Veritasium "How Electricity Actually Works"

[ejeffrey]

The pathological nature of the Two Capacitor Paradox problem leads to the observation that only the Law of Conservation of Charge is satisfied, but the Law of Conservation of Energy is violated.

Law of conservation of energy has never been broken/violated.  There is no paradox related to the two parallel capacitors so that page on Wikipedia is misinformation as anyone can write a wiki page and write whatever it wants.

Of course.  And you absolutely need to stop saying that other people on this thread are claiming conservation of energy is violated.  That is just an outright lie and it isn't ok.  Stop it.

The point of the paradox is to show that the initial setup is non physical.  Postulating zero loss and zero inductance and requiring that you reach a steady state is a non physical setup and the equivalent to dividing by zero.  Any conclusions drawn from it are meaningless.  Your entire premise is based on something that cannot exist and is meaningless.

With no inductance or resistance the voltage is discontinuous and the current flow is infinite.  That is impossible.  If you add any finite resistance or inductance -- even an attohenry everything becomes finite and does not agree with your claim.  This is trivially verifiable analytically, by simulation, or by experent.

[TimFox]

Even "dividing by zero" can be handled analytically with proper limits.
As an example, the famous sinc function
y=sin(x)/x obviously goes to 1 at x=0,
even though both the numerator and denominator are 0 at x=0.

[electrodacus]

Of course.  And you absolutely need to stop saying that other people on this thread are claiming conservation of energy is violated.  That is just an outright lie and it isn't ok.  Stop it.

The point of the paradox is to show that the initial setup is non physical.  Postulating zero loss and zero inductance and requiring that you reach a steady state is a non physical setup and the equivalent to dividing by zero.  Any conclusions drawn from it are meaningless.  Your entire premise is based on something that cannot exist and is meaningless.

With no inductance or resistance the voltage is discontinuous and the current flow is infinite.  That is impossible.  If you add any finite resistance or inductance -- even an attohenry everything becomes finite and does not agree with your claim.  This is trivially verifiable analytically, by simulation, or by experent.

You will be spiking for yourself in regards to energy conservation as it seems many people disagree with that either knowingly or unknowingly like Derek did in the Direct downwind faster than wind video where what he basically presented there was an overunity device so getting more power out than in.

You can not get rid of inductance but you can get rid of resistance.
And you can get inductance as a intermediary energy storage in a normal circuit with resistance and reduce the effect of energy loss as I already demonstrated very close to ideal case of zero energy loss that is possible for a circuit with no resistance.

The two capacitor problem is useful if understood to debunk the main claim Derek made and that is the "energy doesn't travel in wires".
All energy including the initial transient (AC) and the DC after that will travel from the source to the load (lamp/resistor) through wires.
There is absolutely no evidence of Derek's main claim and he's supposed evidence is to show that there is energy arriving at the lamp before the electron wave had the time to travel the entire transmission line about 65ns in his experiment.
The explanation for that small amount of energy he sees before those 65ns is easy to explain if you understand that a transmission line is made of distributed capacitance and inductance elements for the entire length of the transmission line and that the capacitance being an energy storage device is the one responsible for that current seen through the lamp while energy is stored in the two capacitors each side of the lamp.
There is no current flow through a capacitors when charging but there is a current flow in to a capacitor and that just means energy travels through wires as electrical energy is the integral over time of electrical power which is the product of electrical potential and electrical current.
Since current is not possible through the dielectric of a capacitor in this case 1 meter of air electrical energy can only travel through wires.

[IanB]

A question (for Tim, ejeffrey, or whoever):

How do we precisely define the circumstances in which conservation of charge is applicable when analyzing a system, and when it is not?

For example, we may have capacitor A holding a charge Q₀, terminals connected to one side of a black box. A second capacitor B starts out uncharged and is connected to the other side of the black box. The black box draws power from capacitor A and uses it to charge capacitor B. At the end of the process, capacitor A has remaining charge Q₁ and capacitor B has charge Q₂. We can imagine that Q₁ + Q₂ may not always be equal to Q₀.

We know that if the black box just contains a series resistor, then conservation of charge will apply. But if the black box contains a DC/DC converter, then maybe not.

What is the rigorous technical statement about when and where conservation of charge applies? Is it related to the number of closed loops in the system topology? That conservation of charge applies around each loop? But then what about branches where two loops overlap? Or is it that conservation of charge should apply within each separable island in the topology, such as on each side of a transformer?

Or is this just overcomplicating things, and one should just apply KCL around each node in the system and sum over all the nodes?

[hamster_nz]

I would love to hear electrodacus's analysis of the behavior of this circuit when the switch is closed, with V_i of 3 V and C of 3 F, with ideal capacitor, wires and switches, of course, (so no dissipative elements).

[IanB]

I would love to hear electrodacus's analysis of the behavior of this circuit when the switch is closed, with V_i of 3 V and C of 3 F, with idea capacitor, wires and switches, of course, (so no dissipative elements).

It will cause a rift in the fabric of space-time and open up a wormhole to another dimension 

[hamster_nz]

A question (for Tim, ejeffrey, or whoever):

How do we precisely define the circumstances in which conservation of charge is applicable when analyzing a system, and when it is not?

Charge conservation, considered as a physical conservation law, implies that the change in the amount of electric charge in any volume of space is exactly equal to the amount of charge flowing into the volume minus the amount of charge flowing out of the volume.

If we agree that there is no transfer to charge through the capacitors, and there is no charge flowing in from outside of the schematic, then hopefully we can agree that that the charge in the upper half (A) and the charge in the lower half (B) will remain the same before and after the switch is closed.

[electrodacus]

I would love to hear electrodacus's analysis of the behavior of this circuit when the switch is closed, with V_i of 3 V and C of 3 F, with idea capacitor, wires and switches, of course, (so no dissipative elements).

You will have an ideal LC circuit thus current through the inductor will increase gradually until capacitor will be fully discharged so a finite current that will create a finite magnetic field that will then collapse and charge the capacitor all they way back up to 3V.
At any point after you close the switch the energy will be there in an isolated circuit as no energy is lost through resistive loss as there is no resistance and no energy will be dissipated despite the variable magnetic field around the circuit.
But if the circuit is not isolated meaning you have some other loop in range of this and that has resistance then energy will be lost to that eventually all the initial energy will be lost in that magnetically coupled loop. But this will no longer be considered an isolated circuit.
The point is that none of the energy will be radiated away but always be there and if you open the switch at the correct time when all energy is in the capacitor then you will be in the same state as the initial state.
That is how the charging of a capacitor from another using an inductor as intermediary storage works in order to reduce the amount of lost energy during transfer.
Adding the inductor adds no energy to the circuit yet it helps getting very close to ideal where most of the energy is still stored after transfer and not lost as heat.

So no energy is "radiated away" by in this case magnetic field as the electric field exists only between the capacitor plates but that also is not "radiated away"
While in Derek's example there is of course energy stored in magnetic field around the wires it is not what transfers the energy from one wire to the other.
The current in those first 65ns is due to capacitor being charged as current cannot flow through 1m of air at 20V potential (leakage so low that it will not register). Since current flow is not present trough air no energy travels outside the wire let alone all energy as Derek claims. 

[electrodacus]

A question (for Tim, ejeffrey, or whoever):

Just noticed that your profile photo is an illustration of energy traveling through air as there electrical potential is large enough to make the air a usable conductor. Not the case with the 20V in Derek's experiment.

[hamster_nz]

But if the circuit is not isolated meaning you have some other loop in range of this and that has resistance then energy will be lost to that eventually all the initial energy will be lost in that magnetically coupled loop.

Are you saying if there was a suitable loop in range (say with a resistor in it), then all the energy would be lost from this circuit? That all the energy would be transferred through the air to this loop, and it would end up heating the resistor?

[electrodacus]

But if the circuit is not isolated meaning you have some other loop in range of this and that has resistance then energy will be lost to that eventually all the initial energy will be lost in that magnetically coupled loop.

Are you saying if there was a suitable loop in range (say with a resistor in it), then all the energy would be lost from this circuit? That all the energy would be transferred through the air to this loop, and it would end up heating the resistor?

Yes but it may take hundreds of years if the distance of that loop is relatively far.  And that will not be considered an isolated system. The discussion of isolated system can be just theoretical in real tests it will be close enough to an isolated system in order to draw some conclusions depending on what the test wants to show.
It is very similar with a superconductor ring where you induce a current at that remains there indefinitely  https://www.open.edu/openlearn/science-maths-technology/engineering-technology/superconductivity/content-section-2.2
Of course you can not have an ideal capacitor as you can not have a dielectric without losses so the LC even if made of superconductor materials it will eventually dissipate the energy as heat in the dielectric material.

