Cap Dumping

[BobZilla]

Hello forum,
So I wanted to show some basic principals that I have come to understand concerning cap dumping. My methods and opinions are not the only way to be sure and this is not an attempt to prove anything to anyone.

Side note:
Guys can we for once keep all of the mystical babble and smarter than Einstein type rants away from this one? I would be very interested in how real builders take their approach to cap dumping and opinions are most welcome if they are on topic. I hope for this to be a chance for beginners to learn something and veterans to share something,, well hopefully we all learn something ;-)

I find it easier to show in video for the most part rather than try to explain everything in text so as I usually do I am attaching a video.

I will explain a little in text though. What this video is showing is some of the parameters that should be considered when setting up a cap dump.

First I consider the "punch" to the battery. This can be affected by on-time of your dump period, the size of the caps you choose, of course the target battery size is also a factor.

The frequency of the dump should also be considered. Your desired current draw (higher/lower) will be affected greatly by this. The "rest" between dumps is what we are talking about here, 1 sec, half sec, quarter sec, and so on. The over all effectiveness of your charging will depend a lot on the frequency you choose. Too fast and you will waste energy but too slow and you will not get the charge battery up to where it needs to be.

Of course there are many factors to consider and I have not listed all of them but I thought we could start with the basics and go from there.

Here is a little starter video for the thread:
https://onedrive.live.com/redir?resi...nt=video%2cmp4

I am going to make a chart of a fuller run at this but I wanted to show all the adjustments in this video prior to doing an untouched one. In other words I don't want to mess around with the one coming up.

[ZPDM]

Hi Bob,

I've been doing a lot of cap discharging as well recently, in fact my brain is a bit fried from it all. I'll mention a couple strange things I've seen, really I have been reduced to just making and breaking connections by hand to be sure of what I was seeing. So I took a 0.1 uF cap charged to about 18 volts and discharged it into a 470 uF cap. The voltage reading on the 470uF cap would then be 0.0036V. Now of course if you discharge that same cap through either a Bedini set-up or even just throwing a backwards diode across the small cap the ending voltage on the 470uF cap will be a decent bit higher than 0.0036V. However, what should the voltage in the 470uF cap be given that you can calculate the power in the small cap from 1/2cV2. It is just a bit of simple algebra and I must have done it 4 times before I trusted it. The voltage in the 470uF cap should be around 0.268 V, from memory, but again it is simple algebra. This gives an energy transfer efficiency of 1.3%. This is why I was doing things by hand and rechecking math, what the heck 1.3%, along with scurrying around the internet for info on discharging one cap into another it also occurred to me to try discharging the cap twice. In this case the first pulse moved the 470uF to 0.0036, the second pulse to 0.0072. So how about that! The 1/2C will remain constant, if we call the 0.0072 as 1 then the second pulse would be 2 so the energy in the cap would be 1 for the first pulse 4 for the second. the two pulses were twice as efficient in power transfer. And as 3 of 4 units of energy came from the second pulse the second pulse was 3 times more efficient than the first. Hey were up to 3.9% power transfer efficiency from pulse one to pulse two. This improving pattern holds through five pulses that I did by hand. At this point I went and looked at something I must have seen years ago, that is the capacitor charging graph as a function of time, just sitting there right under our noses.

This graph shows capacitor charging per constant units of time
Graph from Learning about Electronics http://www.learningaboutelectronics....r-charging.php
So from 0 to nine volts takes 2 units of time, 0-18V four units, 0-30, 7 units. Knowing the energy in the cap is proportional to voltage squared you can deduce the relative efficiency of how the cap charges at different voltages. Realizing as well that as the cap fills the voltage difference between the DC source and the cap diminishes it is clear that the more efficient charging is not in the bottom third of the caps voltage range it is in the top two thirds top half I am guessing from 50-80% of the caps rated voltage. It is also not a small difference, so as I said this has sort of blown my mind as a whole nother aspect to things. I haven't heard that the charging graph is specific to electrolytics so I am guessing this is a property of dielectric charging. The two questions I have right now are what does the graph look like if you have an input voltage of 100V into a 20V cap. I would suspect it looks the same until you got towards 20 volts after which you would just be wasting an 80V discharge up against a full cap. Similarly does the graph look the same for a 20 Volt input into 100V cap. I.e is what is going on in the cap or is that as long as you haven't exceeded the voltage range of the cap you will see the same sort of graph with 1 or 99 Volts? By the Way, while it doesn't explain the lost power in a two capacitor paradox, I suspect it might account for it, that is it is less efficient to charge the bottom half of the cap.

