 Hi, I'm Zor. Welcome to Inizor Education. We continue talking about alternating current. In this particular lecture, we start basically a new chapter where we will talk about the Ohm's law for alternating current. And that would be related, obviously, with circuits which include not only resistors like in a direct current, but also inductors and capacitors. This lecture is part of the course, Physics for Teens, presented on Unizor.com. I suggest you to watch this lecture from the website. So you go basically to the menu on the left where you choose Physics for Teens. This is electromagnetism part of this course and the chapter is called alternating Ohm's law for alternating currents. Okay. Now it's really kind of involved topic and it has a lot of math which I actually consider to be a good thing and it brings certain precision. So we will talk about different alternating current circuits. We have three major elements. We have resistors, we have capacitors and we have inductors. Which can participate in the alternating current circuits. So I will dedicate this particular lecture to the Ohm's law if the circuits, AC circuits, alternating current circuits, include only resistors and resistor and capacitor in a series. Then the next lecture will be about resistor and inductor and the next one after that would be resistor, inductor and capacitor. So just a little bit more and more complex kind of things, but very much look alike. Now we do know that alternating circuit can include capacitors and inductors and direct current doesn't really go through capacitor. Alternating does. Similarly for inductors direct current basically goes through without any kind of interesting effects. While AC, the alternating current, it creates variable magnetic field and there are certain forces involved in this particular case as well. So it's kind of involved as I was saying but interesting and it's also a good exercise in math because we will solve differential equations and I might actually recall some complex numbers, theory, the Euler's formula, etc. Okay, so let's start. Now my first topic here is in this lecture the AC circuit with only a resistor. Well, that's the easiest topic here. So if you have an alternating current and resistor, now alternating current, this is the generator of alternating current and it's actually reducing the EMF, electromagnetic force, electromagnetic force has sinusoidal oscillations. Omega is the angular speed of rotation of the rotor inside of this generator. So my current goes up and down in some sinusoidal wave. T is obviously time and E0 is a peak EMF, peak electromagnetic force. Which is developed by this particular generator. So all the generators which we are dealing with are actually generating this type of sinusoidal wave of oscillations because it's based on the rotation of the rotor inside this tether. We talked about how generators are actually made, right? Okay, so this is my EMF or voltage if you wish. And this is the only thing which we have in the circuit which is a resistor with resistance R. Okay, so now let's just consider one tiny infinitesimal moment of time between some moment and the moment plus certain infinitesimal increment. Now during this time basically we can consider the electromotive force to be constant and the regular Ohm's law for direct current actually takes effect. And there are no other forces, components, fields which kind of take part in this picture. Because it's just a resistor, which means it just basically resists the electricity and the Ohm's law at that particular moment of time works quite fine without any kind of modifications. So this is the voltage, this is the resistance and this is the current. Now obviously since this is sinusoidal, this will be sinusoidal as well and I of t would be equal to E0 divided by R sine of omega t. So current goes in sync with the electromotive force with voltage. The amperage is in sync with voltage and this is the maximum the peak amperage. So this actually can be represented, well it actually represents basically the equivalent of the Ohm's law for alternating current. Now we also have a concept of effective current and effective voltage if you remember. Now the effective one that would be if we will average square of voltage or average square of amperage and we talked about this before and the peak voltage and amperage are related to the effective one as square root of 2. So if you remember effective voltage would be square root of 2. That's I effective and effective amperage would also be equals to square root of 2. So from this follows that since I0 is equal to E0 divided by R, peak amperage is equal to peak voltage divided by resistance. So effective amperage would be equal to effective voltage divided by resistance, right? Because each one of them is just square root of 2 smaller than the peak. So this is also an equivalent for AC for alternating current equivalent of the Ohm's law. Well, that's it for plane R, plane resistor included into this circuit. That's very easy. So this this formula and this formula, these two formulas represent basically equivalent for AC of Ohm's law for direct current. Nothing here. It's really very simple. But what we now will do we will include the capacitor here and that makes the whole difference. Okay. With capacitor we do have a lot of other things which are happening now. Here we have to go back and recall what we were talking about, about capacitor in the alternating current circuit. We actually had investigated how this type of circuit behaves. So when I was introducing capacitor in the alternating current circuit I did not have any resistor. And we did have whatever the research we have done. You can refer to that lecture. I will repeat a little bit because I will need exactly the same kind of formulas. But