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  #1  
Old 08-03-2017, 03:18 AM
p75213 p75213 is offline
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A Free Energy Idea

I've been doing some electronic tutorials over at allaboutcircuits.com. The idea comes from this lesson: https://www.allaboutcircuits.com/tex...-and-calculus/
At the bottom of the topic there is a circuit showing a neon bulb in parallel with an inductor and the sentence "If current through an inductor is forced to change very rapidly, very high voltages will be produced."
Well I am thinking why not change the bulb for a capacitor and capture the energy? I did some googling and found that the formula for energy of a capacitor is .5CV^2. Also the capacitor voltage would = the breakdown voltage of the inductor which is V = L di/dt. di is large and dt small which gives a large V. The capacitor current is I = C dv/dt. Here again dv is large and dt small which gives a large I.

I'm not sure how this would be done. But why not siphon of some of the energy to repeat the procedure instead of using a battery. The rest could be used to power a load. This should be able to continue indefinitely as the breakdown voltage is much larger than the voltage required to charge up the inductor.

One idea I have is to use 2 capacitors in parallel with the inductor. One with a small capacitance to keep the circuit running and a larger one for the load. The load capacitor would have to use diodes? (I guess) to prevent it emptying back into the inductor. It would also be necessary to employ some switching arangement (transistor?) to obtain the high breakdown voltage.

The idea seems a bit too simple so I wouldn't be surprised if it doesn't work. I've put the topic up at the allaboutcircuits forum and so far nobody has given me a convincing argument why it wont work. Somebody said you can't get out more than you put in. But that's not what the maths (pretty simple stuff. Just what I have mentioned in this post) is telling me.
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  #2  
Old 08-03-2017, 05:26 AM
ricards ricards is offline
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Quote:
Originally Posted by p75213 View Post
I've been doing some electronic tutorials over at allaboutcircuits.com. The idea comes from this lesson: https://www.allaboutcircuits.com/tex...-and-calculus/
At the bottom of the topic there is a circuit showing a neon bulb in parallel with an inductor and the sentence "If current through an inductor is forced to change very rapidly, very high voltages will be produced."
Well I am thinking why not change the bulb for a capacitor and capture the energy? I did some googling and found that the formula for energy of a capacitor is .5CV^2. Also the capacitor voltage would = the breakdown voltage of the inductor which is V = L di/dt. di is large and dt small which gives a large V. The capacitor current is I = C dv/dt. Here again dv is large and dt small which gives a large I.

I'm not sure how this would be done. But why not siphon of some of the energy to repeat the procedure instead of using a battery. The rest could be used to power a load. This should be able to continue indefinitely as the breakdown voltage is much larger than the voltage required to charge up the inductor.

One idea I have is to use 2 capacitors in parallel with the inductor. One with a small capacitance to keep the circuit running and a larger one for the load. The load capacitor would have to use diodes? (I guess) to prevent it emptying back into the inductor. It would also be necessary to employ some switching arangement (transistor?) to obtain the high breakdown voltage.

The idea seems a bit too simple so I wouldn't be surprised if it doesn't work. I've put the topic up at the allaboutcircuits forum and so far nobody has given me a convincing argument why it wont work. Somebody said you can't get out more than you put in. But that's not what the maths (pretty simple stuff. Just what I have mentioned in this post) is telling me.
It really is easy to do the math, but to figure out the RIGHT math.. that is rather hard.. based on my own experience...

You have to get the grasp of what is REAL and what is ABSTRACT.. based from your post.. I would assume you would want to get more V (voltage) and I (Current) based on what I've learned so far they are both ABSTRACT.. what is REAL is.. the Inductor and its Magnetism.. the Capacitor and its Charge..
they are the things that you can manipulate...
meaning if you would want to get more voltage you would either have to increase your magnetism to generate more EMF or charge the Capacitor without using it (open circuit), and if you would want to generate more current you either charge a big capacitor and discharge it to a low impedance coil or low resistance load, or pass your Alternating current in a Step-down transformer (this will lower your voltage).

Voltage can be 2 things :
Electrostatic Potential
Electromotive Force

Current can also mean 2 things:
Conduction current
Displacement Current

"If current through an inductor is forced to change very rapidly, very high voltages will be produced."