Enjoyed your vid, Ciao.

[BobZilla]

The two questions I have right now are what does the graph look like if you have an input voltage of 100V into a 20V cap. I would suspect it looks the same until you got towards 20 volts after which you would just be wasting an 80V discharge up against a full cap. Similarly does the graph look the same for a 20 Volt input into 100V cap. I.e is what is going on in the cap or is that as long as you haven't exceeded the voltage range of the cap you will see the same sort of graph with 1 or 99 Volts? By the Way, while it doesn't explain the lost power in a two capacitor paradox, I suspect it might account for it, that is it is less efficient to charge the bottom half of the cap.

Enjoyed your vid, Ciao.

You bring up some good points. I cannot follow to well on the paradox because I am more of a hands on type and since I am not doing the test I cannot really say much. Some point of view about your last two posed questions however I can say my perspective, not that it is correct.

The first putting 100v to a 20v cap. I think with all of these situations we also need to consider current because without a limiter we would not see any curve at all, I mean it would be there technically but almost undetectable. But what do I think it would look like, almost the same but perhaps a bit steeper. I say that because of the current again. Because of the difference in potential between the 20v and 100v it would naturally want to push more current to make the difference so the chart would probably rise faster until the cap was full and then the extra 80v would still push on the cap until it malfunctioned, the potentials would still be trying to balance.

Now the other way 20v to a 100v cap. I think it would give a "normal" type of curve like the picture shows, perhaps slower but I doubt it matters too much. The 20v cap does not know the 100v cap has it's limit. When they start out it would just see the zero to 20v potential and it would start transferring voltage to try and balance, as the voltage climbs in the empty cap however the potential becomes less and therefor the current becomes less, so slower.

Hopefully that makes sense, sometimes it is difficult to express how I think about unseen things. On a basic level I can agree with what the chart says though. I know from experimenting that the cap does fill faster in the second half of it's range than the first. Those caps I had in the video are 50v caps. I have used them up to 50v in the past but in doing so I have blown out countless FET's and a fer SSR. The density of the power is far greater when you get to the upper end of a cap, almost like a pressure that wants to get out. If I use higher voltage caps and do the same discharge into a battery it is not as "punchy".


Now for the group:

I ran that cap dump using two garden batteries on the front in series and 1 garden battery on the back charging. I did shoot the first part of it in video but I goofed this one up guys. I set the charge last night and went to bed. I thought it would be fine but when I got up and checked it had gone way past the top in the middle of the night and kept going to the point of the primaries getting WAY depleted. When I say what had happened I switched it off immediately. I will have to do this one over and not let it run on.

Here are the charts from that run however. The primary chart is from my CBA in charge monitor mode because at the moment my other meter can only mesure up to 20v so it would be out of range. You can see that things were looking pretty good if I had not slept through the proper time to cut it off.

by the way the timing was set at 247ms OFF, 3ms ON. The first video shows what I mean by that if you didn't see it.





As I said I will have to do this one again.

Edit:
I wanted to add two shots because this is important.

The first shot is showing the dump near the start of the run. Notice the curve is small and the over voltage is high. This is a direct result of the impedance of the uncharged battery.



This second one is toward what would have been the end if I cut it off at the right time. Again notice the same , curve and over voltage. See less impedance translates to more absorption. We see the battery taking more in.



The time is in the lower left in seconds.

[James McDonald]

I know from experimenting that the cap does fill faster in the second half of it's range than the first. Those caps I had in the video are 50v caps. I have used them up to 50v in the past but in doing so I have blown out countless FET's and a fer SSR. The density of the power is far greater when you get to the upper end of a cap, almost like a pressure that wants to get out. If I use higher voltage caps and do the same discharge into a battery it is not as "punchy".

Just to bring in a little theory on the time that it takes for a capacitor to charge to the full applied voltage. A capacitor charges in 5 time constants. On the first time constant the capacitor will charge to 63.2% of the applied voltage. If the applied voltage was 50 volts then the capacitor would charge to 31.6 volts. The second time constant the capacitor will charge up 63.2% of the left over voltage which is 18.4 volts. So at the end of the second time constant the capacitor would have charged another 11.6288 volts. So the total voltage on the capacitor would be 43.2288 volts at the end of the second time constant. By the end of the 5th time constant the capacitor would have charged to 99.99% of the applied voltage. So your observation that the second half of the capacitor charging is faster then the first half is correct. So you need to calculate the capacitor timing first and this is going to depend on its size in Farads and the amplitude frequency of the voltage being applied. So its always safe to stay just below the second time constant. Not too much voltage to blow up the capacitor and not too much stored joule energy in the capacitor to fry your Field Effect Transistor switching circuit.