[hamster_nz]

But if the circuit is not isolated meaning you have some other loop in range of this and that has resistance then energy will be lost to that eventually all the initial energy will be lost in that magnetically coupled loop.

Are you saying if there was a suitable loop in range (say with a resistor in it), then all the energy would be lost from this circuit? That all the energy would be transferred through the air to this loop, and it would end up heating the resistor?

Yes but it may take hundreds of years if the distance of that loop is relatively far.  And that will not be considered an isolated system. The discussion of isolated system can be just theoretical in real tests it will be close enough to an isolated system in order to draw some conclusions depending on what the test wants to show.
It is very similar with a superconductor ring where you induce a current at that remains there indefinitely  https://www.open.edu/openlearn/science-maths-technology/engineering-technology/superconductivity/content-section-2.2
Of course you can not have an ideal capacitor as you can not have a dielectric without losses so the LC even if made of superconductor materials it will eventually dissipate the energy as heat in the dielectric material.

You seem to be saying:

"A portion of the energy is outside of the wires, and can be transferred to completely galvanically isolated circuits. The amount of energy transferred between the two depend on the physical geometry of their physical arrangement - so size, distance, orientation and so on."

Do you get dielectric losses if you use a vacuum as the dielectric in capacitors? What heats up if there is nothing there?

[PlainName]

like Derek did in the Direct downwind faster than wind video where what he basically presented there was an overunity device so getting more power out than in.

You were, and are, wrong about that. You wanna pop back over to that thread and answer the questions you conveniently forgot about?

[electrodacus]

You seem to be saying:

"A portion of the energy is outside of the wires, and can be transferred to completely galvanically isolated circuits. The amount of energy transferred between the two depend on the physical geometry of their physical arrangement - so size, distance, orientation and so on."

Do you get dielectric losses if you use a vacuum as the dielectric in capacitors? What heats up if there is nothing there?

You can only transfer energy while the magnetic field strength is variable so it will not work at DC.
I try to stay on the subject and magnetic field is not involved in what happens in those first few ns when switch is closed.
The line capacity is what induces that current through the lamp/resistor.
Vacuum will also have losses as there is no such thing as real vacuum (But I'm not a physicist so I will not pretend to know all the subatomic particles that may or not pop up or out of existence at random).

Main subject is electrical energy and if it flows through wires or not. Since we are ignoring any super small leakage current through air or even vacuum the main electric current will travel through wires and you need current on top of electrical potential in order to have power with integrated over time will be energy.   

[electrodacus]

You were, and are, wrong about that. You wanna pop back over to that thread and answer the questions you conveniently forgot about?

I will go there and check but you will need to also go there and point to me the question you are referring to as I'm sure there are a lot of them.

[Sredni]

Again, in that simple ideal circuit with a finite-value resistor connecting the second (initially uncharged) capacitor to the first (charged) capacitor, and no additional lossy elements (such as capacitor ESR or leakage) the total energy lost by the circuit itself (into heat) is independent of the resistance value, and my earlier statement holds, that it is the same value in the limit as R-->0.  Of course, the time required to transfer the charge is directly proportional to that resistor value and the limit as R-->0 of transfer time (to a specific settling fractional value such as 1 ppm, or 14 time constants) is zero.  Here, "limit" is the normal mathematical meaning of the term.

My biggest gripe with the R->0 limit is that it necessarily lead to an unphysical model. If charge needs to move from cap1 to cap 2 at a finite distance, the zero transfer time implies faster than light charge motion. Radiation takes you out of this untenable position: the charges move from here to there, but they need to accelerate and decelerate - hence there will be radiation.
The two-cap problem is a model breaker. One way or the other, something's gotta give.

[electrodacus]

My biggest gripe with the R->0 limit is that it necessarily lead to an unphysical model. If charge needs to move from cap1 to cap 2 at a finite distance, the zero transfer time implies faster than light charge motion. Radiation takes you out of this untenable position: the charges move from here to there, but they need to accelerate and decelerate - hence there will be radiation.
The two-cap problem is a model breaker. One way or the other, something's gotta give.

Not sure I understand what you want to say.
If resistance is zero you still have inductance both in connection between the capacitors and capacitor plates. So current and thus the time it will take charges to move from one capacitor to another will be limited.
I do not see the problem with the two capacitors either with resistance or without.

The first electric field in between the two capacitor plates on the discharged capacitor will only happen when the first electron gets to one of the plates and it will not be that electron that jumped first the space between the switch contacts when they got close enough but it will be caused by that forming a cascading wave traveling at the speed of light through wire.

[vad]

So no energy is "radiated away" by in this case magnetic field as the electric field exists only between the capacitor plates but that also is not "radiated away"

Maybe in the world of trivial Spice simulations electric field exists only between capacitor’s plates, but not in the real world.

A hint: add a second ideal inductor close to that ideal inductor that you mentioned and check if you can measure any voltage across terminals of the new inductor during oscillations. Then think about where does that voltage come from. The same question puzzled Michael Faraday about 191 years ago. He found the answer, and kids now learn the Faraday’s law in the 8th grade (well, at least in the country where I went to middle school half a century ago).

Another hint (since you are a Spice fan): by the words “close” and “ideal” I mean coupling parameter’s value 1 in K statement.

PS. Dave ought to add more educational videos to his channel. If people did not study electromagnetic induction in school, maybe Dave can close that gap too…

[electrodacus]

Maybe in the world of trivial Spice simulations electric field exists only between capacitor’s plates, but not in the real world.

A hint: add a second ideal inductor close to that ideal inductor that you mentioned and check if you can measure any voltage across terminals of the new inductor during oscillations. Then think about where does that voltage come from. The same question puzzled Michael Faraday about 191 years ago. He found the answer, and kids now learn the Faraday’s law in the 8th grade (well, at least in the country where I went to middle school half a century ago).

Another hint (since you are a Spice fan): by the words “close” and “ideal” I mean coupling parameter’s value 1 in K statement.

PS. Dave ought to add more educational videos to his channel. If people did not study electromagnetic induction in school, maybe Dave can close that gap too…

There will be a voltage drop across the inductor even if it had no resistance but the inductor has series resistance as parameter in spice and for the example with 47mH inductor that was set at 0.3Ohm. Capacitors and switch also have series resistance included in the simulation.
I'm not a spice fan. Spice is just a tool that if used correctly will provide correct results.
The spice simulations I showed get the same results as the experimental tests.

There is a reason a transmission line in spice is simulated as a series of LC elements as that is the best approximation of what happens and it is confirmed by the results that are not in contradiction to experimental results including the one  Derek made.

[vad]

There will be a voltage drop across the inductor even if it had no resistance but the inductor has series resistance as parameter in spice and for the example with 47mH inductor that was set at 0.3Ohm. Capacitors and switch also have series resistance included in the simulation.
I'm not a spice fan. Spice is just a tool that if used correctly will provide correct results.
The spice simulations I showed get the same results as the experimental tests.

I probably needed to express myself more clear. I asked you to create an ideal transformer by adding the second inductor.

Anyway, you are wrong. Changing magnetic field of the inductor induces changing electric field that in its turn induces changing magnetic field, and so on. And all that radiates into the Universe at a speed of light…

[hamster_nz]

Maybe in the world of trivial Spice simulations electric field exists only between capacitor’s plates, but not in the real world.

A hint: add a second ideal inductor close to that ideal inductor that you mentioned and check if you can measure any voltage across terminals of the new inductor during oscillations. Then think about where does that voltage come from. The same question puzzled Michael Faraday about 191 years ago. He found the answer, and kids now learn the Faraday’s law in the 8th grade (well, at least in the country where I went to middle school half a century ago).

Another hint (since you are a Spice fan): by the words “close” and “ideal” I mean coupling parameter’s value 1 in K statement.

PS. Dave ought to add more educational videos to his channel. If people did not study electromagnetic induction in school, maybe Dave can close that gap too…

There will be a voltage drop across the inductor even if it had no resistance but the inductor has series resistance as parameter in spice and for the example with 47mH inductor that was set at 0.3Ohm. Capacitors and switch also have series resistance included in the simulation.
I'm not a spice fan. Spice is just a tool that if used correctly will provide correct results.
The spice simulations I showed get the same results as the experimental tests.

There is a reason a transmission line in spice is simulated as a series of LC elements as that is the best approximation of what happens and it is confirmed by the results that are not in contradiction to experimental results including the one  Derek made.

If you watch long enough you will see get the general pattern.

