 ###### Jonathan Osbourne

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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# Capacitors in Series

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.

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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.