# Capacitor to store lightning?

Status
Not open for further replies.
Yes that is true; however the "whatever it's attached to" can only have a few ohms impedance to ground, otherwise the lightning will just arc to ground thru the pre-ionized air around the "whatever." If you have a tall, vertical metal rod the lightning very well may hit the top of it and travel down it to the point where your network of dividers and connection wires lead off to your capacitor bank. At that point the lightning will resume its air arc to ground as the impedance of that short path will be less than the inductance of your connector network, and orders of magnitude less than passing thru a voltage divider string made of meg-ohm resisters.

Once the bolt has ionized a channel to a conductive object, like a Franklin Rod, for instance, or any other object in Patent Class 174, Subclass 2, that channel will have the lowest impedance until all the voltage in that part of that cloud has been transmitted. Remember, the clouds that discharge electricity are miles above most land masses, and it requires hundreds of millions of volts to overcome the resistance of the air.

Everyone who has ever gotten a shock from a metal doorknob after walking across a carpeted room has felt lightning on a small scale. Your hands do not even need to be in contact with the doorknob because the voltage you've built up by the friction of your shoes against the carpet can cause electricity to jump across a small gap between your hand and the knob.

As you are so ignorant of the physics of electrical circuits you may not understand that "impedance", Z, is the complex addition of resistance, R, and inductance L. I.e. Z^2 = R^2 + L^2

As you are so ignorant of the material that was taught in my classes,, you may not understand that impedance was mentioned by my teacher and covered by my textbooks.

For the frequencies in very brief duration lightning pulse and wiring network connecting to many physically large condensers, L will normally greatly exceed* R. I.e. L >> R but you ignore it as you falsely think that if component is called a "resister" (or a "capacitor") it's L is zero as it nearly is for DC currents.

Inductance is not a property of an electric current, it is a property of an electric component. This property, measured in Henrys, is large in a coiled wire and small in a straight wire. As I understand it, moving electrons in a wire create a magnetic field that can "impede" the further flow of the current. Whenever two or more wires are next to each other with electrons traveling in the same direction, similar to two cars traveling in the same direction in two adjacent lanes in a highway, the magnetic field would reinforce itself, creating a very large inductance. A store-bought spool of wire would itself have much more inductance than the same wire unwrapped from the spool. A coil of wire, with many wires adjacent to each other, was called an inductor in my classes, and the inductance of this component could be measured, using Ohm's Law and the direct measurements of resistance and current by an ohmmeter and a ammeter.

Every wire will have some inductance, just like every wire has some resistance unless it's supercooled to a temperature that approaches absolute zero. However, the inductance of a straight wire, measured in Henrys, is barely measurable and not likely to be a factor in a resistor divider circuit.

By the way, just in case you or anyone else missed it, let me say this quite explicitly. If I thought I could have obtained a patent for the resistor divider circuit with a cap bank wired in parallel with R11, I never would have mentioned a word of it on this message board, but as long as the patent value is gone, let me say that a more useful resistor divider would consist of only two resistors. Let me name them R1 and R2. R1 would be a water-cooled high-wattage 100 K Ohm resistor and R2 would be a water-cooled high-wattage 100 Ohm resistor. The ratio of the resistances of the two resistors, wired in series, would now be approximately 1,000 to one. Whatever voltage was applied across the series, R2 would get just under 1/1,000th of it.

If a lightning bolt supplied the two-resistor series with 200 MVolts, R2 would get 200 KVolts, a good number, considering the fact that earlier in this thread, I supplied a link to a manufacturer of 300KV capacitors. With two of those in every current branch, the lightning rod could be hit with up to a maximum of 600 MVolts, and that's more than the estimated maximum voltage of all but the rare, positively-charged bolts that come from the tops of the clouds.

One more time. I will send to the U.S. Patent Office some circuitry that I have not published anywhere, in strict accordance with their rules.

Benny

Last edited:
benny Here is a quick crude summary of posts 578 and 579:

Lightning is 5 ton gorilla which does what it wants to, not what you hope it would do. It ain’t never going to climb into your condenser bank thru your resistive divider chains and the inductance of the connection wire network to an array of large condensers.

