 So to go a little further on our practice problem for capacitors in series, we want to look at what happens if I have more than two capacitors. Nye had originally been using these two equations for the equivalent capacitance. And those are good equations, but you'll sometimes see professors use a different one. And it might look like this. And in this particular case, they've taken this equation and simplified it. So you've got C1 times C2 over C1 plus C2. And that equation is a perfectly fine equation, and it works really well for just two capacitors. So for example, with the nice numbers I've been using, a 2 and a 3 Farad capacitor, you'd have 2 times 3 divided by 2 plus 3, which is 6 fifths, which is 1.2. The problem with using this equation is not that there's any problem with the equation itself is that you can only use it for two capacitors. If you have three capacitors, you can't use this equation at all. But if you have my equation, then when you've got more capacitors, it's really easy to extend it. And you just have the 1 over C1 plus 1 over C2 plus 1 over C3. And there's no problem adding as many capacitors as you want in that equation. So let's take a quick example look at this. And I'm just going to start with the 1 over a 2 Farad, a 3 Farad, and a 4 Farad. Put that into my equation, and using this format, I would get 0.923 Farads. Now this also makes sense because as I add more capacitors in series, the total equivalent capacitance should drop. So when I had just the 2 and 3, it was 2.2, 1.2, excuse me, now it's dropped down to 0.923. Just as a further example, what if I have equivalent capacitors? So rather than three random capacitors, I've got three equal capacitors here. So with three equal capacitors, if I do the algebra to rearrange it, what I actually get is this simplified form, meaning if there are equivalent capacitors, the equivalent capacitance is the original capacitance divided by the number of them. So if I had three original capacitors that were each 2 Farads, then it would be 2 thirds for the equivalent capacitance. And as I added more capacitors, it would go from 1 third of the original capacitance to 1 fourth of the original capacitance to 1 fifth of the original capacitance, etc., getting smaller and smaller the more capacitors I add to the situation. So these are just some examples of if you're working with more than two capacitors at once.