it makes really a very important difference, the adding of the resistor. Well, it seems that this circuit should behave like this one because actually it's more or less we're just adding a very simple device in this circuit resistor, which is just like slowing down the electrons or whatever you think about this resistance. But it introduces a significant other detail which we will address. Okay. So the first thing we do is let's consider something which we called voltage drop. Now voltage drop, now this thing is generating certain voltage. This is EMF, voltage if you wish. Now whatever is generated here goes through this circuit and there is a voltage drop on every component because there is a difference in voltage between these two plates and between these two ends of the resistor, right? So let's call this voltage drop Vc and this voltage drop would be Vr. So this EMF is generated, then the voltage drops here a little bit, then here a little bit and returns back. So what it means is that whatever is generated is the sum of these two things and these are all obviously depend on the time. Again, in the direct current when basically the electricity has a constant voltage, these are also constant and the same equation actually is true. Whatever is generated difference in potential should be equal to sum of difference in potential between this and this and then this is this. So at any moment in time we have exactly the same kind of equation. So at any moment in time there is a difference in potential between electric potential between these two ends and that should be equal between this and this plus this and this. Or if you wish the whole sum of these should be equal to zero because this will be then with a minus sign. So the whole sum of all voltage differences between the ends if you will put this and put it on the right with a minus sign. It would be zero. Okay, so that's exactly the same thing. But this is a very important equation which we have to keep in mind. It's logic basically. It's nothing it's not derived from anybody. It's just a concept of voltage drop between ends of this device and this device and this device actually. Since it's a closed circuit, right? Okay Now let's just analyze each part individually. But with VR it's simple. If you have a difference in potential between this and this then as I was saying before there is an ohms law for the alternating current because at any moment in time we don't have anything but the difference in voltage and resistors. So the IR on T this is the current which goes through this is supposed to be equal to difference in potential on the ends of this which is voltage drop divided by R, right? Now let's just think about I will next talk about IC. So the current which goes through the capacitor but it's the same circuit, right? It's a closed circuit which means whatever electrons are coming here they're coming here as well. So I see the current through the capacitor and IR the current through resistor should be exactly the same because it's a circuit. It's a closed circuit. So I might actually drop the whole thing here and that would be exactly the same for this as well as for this resistor, right? Because the same electrons are going through this and this. Okay now let's talk about this one. So from this we can see that the voltage drop is equal to R times I of T, okay? So that's about this one. Now what about this? Well we have to consider what exactly is a capacitor. Some time ago we were when we were introducing the capacitor we were talking about if there is certain voltage between the plates there is a certain charge which is accumulated. So the QC of T is the charge between these plates and it actually grows with the voltage, the greater the voltage between these things. Well the more electrons they actually have to have. So the ratio is a constant which is called capacity. So the capacitor has certain capacity or capacitance whatever you prefer. So this is supposed to be given like R here, the capacitance here should be given. This is the characteristic of the capacitor, right? So this is a constant. The growing of the voltage between these two plates results in the growing of the charge between these two plates. So from this we can see that we can put it this way. This is equal to VC of T. So we have expressed VC, the voltage drop on the capacitor through this and the voltage drop on the resistor through this. This is the characteristic of capacitor, this is the characteristic of resistors. Well but we have these two different. Well they're not so different between let's just recall what is the current. Current is rate of change of the charge. So when this charge between these two plates is changing that's what actually generates the current. It's the rate of change of the charge. So what I would like to say is that I of T is equal to Q derivative. This is the rate of change, right? So derivative of the Q is actually I. Derivative of the charge accumulated on the capacitor's plates is exactly the current which is circulating in the circuit. And this is actually a definition of the current. From here we can actually make some kind of a differential equation for this. E of T which is E0 sin of omega T is equal to VC which is this one which is QC of T divided by C plus VR which is R times Q derivative of C of T divided by no, without any kind of equation. So that's my that's my equation. Now let me wipe out everything else. So all we have to do is solve this differential equation for Q, take derivative of Q and that would be my current. That's how I will connect my current with the EMF to produce something which is equivalent of the Ohm's law, right? So how can I solve differential equation? Well let's just simplify it a little bit. First of all let me just divide it by R. So this thing doesn't