take a complex number, say number 1,  and you multiply it by a second complex number,  you do two things to the first guy.  You rescale it and you rotate it.  So, two operations are done in one shot when you use complex  numbers.  You get a real number that's equal to the length of z1  plus θ2.  So, you get a real number that's equal to the length of  z1 plus θ2 plus θ2.  It adds to θ1.  Likewise, if you divide, if you take z1 over z2,  it is mod z1 over mod z2 times e to the i θ1 minus  θ2, because exponential θ2 is downstairs.  So, multiplication by real numbers is very easy.  You take a real number 4.  You take a real number 5 here.  With complex numbers, you take a number,  you multiply another number, you rescale and you rotate.  That's what I want you to know.  That's going to be very important for what I'm saying.  Okay, now I come to the problem I really want to solve,  which is an LCR circuit driven by an alternating source,  B0 cosine ωt.  It's R.  This is C.  And this is L.  The equation we have to solve is LdI dt  plus Ri plus 1 over C integral I dt is equal to  B0 cosine ωt,  because Q is the integral up to time t of the current.  That's just Q over C.  It's the same equation.  Now, unlike in this problem, where I had no driving  voltage, I have a driving voltage,  and I've also written it in terms of current rather than in  terms of charge, because in electrical  engineering, you don't really watch the  charge in the capacitor.  You look at the current flowing through the circuit.  So, you have to solve this equation.  The question is, can you guess the answer?  That's the only way we know.  You've got to guess.  You've got to try to find a function I of t so that when  you differentiate it, you get a cosine ωt.  When you integrate it, you get a cosine ωt.  So far, we can do it.  Sin ωt will do it.  But when you leave it alone, then also you get a cosine  ωt.  You cannot do that, because sin will become a cosine  and cosine will become a sin.  In the LC circuit, when you didn't have this guy,  you were okay, because if you pick a sin,  this became a cosine, and that also became a cosine,  and you can combine them to get a cosine.  But you cannot have dI dt, I, and integral of I all in one  equation, and all satisfied by trigonometric function.  The only time we can guess the answer is if it looks like v  0 e to the alpha t.  Suppose that was the voltage.  Then, can you guess the answer?  What form the answer will have?  It will have to be e to the alpha t times some number,  because then this will look like e to the alpha t.  That will look like e to the alpha t.  That will look like e to the alpha t.  The e to the alpha t has a great property that whether you  multiply it, whether you integrate it,  differentiate it, or leave it alone,  it looks the same.  Unfortunately, no one's interested in this voltage,  because it's growing exponentially fast,  or if you put a minus sign, it's dying exponentially.  What we really want is this.  So the question is, how do we solve this problem  with cos omega t?  So we're going to use a certain trick.  The trick I'm going to use is the following.  Let me take a general case where this is some function v of t.  Then, but I'm not even assuming v is real.  I'm not assuming i is real, but l and r and c are real.  So something else can be complex numbers.  Now, once you have an equation and you've found a solution,  if you take the complex conjugate of both sides,  they will still match.  Do you understand that?  If two things are equal, the complex conjugates are  equal, because this thing is the real plus imaginary.  That balances the real plus imaginary on the other side.  Complex conjugate just reverses the imaginary terms,  so they will still match when you flip both sides.  So this means if you take l dI star of dt  plus r times i star plus 1 over c integral i star of t  prime dt prime of t equal to v star of t.  So star is the rule by which if v had a real and imaginary  part, you flip the imaginary part.  So that solution for v of t implies a second equation.  In other words, if the voltage v drives a  current i, the conjugate of the voltage  drives the conjugate of the current.  That simply follows from the equation.  It comes from the fact that when I conjugated it,  l, r and c are real.  Do you understand that?  Suppose l were imaginary, or had an imaginary part.  Then I must also put a conjugate on l,  then i star obeys a different equation,  because this had l in it, this had l star in it.  So if everything is real, i star obeys the same  equation when you have a v start driving it.  I'm almost done now.  I want you to add the left hand side,  the left hand side, and the right hand side,  the right hand side.  And what do I get?  Well, if I write it here, you may not be able to see it.  Let me put it here then.  If you add these two, you get l d by dt of  v.  So this is the equation.  I plus i star plus r times i plus i star plus 1 over c  integral of i plus i star dt is equal to v plus v star.  This again comes because the equation is a linear equation.  You can always add the left hand side and the right hand side  to get another problem where the driving voltage is v  plus i plus i star, etc.  But i plus i star is 2 times the real part of i,  and v plus v star is 2 times the real part of v.  Therefore, this implies that l d by dt of the real part of i  is r times the real part of i plus 1 over c times  integral of the real part of i will be equal to the real part  of v.  So let me say in words what the equation means.  If you, by luck, solve an equation with a complex  potential v, and you get a current,  then the answer to the problem where the potential is only the  real part of the actual potential applied,  the current will be the real part of the answer you got.  That's all I want you to know.  If you solve a complex problem, the driving voltage is complex,  which is just a mathematical fiction,  real life you don't have that, then the current that you get  will also be complex, but its real part will be  due to the real part of your voltage.  And you can also show the imaginary part of the current  will be due to the imaginary part of the voltage.  