I'm guessing this is AC.. via a Transformer
but can also be Pulse DC.. via a boost converter..

this is a standard meaning you can see it in almost all power supply... they do not produce more energy.

I would not tell you you can't get more energy out than in, because that's actually our goal too.
it is really good to start with an Idea (even though its not feasible). but an understanding of what your dealing with is necessary to generate the RIGHT IDEA.
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Old 08-03-2017, 08:10 PM
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Originally Posted by ricards View Post
I would not tell you you can't get more energy out than in, because that's actually our goal too.
it is really good to start with an Idea (even though its not feasible). but an understanding of what your dealing with is necessary to generate the RIGHT IDEA.
What do you think a Bedini monopole already does?
Do some homework..
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Old 08-03-2017, 09:16 PM
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I've been giving this some thought. It all comes down to the formula for energy in a capacitor: w = .5CV^2. If we have 2 capacitors both of the same capacitance. One we charge from a 6 volt voltage source at 1 amp. The other one we charge from a 24 volt voltage source at 0.1 amps. In the first case the power is 6 watts and in the second case 2.4 watts. However the second capacitor would have more energy -> 0.5C24^2 joules as opposed to 0.5C6^2 joules. Therefore the second one should also have more power as power is defined as the rate of using energy.
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Old 08-03-2017, 11:38 PM
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You would be wise to pay attention to what Matthew Jones posted. He as well as many of us on this forum have been at this for a long time. You can learn a lot here if you are willing to spend some time reading and studying.
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Old 08-04-2017, 12:30 AM
p75213 p75213 is offline
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Originally Posted by citfta View Post
You would be wise to pay attention to what Matthew Jones posted. He as well as many of us on this forum have been at this for a long time. You can learn a lot here if you are willing to spend some time reading and studying.
I don't doubt that. Getting the time is the hard part. Same goes for everybody I guess. However I'm just tossing this idea around to see if its a goer. It seems like the bedini monopole may do something similar. It's on my to do list while ploughing through the allaboutcircuits site.
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Old 08-04-2017, 06:51 AM
ricards ricards is offline
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What do you think a Bedini monopole already does?
Do some homework..
I do not remember stating Bedini Monopole is not feasible.. maybe you got the wrong Idea... I was talking about the thread starter's Idea of doing things mathematically..

and I don't see the reason why you should show such attitude.
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Old 08-04-2017, 11:35 AM
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You cant get more energy out than you put in?? Just got to love that one. Sounds like someone never heard of nuclear energy, solar panels, or a caveman lighting a piece of wood.

Its ok to throw ideas out there, someone might make a believer out of you one way or an other and save you some time. Sometimes there is no such person, and one experiments according to their instincts. Although it is the slower path, eliminating how stuff isnt, takes one closer to how stuff is.

After a while it may turn out that the second capacitor took 10 times longer to charge, or those circuits posted only swapped amperage for voltage, where as the Bedini SG shows that the magnetic fields and their CMMF's can also be used to produce physical work by turning a motor , also opens up the concept for motor speed controllers to recover alot of the energy used to run the motor, if the right strategy is employed and the motor design is tweaked a bit.
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Old 08-04-2017, 02:51 PM
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Originally Posted by lotec View Post
You cant get more energy out than you put in?? Just got to love that one. Sounds like someone never heard of nuclear energy, solar panels, or a caveman lighting a piece of wood.
That's not quite how it works. A solar panel converts about 20% of the sun's energy that goes in into electrical energy. Just because "you" personally didn't put it in, you are in no way getting more out than goes in to it. Also, the caveman's pile of wood didn't spontaneously appear at his feet. Work and effort is required to gather it. How many years did it take the tree to grow?
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  #10  
Old 08-05-2017, 01:17 AM
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Originally Posted by lotec View Post
You cant get more energy out than you put in??
Just got to love that one.
These guys are locked into the idea that what the schools teach
is the only true science. They can't see beyond so we need to show
them. PHD, BA, MA oh and BS

This statement "You cant get more energy out than you put in" is all
they know. They have never seen anything else mainly because first
of all they think it impossible, next it could be and is right under their
noses and still miss the whole enchilada.

Now that is what I call spirit, call a wondering air head by his real name
a foolish talker. It is amazing to me that most do not see how the
extra is all around them.

