1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Capacitor thermodynamic non sequitur

  1. Nov 27, 2009 #1
    This has probably been posted/asked here before as it seems quite basic, but I can't seem to find a thread on it using the search function.

    According to conventional cap theory (0.5cv^2), two identical caps in series can store twice as much energy as two in parallel (provided the caps in series are charged to twice the voltage as the parallel bank). This seems strange though, is there some subtlety I'm missing here?

    A silly corollary of this line of thinking would be that two caps charged in parallel that are then stacked/erected in series can deliver twice as much energy as the parallel network.
     
  2. jcsd
  3. Nov 27, 2009 #2
    Scroll down on this reference and see the equivalent capacitance Ceq:

    http://www.physics.sjsu.edu/becker/physics51/capacitors.htm

    In theory both configurations could be used to store the same energy. In practice, if the caps are voltage rated for maximum voltage, then stacking two in series lets you double the voltage to double the stored energy.

    I haven't done any analysis since you should be able to do it from this reference and my comment.
     
  4. Nov 27, 2009 #3
    Yes, I should add that I'm treating the capacitor's rated voltage as what sets an upper limit on the amount of energy they can store.

    Your reference cites the theory I'm running with here, which leads to the (seemingly absurd) conclusion that capacitors can be used to store more energy when used in series rather than parallel.
     
    Last edited: Nov 27, 2009
  5. Nov 27, 2009 #4
    I spoke too soon. Let two capacitors have the same capacitance C and maximum voltage rating V. In each configuration let the terminal voltage be Vt.

    Series Configuration:

    Equivalent capacitance: Cs = C/2.
    Terminal voltage: Vt = 2V.

    Substutute into (1/2)*Cs*(Vt)^2 = (1/2)*(C/2)*(2V)^2 = (4/4)*C*V^2 = CV^2

    Parallel Configuration:

    Equivalent capacitance: Cp = 2C.
    Terminal voltage: Vt = V.

    Substitute into (1/2)*Cp*(Vt)^2 = (1/2)*(2C)*V^2 = (2/2)*C*V^2 = CV^2

    So the energy is the same in both configurations, CV^2. This is why you must first do the math.
     
  6. Nov 27, 2009 #5
    Oh sorry, you're right. I did the math before but for some reason I ended up with twice the energy.
     
  7. Nov 27, 2009 #6
    It happens to me all the time.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook