13. LCR Circuits—AC Voltage
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Fundamentals of Physics II13 / 25
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AtmosphäreDrechselnElektrizitätVideotechnikWarmumformenFörderleistungLeistenFlugzeugträgerPlattierenErsatzteilFeinschneidenHochfrequenzübertragungKlemmverbindungMagnetisierungPerlschnurblitzZylinderkopfSchlagwerkAktives MediumTagWolkengattungFeldquantColourElektrische StromdichteVerschiebungsstromMittwochSchwingungsphaseLadungsdichteMagnetspuleDrehung <Textiltechnik>Vorlesung/Konferenz
Transkript: Englisch(automatisch erzeugt)
00:02
So I thought I will go back a little bit to LCR circuits. I think that's one of the more useful things you learn in this course. I want to do it carefully so everyone's on top of that. So remember that the problem you want to solve looks like
00:21
this. That's an AC source. That is a resistor. There's an inductor and a capacitor and that's a circuit. And this voltage we take to be V0 cosine ωt.
00:45
That's the kind of voltage that will come out of any AC generator. The ω is controlled by how the turbine is spinning. That frequency is the same as this frequency. Our job is to find the current in the circuit.
01:01
Now what makes it difficult is that unlike DC circuits, where you take the voltage and divide by some effective resistance, the equations here are not algebraic equations, but differential equations. So we'll write it down, then I'll tell you how we handle it. So the equation for this circuit, you begin as usual.
01:20
You go around the whole loop and you add everything to 0. Then you'll find V0 cos ωt is equal to R times I L times dI dt 1 over integral of I up to
01:41
time t times 1 over C. That's the voltage. Everybody with me on that part? That determines the condition. That determines the current at every instant in time. So if you can solve this equation, you solve for the current. But what's difficult is that it's not an algebraic equation.
02:02
It's not like V equal to I R, where you divide by R, you've got the current. This has got integrals, it's got derivatives. But it turns out that if the voltage is V0 cos ωt, that's a very clever way to solve this problem, which takes it back almost to the good old days of V equal to I R.
02:21
No differentials, no integrals, nothing. That's the magic. That was invented by an engineer from GE called Steinmetz. He put to work some ideas that may seem esoteric because they involve complex numbers, but it makes life tolerable for people doing circuit theory. And here is the trick.
02:41
I ignore this problem and I solve the following problem, purely mathematical device. The problem I solve has a voltage V0 e to the iωt. We all realize that you cannot get a source that does that, but it's pure mathematics. Then the solution to that problem, that voltage I assume
03:04
will drive some current I call I twiddle. So I twiddle is the answer to the problem where the driving
03:22
voltage is not real, but this. We mathematically write the same equation. But now we realize that I twiddle can be complex. I twiddle is not real. It can be a complex number. That's because the voltage is a complex number.
03:40
It's an equation. It's got imaginary and real parts on the left hand side. It'll have real and imaginary parts on the right hand side, and so you will need both to make it work. But if you have solved this problem, I claim you have also solved this problem.
04:00
Why? Because if you take this equation, which is a complex equation, then the real part on the left hand side will match the real part on the right hand side. The imaginary part will match the imaginary part. That's because if you have two complex numbers, Z1, and you say they're equal, you mean x1 iy1 is x2 iy2.
04:22
And they can be equal only if these guys are equal and those guys are equal. The real part has to match, the imaginary part has to. You cannot borrow from the real part and give the imaginary part. They're apples and they're oranges. So you've got to match them separately. Closest analog is a vector equation. You've got a vector equation and you equate two vectors.
04:42
So you say vector F is m times vector A. Then Fx will be m times Ax and Fy will be m times Ay. So there are two equations in one single vector equation, or three in 3D. In a complex equation, there are two equations. So let's take the equivalent of the real part on the left hand side and the real part on the right hand side.
05:01
And what do you get? The real part of this is the most important thing. The real part of this is cosωt. That's because e to the iωt is cosωt i sinωt. The real part is just cosωt. On the right hand side, I should take r times the real part of i twiddle.
05:20
Let me just call that guy i. Then I should take L times the derivative of the real part of i twiddle, which I want to call i. Then 1 over C times integral of the real part of i twiddle. The real part of i twiddle I want to call i. In other words, let the real part of i
05:40
twiddle be called i. Then we notice that this i is exactly the i that we want. It satisfies exactly the equation we want to solve. So we have some good news and some bad news. The good news is that if you solve this equation,
06:03
you have also solved this equation. The bad news is if you cannot even solve the real equation, what makes you think you can solve this complex equation, right? I cooked up another problem. It looks even more difficult, and I say, hey, if you can do that, we can do this. Well, it turns out this complex problem is actually easier
06:23
to solve than this guy. And I try to tell you why it's easier to solve. If you come to this equation, I've told you repeatedly any equation involving derivatives and so on, you guess the answer and you put it in. You try to make a guess.
06:40
I'm trying to guess what kind of function i of t has the property that when I multiply it by r, I should get a cosωt. When I differentiate, I should get something like cosωt. When I integrate, I should get cosωt. That's the only way left and right will match. And neither cosine nor sine will do the trick,
07:00
right? If you pick the sine, then when you differentiate it, you will get a cosine that matches this. When you integrate it, you will get a cosine that matches this, but this will contain a sine. On the other hand here, since it's exponential, if you make the choice i twiddle of t, this i twiddle 0 e to the iωt,
07:24
it is going to work. It is going to work because if you take r times that, you get r times i naught e to the iωt. If you take the derivative of this guy, that's iω times the same current. The integral is 1 over iω times the same thing.
07:40
That's the beautiful property of the exponential that when you differentiate it, it's like multiplying by i times ω. When you integrate it, it's like dividing by ω, and when you multiply by r, it's like multiplying by r. So if you take this assumed solution and put it in that equation, let's see what we get. We get e to the iωt equals.
