Motor consumes no charge ?

[tinman]

Well,so he ended up with more than half the energy in the two caps combined to that of what he started with.
So 2.52 volts across 350F=1111.32 joules of energy in one cap,and none in the other.

The end result was
Cap 1 -1.468v across 350F= 377.129 joules
Cap 2-1.273v across 350F= 283.593 joules.

Total =660.722 joules
If we double this,we have 1321.444 joules.

So yes,he lost less than the half expected.
But there is also one other thing to take into account-->the little motor was running the whole time,and there for was also dissipating power the whole time current was flowing through it,by way of resistive heat loss.

Dont tell the EE guy's,they will start doing back flips.

Great job guys.


Brad

[Dog-One]

Well,so he ended up with more than half the energy in the two caps combined to that of what he started with.
So 2.52 volts across 350F=1111.32 joules of energy in one cap,and none in the other.

The end result was
Cap 1 -1.468v across 350F= 377.129 joules
Cap 2-1.273v across 350F= 283.593 joules.

Total =660.722 joules
If we double this,we have 1321.444 joules.

So yes,he lost less than the half expected.

Yes, I get that we would need about 1.782V to be left in each capacitor to equal the same energy (Joules) we started with.

Okay, screw energy for a moment and just focus on charge:

T_Start
C₁ -- 2.52V -- 882 Coulombs
C₂ -- 0.0V -- 0 Coulombs
Q_Total:  882 Coulombs

T_Final
C₁ -- 1.468V -- 513.8 Coulombs
C₂ -- 1.273V -- 445.55 Coulombs
Q_Total:  959.35 Coulombs

So how in the heck do we gain charge, but lose energy?  Isn't this an entirely "closed loop" system?  Charge just jumps in there as energy is split into two chunks?  If we had a system that took charge in two clumps and put them into one clump, would we gain energy and lose charge?

I need a better way to interpret these results Brad.  What is the correct way to think about this simple system?  Apparently you cannot think of energy like water and capacitors like buckets.  This is so very fundamental, I'd hate to waste this moment and beat my head against the wall for another ten years.

M@

[tinman]

Yes, I get that we would need about 1.782V to be left in each capacitor to equal the same energy (Joules) we started with.

Okay, screw energy for a moment and just focus on charge:

T_Start
C₁ -- 2.52V -- 882 Coulombs
C₂ -- 0.0V -- 0 Coulombs
Q_Total:  882 Coulombs



T_Final
C₁ -- 1.468V -- 513.8 Coulombs
C₂ -- 1.273V -- 445.55 Coulombs
Q_Total:  959.35 Coulombs

So how in the heck do we gain charge, but lose energy?  Isn't this an entirely "closed loop" system?  Charge just jumps in there as energy is split into two chunks?  If we had a system that took charge in two clumps and put them into one clump, would we gain energy and lose charge?

I need a better way to interpret these results Brad.  What is the correct way to think about this simple system?  Apparently you cannot think of energy like water and capacitors like buckets.  This is so very fundamental, I'd hate to waste this moment and beat my head against the wall for another ten years.

M@

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.

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

[Dog-One]

Im not even going to pretend i know how that works.

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

Thank you for giving it a go Brad.

I thought about this all night--charge being the derivative of energy, mathematically.  It kind of seems like energy is stored in a capacitor under pressure, with charge being the medium being compressed.  I suppose there is a way to associate this with an inductor too, but I haven't made any attempt yet to try it.  I did some plots of energy (and charge) transfer between two capacitors, focusing on the break-even point and the curves have some interesting characteristics--much like transferring air between two scuba tanks.

What seems so strange is the actual term "energy".  It's almost as if this is a completely fabricated term, having a very ambiguous meaning, at least in respect to electricity.  We don't actually transfer energy at all, we transfer some medium (charge, voltage, whatever) and in doing so, we can say it took energy to accomplish.  Then of course if it took some amount of time from start to finish, we can say "power" was involved.

Anyway, to quote ol' Ken Wheeler, it's all a bit of a "mind screw".  Hands on the bench is the only way to come to grips with it.

[TinselKoala]

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.



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