In this lecture, we're going to start working on AC circuit analysis. And so the first thing we have to do is talk about two new circuit components, capacitors and inductors. And then, after introducing those, we're going to talk about decomposing signals into their sinusoidal components and the idea of phasors. Then with all of those tools, we can start to do some AC circuit analysis, and then using those techniques, we're going to look at a few simple filters. Low pass and high pass filters built out of resistors, capacitors, and inductors. We'll start by talking about capacitors first. So, capacitors are nothing more than a pair of metal plates that are separated by either a vacuum or maybe some kind of a dielectric material. And there's say a gap, d, in the plates, and on those plates, we can place some charge. So we can put plus charge on one plate and minus charge on the other plate, and you know that if I segregate charge in that way, I get an electric field pointing from the plus to the minus charge, and so there's a, there's an electric field contained inside the capacitor. Now, the electric field in the capacitor gives rise to a voltage difference between the terminals. So this is a device that when I put charge on the two plates, there's a voltage difference between the two terminals. Now, energy is stored in a capacitor in the form of the electric field contained in the dielectric material, and you remember from before that the electric field is the voltage divided by the distance from the the plus charged to the minus charged plates. Or we can rewrite that as the voltage is the electric field times that distance, or the gap in the capacitor. Now the capacitance of this structure Is the ability of this device to store a charge, or the capacity of the device to store a charge. Now, we define capacitance as the number of Coulombs per volts. So, it's Q over V. And capacitance is measured in farads, and so, this device has a capacitance, c, we'll cal it, and it's defined by this relationship, q is c times v. So, if I have, if I take this device and let's say I put one volt of potential difference on the two plates, then there will be some amount of charge stored in the device that is equal to the capacitance times that voltage. So a one-farad capacitor means that if I put one volt pepit, potential difference here, I'll get one coulomb of stored charge. Now a 1 Farad capacitor is a very big capacitor. typical capacitor values go from just a few picofarads, 10 to the minus 12 farads, to maybe a few hundred millifarads, 10 to the minus 3. And picofarad-type capacitors are, you'll see them in various forms in electronic circuits. capacitors with leads maybe a few millimeters or maybe 5 millimeters across. the surface-mount capacitors are typically quite small, maybe 2 or 3 millimeters in size. And so these are on the small end. Now, millifarad capacitors or hundreds of millifarad capacitors are the, the big ones that if you open up a piece of equipment, you see some big capacitors in there that are part of the power supply, typically, and so these may be several inches high and a few inches in diameter. And this is just very large coil, of two metal electrodes, and some a kind of dialectic material, often there's there's a a liquid in the servicing as the dialectic material, and the, these are. Actually, fairly formidable devices and you have to be careful if you charge one up to substantial voltage and, it holds a lot of charge and you can get quit a shock from one of these. So you have to be careful with those. And the other thing, I'll tell you this from personal experience, is that, when you think you've discharged it, say touching 2 terminals with a screwdriver. you better check twice, because I it knocked myself off a chair once by touching the one of the electrodes of a large capacitor that I thought was discharged. I discharged it but it wasn't discharged all the way, and there was still enough residual charge to knock me off a chair. Okay, now I want to look at the current voltage relationship for a capacitor. We gave you the charge voltage relationship for a capacitor which is really just the equation that defines the capacitance of that device. Now let start with Q equals cv that;s just remember that equation that's that's the defining equation for a capacitor and I'm going to take and take the derivative of this with respect to time so dQ dt is current and. Now I'll assume that the capacitance, c, is fixed, it's not changing value. So it's just a constant. So when I take a derivative of this constant times the voltage, it's the constant times the derivative of the voltage. Now, let's use this equation to look at the transient behavior of a circuit with a resistor and a capacitor in it. So what I want to do is find the charge, q, and the voltage across the capacitor as a function of time. Now here's the simple circuit. I just take a battery, connect a resister and a compassitor in series, like this and then I have a switch and I'll say that at T equals zero, I'll throw the switch closed and let's say before I do that I want to make the point that the capacitor is initially uncharged. So there's no charge in the capacitor, and obviously there's no current flowing through the circuit, because this is open. But then at t equals 0, I throw the switch closed, and the battery is then going to force a current through this circuit. And the capacitor is going to charge up and its voltage is going to increase. And we want to know exactly how that occurs in time. I want to find this voltage, Vc, as a function of time. So, to do that, I'll start by using Kirchhoff's voltage law. So. this is right after the switch is closed. in fact, this is true at all times, but the, the voltage across the battery is, if I start down here, and I go up through the battery, I pick up a voltage V0. And then I go through the resistor and I drop whatever the resistor voltage is, then through the capacitor, drop whatever that voltage is and I'm back to zero. So this then, can be rewritten as vr plus vc equals v zero. Now, [COUGH] taking this VR the just using Ohm's law is just i times r so this is just the voltage across the resistor. VC is the voltage across the compasitor I can just go back here and write this v is q over c And so that equals V0. So, I have this equation that I have to solve for Q as a function of time. So, this is a simple differential equation that is not any different form the ones we saw, just previously. when we introduce the exponential function, the one difference is there's a term here on the right that we have to deal with. So, I want to say again that the initial condition of this one, as soon as I close the switch is there is no charge on this capacitor at t equals 0. And what we'll find is that the final state is that the capacitor will be charged up and the current will stop, and so once the current stops, there won't be any voltage drop across the resistor, and the charge, after a long time, it will just be C times 0. Now we'll find this, this is not an assumption, we'll find this from the solution. This is the assumption, the initial condition equals 0.