Kelistrikan dan Kemagnetan
Topics covered: Review for Exam 3
Instructor/speaker: Prof. Walter Lewin
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These are the subjects will be covered during our third exam.
There's no way I can cover all during this review.
Nor can I cover all of them of course during the exam.
I can only touch upon a few of them.
And what I cannot cover today, what I will not cover today, can and will be on the exam.
Let's first look at magnetic materials.
Magnetic materials come in dia-, para- and ferromagnetic materials.
The molecules and the atoms in para- and ferromagnetic materials have intrinsic magnetic dipole moments.
These have always -- they're always a multiple of the Bohr magneton.
Has to do with quantum mechanics.
It's not part of 8.02.
And they are going to be aligned by the external field, I call that um the vacuum field.
And the degree of success depends on the temperature and on the strength of that external field.
The lower the temperature, the easier it is to align them, to overcome the thermal agitation.
And above a certain temperature which we call the Curie temperature, ferromagnet- magnetic material loses all its qualities and becomes paramagnetic and I have demonstrated that during my lectures.
Suppose we have a solenoid and the solenoid has N windings, and the length of the solenoid is N.
And the current I is flowing through the solenoid.
Then the magnetic field generated by that solenoid which I have called the vacuum field, that magnetic field can be derived using Ampere's law, which you see down -- down there.
That magnetic field is approximately mu 0 times I times N divided by L.
If now I put in here ferromagnetic material then I have to include this factor kappa of M or K of M, whatever you want to call it.
The magnetic permeability, and this can be huge.
This can be 10, 100, even up to 1000 and higher.
So you get an enormous increase in magnetic field strength.
Self-inductance is defined as magnetic flux divided by the current I.
That's just the definition of self-inductance.
If the magnetic field goes up by a factor of kappa M then of course the magnetic flux will go up by the same factor and so the self-inductance will go up.
And you may remember a demonstration that I did when I had an iron core which I moved inside the solenoid and depending upon how far I moved it in could we see that the self-inductance went up and when I pulled it out self-inductance went down again.
We have an interesting problem.
I think it is assignment seven, whereby we have iron core here and then we have somewhere an air gap and you may want to revisit that to refresh your memory.
Let's now turn to transformers.
A transformer often comes in this shape.
Let me move it a little bit to the right.
Often comes in this shape which is then ferromagnetic material, to give perfect coupling between the left and the right sides, also increases the magnetic field.
This is the -- let's call this primary side.
N1 windings, index, self-inductance L1.
And here I put in a voltmeter to always monitor that value, I call that V1.
And this is the secondary side.
N2 windings.
Self-inductance L2.
And I put here a voltmeter which always monitors that voltage and I call that one V2.
You can show with Faraday's law as I did in class, in lectures, that V2 divided by V1, let's not worry about plus or minus signs, is N2 divided by N1.
That's a good approximation.
Depends on how well the coupling goes.
It depends on several factors, but you can come very close to this and this means then that if you make N2 larger than N1 then you can step up in voltage, we call that a step-up transformer.
But you can also step down if you make N2 smaller than N1.
Under very special conditions will the power generated on the primary side be all consumed for 100% or nearly 100% on the secondary side.
That is very, very special.
If that's the case then the time averaged power here V1 I1 is the same as V2 I2, here time averaged.
And so as a logical consequence of that you'll find that I2 divided by I1, let's not worry about minus signs, is that N1 divided by N2.
That, however, is not so easy as you may think.
It only can work approximately and I mentioned that on the side in my lectures.
If the resistance here and the resistance there is way, way smaller than the value for omega L.
And we did try to achieve that during one of the demonstrations that I gave on this.
I remember we had the induction oven whereby N2 was 1 and N1 was very large, I don't remember what it was anymore but it was of the order of several hundred, maybe a thousand, and we managed to get a current in the secondary which was huge, which was close to 1000 amperes.
It was enough to melt that iron nail.
And we made every effort then to make sure that the resistance was much, much smaller than omega L.
I think problem 7-1 of our assignments deals with that, and very naively assumes that this is all true.
But you should realize that it is not always so easy to achieve the conditions for that.
So let's now go to RLC circuits there.
Let's take an, uh, system which has a resistor R, it has a self-inductor, a pure self-inductor, L, and a capacitance, C.
AC.
And this driving power supply provides with a voltage, V, which is V0 cosine omega T.
Keep in mind that this can be always be sine omega T of course.
There is nothing special about cosine in life.
The steady state solution, that is not when you turn the thing on but if you wait awhile, you get a steady state solution for the current.
And the current that is going to flow now is V0 divided by the square root of R squared plus omega L minus one over omega C squared times the cosine of omega T minus phi.
And the tangent of phi is omega L minus one / omega C divided by R.
We call this the reactance.
The upstairs.
For which we give often the symbol X.
And so this is also X then divided by R.
And this whole square root that we have here, we call that the impedance.
The units are ohms.
And we call that Z.
And so the maximum current that you can have, the current is of course oscillating with angular frequency omega, the maximum value that you can have for the current, which I call I max, is then V0 divided by Z.
Then the cosine term is either plus or minus 1.
I can plot now this I max as a function of frequency.
So here is frequency and here is I max.
If the frequency is very low or near 0, then this term here becomes infinitely high because the impedance is infinitely high and so the current is 0.
I max is 0.
There's no current flowing at all.
When we go to very high frequencies it is the omega L term that goes to infinity.
And so again Z goes to infinity so again I max goes to 0.
And for other values of omega you get an I max which is not zero and so you get a curve like this which has the name of resonance curve.
This I max reaches a maximum value when the system is at resonance, that's what we call resonance.
And that's the case clearly when the reactance is 0.
Because when the reactance is zero this part vanishes.
And if the reactance is not 0 then the maximum current can only be lower, can never be higher.
And so when X equals 0 you'll find that omega L is 1/omega C and so the -- the frequency for which that happens, I call that omega 0, reminds me that it is the -- the resonance, is one divided by the square root of LC.
When I am at resonance, phi becomes 0.
So there is no phase delay between current and the driving voltage.
They are in phase with each other.
And the value for I max now simply becomes V0 divided by R.
Because the impedance itself becomes R.
Very boring, very simple, you're looking here at Ohm's law.
When the system is at resonance, forget the self-inductance, forget the capacitor, they are not there, they annihilate each other, and so the system behaves as if there were only a resistor, and that's exactly what you see here.
I have here some numbers which you have seen before.
During my lectures.
You can download this from the Web but you have to go back to the lecture when I discussed that.
And you see here s- some numbers for R, L and C and also for V0.
And I calculate for you here the resonance frequency.
I calculate the frequency also in terms of kilohertz.
And here you see the impedance and here you see the reactance.
If I'm 10% below resonance notice that the 1 / omega C term is always larger than omega L.
So your reactance in this case becomes -86 ohms.
The minus sign has of course no consequence for the current because you have an X squared here.
But notice that Z is now almost exclusively determined by X and not by R anymore.
Because the 10 ohms of the R here play no role, almost no role, in comparison with the 86.
Z becomes 87 and the maximum current is one-tenth of an ampere.
When you're on resonance, and that is characteristic for on resonance, the 2 omega L and 1/ omega C eat each other up.
They annihilate each other and so the reactance becomes 0.
So Z now is just pure R.
X is 0.
And so the maximum current in this case is 1.
Because I chose V0 at 10 and I chose R 10.
And then when I'm 10% above resonance then the omega L term is larger than the reactance of the capacitor and accordingly you get a lower current again, about one-eighth of the -- of an ampere.
And so you see this curve being formed in a very natural way and that's quantitative, you see there some numbers.
So now comes the question which of course in practice is very important.
And that has to do with the power that is generated by the power supply.
That power comes out in the form of heat.
Heat in the resistor and so if you time average the power, then the time average value, you can take the -- the voltage of the power supply, multiply that by the current.
You could also take the time average value of I squared R.
Because all that energy will ultimately come out in the form of heat of the resistor.
Either one will be fine.
I've decided to take this one.
So I will get then V0 cosine omega T -- the I becomes V0 divided by Z times the cosine (omega T - phi).
This is the power at any moment in time.
I will do the time averaging a little later.
When I see cosine (omega T - phi), that reminds me of my high school days, cosine alpha minus cosine -- no, cosine alpha minus beta is cosine alpha cosine beta + sine alpha sine beta.
That was drilled into my memory here.
I will never forget that, I think.
And so I will write down here -- my math teacher will be proud of me -- cosine omega T cosine phi plus sine omega T sine phi.
So this is this term.
If I'm going to time average it, I have a cosine omega T multiplied by sine omega T, that time average is 0.
So this term vanishes.
So the time average value of the power, I get a V0 squared, I get a Z here, and now I have here cosine omega T times cosine omega T.
The time average value of cosine squared omega T is one-half.
So I get a two here.
And then I still have my cosine phi there.
And I'm done.
If you like to get rid of this cosine phi, you can do that.
Because remember, the way that phi is defined, the tangent of phi is the reactance divided by R.
You still see it there.
So that means if this angle is 90 degrees that this side must be Z.
That's the square root of X squared plus R squared.
That's this part.
And so the cosine of phi is also R divided by Z.
And so if you prefer that -- there's no particular advantage but if you prefer that you can write down for cosine phi R divided by Z.
And so you get V0 squared times R divided by 2 and now you get Z squared.
And so there you see the power, time averaged power in an RLC circuit.
So now we can look at resonance.
It's always a very special situation.
When we are at resonance, Z equals R.
So you replace this capital Z by R.
And then you find V0 squared divided by 2R.
That's utterly trivial.
You could have predicted that.
It's really Ohm's law staring you in the face.
There is no self-inductance and there is no capacitor at resonance.
So you might as well have treated it as a simple system only with R.
And you find immediately then that answer.
At any other frequency than omega 0, Z would always be larger than R.
You see that immediately here.
And so that means that the average power would always be lower.
So it's only at resonance that you generate the highest power possible.
All right.
Pengembangan Perkuliahan
1. Buatlah sebuah Esai mengenai materi perkuliahan ini
2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini
3. Lakukan Penelitian Sederhana dengan kelompok tersebut
4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat
5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnyaStaff
Visualizations:
Prof. John Belcher
Instructors:
Dr. Peter Dourmashkin
Prof. Bruce Knuteson
Prof. Gunther Roland
Prof. Bolek Wyslouch
Dr. Brian Wecht
Prof. Eric Katsavounidis
Prof. Robert Simcoe
Prof. Joseph Formaggio
Course Co-Administrators:
Dr. Peter Dourmashkin
Prof. Robert Redwine
Technical Instructors:
Andy Neely
Matthew Strafuss
Course Material:
Dr. Peter Dourmashkin
Prof. Eric Hudson
Dr. Sen-Ben Liao
Acknowledgements
The TEAL project is supported by The Alex and Brit d'Arbeloff Fund for Excellence in MIT Education, MIT iCampus, the Davis Educational Foundation, the National Science Foundation, the Class of 1960 Endowment for Innovation in Education, the Class of 1951 Fund for Excellence in Education, the Class of 1955 Fund for Excellence in Teaching, and the Helena Foundation. Many people have contributed to the development of the course materials. (PDF)
Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.
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