SO how many amps will I get if I charge a ONE FARRAD cap up to 12 volts? - Page 1
Forum section: electronics / beginners

author: raspberrypi
author_rank: Frequent Contributor
timestamp: February 03, 2017, 04:16:37 am
content: Need to test the balls factor on some parts but by way of scientific method. My brain broke down when converting coulums to zipzaps...Scariest cap ever!

author: Brumby
author_rank: Unknown
timestamp: February 03, 2017, 04:34:05 am
content: To get any sort of answer for that, we would need to know the ESR of the cap.A better answer would include the maximum rated discharge current. I'd be looking for this on a datasheet.With both these bits of data, you can then plug in the load characteristics and start getting some answers.

author: andtfoot
author_rank: Unknown
timestamp: February 03, 2017, 04:34:17 am
content: I think the actual current delivered is more determined by the resistance (Ohm's law), based on the voltage and internal resistance of the cap (ESR).The time this can be sustained for is based on the capacity of the cap.This seems relevant:http://electronics.howstuffworks.com/capacitor2.htmBy that, it would be 12 Amp/seconds (12A for 1 second, 6A for 2 seconds, etc)

author: BobsURuncle
author_rank: Regular Contributor
timestamp: February 03, 2017, 04:56:49 am
content: The capacitor is holding 12 Coulombs of charge. The rate of discharge, amps, will depend on the equivalent series resistance of the capacitor and the load it is discharging into. Discharged into a constant resistive load the current will decrease exponentially as the voltage in the capacitor drops.

author: sleemanj
author_rank: Super Contributor
timestamp: February 03, 2017, 05:03:40 am
content: My capacitor calculations page may be of assistancehttp://sparks.gogo.co.nz/capacitor-formulae.html

author: neil t
author_rank: Regular Contributor
timestamp: February 03, 2017, 05:19:15 am
content: Quote from: raspberrypi on February 03, 2017, 04:16:37 amI'm guessing you are looking for instantaneous current, perhaps into a car amplifier or something like that.

author: IconicPCB
author_rank: Super Contributor
timestamp: February 03, 2017, 06:57:17 am
content: BobsURuncle gave you the only sensible response.Capacitors store charge. They do not store current.Current is rate of flow of charge.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: February 03, 2017, 07:45:12 am
content: I = V / RIn this case, V=12 (to start with) R is the effective series resistance of your capacitor plus the resistance of the wiring.V will reduce as it discharges.

author: danadak
author_rank: Super Contributor
timestamp: February 03, 2017, 11:02:38 am
content: Q = C x VI = C x dV/dTI in amps, C in farads, V in volts, T in secsSo you know what C is, what desired dV is (12 volts), and pickthe amount of time you want to charge the cap in, dT.Conversely use same equations for discharge. Note these equations arefor ideal capacitor, one with no ESR, but good approximation for whatyou are doing.Also discharge only works as a fixed current if the load is a constantcurrent source. If its resistive use exponential discharge equation.But keep in mind the current will be a f(t) and will decline exponentially.https://en.wikipedia.org/wiki/RC_circuitRegards, Dana.

author: Vtile
author_rank: Super Contributor
timestamp: February 03, 2017, 11:44:03 am
content: You get energy, not amps.

author: raspberrypi
author_rank: Frequent Contributor
timestamp: February 04, 2017, 03:34:19 am
content: Quote from: Vtile on February 03, 2017, 11:44:03 amAre amps not a form of energy? Maybe something was lost in translation.Basically its a car audio cap "Lightning cap Brand" with one of those LED "charge" gauges on top which I'm guessing changes the perceived ESR? I have one of those Chinese 5V to 7kV DC step up transformer cylinders hooked up to it. The goal is to make a continuous spark (a spark that's at the frequency of the device), verse one spark that I get when two 470 uf caps are hooked up to it. I want to make a jacobs latter. I cant seem to find any technical info on this cap so can we guess what the ESR is approximately? What effect would the charge meter circuit have on the ESR (I know it doesn't change the ESR inside the cap but won't it be seen on the output?).

author: Brumby
author_rank: Unknown
timestamp: February 04, 2017, 03:54:33 am
content: Quote from: raspberrypi on February 04, 2017, 03:34:19 amIn discussions like these, you are going to encounter a mix of 'loosely used' and 'scientifically precise' terms. This is one example.Amps is a simple quantity that is measured at a single point in time. You can create a graph of your amps over time by taking lots and lots of measurements and putting lots and lots of points on a chart. You can then join the dots to create a continuous graph.But, the fundamental still applies - Amps are a measure of current at a single point in time. The amount of energy that is transported is measured/calculated over anintervalof time, not apoint.Strictly speaking(and I mean getting into the mathematical definitions), if that point in time is exactly that - a point, then there is no time interval and thus, no energy transported. It's like taking a snapshot of a ball in the air and asking how much kinetic energy it has. You can't do it.This may seem confusing, but you have to understand that there are a lot of members here, with a wide range of experience, background and 'teaching' ability. Be prepared for these - and if you can work out how to fit all the pieces together, you will not only learn more about the subject - but you will also learn how other people express ideas. Both of these make future learning far more productive.

