PhD., University of Maryland
Published author
Jonathan is a published author and recently completed a book on physics and applied mathematics.
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PhD., University of Maryland
Published author
Jonathan is a published author and recently completed a book on physics and applied mathematics.
RC circuits contain resistors and capacitors. Immediately after a switch is closed, completing the circuit, the capacitors behave like wires because they have a potential difference of zero. After the current has been flowing for a long time the capacitors fill and the circuit reaches steady state. In steady state, the resistors in series with the capacitors behave like wires.
So let's talk about RC circuits, now what's RC stand for? Well RC stand for resistor capacity. So unlike just resistor circuits or just capacitor circuits in RC circuits we've got both circuiinfo!!info and we're trying to figure out what's going on. Now in general these types of circuits are very complicated and they require us to solve something called the coupled linear differential equations. But we don't need to do that in order to determine 2 very important kind of asymptotic behaviors of the circuit. We can find out what happens immediately after the switch is closed and we can find out what happens after we wait a very, very long time. Alright so we've got immediate and we've got steady state, so let's look at this circuit as an example. So I've got some resistors and some capacitors, alright here's my switch now I'm going to assume that all of the capacitors are uncharged when I've got my switch open. So initially they're all uncharged and I want to know what happens when I shut the switch okay.
Immediately after I shut the switch all the capacitors are still uncharged because they're not going to charge up instantly. So that means that their charge is 0 so that means the potential difference across the capacitors immediately after I close the switch is 0 they behave like wires. So to do immediately after I close the switch all I need to do is redraw the circuit where I make each of the capacitors just a wire and then I look at it as a resistor network and I just solve it however I solved those. Alright we'll do that in just a second but let's talk after a long time first. After a long, long time essentially what we're saying is all the capacitors are full and so that means that they can take no more current. No longer can current flow into any of the capacitor so after a long time what we're going to do is we're going to take out some sort of fictitious wire cutters and we're just going to cut the wires next to any of the capacitors because the capacitors are essentially full. No more current can flow there, so it really doesn't matter whether or not they're connected.
What that means is that all of the resistors that are in series with the capacitor just behave like wires after a long time. Alright so let's see this they can be a little bit confusing and it's easier to see once you got an example under your bell. Alright so immediately after I close the switch remember what I said all of the capacitors are replaced by wires. So now I just have the following resistor network now this is a really, really simple network especially because look at this I've got a wire connected in parallel with resistors. So that's like you're trying to get from point a to point b and you've got 2 roads that have got a bunch of traffic lights on them and one road that's freeway. Nobody is going to take the traffic light road, everybody is going to take the freeway right it's not like you're trying to stop here you're going from here to here freeway time.
Alright so this called the short circuit, what we've done is we've shorted out 1 ohm and the 5 ohm resistors so that they're not going to participate. No current will go through them. So if we're thinking of these as light bulbs these light bulbs are out, now current will go through here so that means it'll go through the 3 ohm resistor. How much current will go through the 3 ohm resistor? Well 12 divided by 3, 4 amps, so we've got 4 amps that are going like this down feeding both our 15 Farad and our 7 Farad capacitor as those capacitors get fed the potential difference across them will no longer be 0 and that's why this drawing is only appropriate immediately after we close the switch. As this current flows it's going to start filling up those capacitors, so this won't look correct anymore alright but right now initially immediately after we close the switch how much power is this 3 ohm resistor using. Well we'd use i squared r if we want, so we'll have power 4 squared 3 so that's 48 watts. Alright so if it's a light bulb it's fairly bright alright let's do steady state, now what we said was in steady state we can assume that all of the capacitors are full and that means we can just clip all the wires. So we clip here and we clip here because no current can flow into the 15 Farad capacitor so I just clip it.
Alright so what that means is that the resistor is over here, and the single resistor over here which was 5 ohm resistor is really just a wire now because there's no current going through it. Alright so the only circuit that's actually carrying current is now this circuit. Notice that now the 1 ohm light bulb does take current because now the capacitors are full. So that means that this 7 Farad branch which was accessible before is no longer accessible to the current. So that means that the amount of current that is actually flowing through these light bulbs here will be, well it's series right 1+3 is 4, 12 divided by 4 is 3 so that means that I'll have 3 amps going through my 1 ohm and my 3 ohm resistor in steady state after a long time has passed. Alright so how much power is being used? Well the power for the 3 ohm is going to be i squared r, 3 squared times 3 is 27 watts. What about the power for the 1 ohm resistor? Well i squared r so this one is going to be 9 watts, so that means that the total power that's being supplied by the battery after a long time is 36 watts.
Initially it was supplying 48 watts so now it's not supplying as much power and that's because it doesn't have to feed the capacitors anymore. The capacitors have already taken all the power that they need. Alright how much energy is stored in all these capacitors? Well it's really not a difficult question to answer all we need to do is combine them as we have before so we'll start with the 3 and the 7, 3+7 is 10 so we'll have a 15 and a 10 in series and we know that 15 inverse, 10 inverse is 6 inverse so that means that this capacitor network is really just a single capacitor with a capacitance of 6 Farads. Alright if I've got 6 Farads across it potential difference across there has got to be the same as the potential difference across the 1 ohm resistor. Well with 3 amps of current the potential difference across a 1 ohm resistor will be 3 volts. So that means that I've got 3 volts across a 6 Farad capacitor so the amount of charge that I have stored on this capacitor is 6 times 3 which 18 Coulomb's. So we got 18 Coulomb's here and then we got 18 Coulomb's is the sum here. So what we would need to do is find the potential difference across the 15 just by doing q over c right 18 over 15 we'll subtract that from the total 3 and then we'll say "that potential difference has to be the same as these."
Alright I'll let you do that on your own it's not a difficult task but you might want to use a calculator. Alright now there's one other type of resistor capacitors circuit that I just want to mention briefly and that's the situation where we're actually using the capacitor to drive the current. So we start off with the capacitor that is charged not uncharged so now I have a circuit that doesn't even have a battery, it's just using a capacitor to drive everything. So I've a bunch of positive charge here, negative charge here and that's going to discharge so it's going to go around that way. Well how much current initially is going to flow? Well if I want to know the potential difference I'll say it's charge over capacitance. 12 divided by 6 is 2 so this is 2 volts initially, 2 volts over 2 ohms gives you a 1 amp current. Alright now that 1 amp current is only the initial current, so the idea is that initially I have 1 amp but then the potential difference is going to go down because I'm starting to discharge the positive charge is going to go on the other side meaning the electrons are going the other way but we'll ignore that for the moment. So really this current is going to go down in fact it turns out that it goes down exponentially in time like that.
Alright so that's actually what we'll also happen in our previous circuit, remember that we started off over in our circuit with 48 watts going through the 3 ohm resistor. 48 watts so it was bright and then after a long time only 27 watts through the 3 ohm resistor. So we're going to have some sort of an exponential decay from 48 to 27 at the same time our 1 ohm resistor starts off at 0 nobody is going through it because it was short circuited and it ends with 9 watts so that's going to be an exponential increase from 0 to 9 and that's the way that RC circuits work.
Unit
Electricity