If people have any IDEA about extra energy production let them ask
the guys here for years about it and there will be no silly one liners
claiming the impossible.

Good to see you coming around LOTEC

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Old 08-05-2017, 02:30 AM
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Yeah but you didn't address the nuclear option because you can't fit
that into you little one liner caveman nonsense. Where does the energy
come from on the atomic level when smashing atoms, HUH? Where?

You can't explain it so don't even try. The pile of wood analogy is a pile of
what? Never mind.
Perhaps you forgot that I'm not the one who brought up the caveman and wood analogy. You should take up the issue with the other fellow who seems to be on your own side of the argument.

Nuclear energy is harnessed through producing steam to drive turbines. That's not 100% efficient either. You are not getting out more than you put in.

And in that respect, what does it matter if it's the sun, nuclear energy, or a pile of wood? What part of it are "you" supposed to be putting in to be able to claim that you're getting more out, unless you have a generator attached to your bicycle wheel? If you burn a pile of wood then of course you're getting more energy out of it than "you" put in. You didn't make the wood. But I think you know very well that's not what your friend was saying, unless you chose to overlook it in favour of attempting to argue against that which you are simultaneously defending.
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  #12  
Old 08-05-2017, 04:56 AM
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distinctions

I think that a useful distinction for a "free energy" system is one that does not have more output than input, but more output than we have to put in. That obviously implies that there is extra input from somewhere. We can pay a little to leverage environmental potential that is waiting to be put to work so the total work done can exceed what we personally contributed.

The whole point of COP or coefficient of performance is to compare the total work done only to what we personally pay for not including free environmental source potential (free to us).
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Old 08-05-2017, 04:57 AM
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our contribution vs total work done

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Originally Posted by dR-Green View Post
If you burn a pile of wood then of course you're getting more energy out of it than "you" put in.
What dR-Green said.
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Old 08-05-2017, 06:57 AM
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I think that a useful distinction for a "free energy" system is one that does not have more output than input, but more output
than we have to put in. (free to us).
Hey that is a great way to say it Aaron, I never thought of it just
like that but it is true. Like the steam engine for instance, they use to
say they could move a 1000 pounds 1 mile on a pound of coal which in
the beginning was more like 10 pounds of coal with the early inefficiencies.

So it is all in how we look at it. With all of the information releases
you would think that someone would be saying they tapped into the
endless aether by now. Or that they are able to recycle energy by
pumping it around in a circle.

Something, anyone, please, instead all we hear from the so called adepts
is that it ain't so "Can't get the extra" oh well I'll go back to my coils
and then report back later.

I am on #18 coil right now, soon to reach 24. BRB

Keep up the great work.
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Old 08-05-2017, 01:35 PM
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I think you are both right

Its a statement that can be taken both ways depending on how it is worded.

If the statement is " You cant get more out than what goes in. ", there would be no argument from me.
If the statement is " You cant get more out than what you put in ", then I can think of a few.

What gets me is when someone says the second one to someone, and when they are called out about it, they scurry away and hide behind the first one to save face. The people aren't buying it anymore.

One person drags a log of wood 6 miles in the pouring rain and after many hours of rubbing finally gets a small fire going.
A trucker driving through a drought ridden part of the country flicks his cigarette but out the window and nearly burns down half the state.

When it comes to nuclear power, because of the nasty pollution, I sure hope that more energy comes out than what people put into it, otherwise we are in even more trouble than I thought.

Bro Mikey nice to hear from you, been a while .
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Old 08-06-2017, 01:05 AM
p75213 p75213 is offline
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If a coil has been charged with 5 volts and has reached a steady state with 0 volts and 50mA. The power source is switched of, the magnetic field collapses and the coil voltage goes to x volts (greater than 5). What does the current go to? Is it greater than 50mA?

I'm thinking the coil is now the only source of voltage and the resistance hasn't changed therefore there should be an increase in current as: I = V/R. According to this site that is the case: https://en.wikipedia.org/wiki/Voltage_spike "The effect of a voltage spike is to produce a corresponding increase in current (current spike)."

The time constant would be the same in both cases: t = L/R as neither L or R have changed. Therefore there should be an increase in energy as: w = 0.5LI^2. Can that be true? Were getting more out than in.
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Old 08-06-2017, 04:00 AM
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Coil

Quote:
Originally Posted by p75213 View Post
If a coil has been charged with 5 volts and has reached a steady state with 0 volts and 50mA. The power source is switched of, the magnetic field collapses and the coil voltage goes to x volts (greater than 5). What does the current go to? Is it greater than 50mA?