author: Nerull
author_rank: Frequent Contributor
timestamp: February 04, 2017, 07:33:03 am
content: The actual definition of an amp is fairly complicated to explain at a basic level, but essentially it is the number of electrons flowing past a point every second, and is an instantaneous value, like the speed of a car. You can't determine how far a car has driven from just it's speed, and you can't determine the total amount of energy transferred from just amps. You can't even tell how much work those electrons can do from just amps. A capacitor will store a given amount of energy, but that energy could be transferred out of the capacitor in 1/10th of a second, 1 second, 5 minutes, 3 hours, 3 years. Each will give a completely different amount of amps, but the same energy (ignoring losses)

author: EEVblog
author_rank: Unknown
timestamp: February 04, 2017, 08:44:59 am
content: Quote from: Vtile on February 03, 2017, 11:44:03 amHook that cap up to any load and you'll get amps all right. Short it out and you'll initially get the 12V/CAPesr and then it will drop from there.You don'thaveto care about the energy if you don't want to, but of course it still matters because it's the energy capacity that determines how long it power a given load.

author: EEVblog
author_rank: Unknown
timestamp: February 04, 2017, 08:46:49 am
content: Quote from: Brumby on February 04, 2017, 03:54:33 amThere's a term for that, Power.P=V*I

author: Vtile
author_rank: Super Contributor
timestamp: February 04, 2017, 09:10:42 am
content: Good now energy is talked through.For OP, do look up the definitions of Farad unit.source wikipedia.

author: raspberrypi
author_rank: Frequent Contributor
timestamp: February 05, 2017, 05:00:25 pm
content: Quote from: Brumby on February 04, 2017, 03:54:33 amThats how I think of the Heisenberg uncertainty principle. The quicker your "picture" of an object the more you know its location but the less you know its velocity. Time is a measurement of quantity too. Without time there are no quantities, just probabilities or potentials.

author: Ratch
author_rank: Regular Contributor
timestamp: February 05, 2017, 09:57:17 pm
content: Quote from: raspberrypi on February 03, 2017, 04:16:37 amYour question does not make sense for the reasons others have already given. Also, a capacitor does not "charge". It energizes. A capacitor energized to 100 volts has the same net charge on it as it did at 0 volts. For every coulomb of charge added to one plate, one coulomb is deleted from the opposite plate. Capacitors store energy, not coulombs.Ratch

author: mikerj
author_rank: Super Contributor
timestamp: February 05, 2017, 10:28:30 pm
content: Quote from: Ratch on February 05, 2017, 09:57:17 pmFirst time I've ever heard that. A capacitor storescharge, the action of adding charge ischargingand once done it ischarged.

author: Vtile
author_rank: Super Contributor
timestamp: February 05, 2017, 10:37:20 pm
content: Actually it is spring.

author: Ratch
author_rank: Regular Contributor
timestamp: February 05, 2017, 11:09:28 pm
content: Quote from: mikerj on February 05, 2017, 10:28:30 pmWell, it's true a cap does not store charge. When you apply a voltage across a cap, one plate accumulates a charge, and the opposite loses the same amount of charge. The plates will unbalanced, but the net change of the charge will be zero. It takes energy to separate the charge on the plates, and energy is what the cap truly stores W=1/2 C E^2.To say a cap "charges" is using technical slang, which like most slang is not literally true. Sort of like NASA claiming their astros "walk" in space. Even if everyone knows what you mean, it is not a correct description and can be confusing and sometimes lead to mistakes.Ratch

author: IconicPCB
author_rank: Super Contributor
timestamp: February 05, 2017, 11:43:12 pm
content: The electron is dead... long live the hole

author: Dinsdale
author_rank: Regular Contributor
timestamp: February 06, 2017, 01:01:25 am
content: Zero amps.

author: danadak
author_rank: Super Contributor
timestamp: February 06, 2017, 01:05:08 am
content: Not great at Physics, but is there not an equivalency between charge and potential,one does not exist without the other. So if we charge a cap we can talk about it havingand equivalent coulomb charge, or potential, they are one and the same as related,Q = C x V. And we can also talk about the energy stored, and its relationship to charge.Like matter, E = m x c^2, they are inextricably linked to each other. One can talkabout either, both being correct observations.Regards, Dana.

author: Ratch
author_rank: Regular Contributor
timestamp: February 06, 2017, 01:31:53 am
content: Quote from: danadak on February 06, 2017, 01:05:08 amLet me ask a quick question. When you speak about charging a capacitor, what are you charging it with? Please don't say coulombs of charge, because I showed previously that applying a voltage across a capacitor does not increase its net charge. Another question. What do you define as "potential"?Ratch

author: IconicPCB
author_rank: Super Contributor
timestamp: February 06, 2017, 02:50:42 am
content: charging with electrons...the more electrons the more electronegative the notionally unearthed terminalThe electron is impeached... freedom to the hole

author: danadak
author_rank: Super Contributor
timestamp: February 06, 2017, 02:55:12 am
content: If I have a C, both plates connected, then Q = C x V = 0 because V is 0, and V is 0 because nocharge differential exists. Note no voltage on the C, even though there could be an equal amountof excess charge on both plates, or no excess charge. # protons = # electronsIf I then extract some amount of charge from one plate, the other plate unconnected, then I developa V = Q / C. The charge comes from wherever, from the universe of charge, not necessarily the otherplate. So now we have a differential amount of charge on C. Thereby exhibiting a V.Am I missing something ?Regards, Dana.

author: Ratch
author_rank: Regular Contributor
timestamp: February 06, 2017, 03:16:01 am
content: Quote from: IconicPCB on February 06, 2017, 02:50:42 amI thought I made it clear. The same number of electrons are removed from the second plate as are added to the first plate. You cannot give a cap a net charge of electrons.Ratch

author: Ratch
author_rank: Regular Contributor
timestamp: February 06, 2017, 03:28:25 am
content: Quote from: danadak on February 06, 2017, 02:55:12 amYes, you are missing an explanation. How are you going to extract a charge from one plate without giving the same charge to to other plate? Charges, like mass don't just appear and disappear. They have to be accounted for and explained. Applying a voltage across a capacitor will shift some of the charge from one plate to the other for a net charge change of zero, and an energy change of 1/2 C V^2. That operation is energizing a capacitor.Ratch

author: mikerj
author_rank: Super Contributor
timestamp: February 06, 2017, 08:02:37 am
content: Quote from: Ratch on February 06, 2017, 01:31:53 amApplying a voltage (or more accurately a current)movescharge. This process is charging.

author: Vtile
author_rank: Super Contributor
timestamp: February 06, 2017, 08:56:39 am
content: Heaviside explained this displacement as a spring analogy and resisted the charge concept to the end.No I have not read the Electromagnetism 1&2, these are from 2nd hand sources.Interesting discussion shows how much I have forgotten from the little I knew.

author: danadak
author_rank: Super Contributor
timestamp: February 06, 2017, 11:21:22 am
content: Change charge on one and only one plate -1) Apply heat to one plate.2) Radiate/bombard one plate.3) Photo electric effect applied to one plate.Of course then we have the issue the plate much more massive than the electron, so whatis the required escape velocity of the electron to break free of plate entirely. After all thereis gravity to be considered. What is the relative force to break free vs force to extract electronfrom atom, eg. its coulomb and gravitational and atomic forces. So many questions, so little time.....And then we have to consider the electrons in the parallel brane universes, what is goingon with them. And can a brane be considered a dielectric ? I do not know. Are we all justa giant capacitor in the scheme of things ?https://en.wikipedia.org/wiki/BraneThen there is the coupling of the Heisenberg uncertainty principle and the virtual particle froth in theuniverse, where/when does the anti electron get created ? And does the Heisenberg coupling implyelectrons are in violation of signed Coulomb force, that is they are somehow attracted to each other.Ga....h head spinning, do capacitors really work ?Regards, Dana.

author: Vtile
author_rank: Super Contributor
timestamp: February 06, 2017, 12:13:46 pm
content: Atleast it doesn't make it any more clear that the electrons do flow against (so called) the current and the positive voltage side of the resistor is the depletion side of the electrons as there is more negative particles on the negative side that in the positive side which have less negative particles while the current do flow upstream (by convention).It is asconventionalpractical as holding the hammer from the head and using the shaft driving the nails.So we are kind of stupid. We should have left hand rule instead of right hand rule.

author: Ratch
author_rank: Regular Contributor
timestamp: February 06, 2017, 04:11:25 pm
content: Quote from: mikerj on February 06, 2017, 08:02:37 am

author: Ratch
author_rank: Regular Contributor
timestamp: February 06, 2017, 04:15:23 pm
content: Quote from: danadak on February 06, 2017, 11:21:22 amIs that the way you would energize a capacitor capacitor in a circuit?Ratch

author: danadak
author_rank: Super Contributor
timestamp: February 06, 2017, 08:38:31 pm
content: No, but physical laws don't change. I am simply questioning the statement madeearlier that net charge on a cap does not change.Again, just asking questions, not a physicist, if we can't answer these then wehave a deficit in our C working knowledge.Regards, Dana.

author: IconicPCB
author_rank: Super Contributor
timestamp: February 06, 2017, 10:27:02 pm
content: Holy Sheldontalk (TM) ,,,What is it with You ... The OP asked a simple question out of ignorance... what current will he get from... And lo and behold a pissing contest starts.Engineers do not get involved in esoteria... whether it is energy expanded in creating a displacement current or work done in separating "charged" plates against an electrostatic field.. it does not matter.A flow of electrons will take place between two points with different charge density tending towards minimum energy condition along any conductive path which may exist between these two points.How this minimum energy condition may be reached depends on the path. It may be a long or short process but it will certainly not be a steady state process. IT will be a transient process requiring an analytic solution to define it.And in a way of a stright forward no BS response to the OP...current is defined as flow of charge past a point. Charge is carried by electrons in classic definition.. although in modern semiconductor theory electrons coexist with holes ( or absence of electrons).In the old days the Ampere was defined as flow of one coulomb of charge per second.One coulomb of charge was defined as 6.242 x e18 elementary units of charge.An elementary unit of charge is the charge carried by an electron ( deemed to be negative ) or a proton ( deemed to be positive)One elementary unit of charge is defined as 1.6021766208×e?19 coulomb.In case of a capacitor under steady state conditions a plate become charged with excess electrons.One plate becomes more electro negative with respect to the other and conversely one the other plate becomes more electro positive with respect to the first .The increase of electrons on one plate results in a time varying change in electric field. This is known as displacement current. It is NOT a current of moving electrical charges.From late forties onwards an ampere was defined in terms of a force which exist between two current carrying conductors.It is this force which is employed in D'Arsonval meter movement to provide an indication of current flow .BAck to OP s question. The capacitor will provide as much current as the path joining the two plates will carry.It will be a time varying current since the charge density on the plates of a capacitor will vary in time. In order to estimate the current path impedance and duration of current flow need to be specified.

author: Vtile
author_rank: Super Contributor
timestamp: February 06, 2017, 10:34:17 pm
content: It were answered in three first answers. Rest is just for fun.Besides this is really poorly discussed topic atleast on the EE books I have opportunity to get to my hands. There is plates, here is electric field, here is volts between and another is positive another is negative. The type of the model is hardly ever mentioned, nor the application window where it is usefull etc. Even the truth about the plate sizes versus cycletime is forgotten to mention.Then when you start to dig around of it, you discover it is actually a transmission line and everything were a halftruths, assumptions, simplifications.The damn foil roll must be the most misunderstood component of all of them in electric and electronic engineering.

author: free_electron
author_rank: Super Contributor
timestamp: February 06, 2017, 10:54:12 pm
content: how long after charging ?what is the leakage rate ?what is the esr and esl of the capwhat is the resistance and inductance in the wires ?what kind of capacitor ( is it made for pulse discharge ? ) is it fat and short or tall and skinny ?so many factors.

author: danadak
author_rank: Super Contributor
timestamp: February 07, 2017, 02:26:24 am
content: What is it with You ... The OP asked a simple question out of ignorance... what current will he get from... And lo and behold a pissing contest starts.Engineers do not get involved in esoteria... whether it is energy expanded in creating a displacement current or work done in separating "charged" plates against an electrostatic field.. it does not matter.I agree OP probably does not need this. It simply cascaded because of a question raised earlier about net charge.Which brings me to the "esoteria" thought. I had the opportunity to work twice in environments with a team of engineersand PHds in Physics, and lunch time was full of esoteric in the quest for a deeper understanding of phenomena notwell understood. I loved these sessions. These discussions did matter. Even when i worked on bit slice stuff I went tolunch with Mrazek (inventer of TriState, head of 2900 team) and my boss and heads of Hybrids (read analog experts)and we discussed many out of the box kinds of ideas. So yes, if designing an electronics frypan that needs to get tomarket right away, I support your point of view, except at lunch and outside work, where esoteric is a good thing.Regards, Dana.PS : Not a pissing contest yet, no name calling has occurred.

author: IconicPCB
author_rank: Super Contributor
timestamp: February 07, 2017, 05:56:17 am
content: Hello Dana,I dont go for name calling either. I was just a little bit miffed by the swirls in the discussion.

author: NMNeil
author_rank: Contributor
timestamp: February 07, 2017, 03:43:16 pm
content: Funny, in beekeeping circles we have a saying, "Ask 5 beekeepers the exact same question, and you'll get 6 different answers"This seems to also apply to engineers

author: Vtile
author_rank: Super Contributor
timestamp: February 07, 2017, 03:47:09 pm
content: Quote from: NMNeil on February 07, 2017, 03:43:16 pmWell both handle matters that sting and do make a buzz sounds, what did you expect.

author: NMNeil
author_rank: Contributor
timestamp: February 07, 2017, 05:33:45 pm
content: Quote from: Vtile on February 07, 2017, 03:47:09 pmDon't forget the magic smoke, beekeepers and engineers both generate smoke.

author: IconicPCB
author_rank: Super Contributor
timestamp: February 07, 2017, 10:31:58 pm
content: I thought I made it clear. The same number of electrons are removed from the second plate as are added to the first plate. You cannot give a cap a net charge of electrons.RatchOK .. i am now sufficiently agitated to offer Maxwel's work in defence of my position:I refer the reader to the first and second paragraph ofhttps://en.wikipedia.org/wiki/Displacement_currentfor a statement on displacement current and te fact that it is not a flow of chraged particles such as the conventional current or the more correct electron flow.

author: Ratch
author_rank: Regular Contributor
timestamp: February 08, 2017, 01:28:28 am
content: QuoteAnd, how does that factoid about displacementcurrentchange or contradict the statement that capacitors are energized when the charge on their plates are unbalanced?Ratch

author: rsjsouza
author_rank: Super Contributor
timestamp: February 09, 2017, 03:49:31 am
content: The way I learned was that, strictly speaking, when capacitors were connected to a DC voltage source they were able to sustain the potential difference (we called d.d.p. or diferença de potencial) across its terminals in open circuit.We used the term "charge" simply due to the factor a capacitor under a load displays a similar behaviour as a battery - the imbalance of potential causes current to flow to/from its terminals.The lectures on power and distribution systems used to call circuits and its elements "energized".

author: MrAl
author_rank: Super Contributor
timestamp: February 09, 2017, 04:21:53 am
content: Quote from: Ratch on February 05, 2017, 09:57:17 pmHi there Ratch,From a purely physical viewpoint that may be true, but that limited view will just confuse people who are talking about this from a purely electronic viewpoint. Surely you must have suspected something isnt right by now since almost everyone here questions that kind of statement.The difference in the two viewpoints, again, is PURELY PHYSICAL vs PURELY ELECTRONIC.When we talk about the physical nature of the cap itself from an internal viewpoint, we may want to know that the charge is balanced internally. BUT there is a big difference when we view that capacitor from the EXTERNAL world. So here we can say there is an INTERNAL vs EXTERNAL viewpoint difference.The internal view says that the charge is balanced, but the external view sees it as a differential charge, not an absolute charge. What this means is that we could have 1/2 of the total charge on one plate while we have a deficit of 1/2 of the total charge on the other plate, yet when we view this from the external world we have a view of the total charge which comes out to:qTotal=q/2-(-q/2)=q(or we could just say that the total charge is accounted for on just one plate only)This is similar to when we use a plus and minus power supply (plus Vcc and minus Vee) and measure the voltage from Vcc to Vee instead of from one or the other to ground. We dont get zero volts, which would be incredibly ridiculous. We get the difference:V=Vcc-Veeand since Vee is negative if Vcc=5 and Vee=-5 we get:Vtotal=Vcc-Vee=5-(-5)=10 volts.THAT is the way we MUST look at it in the electronics world from the viewpoint of an electrical circuit and it's electrical analysis. If we want to know what the cap is doing alone, we might look at that alone, but in a circuit we MUST know ther total charge equivalent which we almost always just call simply, "The Charge".I dont think i can make it any more clear than that although the more usual view is that we just have charge on one plate that is equivalent to the 'total charge'.You may want to note that there are other times when we have to depart from the purely physical view in the electronics world. We sometimes have to apply different attributes to the same components depending on the actual application. in modern times this would be called "anisotronic" behavior, where we can not analyze the circuit (note this is particular to circuit analysis) without knowing the multifaceted behavior which depends on the particular theoretical application at hand.

author: hamster_nz
author_rank: Super Contributor
timestamp: February 09, 2017, 07:54:21 am
content: When you 'charge' a cap, you are moving charge carriers (electrons) from one of it's terminals to the other.Imagine charging a cap from a battery with a current limting resistor in series. A meter reading amps in series with the positive terminal will read the same as another meter connected in series with the negative terminal.It will start off with the cap appearing to be a dead short (the full battery voltage will be measured over the current limiting resistor), and then quickly fall away as the voltage across across the cap increases.Once fully charged 0 amps will be flowing, and so the current limiting resistor will have zero volts over it - (although you can theoretically never fully charge a cap to the match a voltage source, only get very, very close - the whole exponential decay thingy).And it is possible to add charge to one terminal of a capacitor (e.g a static charge), but that sort of makes a second capacitor between the uncharged terminal and the external environment...

author: Assafl
author_rank: Frequent Contributor
timestamp: February 09, 2017, 09:52:49 am
content: Quote from: Ratch on February 06, 2017, 03:16:01 amIt may appear to be that there are capacitors that can accept a net charge as well. Leyden Jars are good examples of that. The physical mechanisms are the same but diametrically the opposite in quantitative terms. Caps can be tiny, but due to the size of the plates, distance between them and the dielectric constant, can accept a fairly large charge to opposed a reasonable electric field.A Leyden Jar, on the flip side - is tiny, requires a very large electric field to charge it, and seemingly doesn't have an opposing plate - hence the net charge "electrostatic" claim (of course it does - the air around it is charged - that is why the air around it lights florescent bulbs, and is positively dangerous)...One of the rather amazing (if unpleasant) surprises to owning an electrometer is that you realize everything around you (with the exception of good conductors) - including us humans - is a capacitor.

author: danadak
author_rank: Super Contributor
timestamp: February 09, 2017, 11:26:10 am
content: Would not the opposite plate in a Leyden jar be mother earth ?The argument that charge cannot be changed on one plate and only one plate,does that mean every time the charge changes on earth the Space station changesits charge ? After all is not the Space station one plate of a C ?Regards, Dana.

author: Assafl
author_rank: Frequent Contributor
timestamp: February 09, 2017, 12:32:57 pm
content: Earth, air, ions, all the same thing. It is not the sort of jar you'd want to put a finger in... The reason earth is so important is that you cover the Leyden jar with a ground potential shield to keep you safe.Electrons do have to come from an atom somewhere - and that Atom want its electrons back (ions).Earth is important as it is the Human's Potential (usually) - hence will prevent the electrons from deciding to flow through you.So where do the electrons in a thermionic emission come from?NB - If you play around with an electrometer, which is a very sensitive voltmeter that can detect very low amounts of extra (or missing) electrons, you'll see that there are many electrons just "out there". Mainly stuck to an insulator somewhere (such as a piece of dead skin - if you prod yourself with the probe, a tabletop, a piece of paper). Damn things are everywhere.

