A Free Energy Idea

[p75213]

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.

[bistander]

Coil

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

[p75213]

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.

[boguslaw]

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....

[ricards]

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.

[bistander]

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.

[p75213]

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.

[bistander]

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

[p75213]

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."

[bistander]

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.

[ricards]

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?..

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.

I can see who you are.

do you have ESP?

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.

[p75213]

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

[ricards]

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.

[ricards]

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.

[bistander]

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