I'm thinking the coil is now the only source of voltage and the resistance hasn't changed therefore there should be an increase in current as: I = V/R. According to this site that is the case: https://en.wikipedia.org/wiki/Voltage_spike "The effect of a voltage spike is to produce a corresponding increase in current (current spike)."

The time constant would be the same in both cases: t = L/R as neither L or R have changed. Therefore there should be an increase in energy as: w = 0.5LI^2. Can that be true? Were getting more out than in.

http://www.energeticforum.com/attach...ductor-v-i-png

Hi p75213,

Hopefully this example will help you understand. I believe you're talking about the coil (inductor) only so in the example Rs = Rd = coil resistance. Also, the example uses a decay path. When you say "power source is switched of(f)", it implies a sudden opening of the circuit, as in the contacts of a switch separating. As that happens, the gap between the contacts presents a resistance in the decay circuit. The current cannot stop or change instantaneously so an arc or spark will bridge the contact gap and complete the decay circuit until the energy stored in the inductor is used by the resistance of the coil and gap.

Regards,

bi
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Old 08-06-2017, 06:37 AM
p75213 p75213 is offline
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Originally Posted by bistander View Post

http://www.energeticforum.com/attach...ductor-v-i-png

Hi p75213,

Hopefully this example will help you understand. I believe you're talking about the coil (inductor) only so in the example Rs = Rd = coil resistance. Also, the example uses a decay path. When you say "power source is switched of(f)", it implies a sudden opening of the circuit, as in the contacts of a switch separating. As that happens, the gap between the contacts presents a resistance in the decay circuit. The current cannot stop or change instantaneously so an arc or spark will bridge the contact gap and complete the decay circuit until the energy stored in the inductor is used by the resistance of the coil and gap.

Regards,

bi
Hello Bistander,
Actually I was referring to the circuit mentioned in my first post at the beginning of the thread. Here's the link to it: http://www.energeticforum.com/redire...nd-calculus%2F At the bottom of the page you will see a circuit with an inductor in parallel with a bulb. My proposal is to replace the bulb with a capacitor/s.
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Old 08-06-2017, 07:32 AM
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Indeed, very good proposition, though you missed one thing - a coil with a capacitor is a resonant circuit so the oscillation would occur...and what next ?
I have done something very similar (with slight modifications) in the past and got scary results from very little power spent (around 12V at 1A).
Tesla was a master of secrecy , his method of conversion patent is full of statements many still don't understand today....
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Old 08-06-2017, 09:28 AM
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Quote:
Originally Posted by p75213 View Post
If a coil has been charged with 5 volts and has reached a steady state with 0 volts and 50mA. The power source is switched of, the magnetic field collapses and the coil voltage goes to x volts (greater than 5). What does the current go to? Is it greater than 50mA?

I'm thinking the coil is now the only source of voltage and the resistance hasn't changed therefore there should be an increase in current as: I = V/R. According to this site that is the case: https://en.wikipedia.org/wiki/Voltage_spike "The effect of a voltage spike is to produce a corresponding increase in current (current spike)."

The time constant would be the same in both cases: t = L/R as neither L or R have changed. Therefore there should be an increase in energy as: w = 0.5LI^2. Can that be true? Were getting more out than in.
put a diode at the end of the coil that is connected to negative side of the source battery and connect that to the + side of another battery (charge battery) and connect a lead to the - side of the Charge battery to the start of the coil, and every one who has work on John B. works know what happens..

now get more funds.. buy more battery wound larger coils then study the 3BGS..and you will have your idea..

I admit I misunderstood your post.. because of your 2 capacitor idea.
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Old 08-06-2017, 09:52 AM
bistander bistander is online now
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Originally Posted by p75213 View Post
Hello Bistander,
Actually I was referring to the circuit mentioned in my first post at the beginning of the thread. Here's the link to it: http://www.energeticforum.com/redire...nd-calculus%2F At the bottom of the page you will see a circuit with an inductor in parallel with a bulb. My proposal is to replace the bulb with a capacitor/s.
Principle still applies. Charge accumulates on capacitor plates instead of arcing across gap between switch contacts.
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Old 08-06-2017, 09:42 PM
p75213 p75213 is offline
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Quote:
Originally Posted by p75213 View Post
If a coil has been charged with 5 volts and has reached a steady state with 0 volts and 50mA. The power source is switched of, the magnetic field collapses and the coil voltage goes to x volts (greater than 5). What does the current go to? Is it greater than 50mA?

