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Class average on the exam was 55.

Three homework problems were on the exam.

I'm a little puzzled by the fringe field problem.

That was the one homework problem that was graded.

It was graded on your homework, and I went back to your homework scores and they were 90%.

You all did extremely well on that problem.

But on the exam, it was only 40%.

And so perhaps this is telling you something.

And maybe it's telling me something too.

If you turn in a correct solution for homework, it's not very useful if you do not understand what you wrote down.

The understanding, of course, is what matters, not that you somehow, one way or another, get a correct solution.

If I had to judge you only on the basis of exam one and two, forgetting the quizzes, forgetting the homework, then those with 80 or lower would fail the course, as of now.

But of course those who have between 80 and 90 are by no means home free.

They are still in the danger zone.

And anyone who has 95 or even 100, I cannot guarantee you that you will pass the course.

The depends, of course, on how you will do in the future, the third exam and on the final.

Today I want to discuss with you RC circuits.

We already discussed RL circuits.

Now we get RC circuits.

I have here a battery with an EMF V0.

And I have here a switch connecting to a capacitor.

And here a resistor.

And I close the loop.

When I have the switch in this position, the capacitor is going to charge up.

I get a current I going in this direction.

If I call this point A here and this point P and this point S to make sure that we are on the same wavelength, I will call the potential over the capacitor, I will call that V A minus V P.

The potential over the resistor is of course always I times R, is then V of P minus V of S.

The question now is what is the potential over the capacitor doing as a function of time and what is the current doing as a function of time as I charge up this capacitor.

And your intuition will help you a great deal without any fancy differential equations.

It is clear that at T equals 0, the capacitor is not charged.

There is no charge on the capacitor.

And it will take time to charge the capacitor.

So at T=0, you expect that the potential over the capacitor is 0.

If you make T a little larger than 0, you are going to charge up this capacitor, and so the potential over the capacitor will go up, and therefore the current will go down.

And if you wait long enough -- we call that infinitely long -- then the capacitor will be fully charged.

It will have the potential V0 of the battery, and then the current has become 0.

No current is flowing anymore if the battery -- if the capacitor is fully charged.

And so this capacitor is going to charge up, this becomes positive, and this becomes negative.

And so you can construct a, a plot whereby you plot as a function of time here the potential over the capacitor -- we haven't used any differential equations yet.

You know that if you wait long enough you will reach that value V0.

And it's going to build up like this, asymptotically reach that value.

And if C is very large, then the current will be more like this, and if C is very small, then it will go much faster, of course, small C.

The current as a function of time.

In the beginning, the current will be high, but ultimately the current will die down, when the capacitor is fully charged.

So you expect something like this.

So this you can do without any differential equations.

Let's now do it the correct way.

The closed loop integral of E dot dL happens to be 0, which will make Mr. Kirchhoff very happy.

The closed loop integral of E dot dL in this circuit is 0.

Mr. Faraday is happy and Mr. Kirchhoff is happy.

There is no magnetic flux change here; we don't have self-inductances.

The electric field inside the capacitor is in this direction, from plus to minus.

The electric field in the resistor is in this direction, the current is flowing in this direction.

And this battery, which has this as the positive side and this as the negative side, inside the battery the electric field is in this direction.

So if I start at point A and I go around the circuit, then E and dL are in the same direction if I go from A to P, and so I get plus V of C, that's the E dot dL from A to P.

Then I go through the resistor.

Again, E and dL are in the same direction, so I get plus I times R.

Then I come through the battery.

Now the electric field is opposing the direction in which I go, so I get minus V0.

And that is 0.

So that's my differential equation.

And I can write it differently, because I know that I, the current, is dQ/dT, Q being the charge on the capacitor.

And it is only if that number changes, if the capacitor is either charging up or discharging, is there a current flowing.

And in addition I know that V of C is Q divided by C.

That's the definition of capacitance.

And so I can write down for this V of C, I can write down Q divided by C.

For I, I can write down dQ/dT.

So I get R times dQ/dT minus V0 equals 0.

And this is a differential equation in Q.

We have seen an identical differential equation.

It was not in Q, it was in I, but it had of course a completely similar solution.

And the solution to this differential equation is actually quite simple.

I will put it on this board.

Q -- so this is going to be Q as a function of time -- is V0 C times one minus e to the minus T over R C.

If I take the derivative of this Q then I have I, because I is dQ/dT.