08:03
Now allow me to jump one or two steps, because I did it last time. Try to do this in your head. r times i twiddle is going to contain r times i twiddle 0 e to the iωt. How about l times dI twiddle dt?
08:22
When I come with the d by dt, this guy is a constant. This brings me an iω. I write iωl times the same thing. When I integrate, I get 1 over iω times the 1 over c. That looks like that.
08:41
So the time dependence, this is a time dependent problem, but miraculously, the time dependence matches on the two sides, and now we get this equation v0 equals r iωl 1 over iωc times
09:00
i0 twiddle. And our goal was to find i0 twiddle. So i0 twiddle is equal to v0 divided by this complex number z. This complex number z is called the impedance and it's equal to
09:23
r iωl 1 over iωc for this problem. I'm not telling you this is the answer for every problem, but for this problem, life is very simple. You find the current as a voltage divided by a number, just like dividing by r.
09:42
The only trick is there are two catches now. The first catch is this number is not real. So you've got to get used to the fact that you will work with complex numbers. If you're willing to work with them, you get this. The second thing is this is not the current we were supposed to find. What's the current we were supposed to find?
10:01
The current that I wanted was the real part of this i twiddle. That means it's the real part of i twiddle 0 e to the iωt divided by, sorry, that's correct. That's what I want.
10:22
And that's going to be then the real part of v0 over z e to the iωt. We always pick v0 to be a real number. The amplitude and the voltage, we can take it to be a real number. Then we write it as the real part of v0 divided by
10:44
z e to the iΦ. So I should tell you what this is. I'm saying imagine the complex number z plotted this way. This is r. This is iωL minus 1 over i minus i over ωc.
11:01
And this is the complex number z. This is the angle φ. Then you write the same complex number in polar form as the modulus and e to the iΦ. So now you find i of t is the real part of
11:21
v0 over mod z e to the iωt minus φ, and that we know is v0 over mod z cosωt minus φ. So that's the final result. That's the answer to the original question.
11:45
Now you know trigonometry well enough to know that cosωt minus φ is really cosωt cos φ sine ωt sine φ. So the answer cannot be just a cos and cannot be a sine. It's a suitably chosen admixture of sines and
12:02
cosines that does the trick. So if you want to, if you say, I don't want to deal with complex numbers, you can take a guess like this and put it into the equation. And after a lot of manipulation, you will find that it's a φ satisfying this condition. What's the property of φ? tan φ is ωL minus
12:20
1 over ωc divided by R. But the beauty of the complex numbers is it just comes out in one package as the phase of a complex number. And I also told you to think about how you could do this with real numbers. It's not going to be easy to take a voltage which is
12:40
cos ωt divided by anything to get a volt current which is cos ωt − φ. There's nothing you can do to a cosine that will shift its phase. But if you're working with complex numbers, you can take e to the iωt divided by e to the iφ and turn it into e to the iωt − φ. So in the imaginary world, it's very easy to shift the
13:02
phase, because a complex number, when it multiplies or divides another number, rescales and rotates it. So in the complex plane, it's very easy to attach the φ, and at the end of the day, you take the real part.
13:22
So you can imagine doing any realistic problem. Suppose I tell you R is 10 ohms, ω is some 100 π, L is 3 henries, and the capacitance is 2 microfarads, you can find z.
13:41
z for this circuit will be 100. I'm sorry, it will be 10. ωL will be 100π times 3 henries, minus 1 over ωc, which is 100π times 2 times 10 to the minus 6.
14:03
That's some complex number. I forgot the i here. That is a complex number. I don't want to calculate it. It's whatever it is. And then it's got a real part which is 10, an imaginary part which is the algebraic sum of these two. You can actually plot the real values you get here in
14:23
this problem. I'm just going to show you the 10, the other guys, whatever it is, and you can get the phase. That tells you that in this problem, the current will have an amplitude, which is the volts you applied. Maybe you applied 100 volts, and you divide by this absolute value of z. So if you want, I'll complete the last part.
14:46
The absolute value of z, like for any complex number, is r squared omega L minus 1 over ωc squared, right? Any complex number, the absolute value is real squared plus imaginary squared under root.
15:05
So let's write in all its glory the current that we want. I of t is V0 over r squared omega L minus 1 over ωc squared cosine of ωt − φ,
15:23
where tan φ is equal to ωL minus 1 over ωc divided by r. Here is the picture. Here is r, ωL minus 1 over ωc where φ is a complex number, z.
15:42
Okay, so you can imagine now putting numbers and getting what you want. Notice that the current lags the applied voltage. For example, if cos ωt is the maximum at t equal to 0 and φ was 45 degrees, you'll have to wait till ωt is 45 degrees before the
16:02
current reaches the maximum. So it'll be lagging the voltage. But sometimes φ can come more negative. Can you see how φ could come more negative? This angle φ need not be positive. You understand that?
16:22
What would make it negative? Look at the formula there. ωL minus 1 over ωc. Yep, if 1 over ωc is bigger than ωL,
16:42
this will be more negative than positive. The complex number may end up like that. So I've written it for one possible sign. If that flips sign, the φ itself will change sign. It'll become ωt 32 degrees. It can go either way. It depends on who's dominating it, whether the inductor or the
17:00
capacitor is dominating it. So let's look at this answer for one other interesting feature. So I want you to know two things. This trick works only if the voltage is a pure oscillatory function like cos ωt. Then you can write it as a real part of e to the iωt and do this.
17:21
Secondly, the impedance is not a constant. Whereas the resistance is just 10 ohms, impedance varies with frequency. It's a frequency dependent number that you divide the voltage by to get the current. So the consequence of this is the following. Suppose you plot here the magnitude of the current.