What a shame. You won't be able to see any description of my patent circuitry here. If this statement causes you to stop corresponding, I'll be sorry to see you go, because I truly enjoy discussing concepts and unpatentable circuitry with you.

For what it's worth, I happen to think the "brute force" approach would charge up a bank of capacitors with thousands of current branches to deal with the estimated maximum 100 KAmps. If each one of those thousands of branches had thousands of HV caps in them, with voltage ratings of at least 100 KV, it would deal with the hundreds of millions of volts in the estimated maximum negatively-charged bolts. This would also require that the cap bank is attached to the appropriate sort of lightning rod, and located in a place that has, historically, seen many electrical storms, so it would charge up the extensive (and probably expensive) cap bank with a lot of electrical energy. Whether all this would be economical is, of course, debatable, but that's a different question than whether the physics would work in my favor.

Positively-charged lightning bolts would not be a problem because the bank would not use electrolytic capacitors, so they would not be damaged if they were charged with an opposite polarity.

In my mind, the only real questions are how and when to disengage the system from future strikes:

WHEN:
1. after the first strike
2. after the second
3. when the caps are charged up to half of their rated voltage
4. when the caps are charged up to 3/4 of their rated voltage
5. when the caps are charged up to 90% of their rated voltage
6. other

HOW: (No, I'm not going to disclose my circuitry ideas here.)

The circuitry I will send to the U.S. Patent Office will deal with the polarity question. However, the application will not mention the source of the DC electricity that will charge the capacitor in question. The polarity of the DC source will be specified but not the fact that lightning could be the source.

Last edited:
... As you are so ignorant of the material that was taught in my classes,, you may not understand that impedance was mentioned by my teacher and covered by my textbooks.
Of course I don't know what your teachers taught, but you must have sleep thru their lectures as you thought in post 523 that only components called inductors had inductance (until I educated you in post 525):
{post 523}... I'm afraid I'm not familiar with your use of the term "inductance" as used in a cap bank. In the classes I took, the word was only applied to inductors. ...
{post 525, replying to 523}ALL electrical circuits have inductance, even just a straight wire. Inductance makes the current lag the voltage ...
Also with all the investigation of capacitors you claim to have done, How could you not notice that "low inductance" capacitors are much more expensive? …

Again EVERY COMPONENT that has current flowing thru it has inductance. In fact if you know the exact path of the current in the component, you can calculate the inductance. I will not tell you how as that is far above the abilities of one who thinks only components called "inductors" have inductance....
Despite my efforts to teach you that this is not a DC problem, you remain very ignorant and think only in DC terms, as if the resistance and not the inductance is what determines the path lighting will chose:
... let me say that a more useful resistor divider would consist of only two resistors. Let me name them R1 and R2. R1 would be a water-cooled high-wattage 100 K Ohm resistor and R2 would be a water-cooled high-wattage 100 Ohm resistor. The ratio of the resistances of the two resistors, wired in series, would now be approximately 1,000 to one. Whatever voltage was applied across the series, R2 would get just under 1/1,000th of it.

If a lightning bolt supplied the two-resistor series with 200 MVolts, R2 would get 200 KVolts, a good number, considering the fact that earlier in this thread, I supplied a link to a manufacturer of 300KV capacitors. ...
You still have not learned all I have tried to teach you. – You are still neglecting the inductance of R1 + R2 series resistors and the wiring network leading to the many large capacitors, I.e. still throwing out the “L baby” with the wash. For example the IMPEDANCE, Z, of R2 will be much greater than 100 ohms when very short duration pulse is applied, especially if R2 is std wire wound high power resister. Like:
Bigger shown is rated for 130W.

Recall Z^2 = (wL)^2 + R^2
I have used w for the traditional symbol omega, but w = 2 (pi) F where F is the frequency. For lightning pulse F is on the order of 10^6 hertz so for numeral example, which perhaps you can understand:

Z^2 = 10^12 L^2 x 4 (pi)^2 + R^2. I don’t know the size of the resisters you plan, but lets guess their inductance L plus that of the large wire network leading to 1000 or more large capacitors is a milli Henry, 10^(-3)H just to illustrate to you how serious your throwing out the “L baby” is.

