Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Capacitors in Series

  1. Nov 20, 2017 #1
    I understand algebraically that when capacitors are in series, the total capacitance is less than any individual capacitance, but I do not understand this intuitively. How can this be possible? Shouldn't more capacitors equal more capacitance?
  2. jcsd
  3. Nov 20, 2017 #2


    User Avatar
    Gold Member

    That's exactly like saying that more resistors should mean more resistance, but when they are in parallel it is actually LESS resistance, it's only more resistance when they are in series. Similarly, capacitors in parallel means more capacitance and in series means less.
  4. Nov 20, 2017 #3


    User Avatar
    Homework Helper
    Gold Member

    Maybe the energy approach could make it intuitive for you. Compare the total energies stored in two capacitors in both series and parallel connections, keeping the source voltage constant. Which one gives more stored energy?
  5. Nov 20, 2017 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It's easiest to understand capacitors in parallel. The capacitance of a single capacitor is proportional to the area of the plates so putting two capacitors is a bit like increasing the area of the plates of a single capacitor. It's slightly more complicated if the capacitors aren't identical but the general idea is that in parallel the capacitances add together.

    As for series.. Capacitors have an impedance (Z). If you haven't studied impedance yet then think of it a bit like resistance for AC signals. It actually depends on the frequency of the AC signal but perhaps not worry about that for the moment. The impedance of a capacitor is inversely proportional to the capacitance (eg 1/C) so the larger the capacitance the lower the impedance and the better it conducts an AC signal. When putting two capacitors in series the total impedance is the sum of the two individual impedances so Zt = Z1 + Z2. In terms of capacitance that becomes 1/Ct = I/C1 + 1/C2.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted