SO how many amps will I get if I charge a ONE FARRAD cap up to 12 volts?

[danadak]

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 changes
its charge ? After all is not the Space station one plate of a C ?


Regards, Dana.

[Assafl]

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.

[Vtile]

Would not the opposite plate in a Leyden jar be mother earth ?

-snip-

It 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.

[MrAl]

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.

[Ratch]

To the Ineffable All,
     I am going on vacation for a few days.  I will answer this post when I get back.

Ratch

[raspberrypi]

charging with electrons...the more electrons the more electronegative the notionally unearthed terminal

The electron is impeached... freedom to the hole

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.

Ratch

It 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.

I 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.

[Vtile]

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:

Q = C x V
I = C x dV/dT

I in amps, C in farads, V in volts, T in secs

So you know what C is, what desired dV is (12 volts), and pick
the amount of time you want to charge the cap in, dT.

Conversely use same equations for discharge. Note these equations are
for ideal capacitor, one with no ESR, but good approximation for what
you are doing.

Also discharge only works as a fixed current if the load is a constant
current 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_circuit

Regards, Dana.

The 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.

[drussell]

Are amps not a form of energy?

If 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.

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.

For 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.pdf

Mine looks basically like the one on the cover except I built mine with THREE parallel 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's ANOTHER item of mine I should do some videos on.

[rsjsouza]

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. 

[raspberrypi]

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 12
A: 14 amps.

[raspberrypi]

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 12
A: 14 amps.

[Ratch]

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

[hamster_nz]

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

I 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.

[Ratch]

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

[Brumby]

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.

Vtile came closest by pointing out the answer to the original question is very much time dependent.

[Ratch]

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

[Zero999]

The question is badly phrased and indefinite.

I agree but this is to be expected, as it's the beginners section.

Basically its a car audio cap "Lightning cap Brand" with one of those LED "charge" gauges on top

I 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.

[raspberrypi]

The question is badly phrased and indefinite.

I agree but this is to be expected, as it's the beginners section.

Basically its a car audio cap "Lightning cap Brand" with one of those LED "charge" gauges on top

I 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.

For 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.

[Zero999]

For 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.

Then 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.

[yuzuha]

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.

[raspberrypi]

Remember this is a beginners section of the forum so the questions are not always going to make sense or seem concise.

[Brumby]

Remember this is a beginners section of the forum so the questions are not always going to make sense or seem concise.

... 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.

[raspberrypi]

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.

Do 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.

[danadak]

Are amps not a form of energy?

Yes.

http://electricalengineeringforbeginners.blogspot.com/2009/07/voltage-current-power-and-energy.html

Electrons are being moved, they have mass, hence energy equivalence. They
have intrinsic energy as a particle, wave function, and when moved work is
expended.


Regards, Dana.

[Ratch]

Are amps not a form of energy?

Yes.

http://electricalengineeringforbeginners.blogspot.com/2009/07/voltage-current-power-and-energy.html

Electrons are being moved, they have mass, hence energy equivalence. They
have intrinsic energy as a particle, wave function, and when moved work is
expended.


Regards, Dana.

Definitely 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