17:45
Let's call it i0, this whole thing. Let me call it i0. That's the amplitude of the current as a function of frequency for a given applied voltage v0. So v0 is fixed, but when ω varies,
18:00
these numbers vary. When ω goes to 0, you've got a 1 over 0 in the denominator. That's going to beat everything, and 1 over 0 squared in the denominator means the whole thing vanishes. The current starts out as 0.
18:21
That corresponds to the fact that if your voltage had been a DC source instead of an AC source, that's what ω equal to 0 means. The capacitor will charge to some point and then stop the current. That's it. Then it will go up. Then it will come down, because at very large ω, the ωL is going to dominate.
18:41
You get 1 over ωL squared under root. That looks like 1 over ωL. It will fall like 1 over ω at very large frequencies. And it will be the maximum when these two guys cancel each other. See, you've got r squared. If you're trying to get the maximum current,
19:00
there's nothing you can do about r squared. It is what it is. But you can play these two guys against each other and find a frequency, so ωo, so that this is true, and that ωo is just related to the resonant frequency of the LC combination. So at that frequency, the current amplitude will be
19:25
simply I0 maximum will be simply Vnaught over r. It's as if L and C are not there. You've got them there, but they're not there. They're not doing anything. They neutralize each other.
19:42
So you can ask, why put something in that doesn't do anything? Well, first of all, it does something at other frequencies. Only at one magical frequency they go away. But what is interesting is this very sharp response you have. So do you know where that comes into play?
20:05
Resonance, yeah, but which part of your life? Pardon me? Radio. You guys know what that is, because I don't know. You're always carrying some recorded medium. But if you listen to radio, like in the old days, you have this room full of radio signals.
20:22
Everyone wants your attention. All the radio stations are all sending signals right now in this room. And you want to pick just one station that you like. So what happens if that station sends that information at a certain ω'? And if that's all you want, you go to the store,
20:40
buy an LCR circuit with L and C chosen so that Lc is 1 over ω' squared, then you will get a huge response when you get the signal from that station. Now there's another station with a different frequency. You don't want to listen to them, but you have to listen to some of them, because if their frequency is here, your response to that station is not 0.
21:03
It's a lot smaller than this, but it's not 0. And there may be yet another station you can hear in the background, because if you don't want to hear them, you would like to get 0 signal from them, but you cannot kill this function except at one frequency, but you can make it very sharp. If R is very, very small, this function will be
21:23
very large at resonance, because Vnaught over R will be a huge number. It will also be very narrow. So you can make this thing very sharp. And that's really controlled by the ratio of R to L. So other stations do give you weak signal, but you can make them weaker and weaker by picking a
21:43
very finely tuned oscillator. Now the question is, what if you change your mind and you want to listen to these guys? What should you do?
22:01
Can you go and buy one radio for this station and one radio for that station? You know you do something, right? You fiddle with the dial, but what do you think it does? Yep? Changes the capacitance. It's not easy to change the inductance, but it changes the capacitance.
22:20
If you open a radio, but nowadays if you open a radio, I don't know what you will see. It's all glued onto something. But in the old days, when all the parts are big, you can open the radio and look inside, and you have a capacitor which is called a variable capacitor. You show that with an arrow. You can also have variable inductors I suppose,
22:41
but this is very common. Now how do you vary the capacitance? Capacitance you know is ε0A over d. Yep? Student Student change the surface area. That's right. You've got these interlocking plates, and you can have them
23:03
fully overlapped or not overlapped. If you draw the plates like this, it's got many things, and these plates can be either really overlapping and the other plates are pulled out. When you turn the dial, the two things move in and out, and that varies the capacitance.
23:20
That will give you a range of frequencies, and that's the range you can hear. Okay, so that's one thing. Another thing I want to mention is, this solution I wrote down, I of t is V0 over mod z cosine Ωt − Φ.
23:40
It has no free parameters in it. You tell me the time, I tell you the current. Whatever the voltage is, you take that, shift the phase by Φ and divide by mod z. But you know that a second order equation in time must have two free parameters. So where are those free parameters going to come from?
24:05
Have you seen this before in your math, whatever? No? Okay, I'm going to give you a clue, and you have to think about this clue and see what you get out of that.
24:23
V0 cos Ωt equals V0 cos Ωt 0. That's your clue. What can you do with that clue?
24:44
Yes, I will translate. If you have a problem where you apply one voltage, V1, and you get a current I1, and a second voltage V2 and get a current I2,
25:02
just by adding the two equations, left hand side to left hand side and right hand side to right hand side, you can check that V1 V2 drives a current I1 I2. You should do that. If you don't see that, you should do that. Take that equation for any V of t. Get an answer due to V1, call it I1. Another due to V2, call it I2 and add them up.
25:23
Left hand side is clearly V1 V2. On the right hand side, R times I1 R times I2 is R times I1 I2 and so on. So I1 I2 will satisfy the equation when the driving voltage is the sum of the two voltages. It all comes because there's nothing nonlinear. For example, the right hand side at R times I
25:43
squared, then the first equation will have R times I1 squared. Second will have R times I2 squared. When you add them, you'll get R times I1 squared I2 squared, but what you want is really R times I1 I2 the whole thing squared. So if you have nonlinear terms, you cannot add
26:01
solutions, but if you have linear terms, you can add solutions. Therefore, 0 is a hint. It says add the solution for 0 voltage. And you might say there is no solution for 0 voltage, but I will have to remind you that we did solve the problem the other day of an LCR circuit connected to nothing. It had a current flowing for a while, but it won't do it if
26:23
it's completely inert. But if you had put some charge on this guy to begin with, then you saw that the current and the charge, everything starts and oscillates exponentially like e to the minus some number t times some cos ω't minus some other phase. Call it χ.
26:43
And this thing had two free parameters. There's an A here and a χ here. Or you can write it as e to the minus γt times A cos ω't B sin ω't. No matter how you write it, this function, this solution to 0 external voltage is called a
27:02
complementary solution, and the final solution for I is really what I wrote, B over mod Z cos ωt minus φ, plus this complementary function. Because if you take this complementary function and add it
27:22
to the equation, you might think it will mess up the equation, but it won't. Because the I that I wrote down will give me V0 cos ωt. The complementary function, when you take rI LdI dt 1 over cI dt for the complementary part, you will get 0, because the complementary guy obeys the following equation. It obeys the equation LdI dt r times
27:44
Ic 1 over c integral Ic dt is 0, the equation satisfied in that loop. If it has a non-zero solution, you can add it to your equation. See, in elementary calculus, you learn that if you solve an
28:01
equation, say df dx equal to x squared, the answer is x squared over 2, plus a constant. Because when you add a constant, you don't screw up the solution, because the d by dx kills the constant. The question is, what can you add to this differential equation? You can add anything, any function that's annihilated
28:23
by that differential equation. If you put that function in, you get 0 from all the derivatives and integrals. You can add that to anything. Therefore, the true answer is this complementary function. It'll contain these two parameters, a and b, and you pick them depending on the initial conditions.
28:41
In other words, if you took an LCR circuit and took a generator and you hooked it up, from the instant you hooked it up, it won't immediately assume this form that I have here, because, for example, at t equal to 0, maybe the current is 0 in the real problem. This current doesn't vanish, but you should add to this
29:01
function, choose a and b so that the current vanishes at t equal to 0. You can fit the actual experiment to the solution by choosing a and b to match the initial current and say the initial charge in the capacitor. But we don't usually care about this guy. Do you know why we don't talk about this too much?
29:23
Yep? It dies off exponentially. So in real life it doesn't matter. If you really want to know what's happening, in other words, take an LCR circuit, take an AC source, put a switch, keep the switch open. The minute you close it, current will be 0 by continuity. Current cannot jump.
29:40
So this function obviously is not the whole answer. It doesn't vanish at t equal to 0. But this extra part you add in with suitably chosen a and b will fit those initial conditions. But after oscillating for a while, e to the minus something t for long t will just go away. Then it will settle down only to this function.
30:00
This is called the steady state current and this is called the transient current. So transients are important. You cannot ignore them. Maybe a transient will burn your circuit, but it doesn't matter after a long time. If you survive the early time, it doesn't matter for the long time. It's a lot like this course. Okay, all right, so I want to take a minute to
30:25
tell you a little more about the use of these complex numbers. I do it for a lot of reasons. I don't know what field you guys are in. You could be in art history. You just don't know when complex numbers are going to be relevant to what you do, okay?
30:40
You just don't know. But I'm serious. If you're doing engineering, electrical engineering, mechanical engineering, it's all about complex numbers, because everybody writes a differential equation. It's a linear differential equation. You solve it with complex numbers. And I'd like to also show you why sometimes we should not resist the mathematics so much, because you might think in this
31:02
circuit, I don't need complex numbers. After all, in the end, the answer was the cos ωt − φ, and that is cos times some number, plus sine times some number. I can put the combination and fiddle with it till it works. That's true. We're supposed to give you a more complicated circuit. Here's a more complicated circuit.
31:20
I have maybe a resistor, an inductor. Then I come here. I have a capacitor, maybe another inductor. You go like that. Let's call it R1, L1, C2, L3. And this is some Vnaught, cos ωt.
31:44
How are you going to solve this problem by guessing? You have no idea how to guess. You cannot solve this problem by guesswork. But I claim that if you replace this fellow, the Vnaught e to the iωt, you'll be able to guess the answer.
32:01
You'll be able to solve the problem. So I want to take a minute to tell you why that works. So do you know what I'm trying to do now? I'm trying to tell you that an arbitrarily complex AC circuit built of resistors, inductors, capacitors, connected any way you like, add whatever you want.
32:21
If you go back to the differential equation, it'll drive you insane, because there is a derivative of this current, derivative of that current, integral of that current, all coupled in some way. How are you going to solve it? You cannot guess anything cosine and anything sine. It's going to be a madness. But I will show you that if you use complex numbers, you can reduce the problem to something that looks just like
32:42
DC circuit. That's all I want to do. So I'm going to give you the idea, but I don't want to do the whole thing, because I don't want to spend too much time on algebra. For those of you who want to know where it comes from, I want to give you a chance. For those of you who don't want to know exactly where
33:01
everything comes from, in the end I will give you the so-called bottom line, stuff you must know when you forget all this. So let's take that problem. And let me take an easier one than this one, because it's got too many wrinkles in it. But the point I'm making, I can illustrate just as easily
33:21
with this guy. So what did I call it? R1L1C2L3. Let the current here be I1. Let the current here be I2. The current here is I3. And this is V0cosωt.
33:44
Our job is to find those currents, which are all oscillatory functions of time, in magnitude and in phase. If you do that, you're done. So how do we do it normally? What are the fundamental equations for a circuit? Fundamental equations are that at every branch,
34:02
the incoming current should be equal to the outgoing current, which for us means I1 is equal to I2 I3. So I2 and I3 come down here and they join and become I1 again. So I'm going to assume that and write I1 there.
34:20
So I've got two unknown currents, I2 and I3. That's the first thing to understand, how many free independent currents there are. Generally, if it's a simple problem, it's equal to the number of loops you have. That's a loop here, that's a loop there. You can draw many other loops, like that one and that one, but they will not give you any new results.
34:40
So you've got to write voltage equations that says if you go around a loop and you come to where you start, the change is 0. The equation I would write would be V0 cos ωt is equal to, I'm going to take this loop, is equal to R1 I1 L1 dI1 dt 1 over C integral
35:11
I2 dt prime. Back to here is 0. That's one equation. For the second loop, I can go around the other way,
35:24
if you like, going around this inductor, or I can take a loop like this. I just want to write that to show you what it may look like. That equation will say L3 dI3 dt minus 1 over C2 integral I2 dt is equal to 0.
