electricitymagnetism-10

ECE802 – Lec10

[00:00~10:30]

We have often talked about power supply which are the device which maintain a constant potential difference.  Here we have a power supply, potential difference V.  This is to be the plus side, and this is to be the minus side.  I am going to connect this……I have a resistor here, R.  And as a result of this, the current (I) will start to flow in this direction……this direction……this direction……this direction.  So in the power supply, the current flows in this direction.  So the resistance, the current flows in this direction.  In what the direction is the electric field?  Did electric field always runs from plus to the minus potential?  So right here in this resistor, the electric field is in this direction.  From plus, to minus!  But inside the supply, it must also go from plus to minus.  And so inside the supply, the electric field is in the direction that opposite the current.  To some kind of pump mechanism, must force the current to go inside the supply against the electric field.  A border does not all by itself move up to hill.  And so something is needed to push it!  Remember it’s the ventograph, we will bring charge onto the belt, and then we rotated the belt, and the belts forced the charge into the dump (電荷暫存的堆積處).  It has to overcome the repelling force of the dump.  So what have to be down?  So the energy must come from somewhere.  And in the case of the ventograph, it was clearly that the motor keeps the belts running.  In the case of re-inverse, it was I (the machine user; here means Dr. Walter Lewin) to turn the crank.  So I did work.  In the case of common battery is the one that you buy in the store.  It is the chemical energy that provides the energy.  And I will discuss that now with you and demonstrate a particular kind of chemical energy, which is one whereby we have a zinc and a cooper plates in a solution.  So we have here H₂SO₄, and we have here zinc plate, and we have here a cooper plate.  This side will become positive, and this side will become negative.  You will get a potential difference between these two plates.  To understand that, it real takes quantitative mechanic.  That goes beyond this course.  But the potential difference that you get is normally something around 1 volt.  The secrete really is not necessarily in the solution.  Because if you take two conductors, two different conductors and you touch them metal to metal, there will also be a potential difference.  So lets look this now in some more detail.  We have here a pool barrier that the I can flow freely from one side to the other, and we connect them here to the resistor, and so the current is now flowing.   The current is flowing in this direction through the resistor, from a plus side of the battery to the minus side.  That means inside the battery current is flowing like this, and the electric field (E) here is in this direction, from plus to minus.  But also inside the battery, the electric field must be from plus to minus.  So you see it again as we saw here that the electric field is in the opposite direction of the current.  You have here SO₄ minus ions, (SO₄^{– –}), and you have copper plus ions, (Cu⁺⁺), in this solution.  And here you have zinc plus Zn⁺⁺ and you have SO₄^{– –}.  And as current starts to run, SO₄^{– –} ions which are now to the current carrier inside this battery is going form the right to the left.  Now why were as SO₄^{– –} ions travel through the electric field that opposite them?  That opposite their motion.  And they do that because in doing so the engage in a chemical reaction which uses more energy then it cost to claim the electric hill.  And why a current is flowing?  Why the SO₄^{– –} is going from right to the left?  You get fewer SO₄^{– –} ions here.  This liquid here remains neutral, so Cu⁺⁺ must be disappear.  And it precipitates onto this copper bar.  So its likes copper plate.  On this side you get an increase of SO₄^{– –},  therefore you also must get in Zn⁺⁺ because again this liquid remains neutral.  And that means that some of the zinc is being dissolved.  So you get increase in the concentration of the zinc.  So the charge carrier inside this battery, the SO₄^{– –} ions travel through this barrier.  They go from here to here.  So they travel through the electric field that opposite that motion.  And this happens at the expends of chemical energy.  Now in the copper solution becomes very dilute.  Because all the copper has been plated onto the copper.  And when this becomes concentrated, the Zn⁺⁺, then the battery been stopped.  And now what you can do you can run the current in the opposite direction.  So you can running current now in this direction.  You can force a current to run with an external power supply and now the chemical reaction will reverse.  So now copper will go back into the solution so it will dissolve.  And now the zinc will be precipitate it onto the zinc.  And so now if you do this long enough, you can run the battery again the way it is here.  A car battery, is exactly this kind of battery except you have laid and ladoxide opposite side instead of zinc and copper, which also have sulfuric acid like you have here.  And the Nickel-Cadmium battery is well known you can charge that, too.  There is the one that regularly available in the stores.  You can run your flash light with the Nickel-Cadmium batteries.  The symbol for battery that we will be using in our circuits is this.  This is the positive side and this is the negative side.  This is the symbol that symbolize that we are dealing with a battery.  For that this point be B and this point will be A, and here we have a resistor R.  So we have a current going.  The current is going in this direction, the current I.  This could be a light bulb, could be your laptop, could be a hair dryer, whatever that is yours appliance.  If this R is not there, that means that the resistance is infinitely large (¥), that means that the current that is running is zero.  Then the voltage that we were measure over this battery which is V_B – V_A, for which I will simply write down V_b, that voltage we called curl ε.  We stand with EMF, which is an Electric Motor Force.  I will show you that later.  If I put a resistance R in here, which is not infinitely large, then the current will start to flow.  But now we should never forget, that between the point A and B invisible to the numan I, that there is always a internal resistance, which I called little r of i, (r_i).  And so if the current starts to flow, it goes not only through the capital R, but it also goes through to the little r.  And so according the Ohm’s Law,  the ε is now I, times the external resistance plus the internal one (R+r_i).  The voltage that you will measure between point B and A is now going to change.  That voltage according to Ohm’s Law is I*R and so is also the (ε – I*r_i).  And you see it is a little bit lower then the ε.  And the reason is this internal resistance here.  If I short it out this battery, stupid thing to do but if I make R equal zero.  So I take the battery and I just short it out, then the maximum current I_max that I can draw then.  So R is now is zero.  So you can see that the maximum I that you can get is ε divided by r_i.  And V_b, the voltage that you will measure now between point B and A goes to zero.

