Does a battery exert energy keeping a capacitor charged?

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Piotr R
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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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Piotr R said:
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?
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.).
 
berkeman said:
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.).

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
 
Piotr R said:
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
No leakage = no current flow.
 
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
 
Piotr R said:
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
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.
 
Piotr R said:
I guess in this case there's "backwards leakage" from the plate towards the battery
TBH, I don't know what you mean by that... No current flows.
 
berkeman said:
Just keep in mind that all real-world capacitors and batteries have leakage currents.
It would be better to disconnect the battery, once the capacitor was charged. (Ignoring leakage, even through dampness in the air etc)
 
"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?
 
Piotr R said:
"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?
What's an HHO generator? Do you mean a fuel cell? Or just splitting water into O2 and H2 using electricity?
 
berkeman said:
What's an HHO generator? Do you mean a fuel cell? Or just splitting water into O2 and H2 using electricity?

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 because 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
 
berkeman said:
Or do you mean like the MythBusters HHO Generator...


Yes this, haven't watched that video yet, I'm at the cheesecake factory
 
Thread reopened. After a detailed PM conversation, it does not appear that the OP is asking about over-unity.
 
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?
 
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?
 
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Piotr R said:
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 input/output = 15163/10602.7 *100% = 70%
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...
 
berkeman said:
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...

Yeah, I put input/output, and 15163/10602 but i meant the inverse. Thanks.

And in terms of cars, yeah you make a valid point, but for now I'm just looking for the most efficient way to convert electrical energy into hydrogen gas, because there are other uses for them; rocket propulsion being one
 
Piotr R said:
Yeah, I put input/output, and 15163/10602 but i meant the inverse. Thanks.

And in terms of cars, yeah you make a valid point, but for now I'm just looking for the most efficient way to convert electrical energy into hydrogen gas, because there are other uses for them; rocket propulsion being one
Interesting. What are some other ways to get Hydrogen? Do they all start with water?
 
BTW, I think you can increase the efficiency of electrolysis some by adjusting the ionic content of the water. Have you already looked at that?
 
no they don't wall start with water, but water covers 70% of the surface of this planet

there's other ways of getting hydrogen, like through gasification, which is when high temperature steam reacts with methane, or something, but other methods of getting hydrogen haven't been introduced to me in my curriculum, so i wouldn't really know about them
 
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berkeman said:
BTW, I think you can increase the efficiency of electrolysis some by adjusting the ionic content of the water. Have you already looked at that?
i saw that too, yeah.

that's like a final step sort of thing after i design the system, and the fundamentals are finalized i'd probably do control tests to compare different electrolytes
 
Piotr R said:
He produced 1 liter of hydrogen gas, which is 0.044643 moles, because 1 mol H2 = 22.4L

I thought it produce a 1 liter mixture of hydrogen and oxygen? That would be 0.66L (0.03 mole) Hydrogen and 0.33L Oxygen. It could even be less because the mixture was bubbled through water at 91 degrees so the mix is probably not at STP.

I agree with your figure for the input power.

Revising the figures for the output gives..

The output energy (from combusting the hydrogen is potentially) =

(0.03 moles H2) * (237.5 kJ/mol) = 7125J

with the efficiency working out at about

(7125/15163) * 100 = 47%
 
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CWatters said:
I thought it produce a 1 liter mixture of hydrogen and oxygen? That would be 0.66L (0.03 mole) Hydrogen and 0.33L Oxygen. It could even be less because the mixture was bubbled through water at 91 degrees so the mix is probably not at STP.

I agree with your figure for the input power.

Revising the figures for the output gives..
(0.03 moles H2) * (237.5 kJ/mol) = 7125J

with the efficiency working out at about

(7125/15163) * 100 = 47%

Hmm I think you're right, I thought the hydrogen was being separated from the oxygen for some reason.

Either way though, don't you need oxygen for the combustion reaction i.e. 2 H2(g) + O2(g) ->2 H2O(g) + 475 kJ ?

I could be wrong, but that's what I thought
 
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Nevermind I think you're right that it's 47% efficient

Where does all that energy go I'm wondering?

And btw i think the hydrogen combustion energy output is 286 kJ/mol, so the efficiency would be

0.03 * 286 = 8580 kJ

-> 8580/15163 * 100 = 56.59%
 
I think most of the loss goes into heating the water.

In theory you only need something like 1.2-1.4V to split water so running at 12-14V is wasteful..

https://en.wikipedia.org/wiki/Electrolysis_of_water

Electrolysis of water
is the decomposition of water (H2O) into oxygen (O2) and hydrogen gas (H2) due to an electric current being passed through the water. The reaction has a standard potential of −1.23 V, meaning it ideally requires a potential difference of 1.23 volts to split water.

Overpotential[edit]
Real water electrolysers require higher voltages for the reaction to proceed. The part that exceeds 1.23 V[24] is called overpotential or overvoltage, and represents any kind of loss and nonideality in the electrochemical process.

For a well designed cell the largest overpotential is the reaction overpotential for the four-electron oxidation of water to oxygen at the anode; electrocatalysts can facilitate this reaction, and platinum alloys are the state of the art for this oxidation. Developing a cheap, effective electrocatalyst for this reaction would be a great advance, and is a topic of current research; there are many approaches, among them a 30-year-old recipe for molybdenum sulfide,[25] graphene quantum dots,[26] carbon nanotubes,[15] perovskite,[27] and nickel/nickel-oxide.[28][29] The simpler two-electron reaction to produce hydrogen at the cathode can be electrocatalyzed with almost no overpotential by platinum, or in theory a hydrogenase enzyme. If other, less effective, materials are used for the cathode (e.g. graphite), large overpotentials will appear.

Not sure if this has changed but most Hydrogen used to be made from fossil fuel because its cheaper than using electricity. ..

https://en.wikipedia.org/wiki/Hydrogen_production

As of 1999, the majority of hydrogen (∼95%) is produced from fossil fuels by steam reforming or partial oxidation of methane and coal gasification with only a small quantity by other routes such as biomass gasification or electrolysis of water.[/quote]
 
I know the energy on the plates is U = CV^2/2 where U is the energy, C is the capacitance, and V is the voltage.

In this case I'm assuming that the plate configuration creates a parallel plate capacitor

You're saying that some of this energy is heating the water, right? That makes sense. Along with what berkeman mentioned about leakage, I guess that's two sources right there the system is losing energy.. interesting
 

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