35:45
The minus sign comes because the current is assumed to flow this way and your loop goes counter to the current. You're going uphill, so everything should be subtracted, but if you go down, you can write it going down this way. So these are the three equations to solve.
36:01
They're complicated because there are differentials, there are integrals, and you know what. But if I solve the following problem, V0 e to the iωt. So don't bother to write all of this because you're not responsible for this detail. I just want you to know what's going on.
36:21
So I scrapped this problem and I solved this problem, and the current for that I take to be the twiddle. I say, let the i twiddles be the answer to this problem e to the iωt. Can you see that if I took the real part of the first equation, I get the correct equation for
36:43
the physical currents? That if I took the real part of the second equation, everywhere I replace i twiddle by the real part which I call i, and this becomes V0 cos ωt. I get the second voltage equation, and here again you drop the twiddles by taking the real part. That's the equation satisfied by the physical currents i2 and
37:02
i3. So it is still true, even in this complicated problem, that once you solved it, you may take the real part of the voltage and take the real part of the current anywhere. That's the answer to the primordial problem. So why does this make life easy for you? Because now you may take the current.
37:21
I'll just take one of these as an example. Let the current be current i1 twiddle. Let it be i1 twiddle 0 e to the iωt, and i2 twiddle is equal to i2 twiddle 0 e to the iωt, and i3 twiddle is i3 twiddle 0 e to the iωt.
37:46
Make that guess. And what does this mean? Think about it. Go to the first equation. You equate these twiddle guys, and all the e to the iωt's cancel on all the three terms. So you get an equation that says i1 twiddle 0 is equal to
38:04
i2 twiddle 0 i3 twiddle 0. No time dependence anywhere. It looks like a DC equation. And let's go to this equation here. You will find V0 is equal to R times i1 twiddle
38:25
iωL1 times i1 twiddle 1 over iωC. I should have called this C2, iωC2 times i2 twiddle. e to the iωt's have been canceled everywhere.
38:45
So what do you find? This again looks like an Ohm's law problem where R1 is the impedance of this guy, iωL1 is the impedance of that guy. So the whole thing looks like some Z1 times i1 twiddle
39:04
Z2 times i2 twiddle. In other words, let me just write it below this here. The equation we get is V0 is equal to Z1 times i1 twiddle Z2 times i2 twiddle.
39:22
So the problem will look like this. Just put some black box here, Z1, another black box Z2, another black box Z3, join them, bring them out. Call the current here as i1 twiddle 0. The current here is i2 twiddle 0.
39:41
The current here is i3 twiddle 0. And these constant numbers, they have no time dependence, obey the same equations as the real currents did in the real problem. So it's like a DC problem. The only certainty is the impedances are complex.
40:02
And the rule for the impedances is very simple. If you open the box and you find a resistor an inductor, the impedance for that is R1 iωL1. That's it. Apart from that variation, it is just like DC circuit.
40:21
So your procedure, yep, no. If you want, the imaginary part of the solution is the answer to a person whose voltage source is V0 sinωt. Okay, so you are solving two problems, but we believe that it's
40:41
nothing new. If you can do cosωt, the answer to sinωt is the same except you shift by π over 2. Everything looks the same, but it answers that question. Okay, so this is what I want you to know. Given an AC circuit, here's what you do. You make this rule. Whenever you see this guy, it is R.
41:02
Whenever you see this guy, it is iωL. And whenever you see this guy, it's 1 over iωC. Then you can combine inductance, you can combine impedances in series by adding them, or in parallel by doing the reciprocals and adding them,
41:20
and then taking the inverse of that. It's all like DC circuit. What will you have at the end? At the end, you would have found out every current iω0α. α is 1,2 or 3 in this problem, right? You will solve these DC-like equations, and you'll find these DC-like currents, but they will be complex.
41:42
What is the relation of this guy to the actual current flowing through the actual circuit element? The answer is iα of t will be i quiddle α times e to the iωt, then take the real part.
42:03
That's the algorithm. That's the relation between these time independent functions. Because what did you do? You stripped the time dependence. You wrote every current as e to the iωt times a number that doesn't depend on time. It's those numbers we have solved for. But to go back to the original current, first you've got to reinstate the exponential.
42:21
You canceled everywhere, and remember to take the real part, because that was the deal. It's called the real deal. Real deal, take the real part of your answer. How about following? You've solved a very complicated circuit. I don't even know what's going on. Here is one guy, okay? It's part of a big mess. Its impedance is Z.
42:40
There is a current, i quiddle 0, going through it, meaning it's the amplitude. There's a complex amplitude of that. If I ask you, what is the actual voltage across this? If I put a voltmeter here, what will I measure? The answer will be, again, i quiddled 0 times
43:02
Z. That's in the fake Ohm's law calculation. But you've got to remember that these all had time dependence, which I removed. Then you've got to remember that I should take the real part of the answer. That will be the actual instantaneous voltage across that
43:20
circuit element at that time t. But remember, i quilled could be a complex number. The Z inside the box could be a complex number. When you multiply them, you've got lots of complex numbers. You add on to get this guy and take the real part. I would recommend when you come to such problems that every
43:40
complex number like Z, you write as an absolute value times e to the i times some phase. That makes your life easier, because then the whole thing will be the absolute value of this times the absolute value of that times e to the i of ωt plus the phase of this plus the phase of that. And the real part of it is the cosine of ωt
44:01
plus these two phases. You understand? If you wrote it as mod times e to the i times φ1, absolute value of Z, e to the i φ2 times e to the iωt, what is the real part of this crazy number? These are all real numbers. You pull them out, combine the exponentials,
44:22
you'll get e to the iωt iφ1 from this one, iφ2 from that one. So the full current will be cosωt φ1 φ2. So the answer will be mod i0 mod
44:40
Z cosωt φ1 φ2. So I don't mind waiting to give you some time to digest this. You should be able to at least solve simple DC circuit with more than one loop. If you've got only one loop, just do what we did earlier,
45:03
V over Z and all that. But if you've got two loops, you should learn how to handle that problem, because it's very useful. You're going to deal with circuits no matter what you do. Yep, right.