-- James

[Faraday88]

Hello forum,
So I wanted to show some basic principals that I have come to understand concerning cap dumping. My methods and opinions are not the only way to be sure and this is not an attempt to prove anything to anyone.

Side note:
Guys can we for once keep all of the mystical babble and smarter than Einstein type rants away from this one? I would be very interested in how real builders take their approach to cap dumping and opinions are most welcome if they are on topic. I hope for this to be a chance for beginners to learn something and veterans to share something,, well hopefully we all learn something ;-)

I find it easier to show in video for the most part rather than try to explain everything in text so as I usually do I am attaching a video.

I will explain a little in text though. What this video is showing is some of the parameters that should be considered when setting up a cap dump.

First I consider the "punch" to the battery. This can be affected by on-time of your dump period, the size of the caps you choose, of course the target battery size is also a factor.

The frequency of the dump should also be considered. Your desired current draw (higher/lower) will be affected greatly by this. The "rest" between dumps is what we are talking about here, 1 sec, half sec, quarter sec, and so on. The over all effectiveness of your charging will depend a lot on the frequency you choose. Too fast and you will waste energy but too slow and you will not get the charge battery up to where it needs to be.

Of course there are many factors to consider and I have not listed all of them but I thought we could start with the basics and go from there.

Here is a little starter video for the thread:
https://onedrive.live.com/redir?resi...nt=video%2cmp4

I am going to make a chart of a fuller run at this but I wanted to show all the adjustments in this video prior to doing an untouched one. In other words I don't want to mess around with the one coming up.

Sorry...Einstein is in a way the greatest mind block for the development of the Physics of Energy...Einstein himself has patted Tesla on this front...
BTW real learning is done alone..exception bear synchronicity if they happen to work in collaboration...example JB and PL...
Rgds,
Faraday88.

[BobZilla]

Sorry...Einstein is in a way the greatest mind block for the development of the Physics of Energy...Einstein himself has patted Tesla on this front...
BTW real learning is done alone..exception bear synchronicity if they happen to work in collaboration...example JB and PL...
Rgds,
Faraday88.

So did you just come to fart in the room?

[Faraday88]

So did you just come to fart in the room?

Take it as fart.....you will soon know fart smells good at times..

[ZPDM]

Hey Bob,

Won't try to take over the thread but as you mentioned it the capacitor paradox is very easy to understand, so easy that I thought I was missing some other equation. So energy in a cap is defined by the equation 1/2 capacitance X voltage squared. E=1/2cV2

Take a 1 mFarad cap and charge it to 20V. The energy in that cap then is 20*20=400, 400/2 = 200 * capacitance of 1mF = 200 milli Joules

Now take that 20V 1 mF cap and discharge it into a second equally sized 1mF cap. Both caps equalize at 10V. Now consider the energy in each cap
Cap 1: 10*10 =100, 100/2 = 50, 50 x 1mF = 50mJ
Cap 2: 10*10 =100, 100/2 = 50, 50 x 1mF = 50mJ

Energy in cap 1 + cap 2 = 50mJ + 50mJ = 100 mJ

Energy before discharge 200mJ, energy after discharge 100 mJ. 100 mJ of energy is missing, up and vanished like a fart in the wind.

You can't say resistive losses because you are just connecting the two plates of the caps together, so as I said I've heard talk of radiation or some such but I realized today none of that can work either. The defence is that well conservation of charge was retained, i.e you can demonstrate the same number of electrons after discharge as at the start only now they are distributed over two cap plates not one. Well if you have the same number of electrons charge separated only now on two plates instead of one then you didn't have any resistive or other losses did ya? Being an outsider to electrical engineering well it just sort of ticks me off to see something like this. It is basically well sometimes (like all the time) when you do something really simple like discharge one cap into another, you can't have both conservation of charge and conservation of energy so in this case you can only have conservation of charge, and conservation of energy goes away, hehe, then I guess they stick some playdo up their nose and giggle, sorry. This isn't Rocket Science, this is just one cap to another identical cap. Let's pretend you found some losses, (and for a bit earlier today I was thinking maybe the losses are only in electrolytics and they go towards charging the water, i.e cap rebound). Aside from, very sketchily, finding that the same exact behavior seems to be there with thin film caps, okay say you found losses, then you can't have the same number of electrons as you started which you do by conservation of charge, (unless losses are somehow unrelated to charge separation of electrons). So by conservation of charge you have the same number of separated electrons, so no losses. The problem is fundamentally related to definitions. One of your "Laws" is by definition invalid. I am literally ignorant enough in this area now I want to know when does energy in a cap actually = 1/2cV2? Maybe you could do something with that if you weren't throwing half the energy away. Or is the definition just some sort of relic nonsense, some English royalty "arr Maties half of Crown Queen Victoria ... Victoria" Ahh 1/2cV2 how wise my leige.