Now calling 4 (pi)^2 ~35:
Then Z^2 = 10^12x10^(-6) x (35) + 100^2 = 35x10^6 + 10 ^4. I.e. you R2 = 100 ohms is only ~0.03% of the Z^2. Thus R2 is what could be thrown out without further consideration, but certainly not the inductance, which is on the order of 3,500 times larger than R2^2 in the circuit, as you did with your DC only analysis.

As I have already told you even if R1 = R2 = 0 ohms, still no current will flow into your large array of storge capacitors. I don’t know how many you plan but the array of these physically large capacitors to store any commercially interesting energy will cover more than 1000 square meters.

That is you will have large net work of wires. As inductance is only a property of the current path the inductance of this wire distribution network will (when multiplied by 2(pi) F to get the impedance, with F on the order of 10^6 Hz), cause the “path of least resistance” to be an air arc to ground – roughly the same length as your condensers are tall.

Second point: there is no need to water cool R1. – There will not be any current flowing thru it. As both I and Read –only have explained and given real world examples of: lightning will jumping great distances thru the air AFTER it has run down tall metal tower so it will not flow thru R1 + Z2, the correct impedance of R2. It will simple leap to ground thru the pre-ionized air. Also you seem to be ignorant of the fact that almost all high power rated resistors are actually nichrome wire wound on hollow ceramic tube. (The hole thru tube lets them air cool better). Thus actually being a wound coil, they have very significant inductance and impedance for high frequencies.

Yes there is good chance the lightning will strike top of your collector but zero chance any of its current will flow thru the two resister chain, you call R1 & R2.

BTW, why your discussion of lighting high in the air / the cloud discharging/ etc? Was that just confusion smoke screen? What is important is what it it does after having traveled down your metal "attraction tower" I.e. does it flow thru your network of wires connecting to a large array of large capacitors? (No.) Or simply arc to ground? (Yes.)

Read-only told you of the current leaving the metal tower it hit and arcing thru the air despite fact it could have continued down a 2cm copper rod driven 20 feet into the ground with out jumping thru any air. I told you of lightning curling around a strong metal guy wire, which was also very very well anchored in the ground.

Again: Lightning is a 5 ton gorilla that does what it wants to do, not what a DC current would do, or what you wish it would do. Circuit inductance, which you ignore, usually determines the current's least impedance path for pulses as short in duration as lightning. Your DC analysis is nonsense.

SUMMARY Again you show you are totally ignorant about the subject you are posting about - good luck with the patent office. They don't like fools wasting their time. Probably even the secretary can circular file your application if it reflects the same level of understanding you show here in posts.

Last edited by a moderator:
Yes there is good chance the lightning will strike top of your collector but zero chance any of its current will flow thru the two resister chain, you call R1 & R2.

Then you are necessarily saying that in the millisecond after the bolt hits the top of the collector, the two resistors will have more resistance to a ground than any other path, including the air, and that is what I find so confusing.

The only ionized channel that exists at that moment is the one from the cloud to the top of the collector, aka lightning rod. Air, even moisture-rich air, commonly found at the site of an electrical storm, still has quite a bit of resistance. I'm not aware of any measurements of the resistance of air, but I do know that clouds cannot discharge their stored voltage until they've built up a charge of hundreds of millions of volts. This is necessary to overcome the resistance of the miles of air, sometimes even moisture-rich air, that separate the clouds from the potentially conductive spots on the ground that you and I sometimes call people, buildings, and trees.

Then you are necessarily saying that in the millisecond after the bolt hits the top of the collector, the two resistors will have more resistance to a ground than any other path, including the air, and that is what I find so confusing. ...
That is because you have little understanding that it is the high IMPEDANCE at typical lightning frequencies (~10^6 Hz) of the resistors and the network of conductors leading to your physically large (covering more than 1000 square meters) condenser bank array. The impedance of a pre-ionized (by the bolt's flash of UV) much shorter air path to ground is much lower, so that is where and how the lightning current gets to ground. As I thought you would better be able to understand a numerical example, I did crude calculation for you showing that Z^2 was about 3500 times greater than your (R2)^2, I.e. the impedance at lightning frequencies is about 60 times larger than R2's 100 ohms - your DC thinking is useless nonsense.