really be here, okay? Now let me just make certain simpler notation. So Y of T would be equal to Q C of T. 1 over RC would be A and B would be E0 over R. Then my equation would be Y derivative plus AY equals B sin of omega T. So that's my equation where A and B and Y are these guys. How can I solve this equation? Okay, first of all I can actually put the answer right now and refer you to notes for this lecture which contain actually the whole derivation. But I would like just to make some basically some introductory comments about how I can solve this equation. Here it is. It would be nice if this is not just the sum of derivative and basically the function itself with sky-efficient. It would be nice if it would be a derivative of something. So if it's a derivative of something and I know the function on the right, B is a constant, right? Then I would just integrate the right part to get the left. But this is not a derivative of something unless I will do something about it. Now let me just recall something. If you have X of T times Y of T and you take derivative of this, well if you remember the formula for the derivative of the product is XY plus XY or let me just change the order of this. Okay, does it look like this one? Well if I will multiply this by X of T, I will have X of T Y T plus A X of T Y of T, right? So how can I make this one into this? Well if this is a derivative of X of T, if this is X of T derivative, then I will have the same formulas here, right? So I have to find the function X of T derivative of which is A times this function. Well this is simpler. This is X. X equals E A to the power of A T because what's the derivative of this? Well this is X of T, I mean. Well derivative is the same function. It's an exponential function times the derivative of the inner A times X of T. This is the characteristic property of the exponential function, right? So that's exactly what I'm going to do. I'm going to use X of T which is equal to this and multiply both sides by it. So what happens? I will have E to the power of A T times Y of T plus A E to the power of A T Y of T equals B A to the power of A T sine of omega T. So I just multiply by E to the power of A T. Now I can replace this with E to the power of A T derivative, right? Because the derivative is this one. Okay? So let's do that. Okay? And what is this now? Well this is a derivative of the product E to the power of A T times Y of A T derivative. And it's equal to B E to the power of A T times sine of omega T. Now all I have to do is integrate this to get E to the power of A T times Y of T and divide by E to the power of A T, right? So right now I need an integral of this. Now this part again I will not go into the details. I'll just tell you how it can be accomplished. It can be accomplished by converting sine of omega T using the Euler's formula. Let me remind you the Euler's formula. Cos sine phi plus i sine phi is equal to E to the power of i phi. Remember this? Again go back to the complex numbers in the course. On the same website there is a mass for teens course, prerequisite for this one where it's all explained among complex numbers and trigonometry functions. Okay? So basically if you will replace from here you can get the value of cosine. Well let's do this. Minus i, well let's put it this way, plus sine of minus phi equals E to the power of A T, right? Minus. Why? Because cosine of minus phi is the same as cosine of phi, right? So all I did I put minus phi into this formula and I get this one. But now I can just get rid of this sine because the cosine is an even function. This guy goes out because sine is odd function and I have this. From here I can get the value for cosine which is equal to, so cosine of phi is equal to this plus this. It will cancel out divided by 2. So E to the power i phi plus E to the power minus i phi divided by 2. And sine of phi is equal to how to get sine. I have to subtract and I will get 2i. So it will be E to the power i phi minus E to the power minus i phi divided by 2i. So I can replace sine with the difference between two exponential functions and that means that the whole thing becomes, well difference between two exponential functions and the whole integral actually becomes integral of some exponential function. And phi would be in the exponent part of this thing, right? Because if you will multiply it by E to the power of i phi it will have E to the power a t plus i phi, right? So it will be just plain exponential function. So from now on I will just give you the result of integration of this thing divided by E to the power of a t. So I will give you the result for y of t. And again details you can find in the notes for this lecture. So y t the result is equal to b a sine omega t minus omega cosine omega t divided by a square plus omega square plus constant. Since we are integrating, right? There is a free constant here. Now this is y of t which is this one. Now derivative of this is i of t, the current, right? So before we are differentiating this by the time what I can do I can simplify it a little bit. Look what's inside. Let me just divide this thing into two parts. Okay that's the same thing, right? And a divided by square root of a square plus omega square. I will call a sine of some angle phi and omega divided by square root of a square plus omega square. Cosine of phi. Where phi is basically if I will divide one by another I will have a tangent, right? Tangent sine divided by cosine is a tangent of phi. So tangent of phi is equal to this divided a divided by omega. So I'm just introducing a new variable a divided by omega where a is this one and as a result I will have the tangent. Now from this from this I can rewrite this thing using these types of expressions. So what will I have? I don't think I need this anymore. I