So again, and this fries my bacon to point out. If conservation of charge holds, by definition you have not had losses in your system, if you have not had losses then you have to have the same energy out as in. As you have lost half your energy either your definitions are invalid or one or both of your laws are invalid.

I have at this point fiddled around a bit with looking at discharging caps of different capacitance into one another as well as looking at discharging caps say from a high voltage into a small voltage cap, so on so forth. I don't have the full picture, but the take home message from what I've seen is if you charge a cap to say 60 volts with a radiant, the last thing you want to do is discharge it into a single six volt battery, would work better discharging into 12, 24, 36 volts. So, where oh where have I heard this before, yes John Bedini saying line the batteries up in series, water's fine. To give full disclosure, I have to admit that the more I experiment the more I am ending up just wending my way back to things JB said and confirming parts of them.

[James McDonald]

Hi ZPDM --

The energy lost in the capacitor is lost in two forms when two capacitors are hooked together. This is
when one capacitor is charged and the other capacitor has zero charge.

1) Heat -- This is due to the flow of current flowing through the leads of one capacitors to the other capacitor.
2) Electromagnetic radiation -- This is due to the charge being accelerated from one capacitors plate to another
capacitors plate.

-- James McDonald

Hey Bob,

Won't try to take over the thread but as you mentioned it the capacitor paradox is very easy to understand, so easy that I thought I was missing some other equation. So energy in a cap is defined by the equation 1/2 capacitance X voltage squared. E=1/2cV2

Take a 1 mFarad cap and charge it to 20V. The energy in that cap then is 20*20=400, 400/2 = 200 * capacitance of 1mF = 200 milli Joules

Now take that 20V 1 mF cap and discharge it into a second equally sized 1mF cap. Both caps equalize at 10V. Now consider the energy in each cap
Cap 1: 10*10 =100, 100/2 = 50, 50 x 1mF = 50mJ
Cap 2: 10*10 =100, 100/2 = 50, 50 x 1mF = 50mJ

Energy in cap 1 + cap 2 = 50mJ + 50mJ = 100 mJ

Energy before discharge 200mJ, energy after discharge 100 mJ. 100 mJ of energy is missing, up and vanished like a fart in the wind.

You can't say resistive losses because you are just connecting the two plates of the caps together, so as I said I've heard talk of radiation or some such but I realized today none of that can work either. The defence is that well conservation of charge was retained, i.e you can demonstrate the same number of electrons after discharge as at the start only now they are distributed over two cap plates not one. Well if you have the same number of electrons charge separated only now on two plates instead of one then you didn't have any resistive or other losses did ya? Being an outsider to electrical engineering well it just sort of ticks me off to see something like this. It is basically well sometimes (like all the time) when you do something really simple like discharge one cap into another, you can't have both conservation of charge and conservation of energy so in this case you can only have conservation of charge, and conservation of energy goes away, hehe, then I guess they stick some playdo up their nose and giggle, sorry. This isn't Rocket Science, this is just one cap to another identical cap. Let's pretend you found some losses, (and for a bit earlier today I was thinking maybe the losses are only in electrolytics and they go towards charging the water, i.e cap rebound). Aside from, very sketchily, finding that the same exact behavior seems to be there with thin film caps, okay say you found losses, then you can't have the same number of electrons as you started which you do by conservation of charge, (unless losses are somehow unrelated to charge separation of electrons). So by conservation of charge you have the same number of separated electrons, so no losses. The problem is fundamentally related to definitions. One of your "Laws" is by definition invalid. I am literally ignorant enough in this area now I want to know when does energy in a cap actually = 1/2cV2? Maybe you could do something with that if you weren't throwing half the energy away. Or is the definition just some sort of relic nonsense, some English royalty "arr Maties half of Crown Queen Victoria ... Victoria" Ahh 1/2cV2 how wise my leige.