author: Vtile
author_rank: Super Contributor
timestamp: February 09, 2017, 12:49:54 pm
content: Quote from: danadak on February 09, 2017, 11:26:10 amIt depends, in my limited understanding the opposite plate is the one that is reference ground for the thing that put the energy to the Leyden Jar (Edit. To be noted, I'm not familiar with it and my first reference of knowledge actually did omit the grounding information). Through complex network of well charge paths. If you do the experiment on the zeppelin there is no earth ground nearby, question is does the experiment work. If it does the earths soil is not in significant role and if it doesn't the earth is in major role (or something else is going on).The question on the space shuttle is interesting. Since between us standing on the positively charged (atleast in middle of the storm) earth there is negatively charged fields on atmostphere and propably there is also positive fields on ionosphere or upperparts of the atmostphere (since it have been noted that sometimes lightning strikes both up and down from the clouds). It would be interesting to know which side of the space shuttle is in positive potential.To go totally offtangent St. Elmo's fire is really interesting stuff and tales can be found around the world.Edit. PS. The reason St. Elmo's fire is not typically observed is (again with my limited understanding) because the visible part of the phenomenom is caused by plasma which is particles on the fluid (air) and in storms the air is rarely dead still, which prevents visible plasma cloud to form. This feature of plasma is used in some HiPower/HV swithes with rabid airblow to push the arch away from the contacts.

author: MrAl
author_rank: Super Contributor
timestamp: February 09, 2017, 02:32:25 pm
content: Hi again,I think maybe a simpler explanation would go as follows...We start with two metal pots, one on the left L and one on the right R.If we take one C from L and place it into R, L has lost one unit of charge while R has gained one unit of charge.However, note we only actually moved one unit of charge from one place to the other, so we call it 1C.It doesnt matter if we use the same unit of charge or another unit of charge as long when we remove one we immediately add one to the other pot.

author: Ratch
author_rank: Regular Contributor
timestamp: February 09, 2017, 03:12:28 pm
content: To the Ineffable All,I am going on vacation for a few days. I will answer this post when I get back.Ratch

author: raspberrypi
author_rank: Frequent Contributor
timestamp: February 13, 2017, 07:32:26 am
content: Quote from: Assafl on February 09, 2017, 09:52:49 amI made one of those with a 1L liquor bottle with Al foil a copper wire and salt water. What was interesting is that any metal attached to the top (the copper wire that went into the capacitor quickly oxidized. Alligator clips fell apart after a week, and the copper wire turned black then white with green oxide then after a month I touched it and it turned to dust. It must have been picking up the charge in the air then causing the top to turn positive. WAs during a humid maryland summer.I still don't know how many amps my big cap will put out.

author: Vtile
author_rank: Super Contributor
timestamp: February 13, 2017, 09:09:35 am
content: It will put as much amps as you wish, but!Now comes the whole energy / charge sidedrift discussion. Since it stores charge/energy/space lint the outputs current will not be constant but dependaple of time as it discharges by logarithmic curve.If your capacitor have small enough ESR and leads are short and thick you can discharge it with single steelmelting bang if you short circuit it. This can be seen on the Farad SI-unit definition F=As/V even more so when we reorder it like A = (F*V)/s, where A=current, F=capacitance, V=voltage, s=time. This formulae only gives you aproximate answer, since as time changes other values changes aswell. This formula is same as given on the few first answers:Quote from: danadak on February 03, 2017, 11:02:38 amThe reply#7 gives you the answer how you can "program the output", that is with resistance.Now since capacitor is like really poor battery cell or more closely a battery that is depleted already you only get constant current (well same kind of constant as in battery, which is not constant) only if the load current is small compared to the maximum discharges it can deliver. To word it differently, if you do want it to give you a constant current , you have no other option than keep the current draw small (the time constant s is then big) with big load resistance. Or if you are after big current you put in a small resistor for a load, but then both current and voltage will rapidly drop.If you were after a single xA answer no-one can give it for you if there is no knowledge about the loading resistance and even then it is just a good quess, since all other capacitances and inductances will have own effect (time dependaple voltage and current).I don't know your background, but considering your posting elsewhere I think you try to learn these. Then one good resource for circuit theory is "Schaum's outline series - Theory and Problems of Electric circuits" (first published on the 60's), it is more solid package than most internet resources and more problem oriented than most books with the same content, the mathematics might go over your head (again depending your background). It is not many dollars in used condition and the age of print is not a big deal. Your question is answered in detail in Chapter 16 (Circuit transient).Edit. There seems to be also "Schaum's outline series - Theory and Problems of basic circuit analysis" which is more the real basics and all the "ohms law tricks and usage". Combined with these two it makes about 1000 pages of understanding (in best case). Electric circuits is more of AC and transients (like this subject of your topic). These are theory, but it gives you ability to solve problems, not just use presolved problems. Like the teslacoil below, if you understand those two books and some knowledge here and there you can design your own witn pencil and piece of paper.

author: drussell
author_rank: Super Contributor
timestamp: February 13, 2017, 10:19:36 am
content: Quote from: raspberrypi on February 04, 2017, 03:34:19 amIf you think of it as fluid flow, amps are the measure of the flow rate and the voltage is the pressure. It is the same idea whether you are talking fluid or electrons.QuoteFor a Jacob's ladder you need a lot more than 7000 volts and it needs to be continuous. What you really want to build is something like a tesla coil.Many years ago I built "The Big TC" from the July 1964 issue of Popular Electronics, which there now happens to be a copy of online at:http://www.americanradiohistory.com/Archive-Poptronics/60s/64/Pop-1964-07.pdfMine looks basically like the one on the cover except I built mine withTHREEparallel capacitors instead of one (crazy size) or two (insanity size).I still have it but I'm not sure where I could find to fire it up that was far enough away from all of today's tiny, fragile electronics to not blow up every device in sight.If I recall correctly, we figured mine can pump out close to 250,000 volts, I seem to recall that it can arc about 18 inches. The whole thing glows a nice eerie blue when fired up, especially the capacitors!That'sANOTHERitem of mine I should do some videos on.

author: rsjsouza
author_rank: Super Contributor
timestamp: February 13, 2017, 01:22:36 pm
content: Thank you for sending this PDF to the Popular Electronics magazine - it ends a years long search. When I was a kid I had the Brazilian edition (Eletrônica Popular) and was fascinated with Tesla coils the first time I saw this article.

author: raspberrypi
author_rank: Frequent Contributor
timestamp: February 13, 2017, 06:10:45 pm
content: I thought this thread was going to be bit shorter and not involve the space station, I thought it would go like this:Q: How may amps?A: Depends whats the ESR?Q: Its 12A: 14 amps.

author: raspberrypi
author_rank: Frequent Contributor
timestamp: February 13, 2017, 06:12:10 pm
content: I thought this thread was going to be bit shorter and not involve the space station or the answer waiting for someone to come back from vacation, I thought it would go like this:Q: How may amps?A: Depends whats the ESR?Q: Its 12A: 14 amps.

author: Ratch
author_rank: Regular Contributor
timestamp: February 20, 2017, 06:52:25 pm
content: I am back from vacation.A Leyden jar can store a static charge on one of its plates, but it is not a circuit capacitance. A point charge in space cannot be called a capacitor because it has no dielectric. An in-circuit capacitor energized up to 12 volts will have an imbalance of coulombs, not amps. Amps are the rate charge (coulombs) are moved.Ratch

author: hamster_nz
author_rank: Super Contributor
timestamp: February 20, 2017, 11:31:56 pm
content: Quote from: Ratch on February 20, 2017, 06:52:25 pmI would be the first to say I don't understand capacitance very well, but is not the glass of the jar the dielectric?Is not the external environment (often the experimenter's hand) the other plate, which has transferred charge into the jar? It was also common for Leyden jars to have foil on the outside, which was the other plate.

author: Ratch
author_rank: Regular Contributor
timestamp: February 21, 2017, 02:31:01 am
content: The excess electrons on one plate of the Leyden jar will scare off the electrons on the opposite plate, causing the plates to have opposite charges. One plate will have a excess of electrons and the other plate will have a deficiency of electrons. The dielectric keeps the unbalanced charges from combining and neutralizing the plates imbalance. The net charge on the Leyden jar will be zero.Ratch

author: Brumby
author_rank: Unknown
timestamp: February 21, 2017, 05:00:03 am
content: I am surprised no-one has made this point...The answer to the question needs only the voltage the capacitor is charged to, its ESR and the resistance of the load.How many uF doesn't directly come into it - the discharge graph is going to look like this:The only difference is the absolute units on the time axis.Vtilecame closest by pointing out the answer to the original question is very much time dependent.

author: Ratch
author_rank: Regular Contributor
timestamp: February 21, 2017, 04:40:02 pm
content: The question is badly phrased and indefinite. Does the OP mean the amps when energizing or de-energizing? Is it a constant current or a constant voltage energizing/de-energizing the capaqcitor? A larger capacitance will sustain a longer amp rate. So will a greater voltage. ESR is an AC parameter at a particular frequency, and not applicable here. The unit of capacitance is a farad, not "FARRAD".Ratch

author: Zero999
author_rank: Super Contributor
timestamp: February 21, 2017, 06:52:02 pm
content: Quote from: Ratch on February 21, 2017, 04:40:02 pmI agree but this is to be expected, as it's the beginners section.Quote from: raspberrypi on February 04, 2017, 03:34:19 amI hope you realise that those car audio capacitors are an example of audiophoolery?The theory is they reduce the impedance of the wires to the stereo, therefore provide more bass. In practise they make no difference. The car battery already has a very low impedance and the wires to it should be thick enough to make any voltage drop minimal. A very high powered car amplifier will have a built-in boost converter (any amplifier that really delivers over 18W into a 4 Ohm needs to be run off more than 12V) which will have more than enough decoupling capacitors built-in. If adding a capacitor does make any difference, then your battery or alternator are dying or the cable isn't thick enough.