I don't know which condensers you plan to use but probably they are about half a meter tall - That length is easy for lightning to jump and it will.

Last edited by a moderator:
That is because you have little understanding that it is the high IMPEDANCE at typical lightning frequencies (~10^6 Hz) of the resistors and the network of conductors leading to your physically large (covering more than 1000 square meters) condenser bank array. The impedance of a pre-ionized (by the bolt's flash of UV) much shorter air path to ground is much lower, so that is where and how the lightning current gets to ground. As I thought you would better be able to understand a numerical example, I did crude calculation for you showing that Z^2 was about 3500 times greater than your (R2)^2, I.e. the impedance at lightning frequencies is about 60 times larger than R2's 100 ohms - your DC thinking is useless nonsense.

I don't know which condensers you plan to use but probably they are about half a meter tall - That length is easy for lightning to jump and it will.

Let's see how much we can agree on before we start an argument over our disagreements.

Impedance is a property of a component in a circuit, possibly even a bare wire.

Impedance is the sum of the resistance, the capacitance, and the inductance of that component.

In a two-resistor series, the resistance of the series is easily measured, the capacitance (the ability of the resistors to store a DC voltage) is near zero, and the inductance (caused by a buildup of the magnetic field around any component that carries a current) should also be near zero. What does the frequency matter to a pair of resistors anyway? I understand the difference between the performance of a capacitor in an AC circuit and one in a DC circuit, and I understand that lightning is an event with multiple short durations, but at no time during a bolt is there any reverse polarity of the electricity unless a second, positively-charged bolt is discharged from the top of a charged cloud.

Benny

P.S. I use the term "capacitor", common to U.S. electrical design, but you use the term "condenser", leading me to think that you're in the UK. Am I right? In the U.S., "condensers" were used early in the development of automobiles, but the term is now out-dated in the U.S.

Let's see how much we can agree on ...
Impedance is the sum of the resistance, the capacitance, and the inductance of that component. ...
No, that is false. It is not the simple sum. It is the sum using a complex number representation. As I told you:
{FROM post 525}... Inductance makes the current lag the voltage ... Capacitors do the opposite - i.e. the voltage on them lags the current in them.

This is why, if you look at power distribution lines, you will occasionally see some capacitors at the top of the poles. ...
For example, as inductors and capactors have opposite effect on the current / voltage phase relationship the removal of either the C or the L in a series C,L,R series circuit will INCREASE the impedance - hardly what you seem to understand by saying: "Impedance is the sum of the resistance, the capacitance, and the inductance"

I.e. the series string L,C,R has LESS impedance than either of the two components series strings L,R or C,R. In fact the impedance of the L,C,R series string is only R at a particular frequency called the resonate frequency.

I think I should stop trying to teach you. You don't really want to learn, especially any facts than make your dream look silly. Either that or you are a very slow learner.

Let's see how much we can agree on before we start an argument over our disagreements.

Impedance is a property of a component in a circuit, possibly even a bare wire. ....
No. that is false too as impedance is strong function of frequency also. You made two assertions in post 586, both wrong.

Last edited by a moderator:
No, that is false. It is not the simple sum. It is the sum using a complex number representation. As I told you:
For example, as inductors and capactors have opposite effect on the current / voltage phase relationship the removal of either the C or the L in a series C,L,R series circuit will INCREASE the impedance - hardly what you seem to understand by saying: "Impedance is the sum of the resistance, the capacitance, and the inductance"

OK, I went to a website called Answers.com and saw a definition for impedance.

Z = the reactance multiplied by i, which I knew was the square root of -1, and then added to the resistance.
http://www.answers.com/topic/impedance

The website didn't give a good definition for the term "reactance" however.
http://www.answers.com/topic/reactance

So I went somewhere else, helped by Google.

