Motor consumes no charge ?

[Gothic]

Charge is conserved, as is energy. What may  _not_  be conserved is the capacitance value of large capacitors, which may vary somewhat according to all kinds of things, like temperature, pressure, voltage, charge-discharge currents, number of C-D cycles, etc.

Think about it.

I have...

displacement current is a quantity appearing in Maxwell's equations that is defined in terms of the rate of change of electric displacement field. Displacement current has the units of electric current density, and it has an associated magnetic field just as actual currents do. However it is not an electric current of moving charges, but a time-varying electric field.

and as for the term "energy"  to quote Tom Beardon "I didn,t axe you what it does i axed you what it was"

[Gothic]

Awesome replies, keep it up

This is a work in progress...

[Dog-One]

Charge is conserved, as is energy. What may  _not_  be conserved is the capacitance value of large capacitors, which may vary somewhat according to all kinds of things, like temperature, pressure, voltage, charge-discharge currents, number of C-D cycles, etc.

Think about it.

I sure looks like charge is conserved from what I can see in my experiments.  Energy on the other hand, looks to be consumed in the process of moving charge.

As far as mentioning Kenny Wheeler goes.... even a broken clock could be right twice a day... unless it's a digital clock.

So true and I'm glad you mentioned it because surely in your many hours of building and testing have come across a few bread crumbs that would really be helpful to some of us.  If you ever care to toss a few my way where I can find them, I wouldn't mind following a trail.  Getting to a point in my life where working and watching the mountains of BS pile-up just isn't all that much fun anymore.

Awesome replies, keep it up

This is a work in progress...

My, isn't that interesting.  Looks almost identical to what I have on my bench at the moment.


We can't leave out the Russians, so take a peak at this vid and think about converting it to an electric device.
https://www.youtube.com/watch?v=11r3E4eia0U

Hmmm, a charge pump and the faster you spin it, the more charge it moves.  Any chance the capacitors would fill up?

Maybe one of them...
https://www.youtube.com/watch?v=6Ljmd-ygDs8

[Gothic]

Funny how the mind works,  several times I looked over the pdf posted and at the " buzzer"
experiment I always saw 4700 uf capacitors being used, but only c1 and c2 are, c3 is
470...   O.K.  not as hopeful as I was before,  damn.

p.s. I did the experiment with all three caps 400v 2500 uf  and did not get the "overvoltage"
         on c3 as is claimed in the pdf

[TinselKoala]

Im not even going to pretend i know how that works.
C is I x time,where as Joules is watts per second--> 1 joule is 1 watt  second.

No, those two statements are not the same thing. "Joules is watts per second" is wrong. 1 joule is one wattsecond is right. "Watt-second" actually means Watt times second.

Watt=Joule/second, so Joule = Watt x Second. 

The word "per" generally indicates a division operation, as in "Kilometers per hour".... you go 300 km PER 5 hours, how do you find your speed in Km per hour... you divide 300/5 = 60 km per hour.

A watt is a Joule per second. One joule of energy passing your measurement point PER second of time. This is a measurement of power. The more joules of energy per second, the more power.

A Joule is NOT a "watt per second". The only way that statement can make sense is describing a rate of change of power, like if you had a circuit that draws one Watt in the first second, two Watts in the second second, three Watts in the third second, and so on... this would be a change in power of one watt per second. Think of turning up a Variac slowly with a light bulb on the output.

You have "C= I x Time" while correct, is kind of misleading. Yes, I (current in Amperes) is equal to Coulombs per second, or rearranging as you have it Coulombs = Amperes x Seconds .... but you are using "C" here to mean Coulombs, where in other places in this discussion "C" is commonly used to represent Capacitance in Farads.
For example we say Q=CV when we mean Coulombs of charge = Capacitance times Voltage.
But then we also use "Q" as a symbol for the "quality" factor of a coil or resonant system, where it has nothing to do with Q= charge in Coulombs.

Looks odd when we look at energy stored in cap's
Example
1 volt over 1 Farad= 1 coulomb of charge,but only 500mJ of energy
2 volts over 1 Farad =2 coulombs of charge,and 2 joules of energy
3 volts over 1 Farad =3 coulombs of charge,but now the energy is higher at 4.5 joules.

As we go on,the charge always is the same as the voltage across the cap. If you double the size of the cap,you double the value of charge and energy.
So if we had 3 volts across a 2 Farad cap,we have 6 coulombs of charge,and 9 joules of energy.

I am not sure why at a low voltage,the charge is a higher value than the joules value,and then they switch places as you raise the voltage,where the joules of energy becomes more than the charge value.

I have never had the need to understand or study this situation-Joules V charge,so i can t offer any help there.


Brad

The energy, in Joules, on a capacitor goes linearly with capacitance but goes as the square of the voltage: E=(CV²)/2
So if you double the voltage the energy goes up by 2² or 4 times, but if you double the capacitance the energy goes up by 2.

The charge Q in Coulombs on a capacitor goes linearly with voltage and capacitance: Q=CV, or rearranging algebra, C=Q/V or V=Q/C. So of course for a 1 F capacitor, V=C numerically.

Taking both sets of relationships together you can get the full set of numerical values relating E (Joules), C (Farads), V (Volts), and Q (Coulombs).