Work done by the battery?

  1. Can someone give me an idea how to compute or how to get the work done by the battery if there is a circuit given with resistor and a capacitor?:grumpy:
     
  2. jcsd
  3. Hootenanny

    Hootenanny 9,678
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  4. Hurkyl

    Hurkyl 16,089
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    Power relates to work, right? ...
     
  5. Hootenanny

    Hootenanny 9,678
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    What are you getting at?

    ~H
     
  6. Hurkyl

    Hurkyl 16,089
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    Presumably he can figure out the power dissipated by the resistor, and get the amount of work done through that component of the system.
     
  7. Hootenanny

    Hootenanny 9,678
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    Yeah, he could. But it also takes work to place charge on the capacitor plates.

    ~H
     
  8. Hurkyl

    Hurkyl 16,089
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    Right. But you already covered that part, and I didn't have anything to add to it.
     
  9. Hootenanny

    Hootenanny 9,678
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    Oh right, I thought I was missing something, like you could just work out the problem by considering power dissapated by the resistors. Thanks

    ~H
     
  10. Matter of fact you could!
    The energy dissipated during the charging of a capacitor equals the energy stored in the capacitor at the end of the process.
    So the total energy that battery gave away is

    [tex] U = \frac{Q^2}{C} [/tex]

    It's more complicated to consider it from the resistor side than from the capacitor side I suppose.
     
    Last edited: May 21, 2006
  11. Hurkyl

    Hurkyl 16,089
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    It's clear that when the battery is disconnected, the power subsequently dissipated by the resistor will be exactly the energy stored in the capacitor.

    But I don't see why there should be such a trivial relationship between the power dissipated by the resistor with the power stored in the capacitor during the time the capacitor is charging!
     
  12. It's a just a variant of Kelvin's rule applied to electrical circuits.
    If you don't trust me, write down differential equation ,solve it in current :wink: , and take the integral of [tex] i^2 R dt [/tex].
    Ratio of two energies will be always the same regardless of electromotive force of the source [tex]E[/tex],resistance [tex]R[/tex] of the circuit, and amount of capacitance [tex]C[/tex] in the circuit!:smile:
     
    Last edited: May 21, 2006
  13. I would reason like this:

    Assuming the battery voltage is constant:

    [tex]\text{Work by the battery} = \int V_{battery} I dt = V_{battery} Q[/tex]​

    and since at the end of the charging

    [tex]Q = C V = C V_{battery}[/tex]​

    we get

    [tex]\text{Work by the battery} = C V_{battery}^2[/tex]​

    (I assumed the capacitor has no charge at the beginning)
     
    Last edited: May 21, 2006
  14. Meaning the energy that occurs in the capacitor is also the same as the energy use by the battery?

    And it goes like this?

    [tex]E=\frac{1}{2}qV = \frac{1}(2}CV^2 = \frac{q^2}{2C} [/tex]

    Am I right?
     
  15. I don't CARE what work the battery is doing, the battery needs to get its smelly arse back to Mexico!!!! :tongue2:
     
  16. No!You are missing the point.
    The energy that is delivered by the battery is twice as high as the energy of the capacitor.
    Other half is dissipated by the resistor during the charging as the heat.
    You have a simple energy bilance :
    [tex]W_{battery}=\int_{0}^Q Vdq=QV=W_{dissipated}+U_{capacitor}=W_{dissipated}+ \frac{VQ}{2}[/tex]
    Maybe I should emphasize that prior to the stationary state established,more energy is spent as a Joule heat than being stored in the capacitor. Also easy to show that.
     
    Last edited: May 22, 2006
  17. Hi,

    I think something is still missing in this thread: what is the energy stored in the capacitor, and how this can be proved/calculated.

    My suggestion on how to do that:
    once the capacitor is charged, calculate how much energy it will dissipate in the resistor if the battery is removed and replaced by a shortcut. The calculation is similar to the previous work done by the battery except that the voltage on the capacity will decrease during this process. The outcome is then the 1/2 factor.​
    But I think there is a better way to proof the formula for the capacitor energy storage. I don't remember, help me.

    Michel

    PS:

    The fact that the work done goes in equal amounts to dissipation and to storage in the capacity is strinking. Is it related to a more general rule?
     
    Last edited: May 22, 2006
  18. energy=1/2q^2/c= 1/2C^2V.... Can I ask, do you know what is the formula for the work done by the battery? Thanks.
     
  19. Techno you mean, if the energy in capacitor is 20 joules, the energy in the battery is 40 joules? Am I right?
     
  20. That's right.
     
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