So again, and this fries my bacon to point out. If conservation of charge holds, by definition you have not had losses in your system, if you have not had losses then you have to have the same energy out as in. As you have lost half your energy either your definitions are invalid or one or both of your laws are invalid.

I have at this point fiddled around a bit with looking at discharging caps of different capacitance into one another as well as looking at discharging caps say from a high voltage into a small voltage cap, so on so forth. I don't have the full picture, but the take home message from what I've seen is if you charge a cap to say 60 volts with a radiant, the last thing you want to do is discharge it into a single six volt battery, would work better discharging into 12, 24, 36 volts. So, where oh where have I heard this before, yes John Bedini saying line the batteries up in series, water's fine. To give full disclosure, I have to admit that the more I experiment the more I am ending up just wending my way back to things JB said and confirming parts of them.

[ZPDM]

Hi ZPDM --

The energy lost in the capacitor is lost in two forms when two capacitors are hooked together. This is
when one capacitor is charged and the other capacitor has zero charge.

1) Heat -- This is due to the flow of current flowing through the leads of one capacitors to the other capacitor.
2) Electromagnetic radiation -- This is due to the charge being accelerated from one capacitors plate to another
capacitors plate.

-- James McDonald

Hi James,

Sorry neither answer flies as can be demonstrated in numerous ways. 1) Heat, I have discharged small caps say 1000 times/sec continuously, I should double check, but I don't see any heating per IR thermometer. b) I have discharged a small 0.1 uF cap into an empty 470uF cap. The first discharge showed almost 99% losses, the second discharge 97% losses. As voltage difference has hardly budged between the two the difference in energy transfer is too great to be accounted for by resistive heat losses. 2) EM radiation is a function of magnetic flux. There is magnetic flux off a discharging cap but it would be increased by the addition of an inductor. So if this was a significant source of loss an inductor would make things worse whereas depending on the set-up an inductor either has no effect or improves things. Also, and here I am simply asking not trying to defend my position, I realize now that Maxwell is the theoretical genius behind so much of this, where in Maxwell's equations can an argument be made that EM radiation would be a significant loss? Further, EM radiation is not undetectable, if one is losing half ones' energy to EM radiation in cap discharges, where is the research, the analysis in the literature of this radiation?

Lastly and most importantly as noted above conservation of charge is retained in the cap discharge scenario, that should sink in, it took a awhile for me. So for the fun of it let's calculate the starting and ending electrons

1 Coulomb = 1 Farad x 1 volt

Cap before discharge while at 20 Volts .001 Farads, 20*.001= .02 Coulombs

Two caps after discharge 10 Volts
Cap 1: 10*0.001 = .01 Coulombs
Cap 2 10*0.001 = .01 Coulombs

Cap 1 + Cap 2 = 0.02 Coulombs

1 Coulomb = 6.242*10e18 electrons

Number of electrons before discharge 0.02* 6.242*10e18 = 1.24 *10e17, number of electrons after discharge = 1.24*10e17

So of course if you had any losses the number of charge separated electrons would have had to be fewer and per Kirchoff's second Law the voltages of the two caps would have had to settle at something less than 10 volts each, they didn't so again by definition there were only unmeasurable losses. Unless you want to argue you lost half your electron charges to heat, EM etc, but they somehow reappeared. As I said the problem is right there in the definitions. I do feel confident if I discharge a cap into a strictly resistive load I will see the E=1/2cV2 behaviour. This leaves Conservation of Energy as the most likely problematic part.
Lastly, this property of varying efficiency of energy transfer in relation to the relative voltage difference between a cap and the charging source seems to be a universal property of electrostatic charging across a dielectric, put another way the capacitor charge curve applies consistently to different types of capacitors and over varying voltage ranges. You know one last thing, even if you are right and maybe you are, if we accept that there are these losses from EM radiation, heat, etc. I think we would both agree we would still want to work in the area of the charge curve where energy transfer is most efficient.

[James McDonald]

Hi James,

Sorry neither answer flies as can be demonstrated in numerous ways. 1) Heat, I have discharged small caps say 1000 times/sec continuously, I should double check, but I don't see any heating per IR thermometer. b) I have discharged a small 0.1 uF cap into an empty 470uF cap. The first discharge showed almost 99% losses, the second discharge 97% losses. As voltage difference has hardly budged between the two the difference in energy transfer is too great to be accounted for by resistive heat losses. 2) EM radiation is a function of magnetic flux. There is magnetic flux off a discharging cap but it would be increased by the addition of an inductor. So if this was a significant source of loss an inductor would make things worse whereas depending on the set-up an inductor either has no effect or improves things. Also, and here I am simply asking not trying to defend my position, I realize now that Maxwell is the theoretical genius behind so much of this, where in Maxwell's equations can an argument be made that EM radiation would be a significant loss? Further, EM radiation is not undetectable, if one is losing half ones' energy to EM radiation in cap discharges, where is the research, the analysis in the literature of this radiation?

Lastly and most importantly as noted above conservation of charge is retained in the cap discharge scenario, that should sink in, it took a awhile for me. So for the fun of it let's calculate the starting and ending electrons

1 Coulomb = 1 Farad x 1 volt

Cap before discharge while at 20 Volts .001 Farads, 20*.001= .02 Coulombs

Two caps after discharge 10 Volts
Cap 1: 10*0.001 = .01 Coulombs
Cap 2 10*0.001 = .01 Coulombs

Cap 1 + Cap 2 = 0.02 Coulombs

1 Coulomb = 6.242*10e18 electrons

Number of electrons before discharge 0.02* 6.242*10e18 = 1.24 *10e17, number of electrons after discharge = 1.24*10e17

So of course if you had any losses the number of charge separated electrons would have had to be fewer and per Kirchoff's second Law the voltages of the two caps would have had to settle at something less than 10 volts each, they didn't so again by definition there were only unmeasurable losses. Unless you want to argue you lost half your electron charges to heat, EM etc, but they somehow reappeared. As I said the problem is right there in the definitions. I do feel confident if I discharge a cap into a strictly resistive load I will see the E=1/2cV2 behaviour. This leaves Conservation of Energy as the most likely problematic part.
Lastly, this property of varying efficiency of energy transfer in relation to the relative voltage difference between a cap and the charging source seems to be a universal property of electrostatic charging across a dielectric, put another way the capacitor charge curve applies consistently to different types of capacitors and over varying voltage ranges. You know one last thing, even if you are right and maybe you are, if we accept that there are these losses from EM radiation, heat, etc. I think we would both agree we would still want to work in the area of the charge curve where energy transfer is most efficient.

Hi ZPDM --

The heat losses in the capacitor charge potential would be due to copper leads on the parts have resistance. The lead length of two capacitors being hooked in parallel is say like 4 inches per plate maximum unless they have a 18 inch to 24 inch gator clip wire between them. Depending on the thickness of the wire there will be heat losses in the wire itself. The second thing that is not talked about much in electronics theory is the fact that every capacitor has inductor properties and every coil has capacitor properties. As far as the EM radiation losses go I have some books with Maxwell's information in them. I will look up some of the EM information and report back to you what I find out.

-- James

[Faraday88]

Hi ZPDM --

The heat losses in the capacitor charge potential would be due to copper leads on the parts have resistance. The lead length of two capacitors being hooked in parallel is say like 4 inches per plate maximum unless they have a 18 inch to 24 inch gator clip wire between them. Depending on the thickness of the wire there will be heat losses in the wire itself. The second thing that is not talked about much in electronics theory is the fact that every capacitor has inductor properties and every coil has capacitor properties. As far as the EM radiation losses go I have some books with Maxwell's information in them. I will look up some of the EM information and report back to you what I find out.

-- James

For God sake please do not discriminate between Power and Energy if you are considering only the spatial (Flat Space -Time) aspect of Energy...Energy dissipation is power..
you shall run in circles for sure if you do not understand this..

[ZPDM]

Hi ZPDM --

The heat losses in the capacitor charge potential would be due to copper leads on the parts have resistance. The lead length of two capacitors being hooked in parallel is say like 4 inches per plate maximum unless they have a 18 inch to 24 inch gator clip wire between them. Depending on the thickness of the wire there will be heat losses in the wire itself. The second thing that is not talked about much in electronics theory is the fact that every capacitor has inductor properties and every coil has capacitor properties. As far as the EM radiation losses go I have some books with Maxwell's information in them. I will look up some of the EM information and report back to you what I find out.

-- James

Hi James,

Okay, we can have it your way, half of your charge is lost to heat/radiation, conservation of energy holds, conservation of charge doesn't. I really don't want to philosophize further on Bob's thread as he specifically requested this not happen. I started a thread on this which would be appropriate if we want to discuss more. Would apologize for the tone of my post here, while as I mentioned in the back of my head I thought things didn't add up I only trusted this enough to put out a snarky post after reading what an Hewlett Packard engineer had to say on the topic.