will have this thing remains. So b divided by square root of a square plus omega square. Now this would be sine times sine minus cosine times cosine which is a formula for cosine of an sum of two angles with the minus sign, right? So I will have sine phi sine omega t minus cosine phi cosine omega t. That's what I will have, right? a divided by this would be sine and omega divided by this would be cosine. Okay which is equal to b divided by square root of a square plus omega square times. Now what is this? This is minus cosine of omega t plus phi. So that's what my y of t is, plus some kind of free constant which we don't care right now because we will take the derivative anyway from this from y to get the current. Well that's what it is. Now let's take derivative now. So i of t would be derivative of this, right? The y of t is the charge. Rate of change is the current. So this is b divided by square root of a square plus omega square. Derivative of cosine is minus sine, but another minus would be the plane sine of omega t plus phi and times the derivative of inner function which is omega. So that's what we have derived this. Now let's go back to original constants. Instead of a and b we have to substitute whatever is necessary. So equals to b is e0 omega divided by r square root a square 1 over r square c square plus omega square times sine of omega t plus w equals 2. Okay if we will put r inside we will have this r square and r square, right? So r square would go here. Now if we will take omega out of this I'll have to put omega square here, right? In the denominator. So what I have as a result as a result I have r square plus 1 over, so let me rewrite it, r square plus 1 over c square omega square, right? That's what we have. And now omega cancels out and that's my result. So if you recall this thing 1 over c omega is called reactance of the capacitor and it's actually measured in ohms. So it's the capacitor's equivalent of the resistance. So capacitive reactance, capacitive reactance is functionally equivalent to resistance for resistor. We talked about this in the previous lecture. So instead of this I will put much more convenient xc. This is the resistance and this is the capacitive reactance and this is the main formula which basically gives us the ohms law. This is i of t. So obviously this is i0 the peak of the sinusoidal oscillation. So we have the peak is equal to peak divided by this r square. It's kind of sum but it's not just a sum. It's a quadratic sum of the resistor's resistor and capacitors' capacitive reactance. They are the same functionally, the same kind of characteristics. Whatever is for resistor, this is for capacitor. Well obviously since it's plain sinusoidal with the same, by the way, the same frequency, you can go into effective because they are only a square root of 2 smaller than the peak. So these two kind of representing the ohms law equivalent for ac formulas. And in this case this square root of r square plus xc square plays the role of a well resistance of the whole circuit which contains a resistor and capacitor. This is the resistor's resistance. This is the capacitor's reactance. But what's interesting is this phi. What is phi? Phi we were talking about. Phi was arc tangent of what? A divided by omega. Okay, instead of A we can put 1 over rc. So it would be this. Now 1 over omega and c is our reactance. So I can put instead of this, I can put this. Now it looks much better, right? So you see this is xc and this is xc which is actually 1 over omega c. Now this is the phase shift. So you see if my initial oscillations are like this the oscillations of the current are shifted. So the graph of this is just shifted by phi to the left whatever the phi is. And phi is some kind of a, now as a very quick check what if there is no r, no resistance. If you remember we were talking about this problem when I introduced the capacitors in the AC circuit. And we had very similar formula when the peak current was e0 divided by xc. Let me just remind you. The formula without resistor was i of t is equal to e0 divided by xc times sine of omega t plus pi over 2. Now that formula was derived when I introduced the capacitor in the AC, in the AC circuit. Now we have slightly different formula if there is a resistor. But what if the resistor doesn't exist that this formula corresponds to this? If r is equal to 0, well let's think about it. If r is equal to 0 then this would be square root of xc square which is this which is correct. Now if r is equal to 0 then this is infinity. And what is the angle tangent of which is equal to infinity? That's pi over 2, 9 to the degree, right? So that corresponds. My phi would be equal to, in this formula would be equal to pi over 2. So it corresponds to whatever we have derived previously for plain capacitor without any resistor. So resistor gives a different shift. Capacitor by itself gives the shift by pi over 2. But if there is a resistor as well then the shift would be different. It's kind of hard to explain for me why it happens from purely physical standpoint. But whatever it is it is. Maybe you can think about it like, I know it's slowing it a little bit. I don't know how to explain it from the physical standpoint. But addition of the resistor shifts the phase from pi over 2 to something other which in case of no resistor would actually be pi over 2. But in case of there is a resistor it would be a different angle. Well that's it basically. I do suggest you to read the notes for this lecture. They are much more detailed and the differential equation is fully solved, explained. So it's a good exercise in math as well. And well you know my opinion you can't go into physics without math. So on this note I thank you very much for listening to me. Read the notes for this lecture and good luck.