So I, which is dQ/dT then becomes V0 times C, I get a minus sign, I get another minus sign, then I get one over R C, and then I get e to the minus T / R C.

So the two minus signs eat each other up and I lose one C here.

And so I get that the current as a function of time is V0 divided by R times e to the minus T divided by R C.

So that's the curve at the bottom.

And the potential over the capacitor is now very simple, because the potential over the capacitor is Q / C.

And I already have Q here, so I simply have to divide this C out.

So I get V0 times one minus e to the minus T / R C.

And so that is the upper curve.

And so we can now make a small table and we can look at various values for T.

Maybe I should do that here on the blackboard, because I don't want to erase anything yet.

So we have T here, we have I, and we have V of C.

And when T is 0, you go to your equation of I, so this is one.

So you have a current V0 divided by R.

And your V C -- V C of C is 0.

You can see that.

If T is 0, you get 1 - 1 -- oh sorry, I have to go here.

You get 1 - 1, so you see indeed that the potential over the capacitance is still 0.

If you wait long enough, then the current must go to 0.

That exponential function goes to 0 if you wait long enough.

And your potential over the capacitance then reaches V0, which is exactly consistent with our solution.

If you wait a time RC, that's called the time constant of this circuit.

In the case of the current, it's also called the decay time of the circuit.

Then of course your current is one over e times V0 / R.

And one over e is a roughly .37, 0.37.

So your current is down to 37 percent of what it was at the beginning.

And after a time R C, the potential over the capacitor is one minus one over E times V0, and that is then about 0.73 -- uh, .63, 0.63, 63 percent.

In other words, if I go back here to my plot and if I draw a line at time R C, then this value here is 37 percent of the maximum and this value here is then 63 percent of the maximum.

So it's in -- the solution is rather obvious, very intuitive.

And the R C times can vary an enormous amount, as you can imagine, depending upon the values for R and C.

If we have here an R and a C and we want to know what the R C time is, convince yourself that the product of R and C indeed has units of seconds.

Remember we had L over R before, which had also units of seconds.

R C also has units of seconds.

If you have R equals 1 ohm and C is 1 microfarad, then the R C time is only a microsecond.

But if you have R equals 100 megaohms and you have this 1 millifarad, then this is 10 to the 5 seconds, which is longer than a day.

And that would mean that it would take you even three days to reach 95 percent of V 0.

After three days, you would still have only 95 percent of the potential difference of the capacitor of the maximum value that you can get.

Now what I want to do is I make a change here.

I have here a conducting wire.

I call this position 1 of the switch.

And I'm going to put the switch in this position, position 2.

And I can do that without any danger.

Remember, I waited until this capacitor was fully charged.

There was no current running.

And so when no current was running, I can quietly take the switch and put it in this position.

And what is going to happen now is of course this side is positively charged and this is negatively charged, so now you're going to get a current which is going to run counterclockwise, in opposite direction.

And what is going to happen is the capacitor is discharging now.

And the resistor will dissipate the energy that is in the capacitor.

The one half C V squared energy stored in the capacitor is going to be dissipated in the resistor in terms of I squared R, in terms of heat.

And if you wait long enough, the current will become 0.

So it should be obvious what's going to happen.

If I return to this curve here, if I redefine my time equals 0, and if this is the moment that I put the switch in position 2, then I expect that the capacitor will discharge.

You get a curve like this.

And I expect that the current, which now becomes negative -- it reverses direction and I call that negative.

And so the current will come like this.

And if you wait long enough, of course, the current will aga- again become 0.

You have discharged the capacitor.

So if you want the formal solution, you have to go back to the differential equation.

And you take this term out, because it's not there.

And now you have to solve this differential equation again, which is now utterly trivial.

And I would like you to solve that differential equation.

You couldn't have an easier one.

I will give you the solution to I as a function of time, and you then will come up with this part.

I as a function of time is exactly the same as this except with a minus sign, provided that I call this T equals 0.

So I redefine the 0 time.

And so I as a function of time is the equation that you have here, but now with a minus sign.

So you get an exponential change again, but the current has flipped over.

I can demonstrate this to you.

I have a electronic switch, so I go between one and two -- every four milliseconds I throw the switch.

And so what I have is, as a function of time, what we call a square wave.

So this is my battery.

And this time here, from here to here, is eight milliseconds.

This is time.