45:21
That's what I said. That's correct. So let us ask the following question. His question was, what if I like to know the current coming out of this guy, right? Everybody with me? That was his question. So we do it as if everything was real. First we combine these two guys into a single impedance, which will be Z2Z3 divided by Z2 Z3,
45:43
just like you combine resistors in parallel. To that guy, I will add Z1. That whole thing is the impedance seen by this thing. Divide the voltage by this total effective impedance. That will give you the current coming here.
46:01
Yes? Take that current and multiply it by Z1 times e to the iωt. You take the real part, that's instantaneous voltage on this guy. Here's one more thing. When that current comes here, how is it going to branch between these two? In DC circuits, if this was R2, this is R1, and the current comes here,
46:23
its propensity to go on this side is proportional to that resistor, because if that is more resistive, it's more likely to go here. So the fraction going to this side will be R2 divided by R1 divided by R2, the fraction going to this branch will be R1 divided by R1 R2. You replace all of the Rs with Zs and you get the same
46:42
result. So you can do everything like before, except you've got to get used to complex numbers. Here's the problem. Take the circuit, assign to each element these impedances, add them like in the old days, combine them like in the old days, solve for the current. You get the physical current, multiply it by e to the
47:03
iωt and take the real part, and to find the drop against any element, take the current there times the impedance times e to the iωt, then take the real part. That's it. I cannot say it anymore, because I don't know another way to say it. You've just got to do problems so you get a feeling for how it's done.
47:22
I wanted you to understand sometimes you take a problem, you embed it in a bigger family of problems that looks more difficult, but actually easier. It's even true there are many integrals you try to do which are very difficult, but if you think of it as an integral in a complex plane, sometimes that problem is easier to solve than the real integral.
47:41
So this happens a lot. You take a problem, you generalize it. Sometimes a generalized problem is easier than the original one. Now I'm going to give you one counterexample. Let me see where to pick it. This rule about taking the real part at the end of the day
48:00
fails. So far the rule was do everything with a complex thing. At the end of the day, when you've found something, you take the real part. Here is a place where it does not work. So let me tell you where it is. Take the LCR circuit where we found that the voltage was
48:24
v0 cos ωt and the current was v0 over absolute value of z cos ωt − φ, where φ is all that stuff. I don't want to repeat it. What is the instantaneous power generated by the power
48:41
supply? Instantaneous power for any voltage source is equal to voltage at that instant times the current at that instant, and all these are real voltages and real currents. I'm not doing any games right now. This is the real thing. So what do you get here? You get v0 square over mod z cos ωt
49:04
times cos ωt − φ. That's what it is, right? One voltage going like cos, and the current is going like cos ωt − φ. So I want to write this out for you as follows.
49:21
This is cos square ωt cos φ plus sin ωt cos ωt cos φ. This is just some high school trig identities. Cos of A B is cos A cos B plus sin A sin B.
49:41
So what do you notice? Cos φ and sin φ are some constants, but these are functions of time and they're all oscillating. The reason they're oscillating is that when you drive a current in these circuits, sometimes the L and C are drawing energy from the source. Sometimes they're giving it back.
50:01
So sometimes it's not monotonic. It's not always drawing energy. Sometimes it's taking it, sometimes it's giving it back. That's all the oscillatory terms. So what one likes to study is called the average power over a full cycle. That means you take this time dependent function, integrate it over time over a full cycle at frequency ω,
50:24
and you divide by the time and ask what you get. So let me write the obvious parts, B zero square over mod Z. Cosine square is a positive definite number. Its average over a cycle is ½.
50:41
I don't know how you want me to show this. One is to say cosine square θ is 1 cosine 2θ over 2. And if you integrate that guy over a full cycle, cosine 2θ is periodic, gives you 0, you get ½. Sine ωt cos ωt, if you put a 2 and divide by 2, is proportional to sine 2Ωt.
51:03
That guy completes two cycles in one period. Its average is definitely 0. So the only thing that survives then from all of this is the cosine phi term and the 1 over 2, because the average of cosine square is half. So the average power in the circuit looks like 1 over
51:24
2Ωt over 2 mod Z cos phi. That's the correct result. In fact, we can see it in another way. Let's write it as V0 over mod Z square times mod Z cos phi over 2.
51:43
What is mod Z cos phi? Can you tell from some picture I drew of Z? I don't know if I have any pictures. I've hidden all of them, but here is what Z looks like. That is Z. That's phi. This is r, so ωL minus 1 over ωc.
52:01
What is mod Z cosine phi? Yes? It is just the resistance. So this is a fancy way for r. So this is just 1 over 2. If you want, it's just the amplitude of the current squared times r. That means that real energy loss is taking place only in the resistor.
52:21
You can write it as mod Z cosine phi if you like. The 1 half is new to AC circuits. In a DC circuit, it's just I squared r because it's constant. Here it's oscillating like a cosine squared, and the average of that is 1 half. So what people sometimes like to do is define something called the RMS current, which is I0 over
52:41
root 2, and they define the RMS voltage, which is V0 over root 2. Then the whole thing looks like I0 I rms squared times I rms squared times I rms squared.
53:02
Now RMS means root means squared. That's the reason why for a trigonometric function it's got a number which is 1 over root 2 times the amplitude, but I don't want to go into that because we don't use it again. The main thing to know is, if you want, you can redefine a new quantity, V0 over root 2 and I0 over root 2, so that these factors of 2 disappear,
53:22
and things look like the good old days of resistive circuits. So very often when they say this is 110 volt power supply, they really mean the RMS value 110 volts. That means V0 over square root of 2 is 110.
