As you asked someone else a few messages back, "I need to ask what is your qualification." Seems that literally everyone is wrong except for you, so what is your qualification for having the definitive knowledge about energy?
QuoteI guess those two are professors at MIT and same as the other professor at University of California Alex have a poor understanding of energy storage.
Tim
I will need to correct this as it is unfair since I have no read the book written by them. In case of Alex I have seen his inability to understand energy by being unable to explain how that vehicle works and win the bet with Derek.
As you asked someone else a few messages back, "I need to ask what is your qualification." Seems that literally everyone is wrong except for you, so what is your qualification for having the definitive knowledge about energy?
Circuit theory isn't sufficient to solve the two-capacitor problem. We need to take the electrodynamic behavior of the system into account.
So let's do the thought experiment of connecting a charged and a discharged capacitor in parallel at t=0s. We assume that all the conductors are perfect and that the two capacitors are of the parallel plate type with vacuum between the plates. The two capacitors can be modelled as two lossless transmission lines (see for instance House and Melcher example 14.2.1). After the two capacitors are connected in parallel, an electromagnetic wave will bounce back and forth between the two capacitors. The amount of energy stored in each capacitor will fluctuate, but the total amount of stored energy in the two capacitors will remain the same.
If we go to the next level and take electromagnetic radiation into account, then al the energy will be radiated into space over time.
"Photons do not looz energy over time (ignoring things which rob energy from a photon)."
Tautological, yet mis-spelled.
Photons are emitted and absorbed all the time. In Compton scattering of x-ray photons, the photon loses a fraction of its energy.
When radioactivity was first discovered, around 1900, three types were identified: alpha, beta, and gamma.
They were initially named in order of their penetrating power through matter.
Since their first discovery, it was found and demonstrated experimentally that beta particles were negatively-charged electrons, and gamma rays were uncharged photons.
In 1900, Becquerel determined that the charge/mass ratio of beta particles equalled that of electrons in cathode rays.
Around 1914, it was demonstrated that gammas are electromagnetic radiation, i.e., photons.
How, then, can electrons be photons and vice-versa? Electrons have charge and mass, photons have neither charge nor mass.
It certainly deserves to be separated because the physical behavior is quite different:
Separated.
Since the 19th century, everybody thought that everything electric and magnetic, from DC to cosmic rays, through radio waves, heat, light, ultraviolet, X-rays and whatnot, is the manifestation of the same freaking physical phenomenon.
Now you're saying that they are different. I wonder why the Nobel Committee has not noticed you yet.
QuoteYes I can. Energy flows in wires and reappear in a lightbulb/engine/LED.QuoteYes energy is transferred from the battery to the short, through the wire.If you say so, it must be true.
The circuit theory is perfectly capable in solving the two parallel capacitor problem.
There is no such thing as conservation of charge. The conservation of energy is a law.
I looked probably at all forms of energy generation and energy storage so maybe that gives me a better understanding of energy.
You lack fundamental understanding though.
So photons form a loop, and gain mass to become electrons.
Whence comes the charge?
I'm glad to see you admit the possibility of electrons drifting: TI and the other semiconductor manufacturers can now continue their processes.
Electrons accelerating emit electromagnetic radiation: see antenna theory and synchrotron radiation (both of which work).
Photons interacting with electrons can increase the energy of the electrons: see atomic structure theory (not the archaic "orbits" you keep harping on) and spectroscopy (both of which work).
The entire discussion is about energy traveling through wires or outside of wires. Same can be summarized as energy traveling through a capacitor (not the leakage through dielectric) that is basically what Derek will say vs energy flows in and out of the capacitor as capacitor is an energy storage device and that is what I and everyone that has correct understanding of reality is saying.
Oh they are both implied by Maxwell's equations.
But because their physical behavior is quite different, they are called differently. Much like DC current and gamma rays are called differently. Duh.
You don't call wires, capacitors, coils and antennae "Maxwell stuff" do you?
Well you say energy flows through vacuum just to mock engineers, why can't I say it flows through wires?
I proved that energy travels through wires in multiple ways sumarized in this post https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4156462/#msg4156462
Let me know what part of that is wrong.
The entire discussion is about energy traveling through wires or outside of wires. Same can be summarized as energy traveling through a capacitor (not the leakage through dielectric) that is basically what Derek will say vs energy flows in and out of the capacitor as capacitor is an energy storage device and that is what I and everyone that has correct understanding of reality is saying.Is that statement compatible with an LED + resistor between two physically small capacitors emitting light for a few seconds - a physical demonstration that energy can flow through the dielectric of the capacitors, even though electric charge can't.
If a glowing LED doesn't suffice, how about you put a third identical capacitor in the middle (so three capacitors in series). apply 12V. Then remove the middle capacitor, walk it to the other side of the room and measure the voltage on it with a DMM. It will read 4V, and will have 1/2 C V^2 Joules of energy in it. That energy is far higher than a microamp or so of leakage current count account for.
Where did it this energy come from? How does that energy get into the middle capacitor if energy can only flow in wires and not through capacitors?
Maybe your definition of "energy traveling through [a capacitor]" doesn't match mine?
Acutally, what is your definition of "energy traveling through" something?
I just did it. From a contradiction you can prove whatever, remember? So the flaw of your entire reasoning lies in the fact that you don't understand the most fundamental tenets of electricity. Go back to the books and if you have any question, we're here to help.
That was energy through a leaky capacitor (as all electrolytic capacitors are). But if you are referring to initial high current through LED as you seems you do that is due to capacitors charging but no current passes through capacitor. Current flows in or out of the capacitor.
But the fact that you measure 4V means that all energy that flowed in that circuit with 3 identical capacitors in series supplied by 12V power supply flowed into the capacitors and not through them.
Maybe is a matter of understanding the definition of flowing in or flowing through.
Like again an analogy with limited scope just so we batter define flowing through vs flowing in/out of.
Water flows in a bucket and what will mean for water to flow through a bucket will be a bucket with a hole on the bottom so water can flow through.
Those electrolytic capacitors have a leakage current same as a bucket with a very small leak (the bucket is still useful to transfer liquid from some part to another but no long storage).
The capacitor made by the long transmission wires are more like a small coffee cup so low capacity but very well build with no measurable leakage.
Also back to those two identical parallel capacitors.
"Energy through a leaky capacitor"? What does "leaky" mean in the context of energy?
How long after removing from the circuit do you want me to leave the middle capacitor alone before I measure it? If it is 'leaky' and the energy is there because of such leaks, then it should self-discharge in a short period of time.
C
+---||---+--o Vr
|+ |
Vs R
|- |
+--------+
_|_
I do not see how any engineer will be able to contest the above.
"Energy through a leaky capacitor"? What does "leaky" mean in the context of energy?
How long after removing from the circuit do you want me to leave the middle capacitor alone before I measure it? If it is 'leaky' and the energy is there because of such leaks, then it should self-discharge in a short period of time.
Leakage is an undesired effect of real capacitors and has nothing to do with anything related to the main question.
Energy flows in capacitors not through capacitors except for that pesky small current of few uA typical.
When you have a few mA or A of current in a loop with a capacitor in series then you have current going in to the capacitor not through it.
I sort of get frustrated and annoyed of explaining what in/out and trough means.
Se the problem above post #316 as that is proof energy only travels through wires not through the space around the wire or capacitors.
Wires transfer charge. Space transfers energy.
If current "goes into" a capacitor and not "through" it then would you agree Vr = 0 for an initially discharged capacitor and voltage step Vs?Code: [Select]C
+---||---+--o Vr
|+ |
Vs R
|- |
+--------+
_|_
Tim
If current "goes into" a capacitor and not "through" it then would you agree Vr = 0 for an initially discharged capacitor and voltage step Vs?Code: [Select]C
+---||---+--o Vr
|+ |
Vs R
|- |
+--------+
_|_
Tim
Electrons entering on one side of the capacitor to increase the density of free electrons means that electrons will need to leave the other plate.
But this energy is not used to do any work but stored and can be retrieved later to do work.
Say Vs is 3V and capacitor 1F the value of the resistor is irrelevant.
The capacitor will have a voltage potential of zero at the start so all 3V will drop across the resistor (will say resistor value is large enough that voltage source and capacitor DC resistance is low enough to to be of importance).
The charge current will start high then drop as the capacitor voltage increases until it gets to zero when voltage across the capacitor equal Vs (3V) then no current will flow so no energy will travel through the circuit (we ignore the capacitor small leakage current).
So now no energy can be transferred from Vs to resistor as current is basically zero.
Charging was inefficient but energy is stored in capacitor and can be used.
At all time no current (except that annoying leakage) has flown through the capacitor.
The way you charge a capacitor is by moving electrons from one plate to another but that is not done through dielectric but externally through wires.
There are now 4.5Ws of energy stored in the capacitor even if the power supply needed to deliver 9Ws as the other 4.5W was lost as heat on the resistor.
If you had a constant current power supply then no current resistor will have been needed for current limiting and power supply will have delivered 4.5Ws and all of that 4.5Ws will have been stored thus no work done at all / no heat loss.
Is like charging the capacitor with a linear regulator vs a DC-DC charger.
If you agree with the above (current flows in to capacitor as it is being charged and current flow stops when capacitor is fully charged) then you agree that energy is not flowing through the capacitor but in to capacitor.