(The first sentence)"Reactance, denoted X, is a form of opposition that electronic components exhibit to the passage of alternating current (alternating current) because of capacitance or inductance.​
http://searchcio-midmarket.techtarget.com/definition/reactance

Remember, any one bolt of lightning is not a source of AC, it's a source of DC. Any current flow from the earth (or an object on it, like a tree or a building) to a cloud will only happen if the cloud produces one of those rare positively-charged bolts.

That means that this next sentences do not apply due to the fact that lightning bolts are a DC source:

When alternating current passes through a component that contains reactance, energy is alternately stored in, and released from, a magnetic field or an electric field. In the case of a magnetic field, the reactance is inductive. In the case of an electric field, the reactance is capacitive. Inductive reactance is assigned positive imaginary number values. Capacitive reactance is assigned negative imaginary-number values.​

The page didn't have any definition for reactance as used in a DC circuit.

... OK, I went to a website called Answers.com and saw a definition for impedance.

Z = the reactance multiplied by i, which I knew was the square root of -1, and then added to the resistance.
http://www.answers.com/topic/impedance

The website didn't give a good definition for the term "reactance" however.
You probable know very little about complex numbers. Adding them is easy but multiply is a little tougher, unless you know their exponential representation.

I'm glad to see you are trying to learn. I can help you better understand reactance (than your link did): Reactance contrast with Resistance. Resistive elements do not make the AC voltage and current shift wrt to each other in phase. I.e. if the peak of voltage and peak of current occur at the same time when coming to a resister, they still will have their peaks at the same time after passing thru the resistor but not if passing thru reactive element.

Recall I have twice now told you the current will lag the voltage when passing thru an inductor and the other way round if passing thru a capacitor. This shift of phase is what reactance is all about. Unlike resistive elements, reactive elements do not directly cause any loss of energy. In post 525 I told you that reactive elements do cause a less than unity power factor, which power companies hate, so they put capacitors on the tops of some of their poles to try to balance out the inductive reactance of the long wire.
...
Remember, any one bolt of lightning is not a source of AC, it's a source of DC. ...
Your second link told you: "Reactance also occurs for short intervals when direct current is changing as it approaches or departs from steady flow (e.g., when switches are closed or opened)." lightning is about as far from "steady DC flow" as it gets. Lightning is much more like AC than DC.

If you knew Fourier analysis, you would understand that even a single pulse of positive voltage only, for example a square pulse, is really made of many AC components. This is the case for a brief pulse of lightning with current always flowing in the same direction. The inductance and capacitance of the circuit will still cause reactance. The current does not need to reverse direction - It will be "AC like" (have reactance) even if only flows in one direction. This is why in my numerical example for you I assigned the lighting's unidirectional pulse an AC frequency of 10^6 Hertz. Steady DC (lasting significantly longer than several time constants of the circuit), will not have any reactance.

Search for "Fourier analysis of a square wave" and you will see the sequence of AC frequencies it really is.

I know all this is hard for your to understand - it is just that you have not had much education in complex numbers, Fourier analysis, etc, I assume, (instead of that you are just too dumb to understand what you have been exposed too).

Last edited by a moderator:
BennyF #581:

Would be much simpler to solve the world's energy problems if we ALL just shuffled around on our carpets and occasionally poked our finger in a light socket! Tee Hee!

Brevity and Levity - cornerstones of the Universe!

wlminex

There's no argument. Capacitors block DC. The lightning is not going to find it an attractive path, because it ISN'T a path to Earth.

Yeah, how're you going to get the lightning to go inside the capacitor and not just melt it Benny?

Yeah, how're you going to get the lightning to go inside the capacitor and not just melt it Benny?

Phil has just as much inability to phrase a question as the guy who started this thread. One asks how to store "lightning", and the other says that capacitors block DC.

Capacitors store static electricity, up to their voltage ratings, give or take. Any circuit that has a DC source and a capacitor can be arranged in such a way as to put electricity inside the capacitor. If I have to explain this to you, I will quickly regret asking you to come back to this board.

If you're asking me to show you the circuitry I intend to send to the U.S. Patent Office, then I guess I'll have to repeat what I said to others over the past week or so. It ain't gonna happen. In fact, any circuit ideas I mention on this board or anywhere else, in accordance with the strict rules that the patent office requires of any applicant, are in my opinion, unpatentable, and the reason is usually because somebody else thought of them first, but possibly not for use in collecting the extremely high amount of electrical energy that is found in lightning.

For your benefit, Captain, I have recently discussed two very different kinds of circuits that I think could be used for collecting electrical energy from lightning, and please allow me to express some humility at my initial claims of being able to store "voltage". I now understand that voltage is simply a measurable property of electricity, including the electricity found in lightning.

The first circuit design was what I called "the brute force approach". I envisioned and discussed here, in this thread, a very large current divider network, with a thousand current branches to divide up the tens of thousands of amps in the typical lightning bolt. Each and every branch would include thousands of HV capacitors, which would divide up the DC electricity among them equally because all the caps would be identical and rated at least 100 kV. If there enough of these in every branch, each and every current branch would have a combined voltage rating that exceeds the approximately 500 MV of voltage in the typical negatively-charged lightning bolt.

Positively charged bolts are rarer, but can have double the voltage, but if the caps are non-electrolytics, and if each branch has a combined voltage rating of at least 1 GV, then the caps should not be overcharged.

The second circuit design I proposed, also unpatentable in my opinion, which is the ONLY reason why I discussed it at all, was to use a pair of high-wattage resistors, wired in series and acting as a resistor divider. One resistor, with a much higher resistance than the other, would convert the bulk of the electrical energy to heat, which would be carried away by a cooling system. The other resistor would have a scaled-down version of the cap bank wired in parallel with it, using only one or two of the 100 KV caps in each of the current branches.

Just for your benefit, Captain, I have been to the website for the US Patent Office many times. I have seen the procedure for applying for a US Patent. I have seen the requirements that they demand of all applicants, including the requirement that any application be unpublished. A US Patent is a high priority for me. I have been working alone on my application for over a year now. The circuitry I intend to send them is NOT anything I've discussed here or anywhere else. I will not jeopardize my chances for getting a patent by violating the strict rules that the US Patent Office sets down for all applicants.

The question has come up many times. Do I know enough about lightning to be able to make it work? I have two answers, because it's really two questions.

Q1. Will the circuitry I send to the US Patent Office collect the electricity from lightning?
Answer: My application will say that it's intended to charge a capacitor. It won't mention lightning at all, and the source of the DC voltage won't even be specified. I have looked at every issued patent in the Patent Office's numbered class and subclass for charging caps, and I am satisfied that my circuitry isn't covered by any other patent.

Q2. Do I know enough about lightning to discuss it intelligently on this board?
Answer: Maybe and maybe not. Learning never stops. I did graduate from an electronics school 30 years ago, one which taught electricity before anyone learned electronics. The principles I learned are still valid. The components still do what they did back then. All that has really changed is the introduction of some new technology and the quality of the existing technology.

The transistor was invented in 1948 and didn't come into wide use for a few years after that. Integrated circuits were developed in the 1960s, but they're not made to handle hundreds of millions of volts or a tens of thousands of amps. The older technology is better suited for this task, as long as the specs are appropriate for the size of the problem, so any resistors that are used in my circuits would have to have high wattage ratings, and any caps I use must have high voltage ratings. The circuitry I intend to send to the US Patent Office doesn't need to have any specifications of wattage or voltage ratings, and I don't even need to specify the wire gauge for any of the wires I will use because the Patent Office deals with theoretical concepts, not real-life components that can break, burn out, or have manufacturing defects.

Benny

Last edited:
Phil has just as much inability to phrase a question as the guy who started this thread. One asks how to store "lightning", and the other says that capacitors block DC.

Capacitors store static electricity, up to their voltage ratings, give or take. Any circuit that has a DC source can be arranged in such a way as to put electricity inside the capacitor. If I have to explain this to you, I will quickly regret asking you to come back to this board.
Benny

Benny, FFS dude. Yes you can apply a potential difference across a capacitor and charge it, because the WIRES make that happen.

It is a different proposition when there are no wires.

And capacitors DO BLOCK DC. Get that in your head, learn a fact. CAPACITORS BLOCK DC. "Capacitors are widely used in electronic circuits for blocking direct current" (http://en.wikipedia.org/wiki/Capacitor)

Wikipedia is useless as an authority for anything, because every word in an entry can be changed by the public.