[10:31~20:44]

[20:45~30:34]

In case that happened, I will walk out of here (student laughing!) and I advice you to do the same.  Are you ready?  Ok! I go now.  You see, that is what happened!  It is a very high current!! And when you do this too often to your battery, then they are not going to live very long.  They don’t like it.  But I were joking when you work on your car that you should avoid this because I have seen it happened that wrench actually weld onto the terminal.  Electric company charges you for energy.  They don’t care about the power how many joules you used per second but they care about how much energy you using.  So they will charge you then for joules you think.  That’s energy.  How every if you look at your bill, You were been charged for kilo-watt-hours (kWh).  A kilo is a thousand, and an hour is 36,000 seconds.  So the unit of energy for which they charge you is this in joules (3,600kJ).  Two thousand watts, cooking plate, you run for 2 hours that is 4kW an hour.  They will probably charge you for 10 cents per kilo-watt a hour.  For the same amount of money you could run a 100W light bill for 40 hours.  Again, that will be the same 4kW a hour.  Or you could brush your teeth with your electric tooth brush for about 1000 hours.  Now I want to take a look with you at a network which consists of resistors and batteries.  This is the kind of stuff that you see on the homework assignments, and perhaps…… on exams.  So now we start out with a very modelest circuit.  Here we have a resistor R₁, here we have a resistor R₂, and here R₃, and then we put a battery in here, and we put a plus sign say on the left, plus side and minus side, and let the potential difference of this one is V₂.  It is real the EMF (ε).  But I will ignore any kind of internal resistance of the battery.  It is completely negligible in this problem.  And here I put also a battery, and this be the negative side and this be the positive side, and let the potential difference be V₁.  And so imaging you know V₁, V₂, R₁, R₂, and R₃.  But what I am going to ask you is what is I₁, what is I₂, and what is I₃.  I want the magnitude and I want the direction.  When you look at this, it by all means is obvious that the current in this resistor will be to the right or to the left.  That is by all means obvious.  Depends on the values of V₁ and V₂ and on the resisrtors.  The basic ideas behind solving this problem are the keys what we called Kirchhoff’s Rules.  Kirchhoff’s first rule is that the closed loop integral over closed loop of E×dl is zero, which we seen that before.  I don’t know why Kirchhoff get a credit for this.  It is always is the case when we dealing with conservative field.  We start at the particular point, you go around E×dl, you back the same potential you were before, so it must be zero.  As long as you deal with a conservative field.  But that is the first rule.  And you can do this closed loop anywhere.  You can even do it here, it still be zero.  You can do it here, also zero.  You can do it there.  No matter where you do it that closed loop integral must be zero.  Now there is a second key of Kirchhoff’s rule.  And that is what we called Charge Conservation.  If there is a static situation, then independent of which junction you go to, the current that flows in must flows out.  Can’t have a pirate of charge.  That’s the second rule.  And I gave you a problem 3-7 to work out.  And you can look into your book how that is done.  However, I am going to work on this with in a slidely different way then the book is doing it, which I persuade it likes better.  But it may confuse you.  So I warm you in advance.  You may not want to use my method at all.  What I do is the following.  I said ok, I assume that there is a closed loop current here I₁, and that is a closed loop current here I₂.  Weather I make them clockwise or counterclockwise, not important.  I could have chosen one clockwise, the other counterclockwise.  Not important.  However once I choose the direction, it has a consequence as you will see.  And that’s all that running.  One current likes this, and one current independently like that.  If I assume that then I automatically……automatically I am obeying the second rule because the current that goes around is charge conservation, right?  There is no charge palling out.  So the second rule of Kirchhoff is really obeyed.  So now I go to the first one, and I can start now at any point in that circuit and go around.  I can go around clockwise, I can go around counterclockwise.  There makes no difference as long as I return to the same point that integral E×dl must be zero.  I am returning the same potential.  What is the integral of E×dl in going from point 1 to point 2.  Well, that is the potential difference between point 1 (V₁) and point 2 (V₂).  And so let us start here and let it goes around.  And we have to adopt the certain convention mainly if you go up potential and when you go down the potential.  Again, you are free to choose the sign convention.  But I will say when I go up potential, I give that the plus sign, and when I go down in potential, I give that the minus sign.  So I start here.  I could have started there……I could have started there……make no difference.  As long as I don’t start here that make no sense.  So I start it here and I go around like this.  So right here, I go down in the potential V₁, so I get –V₁.  Now I go with current I₁ in the direction from left to right.  So that means that the potential here must be higher then there.  V = I*R.  Potential here must be higher then there.  So I go down in potential so I get –I₁*R₁.  Now I go through R₃.  This current I₁ is going down.  And this has a higher potential then here so I go down in potential so I get –I₁*R₃.  But, I have independently a current I₂ which is now coming toward to me when I go down.  And so if it comes toward me, that current will give me increasing potential.  This will have has a higher potential then this.  For this current to do this.  So now I climb up the potential hill.  So I get now plus I₂*R₃.  Oh~o~~ look what I did, I wrote down capital R.  Clearly I meant R₁.  There is no capital R in the whole problem.  Sorry for that!  You should read this as –I₁*R₁.  I am back where I was because this wire has no resistance.  So I am back where I was.  So this is zero.  One equation with two unknowns, I₁ and I₂.  So now lets go around this one.  We can go clockwise, we can go counterclockwise makes no difference.  Lets start here.  And I go in this direction once arou