I'm thinking the coil is now the only source of voltage and the resistance hasn't changed therefore there should be an increase in current as: I = V/R. According to this site that is the case: https://en.wikipedia.org/wiki/Voltage_spike "The effect of a voltage spike is to produce a corresponding increase in current (current spike)."

The time constant would be the same in both cases: t = L/R as neither L or R have changed. Therefore there should be an increase in energy as: w = 0.5LI^2. Can that be true? Were getting more out than in.
Somebody said on another forum that when the circuit is open R becomes large and therefore I becomes very small - I=V/R. However the inductor and capacior/s form their own closed circuit where the resistance is small and so I believe my logic still applies.
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Old 08-06-2017, 10:05 PM
bistander bistander is online now
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Somebody said on another forum that when the circuit is open R becomes large and therefore I becomes very small - I=V/R. However the inductor and capacior/s form their own closed circuit where the resistance is small and so I believe my logic still applies.
The DC resistance of a capacitor is extremely high often considered infinite. The capacitor will have an impedance to AC, but essentially blocks DC. As the voltage across the capacitor's terminals changes it will store or deliver charge.

bi
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Old 08-07-2017, 12:52 AM
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The DC resistance of a capacitor is extremely high often considered infinite. The capacitor will have an impedance to AC, but essentially blocks DC. As the voltage across the capacitor's terminals changes it will store or deliver charge.

bi
Hi Bistander,
Just to make it clear. A capacitor has infinite resistance to DC when it is charged. However until it is charged current flows to charge the plates. https://physics.stackexchange.com/qu...-of-capacitors "In the moment that you turn on the current in the circuit (as "closing" the switch in the picture), current runs as if there was no blocking (corresponding to 0 resistance). Charge flows to the plate and pushes the same amount of charge away from the other plate. From the outside this looks as if it was just a wire."
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Old 08-07-2017, 01:16 AM
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Hi Bistander,
Just to make it clear. A capacitor has infinite resistance to DC when it is charged. However until it is charged current flows to charge the plates. https://physics.stackexchange.com/qu...-of-capacitors "In the moment that you turn on the current in the circuit (as "closing" the switch in the picture), current runs as if there was no blocking (corresponding to 0 resistance). Charge flows to the plate and pushes the same amount of charge away from the other plate. From the outside this looks as if it was just a wire."
Yes, for how long? So the instant that current would start to flow it puts a charge on the plates and then there is a capacitor voltage opposing the applied voltage and limiting the current. That was included in my statement about storing charge.
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Old 08-07-2017, 01:18 AM
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You are a sloppy slapstick antagonist. Making simple sentences in an
effort to try to reverse your portrait image, trying to stay neutral or
politically correct while injecting doubt.
another attack what did I ever do to you?..

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Originally Posted by BroMikey View Post
I realize how mad you must be that what appears to be tiny additions or
variations could have an overwhelming impact on electric system.
Isn't it really like that? an addition of a diode would make a simple circuit function differently, I don't know what you are picturing on your head bromikey.

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I can see who you are.
do you have ESP?

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Originally Posted by BroMikey View Post
Is it feasible that you should poke fun for no reason? What is your
job title? Are you ready to build? Or just post?

The fellow who made this thread has heard all of the negatives about
how FE must be a hoax, maybe you could help him out by showing us
all how you know the extra does exist.

You see I am open for your input at the risk of being steps on, so what
about it? You got anything at all?
It doesn't mean If I don't post I don't build. I'm not the "kiss and tell" type bromikey... I know this a forum and this is a discussion, and its open to the public and I'm part of it.. I have the rights to choose what to share.. and my rights as a member to post my opinion is the same as yours.. If you do not like my opinion, you can block me as you wish, I do not care.

you want to know what I think where the Extra is coming?.. I don't even think there is any..

you see bromikey.. I do not see Energy as a Substance, or a juice or some sort of "something" in order to do work.. like I did before...now I don't even think Energy actually exist.. you cant see it, you can't feel it, taste it, or touch it. but you can perceive it.. so it must be real right??.. Energy is such a complicated thing..

