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Capacitor combination Question

  1. Jul 8, 2013 #1
    1. The problem statement, all variables and given/known data

    A circuit contains a single 250 pF capacitor hooked across a battery. It is desired to store three times as much energy in a combination of two capacitors by adding a single capacitor to this one. How would you hook it up, and what would its value be?

    2. Relevant equations

    PE = (1/2) CV^2

    3. The attempt at a solution

    PE(final) = 3PE(initial)

    so then:

    (1/2)CV^2 = 3 * (1/2) CV^2

    Not sure where to go from here? I would think that if it is three times as much energy, then it should be 3 times the capacitance 3C = 3 * 250 pF = 750 pF but this isn't correct
  2. jcsd
  3. Jul 8, 2013 #2


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    Staff: Mentor

    I would call the initial capacitor C1, and the cap you add could be C2. That will make your equation make more sense.

    And you have the right idea, but the question asks what size capacitor should you *add* to make the total energy storage 3x the initial storage on C1. How does that change your answer? :smile:
  4. Jul 8, 2013 #3
    Ok so then it's:

    (1/2)C2V^2 = 3 * (1/2)C1V^2

    When you say "add", does this mean C2 will be in parallel with C1?

    OOH....so if it's parallel, then It's C1 + C2 but is C1 = C2? If it is, then C = C1 + C2 = 250 + 250 = 500 pF

    But how do you know that both capacitors are 250 pF?
  5. Jul 8, 2013 #4


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    Staff: Mentor

    If you add C2 in parallel, what should you have written instead on the lefthand side (LHS) of your equation:

    "(1/2)C2V^2 = 3 * (1/2)C1V^2"

    Your LHS is incorrect as written.
  6. Jul 8, 2013 #5
    Do you mean LHS for C2 = C1 + C2?
  7. Jul 8, 2013 #6


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    Staff: Mentor

    Yes, the LHS should be the sum of C1+C2, not just C2. Your intuition has been correct all along. What value do you now get for C2?
  8. Jul 8, 2013 #7
    Ok I got it:

    (1/2)*(C1 + C2)*V^2 = 3 * (1/2)C1V^2

    C1 + C2 = 3C1
    C2 = 3C1 - C1 = 2C1 = 2 * 250 = 500 pF

    Thanks! :)
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