author: raspberrypi
author_rank: Frequent Contributor
timestamp: February 22, 2017, 01:01:32 am
content: Quote from: Hero999 on February 21, 2017, 06:52:02 pmFor small systems yes, but if you have ever built a big car audio system it makes a huge difference. I had two big Kicker amps from back in the day (They were made in Stillwater OK, and had a "birth certificate" of the power output, not the Chinese junk they sell with the kicker brand in Bestbuy nowadays). These amps used massive amounts of power. They were running dual 15" subs and 10 high draw in car speakers. Each had 60Amp fuses on each channel. Running a 60Amp alternator couldn't power them and the car head lights would dim out and some times even stall the engine at idle. I added two 1Farad caps to the system with short runs of 4 ga wire and it completely fixed the problem. I agree they are not going to "add more bass" to your best buy system if your alternator is already keeping up but they do have their use for crazy 100+db systems.

author: Zero999
author_rank: Super Contributor
timestamp: February 22, 2017, 11:31:01 am
content: Quote from: raspberrypi on February 22, 2017, 01:01:32 amThen it sounds like either your battery was on its last legs (not helped by the undersized alternator) or the wiring was poor. The internal resistance of the battery should be under 0.01Ohms to reliably start the engine. 60A isn't much for a car battery, given the starter motor will use several times the current. With a decent battery, the voltage drop at 60A should be under 0.6V, which shouldn't make much difference.Adding a monster decoupling capacitor was just masking the symptoms, rather than solving the problem.

author: yuzuha
author_rank: Contributor
timestamp: February 22, 2017, 12:49:31 pm
content: Reminds me when I used to use an SCR and a bathtub cap to fire up an auto ignition coil, which I then fed through a stack of 30KV television diodes to charge up a capacitor I made from windows and aluminum foil Capacitance = dielectric constant x permittivity of free space ( 8.854 x 10 to the -12 farads/meter) x Area of a plate / thickness of the dielectric (using meters for area and thickness) dielectric constant is 1 for a vacuum, pretty close to 1 for air and about 7.5 for window glass (or leyden jars) window glass breakdown voltage is about 9.8KV/mm so you need to make sure your glass is thick enough and large enough that the plates won't arc over the surface of the glass. Think of a bug zapper that could vaporize the end of a hot dog with a bang.In fluid terms, I think of a capacitor as two hemispherical pipe fittings with a rubber diaphragm bolted between them.====<|>==== current flows in and depending on the voltage/pressure bends the diaphragm ====<)>=== where the energy is stored in the diaphragm.

author: raspberrypi
author_rank: Frequent Contributor
timestamp: February 22, 2017, 05:27:43 pm
content: Remember this is a beginners section of the forum so the questions are not always going to make sense or seem concise.

author: Brumby
author_rank: Unknown
timestamp: February 23, 2017, 03:59:33 am
content: Quote from: raspberrypi on February 22, 2017, 05:27:43 pm... and responses need to keep that in mind as well. A vague question would indicate it would be rather inappropriate to respond with design concepts that are too advanced for the level indicated by the question.

author: raspberrypi
author_rank: Frequent Contributor
timestamp: March 10, 2017, 04:37:38 am
content: Quote from: yuzuha on February 22, 2017, 12:49:31 pmDo you have pictures? Whats a bath tub cap? I'm guessing the windows were stacks of glass with foil n them?When I made a layden jar from saltwater it didnt seem to make a huge difference when I connected it up to my $5 7,000kv (yes they actually said it makes seven million volts in the description) ebay step up. But did work as a cap when I put lower voltages in it. But it did charge it self so much that any metal connected to the top would quickly oxidize and corrode apart. 1" alligator clips would turn to rust in a week in open air. No effect on the alligator clips attached to the foil on the outside end. So it would charge in one direction. I'm guessing the top electrode that went into the salt water was positive.

author: danadak
author_rank: Super Contributor
timestamp: March 10, 2017, 01:03:58 pm
content: QuoteYes.http://electricalengineeringforbeginners.blogspot.com/2009/07/voltage-current-power-and-energy.htmlElectrons are being moved, they have mass, hence energy equivalence. Theyhave intrinsic energy as a particle, wave function, and when moved work isexpended.Regards, Dana.

author: Ratch
author_rank: Regular Contributor
timestamp: March 10, 2017, 02:08:23 pm
content: Quote from: danadak on March 10, 2017, 01:03:58 pmDefinitely not. Amps are the rate of charge movement, not energy. You have to know the voltage before you can calculate the power. You would not say that velocity is energy either, would you? You have to know the mass before you can calculate the kinetic energy, right?Ratch

author: raspberrypi
author_rank: Frequent Contributor
timestamp: March 12, 2017, 12:49:46 pm
content: Quote from: Ratch on March 10, 2017, 02:08:23 pmAlso you have to take into account time. watts isn't energy until its been some amount of time, other wise its just potential to do work. Like asking how many kilowatts are in a gallon of gasoline.

author: yuzuha
author_rank: Contributor
timestamp: March 13, 2017, 08:00:58 am
content: Quote from: raspberrypi on March 10, 2017, 04:37:38 amSorry no pics that was back in the late 1970s. A bathtub cap is a sort metal can, often rectangular, that is filled with oil and range from the size of a walnut to several hundred kg in size and with voltage ratings from a few hundred to thousands of volts. Think they got the nickname from the smaller rectangular ones used in Ham radios. Yes my caps were panes of glass about 1 meter square with a 10-15cm margin around the edges.