### Learn math, science, English SAT & ACT from

high-quaility study
videos by expert teachers

##### Thank you for watching the preview.

To unlock all 5,300 videos, start your free trial.

# Capacitor Circuits

###### Jonathan Osbourne

###### Jonathan Osbourne

**PhD., University of Maryland**

Published author

Jonathan is a published author and recently completed a book on physics and applied mathematics.

**Capacitor circuits** are circuits which flow until the capacitors are full. Once the capacitors are full, it is called steady state.

So let's talk about capacitors circuits. Alright what is a capacitor circuit? Well it's a circuit that consists of a battery and a bunch of capacitors. Now we're going to start with these capacitors uncharged and notice the way that I've written this. I've got a little switch so that current can flow until I close the switch, alright as soon as I do that, current is going to flow and start charging up the capacitors. Remember that the purpose, the job of a capacitor is to store charge. So that current flows up into the capacitor and it can't jump across. But what will happen is this guy will start being charged positive and then that will scare all the positives away from the other side so we get negative there. The positives will then go down and charge here and here. And then that will scare the positives away so we'll get some charge there. And then we've got all the negatives on the negative side of the battery and all the positives on the positive side. So what's going to happen is, there will be current that flows until the potential differences across these capacitors balance the potential difference across the battery and then no more current will flow. So in a capacitor circuit, current flows until the capacitors are full, they can't eat anymore so no more current can be forced into them. This is called steady state because there no longer are any changes.

Alright now aside from that kind of qualitative understanding the difference between a capacitor circuit on the one hand and a resistor circuit on the other. Aside from that we solve everything basically the same way. We look at our network and we say "how can I combine these capacitors either in series or in parallel to make it a simple circuit that consist only of a battery and a single effective capacitor?" And then what we do is we reconstruct the circuit, the original circuit just by going back through our analysis. Alright so let's go ahead and do this, the first thing it's obvious these two 4 Farad capacitors are connected in series so that means that we have to combine them in series. Remember in series capacitors add in reciprocal, now this is especially easy when as in this case it's the same resistance because we'll do one forth plus one forth and that will give us two forths. So when we flip the back up-side-down, it's going to be 4 over 2. That happens anytime you have 2 capacitors that have the same capacitance connected in series. The effective capacitance is just one half.

Alright so we'll write up the circuit as follows; just like that this is the series combination of the two 4 Farad capacitors so it gives us 2 Farads. Over here we got 8, here was 15 and over here our battery is 10 volts alright easy enough. What's next well the 8 and the 2 are clearly connected in parallel, now the nice thing about parallel with capacitors is that you just add. So we'll have, this is 10 volts, this is 15 Farads and this over here 10 Farads. Alright now these 2 guys are in series, so let's combine them in series again it's 1 over 15 plus 1 over 10 and it turns out that, that ends up giving you 1 over 6. Alright so that means that this circuit really is nothing more than this circuit 6 Farads, 10 volts. Alright now we need to know how much charge is on our effective capacitor. Well potential difference is q over c so q must be c times potential difference 6 times 10 gives us 60 Coulomb's and now we'll just reconstruct the circuit. I'll show you how to do this and then we'll look at the answers.

Alright so in series charge is the same, so that means that both of these capacitors have 60 Coulomb's alright now 60 Coulomb's with 15 Farads potential difference is q over c so this guy is 4 volts what does that mean this one is supposed to be? Well he's supposed to be 6 volts because I got 10 altogether 4 here, so I got 6 left over. Another easy way that we could that is just q over c again 60 over 10 gives 6. Alright so now I've got my 15 all covered 60 Coulomb's, 4 volts. So how I'm I going to reconstruct my 8 and my 2? Well they combined in parallel and remember that in parallel potential difference is the same right? So that means that the potential difference here across both the 2 and the 8 must be 6 volts. Alright then we do c times v and that's going to give us the charge 2 times 6 is 12, 8 times 6 is 48 alright so that's the way that it goes.

Notice of course that 48 plus 12 gives us the 60 that was on the 15 Farads after that we'll break these 2 back into the 2 force. I'll let you do that one on your own. Let's go ahead and look at the answers so I'm going to put them in a table just like this capacitance in Farads, potential difference in volts, charge in Coulomb's. Now as before I've got the 4 guys that I had on the original circuit but then at the bottom I've got an effective capacitance the 6 Farad guy. Alright now there is a check just like there is resistor circuits that checks all of our Math just to make sure that we get everything correctly. And that check is by asking how much energy is stored by all of our guys right? So remember energy you can get just by doing q squared over 2c or one half c v squared or one half qv. So doing that in each of these cases gives 18 joules of energy by both of the 4 Farad capacitors 144 joules for the 8 Farad and 120 for the 15.

Alright now if I add all these up what I end up with is 300 joules. That's how much energy total is stored in my capacitor circuit once all the capacitors are full. They can't take anymore current, so that means that, that 300 joules has got to be stored in my effective capacitor alright let's go through and do it. The potential difference across the effective capacitor is just that 10 volts that we had across the battery. To get the charge in Coulomb's we do capacitance times potential difference. And then to determine the amount of energy stored we need to do one half cv squared, so one half times 6 is 3 times 10 squared 300. Since those two numbers agree it means pretty much that we didn't make any mistakes. And that's capacitor circuit.

Please enter your name.

Are you sure you want to delete this comment?

###### Jonathan Osbourne

PhD., University of Maryland

He’s the most fun and energetic teacher you’ll ever meet. He makes every lesson sound like the most exciting topic ever so you never get bored when he's teaching.

#### Related Topics

- Conservation of Charge - Electric Charge 34,658 views
- Charge Transfer - Electroscope 22,997 views
- Electric Force - Coulomb's Law 26,590 views
- Electric Fields 23,930 views
- Electric Potential 24,720 views
- Electrons - Quantization of Electric Charge 20,331 views
- Conductors 12,523 views
- Insulators 10,403 views
- Electric Current 22,644 views
- Resistance 12,695 views
- Potential Difference 22,661 views
- Ohm's Law 18,301 views
- Electric Circuits 16,844 views
- Resistors in Series 14,563 views
- Resistors in Parallel 16,131 views
- Resistor Circuits 13,778 views
- Power 8,536 views
- Capacitors 16,834 views
- Capacitors in Series 12,852 views
- Capacitors in Parallel 11,204 views
- RC Circuits 21,150 views

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete