So the resistors in series, we know how they behave. R sub s equals R1 plus R2. Piece of cake. But in parallel, we have to add their inverses. One over Rp equals 1 over R1 plus 1 over R2. Here we added resistances, here we, in fact, added currents, and that led to the inverse of the resistor. For capacitors, we have exactly the opposite. So for series, we add the inverses, and for parallel we add the direct values. And so you can see the symmetry here, right? Those two behave the same, and those two behave the same. Resistors and capacitors behave exactly opposite in series versus parallel. So let's say we do the following. Let's- to start with, let's just say we do this. We take a battery, we have a switch and we have a capacitor. This device right here is just a switch, a break in the wire, I'm gonna close the switch. When I have the switch open like this, what is the current I in the circuit? Switch is open so there's no connection in this wire right here. What's the current I in the circuit at that point? What is it ninety percent of the time when I ask you a question? >>Zero. >> Zero, right? I is zero. There's no current flowing. Okay. Why is there no current flowing? Because the wire is broken. Those electrons that are trying to move through get to the end of the wire and they say 'I can't go any further. I'm gonna stop.' Alright but now we close the switch. So the switch is now closed, here's our capacitor. Is there I in the circuit? And to be fair, let's say we're looking right after we close the switch. We didn't have any current, and now we're gonna close that switch which connects the wire, is there going to be current in that circuit? What do you guys think? Yes. There is going to be current, because it's got to charge up the capacitor. It is trying to get some charge on that side of the capacitor, some charge on that side of the capacitor. So there has to be movement of charge. If there's movement of charge, there's current. Alright. Now we wait a long time. The switch has been closed for a while, a long time later. Is there any current flowing in that circuit? What is the current flowing in that circuit if we wait for a long time? Somebody wants to say it, they're on the- zero! Okay, it's zero again. Why is it zero? I mean, I've made this connection here right I have a voltage in this battery that wants to push current around, and yet we're saying if we wait a long time it goes to zero. Why is that? Because there is a gap right there. The capacitor is two parallel plates. And so if you wait until those plates get charged up, you can't put any more charge onto those plates, and that can't jump across from one plate to another because there's a gap there. There's no connection, literally it's a parallel plate, a space in air or some other material, and then there's another plate, another piece of metal and so it can't jump across that gap. So it goes for a while charging up the capacitor, But then later on it is fully charged. and so the current goes to zero. It's kind like when you charge up your batteries right? a rechargeable battery is very much like a capacitor. You put it in your charger, you close the switch, current starts flowing, After a while your battery is fully charged, there's no more current flowing you take it out, and you use it. Then you drain all the energy out, you do that again at the end of the day. Okay, let's do it now with a real resistor and circuit and see how it behaves. So let's draw the battery, we've got our switch, and then we're going to put a capacitor right there, and we'll put a resistor right next to it. Both of these are coming down and back to the negative side of the battery. Now when the switch is open there's no current flowing of course because there's a break in the wire right there. So as soon as we close the switch something's going to happen. The switch is closed, current starts to flow, and we can think about the charge on the capacitor. What is the charge on the capacitor? Well, we know that. Some Q naught is equal to C times V naught. If we open the switch after the capacitor is charged up, let's see if we can figure out what's going to happen. And I guess we call this initial voltage V naught on the battery, and you'll see why we did that in a second. So when we open the switch, The current in this part of the circuit, of course, stops flowing, but there is a whole bunch of positive charge on that side of the capacitor, and a whole bunch of negative charge on the other side of the capacitor. And so now the capacitor, in fact, looks like a battery and can drive current I through this part of the circuit. It's attached to a resistor and so it can drive current around like that. And now let's think about what happens to this charge on the capacitor. As you start to power this current around this loop, the charge on the capacitor decreases. And you can write down a nice little differential equation to understand exactly how the charge decreases on the capacitor, but like a lot of things in physics you might guess, it's got to be exponential. And so that is in fact what happens. Charge on the capacitor is where it started, which we called Q naught, and then we have an e to the minus T over R C. Or, Q naught e to the minus T over tau. Tau is called the time constant, not to be confused with torque right we use tau for torque, unfortunately, but now, tau is just the time constant of that circuit. And it's equal to R C. So if you think about what the charge on the capacitor looks like as a function of time, we can draw it up. It started at t equals zero and a value of Q naught, that's how much charge was on our capacitor. But then it started to discharge and it decayed exponentially and so it fell like this. And what's the shape of that curve? It's exactly that. Q naught e to the minus t over tau, where tau equals RC. Tau is the one over e point. It's how long does it take for that charge to drop to 1 over e it's value Right, e is a natural number 2.7 blah blah blah blah blah, 1 over e is gonna be somewhere about there. This is the 1 over e point Q naught over e and you can see if I put t equals tau here, I get e to the minus and so that becomes a Q naught over e, that's the 1 over e point. And this is how you characterize RC circuits. If you have a very big resistor R, then your time constant is much bigger and this thing decays less steeply. It takes longer to get rid of the charge. That makes sense, right? You're trying to push it through this resistor, that river full of rocks, and it's hard to push it through there. But if R is very small then this thing will decay very quickly, you will discharge the capacitor very quickly. Likewise if you have a very big capacitor C, then you have a lot of charge that you're trying to move and so again this thing will decay very slowly. But if it's a small capacitor C then it's not much charge that you have to get rid of and it will decay much more quickly than that. Okay, so the shape of that curve is dictated by the time constant which depends on the values of R and the values of C.
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Solving Capacitor Circuits
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