This is just the principle of superposition.  So here is the trick we are going to use.  We are going to go back to the original equation,  LdI dt plus Ri plus 1 over c integral I dt  equal to v0 cos omega t.  And we're going to solve a new problem.  We're going to solve a problem where this is v0 e to the i  omega t.  And the current for that I'm going to call some i twiddle.  It's a complex current because the driving voltage is a  complex voltage.  So think of v e to the i omega t as the v in the previous  problem.  The actual problem I was given had v0 cos omega t,  but you know that v0 e to the i omega t is v0 cos omega t  plus i times v0 sin omega t.  Consequently, this potential has a real part,  a cos sign, an imaginary part which is a sign,  and the current will have a real and imaginary part.  And the answer to our original question is the real part of the  answer to this question.  This is what you should think about this point,  you should understand.  So I've manufactured a problem with an unphysical,  complex voltage, and all I know is that at the  end of the day, if I take the real part of the  current, I'll get the answer to the real part of the potential,  which happens to be the actual potential.  This is not cos omega t.  So why would I do all this?  Why would I take a problem that's bad enough with a  cos sign and turn it into a complex exponential?  I think you can sort of guess what the reason is.  The reason is that it may be complex, but it is still an  exponential.  That means its derivative is going to look just the same.  That means I can make a guess that the current i twiddle is  proportional to e to the i omega t.  So, you make the assumption that I twiddle of T is some  constant I twiddle times e to the iωT.  Make that assumption and put it in this equation and see what  you get.  You will find you'll get iωL times I twiddle plus R times  I twiddle plus 1 over iωC times I twiddle times e to the iωT  times V0 e to the iωT.  This is because if you assume the current is an  exponential, every derivative is equal to an  iω and every integral is equal to 1 over iω and  multiplying is just multiplying.  So, d by dt is got replaced by iω and you get this equation  for I twiddle.  That means you will satisfy the equation provided I  twiddle times all of this is equal to V twiddle.  And I'm going to write this as Z times I twiddle,  equal to V, where Z is the name for all the  guys multiplying I twiddle.  So, let's start with that here.  I wrote IωL plus R plus 1 over iωC,  I twiddle, e to the iωT is V0 e to the iωT.  Everybody get that?  That's because differentiating on an exponential is just  multiplying by the exponent.  So, this is a complex number, R plus i times  ωL minus 1 over ωC.  I want you to know that i in the bottom is like minus i in the  top.  And this is called the impedance,  Z.  It's a complex number.  You can visualize the complex number as follows.  It's got a real part which is R, got imaginary parts,  I twiddle minus 1 over ωC, and the impedance Z is a  complex number with some modulus and some phase.  If you know the real and imaginary parts,  you can construct the modulus in the phase like that.  You want this, the modulus of Z.  Therefore, canceling this, I find I twiddle is equal to  V0 over Z.  So, this becomes V0 over the modulus of Z times e to the  iφ, where φ is its angle defined by tan  φ equal to ωL minus 1 over ωC divided by R.  So, don't worry too much about writing all of this down,  because this you're going to find in every book.  There's nothing novel or different.  So, this is the formula for I twiddle.  The formula for I itself, you remember,  is equal to the I twiddle of 0.  I'm sorry, this I twiddle of 0.  Did I define it that way?  Yeah, I should write it, I'm sorry, I should write it as  I twiddle of 0, with a subscript 0 there,  because the full I twiddle has got e to the iωt in it.  So, this looks like V0 over mod Z times e to the iωt  minus φ.  But this is not the physical current, because it's the  result of a complex voltage.  The physical current is a real part of this,  without the twiddle, that we write as real part of  the I twiddle.  Where do I get the real part?  V0 and absolute value of Z are both real.  The real part of e to the i something is cosine of something.  And that's our answer, that's the final answer you get.  This tells you that the current has a magnitude,  which is V0 over the absolute value of Z,  and it's got a phase φ, by which it is behind the  driving voltage.  The magnitude of Z, for as in any complex number,  if someone says what's the magnitude of Z,  it is simply square root of real part square,  plus imaginary part square.  So, I want you to think about the magic of complex numbers.  So, the final answer has a current,  which is a cos ωt minus φ.  Your driving voltage is a cos ωt.  There's no way you can divide the voltage by anything to get  that current.  But in the complex language, the complex current is the  complex voltage divided by the complex impedance.  How does that happen?  That happens because the impedance Z has got a magnitude  and a phase, therefore, if you had a  voltage V0 e to the iωt, and you divide by mod Z e  to the i φ, that becomes e to the  minus i φ upstairs, you're able to change the  magnitude and you're able to change the phase by dividing.  So, if you want to take the voltage and you want to  rescale it and shift its phase all in one shot,  you can do it if you divide by a complex number,  because a complex number will rescale this V and also  the i φ in the same phase.  And then you take the real part.  In other words, cos ωt divided by nothing will give  you cos ωt minus φ.  But e to the iωt, when divided by e to the i  φ, will in fact give you e to the  ωt minus i φ.  So that phase shift you can produce only by dividing the  complex number.  So, I'll come back and explain to you a little bit more on how  you do circuit problems using the complex impedance.  But I think it'll be helpful for those of you who don't know  complex numbers to get used to this part of it.
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