"more energy out than in" is a metaphor we used to describe a phenomenon of a device "tapping" in with the "Natural Flow" of nature... what is it?! In my current state of understanding I cannot answer that.. but I know I'm on the right mindset..

just to be fair.. where do YOU think this "extra" your talking about is coming from?..
also.. what do you hope to achieve in your project?..

I do mine for research and I keep it to myself because I don't want people to misinterpret me "showing off" my work..

I'm in this forum (and many others) to interact with people with the same view, share my thoughts and experiment, I've found few. and your not one of them.
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Last edited by ricards; 08-07-2017 at 02:55 AM.
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  #27  
Old 08-07-2017, 01:36 AM
p75213 p75213 is offline
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Quote:
Originally Posted by bistander View Post
Yes, for how long? So the instant that current would start to flow it puts a charge on the plates and then there is a capacitor voltage opposing the applied voltage and limiting the current. That was included in my statement about storing charge.
The instant current starts to flow current is at a maximum and voltage at zero. Its only after the capacitor is fully charged that voltage is at its maximum.
RC Charging Circuit Tutorial & RC Time Constant
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  #28  
Old 08-07-2017, 02:35 AM
ricards ricards is offline
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Quote:
Originally Posted by p75213 View Post
The instant current starts to flow current is at a maximum and voltage at zero. Its only after the capacitor is fully charged that voltage is at its maximum.
RC Charging Circuit Tutorial & RC Time Constant
bistandar is correct,
the moment you close the switch, current is max, current gradually decreases as voltage builds up on capacitor plates, its not something like give and take type of situation. also the reason why a sinewave is formed on an oscilloscope and not a triangular shaped.
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  #29  
Old 08-07-2017, 03:18 AM
ricards ricards is offline
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Quote:
Originally Posted by BroMikey View Post
I agree, don't be a show off, but there is nothing wrong with "putting on"
in other words "put on a happy face" people here get tried of hearing
said that it is impossible. Sharing is good. Being braggadocios has been
your error in the past?

Where I think the energy is coming from? You wouldn't agree anyway
so you have asked another loaded or foolhardy question that I refuse to
answer.

Stop raining on our fire and start your own downer thread if you are
against getting the extra. You want to mock others who are wanting
and getting the extra then post ugly non sense?

Go away and start a thread with your name on it tell up who and what
makes your view valid, don't go around doodling on everyone elses
conversation.

By all means have your opinion, but have it with restraint and consideration
for others. Trust me I am glad you have rejected me. Because I
am getting the extra, won't tell you and am doing my best to help others
who are open to understand.

Be nice or I will show you some spirit. This is as nice as I can put it.

The man who wants to talk in this thread should be respected as he has
some encouraging and stimulating postulations. You hate that? Hate it
on your own thread.
your starting to sound like the guy with the part G.. be careful..

and I really do not know where you got your Idea of me saying its "Impossible", the very reason I'm here clearly states my belief its possible..
please do not pick words out of my statement (like matthew jones did), instead try to understand what I'm trying to say to the TS.

as I have said, this is a forum Open to public, and I'm not raining fire, you are.

perhaps you didn't notice all your post are off topic.. please do not abuse the liberty given to you by the owners of this forum to linger and attack people you do not share the same view,
Yes! the TS deserve the respect.. so Am I.. .. so I suggest you stop calling me names and spread hate, your not getting anything by doing of that.
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Last edited by ricards; 08-07-2017 at 03:20 AM.
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  #30  
Old 08-07-2017, 03:21 AM
bistander bistander is online now
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Quote:
Originally Posted by p75213 View Post
The instant current starts to flow current is at a maximum and voltage at zero. Its only after the capacitor is fully charged that voltage is at its maximum.
RC Charging Circuit Tutorial & RC Time Constant
You're talking about this circuit except you have replaced the bulb with a capacitor, right?



So when the switch is closed there is 6V across the coil and cap. The current through the coil is 6V/R where R is the coil resistance. When the switch is opened the coil current remains the same because it can not change instantaneously. Neither can the voltage on the capacitor.

bi
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