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I Does a battery exert energy keeping a capacitor charged?

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  1. Apr 15, 2017 #1
    I am researching whether or not there is an equation that calculates energy used by a DC battery to keep a capacitor charged over time in a capacitor circuit.

    I know a capacitor in a circuit acts as a break when it is fully charged, but does the battery continue using energy to keep the capacitor charged once it is fully charged?
     
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  3. Apr 15, 2017 #2

    berkeman

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    Welcome to the PF.

    It depends on the type of capacitor. All real capacitors will have a small "leakage current", which the battery has to supply to keep the cap from self-discharging over time. You can do a search on capacitor leakage current to see what typical values for different types of caps (electrolytic, tantalum, ceramic, etc.).
     
  4. Apr 15, 2017 #3
    The type of capacitor I'm talking about is a parallel plate capacitor C = keA/d where k is the dielectric constant, e is 8.85*10^-12, A is the area of the plates, and d is the distance between the plates.

    Assuming no leakage, I want to know in a theoretical sense whether or not a capacitor continues to need the battery to "push" protons onto its plate.

    Intuitively, I assume that the battery has to exert energy to keep the protons pushed up against the capacitor plate, because the protons are attracted to the electrons on the opposite plate, and the "path of least resistance" is back through the wires and back through the battery.

    I'm looking for an equation that depends on time, emf, capacitance that calculates the energy needed to keep the capacitor fully charged in the form of E(t) = ... where E(t) is the total energy exerted keeping the protons on the plate at time t
     
  5. Apr 15, 2017 #4

    berkeman

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    No leakage = no current flow.
     
  6. Apr 15, 2017 #5
    So a battery hooked up to a capacitor will run out of energy no faster than a battery hooked up to nothing? I guess in this case there's "backwards leakage" from the plate towards the battery
     
  7. Apr 15, 2017 #6

    berkeman

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    For the ideal situation you are asking about, that's correct. Once the cap is charged up, you can disconnect it from the battery and both will stay at that final voltage.

    Just keep in mind that all real-world capacitors and batteries have leakage currents. That's why real battery storage life is not infinite, and why real capacitors don't stay charged up forever.
     
  8. Apr 15, 2017 #7

    berkeman

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    TBH, I don't know what you mean by that... No current flows.
     
  9. Apr 15, 2017 #8

    sophiecentaur

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    It would be better to disconnect the battery, once the capacitor was charged. (Ignoring leakage, even through dampness in the air etc)
     
  10. Apr 15, 2017 #9
    "backwards leakage" means leakage from one plate towards the other passing through the battery

    I'm asking because I am trying to figure out where energy is being used in an HHO generator, which essentially uses parallel plate capacitors in parallel.

    According to this discussion the efficiency of generating oxyhydrogen using the HHO method should be very high given that after charging plates, the only energy usage is keeping the plates charged because of leakage

    Does anyone know the efficiency of HHO generators btw?
     
  11. Apr 15, 2017 #10

    berkeman

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    What's an HHO generator? Do you mean a fuel cell? Or just splitting water into O2 and H2 using electricity?
     
  12. Apr 15, 2017 #11

    berkeman

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    Or do you mean like the MythBusters HHO Generator...

     
  13. Apr 15, 2017 #12
    I'm not sure what you mean by fuel cell, but if your talking about converting hydrogen and oxygen until electricity like Honda's hydrogen car's engine does then no.

    There's a lot of content on the internet about it I'm not sure which link to give you cuz I'm not referring to anything in particular. Maybe mythbusters? Although I'm on my phone right now and can't figure out how to get a working link.

    I found this on eBay
    https://www.google.com/search?q=hho...&ved=0ahUKEwj0qtPivqfTAhXk1IMKHfzBDAIQgjYIjgU
     
  14. Apr 15, 2017 #13
    Yes this, haven't watched that video yet, I'm at the cheesecake factory
     
  15. Apr 15, 2017 #14

    berkeman

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    Thread closed temporarily for Moderation...
     
  16. Apr 15, 2017 #15

    berkeman

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    Thread reopened. After a detailed PM conversation, it does not appear that the OP is asking about over-unity.
     
  17. Apr 15, 2017 #16
    To clear things up, I know 1 mol of water requires 237kJ of energy to dissociate into hydrogen and oxygen plus 49kJ to overcome the change in entropy of the reaction, and I know you get 237+49=286kJ of energy back from combusting the hydrogen and oxygen back into water.

    I'm trying to dissect how the HHO generator is working, because I go to New Jersey Institute of Technology, and I have a capstone engineering course I have to take in a year or so, which is sort of like an undergraduate thesis, and I'm trying to figure out what topic I want to pursue.

    I have an alternative, but similar proposition to approach electrolyzing water, but before I go and tell my professor it's what I want to work on, I want to have a decent understanding of the topic.

    I have a few questions about the HHO generator, which if you want to know more about I recommend watching

    One question I have is how efficient these generators are, and to calculate that simply divide energy output by energy input and multiply by 100.

    The king of random channel that I linked claims 5 Liters per minute, but even accounting for the fact that there are 55.5 moles per liter of water, don't we need to know the resistance in the system to find the heat dissipation?
     
  18. Apr 16, 2017 #17

    CWatters

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  19. Apr 16, 2017 #18
    yeah, i'm doing that now. i just discovered a mole of hydrogen fills up about 22 Liters of space, which goes against what I initially assumed.
     
  20. Apr 16, 2017 #19
    So i just watched this guy test a Punch 5.0 HHO generator, which claimed near 100% efficiency

    and I used his voltage, current, and time measurements to calculate the energy input. He produced 1 liter of hydrogen gas, which is 0.044643 moles, because 1 mol H2 = 22.4L

    The Input energy = V*i*t= (14.13V)(9.845A)(109s)=15163J

    The output energy (from combusting the hydrogen is potentially) = n*H = (0.044643 moles H2) * (237.5 kJ/mol) = 10602.7 J

    The efficiency is therefore output/input = 10602.7/15163 *100% = 70%

    I wasn't sure whether I should use 237.5 kJ/mol, or 286 kJ/mol for the combustion output energy of hydrogen gas, but in either case, if I was supposed to use 286 kJ then the efficiency of the Punch 5.0 is 84.2%

    Did I do this right? thoughts?
     
    Last edited: Apr 16, 2017
  21. Apr 16, 2017 #20

    berkeman

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    Efficiency is Output / Input, so I'm not sure how you got the right answer with your math, but whatever.

    Efficiency = 10602 / 15163 = 0.699 = 70%

    So it's better to use the electricity directly if possible, instead of storing the energy as Hydrogen. If you have a hydroelectric dam and want to power your car with it, then the efficiency difference between using batteries for storage versus Hydrogen and fuel cells is worth considering...
     
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