00:01 Let's end with a good example on how to use capacitors and this question asks if we have a 12-volt battery connected to 3 different capacitors, where these 3 capacitors are in parallel with their 3 capacitances of 1, 2 and 3 microfarads, how much total energy is stored on the capacitors at equilibrium and then we change it a little bit by adding a dielectric. 00:22 Teflon turns out to be a common dielectric. 00:24 We add this dielectric which has a dielectric constant of 2 and put it just in the 1 and 2 microfarad capacitors and then we ask what is the new total charge stored in all 3 capacitors. 00:37 So give this a try, we have equations now for the capacitance. 00:40 The capacitance with a dielectric included as well as for the energy stored and see what you get. We're gonna try that here and see what happens if we have these 3 capacitors. 00:50 Again, what's going on here is we have some battery. 00:53 So be the positive side of our battery and this is connected to 3 capacitors that are all in parallel. 01:00 So we have a capacitor here, a capacitor here, and a capacitor here. 01:05 So what's happening here, since we have 3 capacitors in parallel, where we have 1 microfarad, 2 microfarads and 3 microfarads. The total capacitance once again for capacitors in parallel will simply be the sum of these three. So this is the total capacitance. 01:27 In this case, the total capacitance is 6 microfarads. So now, we're almost done. 01:33 We have an equation for the energy stored in a capacitor. 01:37 It is 1/2 the capacitance times the voltage squared and we can always see how this comes from the equations starting with the charge may also had the equation 1/2 the charge times the voltage. 01:51 But we also saw that the charge Q is equal to the capacitance times the voltage. 01:56 And so there's always 2 equations or two ways to write the energy in the capacitor. 02:00 One is 1/2 Q times V, the other one is 1/2 Q which is CV times V which is 1/2 the capacitance times the voltage squared. 02:11 So certainly, always be aware that there are be more than one way to write the energy stored in a capacitor. So now that we know the voltage and we know the total capacitance which we just found here, we can say that the energy stored is 6 microfarads times the voltage which were already given in this problem is 12 volts squared. 02:31 And so now we have 1/2 times 6 microfarads times 144 volts squared as our unit. 02:42 Now here's where we have to be a little bit careful. 02:45 We can put all of these numbers together and see something like we could take the 1/2 and apply it to the 144, for example, or 1/2 times 6 and C3 times 144 and again we have units of microfarad times volts squared but if we simplify this and we get an answer, so hopefully you get something like I got which 432. 03:08 We have to be careful with our units here. 03:10 We know this unit here farads and volt squared is our energy unit, that's joules. 03:15 But we have this micro term here and don't forget that micro is simply a way of saying 10 to the minus 6th, one-one millionth and so this would be 432 microjoules. 03:26 This is the amount of energy stored in our 3 capacitors. 03:30 For the second part of this problem, we're going to do one small difference which is that we're going to add a dielectric to a couple of our capacitors. 03:38 So we're going to do the exact same type of analysis but we're going to have a different capacitance now, not for all but for some of the capacitors that we have. So let's see what that looks like. 03:47 In this case, what we have is a dielectric just added to the 1 and 2 microfarad capacitors which is going to change our analysis but only slightly so now instead of 1 microfarad and 2 microfarads with our 3 microfarads we now have the double value because we have K being equal to 2. 04:06 So now we have 2 instead of 1. We have 4 instead of 2 but we do not have any change to our 3 microfarad capacitor because we did not add any dielectric here. 04:18 So now, we have a slightly different total capacitance which is now 9. 04:23 So using this new value of 9, we can see that the energy stored in our capacitor is going to follow the exact same logic except instead of 3 times 144, we'll get 4.5 which is half of 9 times 144 and then again the units will be the same. 04:45 We still have microfarads times volt squared. 04:47 And if you get the same thing I got, you'll see 648 and again we have microjoules as our units. 04:54 So this is an example of how we could do a capacitance problem both with adding capacitors in parallel in this case as well as adding dielectric into a capacitor and see what happens to a quantity relating to capacitors like the energy. This also concludes our discussion of capacitors. 05:13 And now that we have both resistors and capacitors discussed, we're ready to discuss some more complicated circuits and see how we can put all these together. 05:21 Thanks for listening.
The lecture Capacitance Example by Jared Rovny is from the course Circuit Elements.
If we insert a dielectric material with K = 2 into a capacitor with C = 6 µF, what is the new value of the capacitance?
A 16 V battery connects to 3 capacitors in parallel with capacitances of 2 µF, 3 µF, and 5 µF. We insert a Teflon with dielectric constant K = 2 in the 2 µF and 3 µF capacitors. What is the total energy stored in the capacitors?
A 10 V battery connects to 3 capacitors in parallel with capacitances of 1 µF, 3 µF, and 4 µF. How much total energy do the capacitors store?
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