53:41
V0 itself is 110 root 2, so the actual voltage goes up and down from 110 root 2 to 110 root 2. Because for computation of power, the peak amplitude doesn't control it. So it's this RMS value that controls average power consumption.
54:01
Now, there is a way to get the power using the complex numbers, if you like, but I'm going to assign that as a homework problem rather than do it here. But I want you to think about why does it fail here? Why is it that when you come to the power,
54:22
this is not simply... So what I'm telling you is the power is not simply the real part of I twiddle T and V twiddle T, but V twiddle is the one with V0 e to the iωt. Why is it that taking the real part of this guy doesn't work?
54:41
Everywhere else we took the real part of the current, we got the right answer. The real part of the voltage, we got the right answer, but not when you take the real part of this. Yes, if you want, he's saying you're multiplying, but not adding. So let me say the following.
55:01
Suppose you had a complex number z1, it's x1 iy1, and another complex number z2 is x2 iy2. If you take the real part of z1, you get x1. You take the real part of z2, you get x2. But suppose you wanted x1 x2 for some reason.
55:23
That is the real part of z1 times the real part of z2. That is not equal to the real part of z1 z2, because the real part of z1 z2 has got x1 x2. That's pretty obvious, but it also got minus y1 y2, because when these two imaginaries multiply,
55:42
they can contribute a real part. But what you wanted is the analog of this one. So you can fix that. I mean, it's not that you can never extract x1 x2. The correct solution is x1 is z z star over 2, x2 is z2 z2 star over 2.
56:04
Then you can multiply it all out and see what happens. Then you can write the answer in terms of these guys if you like, but I'd rather just leave it this way, because it's not going to be used extensively. But I just want you to be careful. This is an example of something that is quadratic in
56:21
the interesting quantities. You cannot take the real part at the end of the calculation, only for things which are linear. At every time, if the real and imaginary parts do not talk to each other throughout the whole calculation, you can follow them through and at the end take the real part as what's interesting. But if you're going to multiply two things where you
56:41
have added on a complex part to this guy and a complex part to this guy, in that product you've added on something real, and that was not your intention. So you've got to take that out. That's what makes it complicated. Therefore, I tell you that when you do these AC circuits, when it comes to power, forget any of the gimmicks you learned. You will be usually asked to find the power only for LCR
57:03
circuit, the single loop like that. Then just go back to this one. It's just I squared R written in this language. And cosine Φ is called the power factor. Okay, so this completes one chunk of the course.
57:24
I'm really going to the finishing line for electromagnetic theory. There's only one big stuff left. Then we'll do optics and then we'll do quantum mechanics. Now, the nice thing about quantum mechanics is that you don't have to worry about whether you will get it,
57:41
because nobody gets it. My idea is to take a class where only I don't get it and turn it into a class where everybody don't get it, okay? That's the plan. When I say get it, I think I talked to one of you guys who called me and said something which is quite true. They said, in the first part of the course,
58:02
you could visualize what you were doing. You know, the mass is rolling down and things colliding. Electricity and magnetism, you're not afraid of the math, but we don't have an intuitive feeling for these things. So I agree, but if you work long enough with electricity and magnetism, you get used to them.
58:21
The only things born with the knowledge of electricity and magnetism are some creatures like ducks. Apparently ducks can feel the earth's magnetic field, and they're just going through and they know which way to go. They don't solve Maxwell's equation, but they know how to travel. And some bees are sensitive to the polarization of light.
58:41
So they do respond to certain things in an intrinsic way. People don't. So you have to get used to it only by doing more problems. But in quantum theory, any intuitive feeling you have is actually detriment, because it will get you in trouble, because nothing is the way you imagine it. So no one has an advantage over anybody else. So it's better not to be well informed.
59:02
So you may be saying, hey, here is the course for me, right? So the less you know, better off you are in quantum mechanics. So don't worry about that. It's a very strange world, and I want to give you an introduction to that. Anyway, there are the two things. So what's left now is just electromagnetic waves, and we're sort of building up to electromagnetic waves.
59:21
So I'm going to start by writing here the equations that we know about electromagnetism as of now. So the surface integral of the electric field on any surface is the charge enclosed divided by ε zero.
59:41
That's called Gauss's law. The line integral of the electric field on a closed loop used to be zero when things are static, but when things are changing with time, is the rate of change of the magnetic flux. That's it.
01:00:00
The surface integral of the magnetic field, B⋅dA, is just 0 all the time, because there are no magnetic monopoles. There's nothing from which lines come out. So if you integrate any configuration, the lines don't start anywhere, so whatever enters the surface leaves the surface. So surface integral of B is always 0. B⋅dA is 0.
01:00:21
Line integral of B⋅dA, B⋅dL is μ0I. That's the old Ampere's Law. These are the four equations we have. Now it turns out this is not still the end.
01:00:42
It's not the end. There's one more fiddling you have to do. That's the last part I want to talk about. Remember, this itself came from new experiments where you started moving magnets and so on, found a current whenever there's a changing magnetic flux. That's the induced electric field due to changing flux. So here is the thought process that led Mr.
01:01:05
Maxwell to a little paradox. So here is some circuit. We don't know where it begins and ends. We don't care. Here's a circuit where there's a current flowing through a capacitor. When I say flowing through a capacitor, I hope you know what it means. I want you to be very clear about this.
01:01:21
Suppose you're connected to some AC source. I want you all to know what's going on. Nothing really flows through the gap of the capacitor. It's not possible. What happens is, for a while, plus charges rush here and minus charges rush the other plate, because that's the polarity of my AC source. Later on it's reversed, then minus charges go here
01:01:41
and plus charges go there. So there is an alternating current in the circuit, but there is no current flowing right through in one sense. But there's nothing to keep it from going this way and that way and this way and that way, because they don't have to jump the gap to do that. Yep, it'll jump the gap if the, here's what happens.