That said, I'm glad that you finally realized the difference between a cap and a broken wire. It sure took you long enough.

Wikipedia is useless as an authority for anything, because every word in an entry can be changed by the public.

OK Benny, you show me a source that says Capacitors conduct DC. But you won't be able to, because it's not true. The Wikipedia article is correct, and you are just slinging mud, just like you did, even after Mac and I showed you our qualifications, something you ave failed to provide yourself.

That said, I'm glad that you finally realized the difference between a cap and a broken wire. It sure took you long enough.

You made that comparison, and I questioned you on it. Short memory eh?

OK Benny, you show me a source that says Capacitors conduct DC. But you won't be able to, because it's not true.

Did I ever say that capacitors "conduct" DC? Either show me one message I've ever written that said those exact words or admit that you misquoted me.

What I've said, time and time again, was that capacitors store static electricity, which they do. Because of that fact, they behave differently than a broken wire, they behave differently if AC and DC are applied to them, and the voltage ratings matter for a capacitor. There are no voltage ratings for broken wires.

your talking about storing all the electron discharge from a single bolt of lightening, which if you tapped it out slowly would make a good amount of power for a town for a month or so.

?? A lightning bolt is about 5 million joules, or about 5 million watt-seconds, or about 1.5 kilowatt-hours. Your house takes about 24 kilowatt-hours a day. So you could power your house for a little over an hour; that's it. You'd need to have every house in the town hit by lightning once an hour all year. It would be hard to sleep.

?? A lightning bolt is about 5 million joules, or about 5 million watt-seconds, or about 1.5 kilowatt-hours. Your house takes about 24 kilowatt-hours a day. So you could power your house for a little over an hour; that's it. You'd need to have every house in the town hit by lightning once an hour all year. It would be hard to sleep.

It looks like you haven't read the rest of this thread. I said many months ago that some of the electricity would go into an chemical reaction called water electrolysis, in which water is separated into hydrogen and oxygen, a process that some high-school students do during a routine part of their classes. Technically, using the terminology of chemistry, this is an endothermic reaction. This means that it won't happen unless you add some energy.

Typically, two electrodes are inserted into a container of purified water, and a DC current is applied to the two electrodes. If the voltage and current levels are ideal, depending mostly on the chemical composition of the metal in the electrodes, you'll see bubbles coming from each electrode. One produces of gaseous hydrogen and the other produces gaseous oxygen.

The electricity that is stored in a bank of capacitors could be used as the DC voltage source for this reaction. The chemistry is well-understood, even when the water is at a high-temperature, which can reduce the overall energy requirement for the reaction to proceed.

A link to a 2005 paper produced by the Idaho National Laboratory:
http://www.inl.gov/technicalpublications/Documents/3169893.pdf

Earlier in this thread, I said that part of the electrical energy would go into the reaction and the rest would be converted into AC for use by the company office. Current prices for electricity may make this uneconomical, but electricity prices have been rising, so by the time the US Government gives me my patent, the prices may be so high that extracting electricity from lightning may in fact be competitive with the electricity that is available from a wall outlet, especially after you factor in the costs of converting AC to DC.

Last edited:
Did I ever say that capacitors "conduct" DC? Either show me one message I've ever written that said those exact words or admit that you misquoted me.

I haven't quoted you, have I? I merely pointed out that capacitors block DC, and that lightning is looking for a path to Earth, so will not pass through a capacitor UNLESS they can overcome the breakdown voltage of the capacitor, and then all you'll have is a fried capacitor.

It really is that simple, below the breakdown voltage the lightning will not strike your apparatus. Over it, and your apparatus is broken.

What I've said, time and time again, was that capacitors store static electricity, which they do. Because of that fact, they behave differently than a broken wire, they behave differently if AC and DC are applied to them, and the voltage ratings matter for a capacitor. There are no voltage ratings for broken wires.

Let the broken wire thing go, you keep bringing it up as if it's relevant, it isn't.

You cannot catch lightning in a box Benny.

Status
Not open for further replies.