PhD., University of Maryland
Jonathan is a published author and recently completed a book on physics and applied mathematics.
Like currents in series, charge is the same when capacitors are connected in series but the potential difference adds. Unlike currents in series, effective capacitance is found by (1/effective capacitance)=(1/capacitance 1) +(1/capacitance 2). Effective capacitance is smaller than that of any of the individual capacitors.
Let's talk about capacitors in series, now as always series can be thought of as just meaning no choice, it's just one road available to the charge. So when I have capacitors in series I draw them just like this, capacitors look kind of like batteries except both of the bars are the same length. Alright, so we've got a charge on this first capacitor and it's plus on that side that means it's minus on that side. Now here's the idea as I was charging up this series combination of capacitors all the charge that came on c1 had to be taken from c2 because there was no choice. The charge if it's coming around this it's collecting here well it's got to come around like that. So charge is the same when I connect capacitors in series. The potential difference will add, now this is just like what happens with resistors, when I have resistors in series the current is the same and the potential difference adds.
Here the charge is the same and the potential difference adds, alright as always when we've got a capacitor or a resistor network, what we really want to do is boil it all down to just one capacitor and ask what's the capacitance of that capacitor in order to have the same potential difference? Well clearly the same charge is going to be stored on this effective capacitor and the potential difference across it got to be the same as the potential difference across this combination. Alright so we'll do, delta v, potential difference equals potential difference across the first one plus potential difference across the second one. Now with capacitors potential difference is charge divided by capacitance. So we'll have q over capacitance series equals q over capacitance 1 plus q over capacitance 2.
At this point since the charge is the same, we're going to cancel him out and that gives us the following expression for adding capacitors in series. Notice that I can't just add the capacitors they add in reciprocal so this is exactly the same thing that we saw for resistors added in parallel. So parallel and series are going to switch places as far as adding capacitance is concerned. Alright so just like with resistors added in parallel when I add another capacitor, the overall capacitance decreases alright so you add another capacitor your overall capacitance goes down it doesn't go up. And you can just think about this in terms of if I had these guys and I didn't have the one in the middle then you can kind of think well jeez the plus and the minus are closer together. So it's like reducing that distance, so then when I add in the capacitance in the middle it's like pushing it apart. So that means that the capacitance goes down not up and that's something very counterintuitive but it's easy to get messed up with.
Alright now adding these things is a little bit of a hassle, because you've got to do fractions if you're going to do it by hand. So there are 2 things that people generally do, first thing if it's only 2 they don't write the formula that way they write it this way. So you just multiply, you add and then you divide. Alright, that might not be something that's easy for you to remember but it will work when there's only 2. The way that I've often really pushed with my students is this way. Suppose that I've got 3 of them that I want to add, so here's what I'm going to type in my calculator 3 negative 1 plus 2 to the negative 1 plus 1 to the negative 1 and then equals or execute or whatever it is on your calculator and then answer to the negative 1 equals. Now if you do that in this case what you'll end up with is 0.5454 Farads and one thing to keep in mind about this, is that this number is smaller than any of these 3. It will always come out that way if you do it right, if you forget to do this last piece it will come out wrong and you'll immediately see that and you won't just pick that choice and go on to the next problem. So when you add capacitors in series the net capacitance always goes down. It's always less than anyone of these capacitors that you've got. Alright and then the last thing to think about in terms of capacitors connected in series is that the smallest capacitor will store the most energy. Okay and that's because charge is the same and energy is q squared over 2c so when c gets small, energy gets big. Alright those are capacitors in series.