- Nobody is allowed an electric field, except in the dielectric of capacitors

- Nobody is allowed a magnetic field, except in for in inductors or transformers

- Nobody is allowed to have magnetic or electric interaction between wires, but transmission lines are allowed

... leading the result that if you deny any existence of the electric and magnetic fields and their ability to actually transfer the energy, then it is a fait accompli that energy flows only in the wires.

The argument from electrodacus all along is that Lumped Element model the "level 0" description of electrical reality, and the underlying physics (e.g. electro magnetic field simulations) are misguided, do not reflect reality, and can be ignored, because against intuition they show that the energy does not flow in the conductors.

[IanB]

There is an interesting thought experiment which was hinted at earlier in the thread I think.

You are allowed to observe two wires floating in free space, which stretch out of sight in both directions. There is nothing near either wire except you and any instruments you care to have about your person.

Each wire is identical in all physical respects, and each wire is carrying exactly 1 amp DC (which you can verify by measuring the magnetic field, or by measuring the voltage drop along a short length of the wire, or by measuring the heat being radiated).

One wire is transmitting 10 W and one wire is transmitting 1000 W. You are required to determine, without interfering with the wires in any way, which one is which. Can it be done?

[electrodacus]

I probably needed to express myself more clear. I asked you to create an ideal transformer by adding the second inductor.

Anyway, you are wrong. Changing magnetic field of the inductor induces changing electric field that in its turn induces changing magnetic field, and so on. And all that radiates into the Universe at a speed of light…

Are you trying to say magnetic energy going in to creating the magnetic field around a wire or inductor (same thing) will be energy lost ?
Because that is not according to any evidence.
That magnetic field is conservative so when you disconnect the circuit all that energy will be put back in to the circuit.
It seems you are not the only one having this incorrect view about an inductor/wire.
If you looked at any of my simulations maybe the most relevant is that simulating a transmission line you will see that all energy is accounted for.
The one where switch is turned ON for just 30ns is particularly relevant for your concern.

 

[electrodacus]

If you watch long enough you will see get the general pattern.

- Nobody is allowed an electric field, except in the dielectric of capacitors

- Nobody is allowed a magnetic field, except in for in inductors or transformers

- Nobody is allowed to have magnetic or electric interaction between wires, but transmission lines are allowed

... leading the result that if you deny any existence of the electric and magnetic fields and their ability to actually transfer the energy, then it is a fait accompli that energy flows only in the wires.

The argument from electrodacus all along is that Lumped Element model the "level 0" description of electrical reality, and the underlying physics (e.g. electro magnetic field simulations) are misguided, do not reflect reality, and can be ignored, because against intuition they show that the energy does not flow in the conductors.

The lumped element model will not have been used if it did not provide accurate predictions of what happens. And it is accurate because it represents reality just a reduced form of that in terms of resolution.

Where is the electric field inside a charged capacitor transferring energy ?
The switch itself is also a capacitor and there is an electric field before closing the switch so how come you need to actually close the switch to transfer energy.

[Sredni]

Not sure I understand what you want to say.
If resistance is zero you still have inductance...

No, I was talking of a circuit where the only dissipation mechanism is that through Joule loss. The reasoning is that the loss is the same irregardless of the value of R, so all dissipation is accounted for when R= 1 ohm, or 0.1 ohm, or 0.00001 ohm or... and then we take the limit for R->0 and say that the loss is still the same - so it is all accounted for by ohmic losses even when R->0.
The first paper in my list above (the paper that was posted earlier by someone else here) avoid this problem by computing the radiation loss contribute considering NOT and oscillating LC circuit, but simply the radiation associated with the accelerated charges. It shows that for a circuit with IIRC a diameter of 10cm but still negligible self-inductance, the radiation loss becomes relevant only when the resistance of the loop falls under a handful of microohm (I don't have it at hand now, but the numbers are in that ballpark).

So, ohmic loss till a certain value of resistance, but under that it's radiation that takes over. At first they share the losses, then radiation becomes dominant and account for nearly all losses.
The current does not oscillate: the solution is a decaying exponential - and with a more advanced model - there are also solutions where the current dies off in a finite time (paper 2 or 3 in the list above).

No more paradox of the missing energy (it is extracted from the circuit)
No more paradox of ohmic loss with R=0 (it is taken care of by radiation)
No more need for an inductance that makes the circuit oscillate at a frequency 1/Sqrt[L Ceq].