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Confusing capacitor question

  1. Apr 26, 2006 #1
    I have a question that has confused me a little,

    A 10 micro-farad capacitor is fully charged by a 24v battery.

    The question asks how much charge is stored and how much energy would be stored.

    I think i have found the answer to these using,

    Q=CV and E = 1/2 CV^2 respectively

    but then it goes on to ask,

    how much energy would be required from the battery to charge the battery to 24v.

    Explain why the answers are different
     
  2. jcsd
  3. Apr 26, 2006 #2

    Doc Al

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    Staff: Mentor

    What's the energy supplied by the battery in delivering the charge Q?
     
  4. Apr 26, 2006 #3
    so is there an equation to find this?
     
  5. Apr 26, 2006 #4

    berkeman

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    "Explain why the answers are different"

    The times I've seen this question, they usually want you to think about what happens in a real situation when you charge a capacitor from a real voltage source. If the voltage source had zero output resistance and you connected an ideal capacitor to it, what would the initial charging current be?

    So instead, draw an ideal voltage source, and a series output resistor that connects the source to the capacitor. Start with a resistance of something like 10 Ohms. Now calculate how much energy is dissipated in that resistor over the course of charging the capacitor up to the source voltage. Change the output resistance to something like 1 Ohm, and recalculate the energy dissipated by the resistor. See a pattern? Does this lead you to the answer to the question above?
     
  6. Apr 26, 2006 #5
    how do i calculate the energy dissipated?
     
  7. Apr 26, 2006 #6

    berkeman

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    What is the relationship between energy and power? What is the equation for the power dissipated by a resistor, in terms of the resistance and the current through the resistor?

    When you connect the voltage source to the capacitor via that resistor, an initial charging current will flow, and that current will keep getting lower exponentially over time as the cap voltage ramps up (also exponentially) to the source voltage. Write the equation for the current into the cap as a function of the source voltage, capacitor voltage, and resistance, and then write the equation for the instantaneous power being dissipated by the resistor. Once you have the equation for the instantaneous power, what do you need to do to add all the powers up to get the total energy dissipated during the charging process?
     
  8. Apr 26, 2006 #7
    if i have calculated the charge and energy stored in the capacitor can i then use,

    energy dissipated = 1/2 QE

    to find the energy lost as heat, therefore the energy needed to charge the battery to 24v is double the energy stored?

    or would the energy needed be, the energy stored plus the energy lost?
     
    Last edited: Apr 26, 2006
  9. Apr 26, 2006 #8

    berkeman

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    I think double is the correct answer, but I don't see how you got it. Plus, you need to be able to explain why it's double. Where did the extra energy go?
     
  10. Apr 26, 2006 #9

    berkeman

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    Yeah, I redid the math, and you lose the same amount of energy in the series charging resistance as you end up with in the storage cap. But it took me half a page of exponentials and an integral to prove it. You should be able to write those same equations with the hints provided above. And be sure to give PF a little credit on your homework answer sheet :rolleyes:
     
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