01:02:11
It depends on what's in the medium. If you put air, for example, the air molecules are all neutral. So there is no way for the charges to ride them and go to
01:02:24
the minus charge, to the minus terminal. But if the electric fields, this will certainly produce electric fields. If they're so strong, they'll rip out the plus and minus charges. Then suddenly you've got a bunch of free carriers. Then the minus charges can immediately go and neutralize this one, plus can go and neutralize that one.
01:02:42
That's like a lightning strike inside the little world. That's what happens to us during a lightning. We are in a capacitor. The lower plate is the earth, upper plate is the clouds, and after a while there's a heavy charging. At some point they cannot take it anymore and they ionize the air and they create a little path, and the current flows through that and the clouds are
01:03:02
discharged. Okay, so here is the problem that Maxwell had. He says, let's look at this equation, B⋅dl in a closed loop equal to μ0I, where I is the current passing through any surface with that loop as a boundary.
01:03:22
Remember that? So if you draw a loop here like that, then some current is crossing that shaded surface. In the line integral of B, you know B fields go around the current, and the line integral of B around the loop will be the current crossing that shaded
01:03:42
surface. That's how we get Ampere's Law. But now you say, that's not the only surface with that loop as the boundary. I should be able to draw any surface with the loop as the boundary. So you say, let me take that surface. You're still okay, because whatever current passes
01:04:02
this phase also passes through that phase. The law is still good. So giddy with success, you say, why not this? That's my new surface. It goes all the way around one of the plates of the capacitor and comes back.
01:04:20
Now I have a problem. I have a problem because there is no current passing that surface. There's nothing going on between the plates in terms of current. If I draw an even bigger one, like this one I'm again okay, because now that same current is passing through this one. It's this surface in between where things don't work.
01:04:44
You follow that? We have a crisis where if you say it's μ0I and you say it's on any surface with this loop as the boundary, this is the boundary of the surface, then this kind of surface, half in and half out, has a problem. So what will you do? You realize you have to modify your equations.
01:05:07
It's quite often how people modify equations. Sometimes they do experiments. Sometimes they do thought experiments. Einstein loved doing this thought experiment, which are called Gedanken experiments. You don't really do the experiment, but you say, if I did this, what will happen?
01:05:21
That's a little paradox. And you have to modify your theory. There are two reasons to modify the theory. One is experiments tell you it's wrong. Other is theory tells you there's a problem. So you have to add something to this. That something I add should not have any contribution from this
01:05:41
surface, but on this surface it should make exactly the same contribution as the physical current made on this surface, so that no matter which surface I take, I get the same answer. And I'm going to do that by the fault. There are many ways to do this. If you know more math, there are more ways to get this, but this is the one that's good enough.
01:06:01
I'm going to do the following. I'm going to rewrite this as follows. You see, in the region between the plates, I have no current. I agree. I give up. I do have something between the plates I don't have in the wire. You know what that is? Pardon me? You have an electric field.
01:06:21
There's something non-zero in between the plates. So I'm going to write this μ0I somehow in terms of electric field between the plates, and let's see how it goes. So I'm rewriting the very same term. I is dQ dt. Q is the charge in the capacitor. And let me write it as Q over ε0 d by
01:06:43
dt. And let me write it as μ0 ε0A d by dt of Q over A divided by ε0. That is equal to μ0 ε0A times d by
01:07:02
dt of the electric field between the plates. Why is that? Because electric field is ε over ε0, and ε is just Q over A.
01:07:21
Now bring that A in here, and you get μ0 ε0 d by dt of electric field times A. But that we can write as μ0 ε0, the surface integral of the electric field over the surface, and that's what we call μ0 ε0
01:07:42
electric flux d by dt. Therefore, my modified law is going to be v dot dl equals μ0 I μ0 ε0 d by dt of phi electric.
01:08:04
That's the bottom line. You can fill in these blanks if you have trouble writing and listening. I'm just saying go to the region between the plates. Once you start with dQ dt, turn Q into the sigma, the charge density on the plates. That's simply related to the electric field. Then you'll find this is simply the electric flux rate of
01:08:22
change times μ0 ε0. This extra term is called the displacement current. I don't know why it's called that name, but you know that you have to put that extra term.
01:08:42
But now look at this equation with that extra term. Notice it works everywhere. If you took a surface like the first one I took that slices through the wire here, there's no electric field anywhere inside a perfect conductor, or anything that's going to be no electric field, then μ0 I is going to contribute.
01:09:03
If I go to the region between the plates, there is no I there, but that's the rate of change of electric flux. Therefore, the last Maxwell equation, the Ampere's Law, is going to be modified with an extra term, μ0 ε0 dΦ electric
01:09:21
dt. And it has a nice symmetry now that the line integral of the electric field is the rate of change of magnetic flux. The line integral of the magnetic field contains the rate of change of electric flux. It's got other stuff, but it's got that extra term. So I'm going to come next time and start looking at these
01:09:42
equations. And remarkably, these equations imply there are electromagnetic waves. Without the benefit of any charge, without the benefit of any current, there will be electric and magnetic fields. That's the interesting part. We know electric fields can become produced by charges,
01:10:01
and magnetic fields can be produced by currents. But I'm saying go light years from everything. No ρ, no charge density, no current density. There will be non-zero E and B just moving on their own, and you won't understand how that happens. First of all, it's amazing it's predicted by the stuff. So basically, this is a very important day in your life.
01:10:21
Because now you know all of electromagnetism is that equation plus that equation. This is it. There is no more stuff any of us knows, at least in classical theory. In quantum theory, there's new stuff. Classical electromagnetism is only that. So you don't have to pack your head with all kinds of
01:10:41
results. You can derive everything from these. Okay, so let's do that on Wednesday. Thank you.