Veritasium "How Electricity Actually Works"

[electrodacus]

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?

You will need to clarify what you mean it transmits 10W and 1000W. To where and in what form ?
I'm assuming you mean this are two isolated systems each with its own source but then where is the return wire ?

As you mentioned wires are identical and both cary 1A the difference will be the electric field between those wires and their respective return wires.
As far as wires are concerned the loss as heat will be the same IR

There may be no load just one wire 100x longer than the other one and so a higher voltage supply in order to be able to push 1A through a higher resistance wire (higher resistance due to extra length).
This is how my house heating system is designed as I have loops of 80m of 18AWG wire connected to about 30V DC and so the wires are the heating elements embedded in the concrete floor witch acts as thermal mass storage.

[electrodacus]

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].

I'm not aware of such an effect and it will not make sense as it will be a discontinuity for the zero ohm case (superconductors) where a current induced in a superconductor ring will flow forever with no loss so magnetic field around the ring and the current is maintained with no losses of any type.

But say that what you claim is true how will this change the fact that energy transfer is through wires ? Derek's example had a 1 or 1.1KOhm resistor as the load so total circuit resistance was fairly high so even with your claim energy will still travel through wires.
If energy did not travel through wires then wires will not have been needed or at least not a closed loop.

The loss is not the same if you introduce an inductor as an intermediary energy storage device.
The inductor is ideal for this job as when you apply a voltage across the current will start slowly to increase as energy is stored in the magnetic field created.
So if you connect a 0.1Ohm resistor across a 3V capacitor the current will be 3/0.1 = 30A * 3V = 90W of power loss as heat while an inductor with same resistance 0.1Ohm will just slowly increase current as it stores that energy and not waste it as heat then you just connect the charged inductor across a discharged capacitor and transfer all that stored energy to that capacitor.
You can get 90% efficiency with two identical capacitors and inductor vs just 50% when paralleling the capacitors directly or through an additional resistor.
When you look at the numbers for both just capacitors or capacitors and inductor all energy is accounted for nothing lost through radiation.
So I will like to see what equation have you used where radiated energy loss is present in an isolated system. 

[IanB]

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?

You will need to clarify what you mean it transmits 10W and 1000W. To where and in what form ?

Electrical power, from a source, to a destination.

I'm assuming you mean this are two isolated systems each with its own source but then where is the return wire ?

The return path is far away and out of sight.

As you mentioned wires are identical and both cary 1A the difference will be the electric field between those wires and their respective return wires.

If you look at the electric field, this is outside the wire. But you say the all the power is carried inside the wire, so doesn't that mean the electric field outside the wire is not important? If it is not important, why do you wish to consider it?

[hamster_nz]

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?

You will need to clarify what you mean it transmits 10W and 1000W. To where and in what form ?
I'm assuming you mean this are two isolated systems each with its own source but then where is the return wire ?

As you mentioned wires are identical and both cary 1A the difference will be the electric field between those wires and their respective return wires.
As far as wires are concerned the loss as heat will be the same IR

There may be no load just one wire 100x longer than the other one and so a higher voltage supply in order to be able to push 1A through a higher resistance wire (higher resistance due to extra length).
This is how my house heating system is designed as I have loops of 80m of 18AWG wire connected to about 30V DC and so the wires are the heating elements embedded in the concrete floor witch acts as thermal mass storage.

I assume IanB is talking about something like this schematic attached (of course this is one possible hypothetical arrangement, in your test cell you can only see the two wires).

If you want to pedantic about the resistive loss in the wires, at the source the voltages are adjusted until 1A is flowing, so the 10W and 1000W is being delivered to the loads.

[T3sl4co1l]

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?

For they were, all of them, deceived, as another wire was forged, one to rule them all, and at infinity, bind them [read: circulate their currents].




Tim

[electrodacus]

If you look at the electric field, this is outside the wire. But you say the all the power is carried inside the wire, so doesn't that mean the electric field outside the wire is not important? If it is not important, why do you wish to consider it?

Electric field will be directional towards the return wire so if you knew where the return wire is and at what distance you will be able to make a distinction between the two.
If as you say you have no clue where the return wires are and how far you will not be able to distinguish between the two super long wires.
It is irrelevant as far as wires are concerned if they transport 10W or 1000W as both transport 1A and are identical same resistance per unit of length.
So you can have say a 10Vdc 1A trough say a 10km long wire loop and 1000Vdc 1A through 1000km long loop but all you see is maybe 1km or so of wire from each and there will be no way to distinguish them.

And yes all energy is carried inside the wire in this particular example it will be 1W per km of wire all lost as heat.   

[timenutgoblin]

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.

The paradox exists if only the DC steady state is considered and not the AC transient state.

Did you even looked closely at the equations you posted from wikipedia to see how absurd and stupid they are.

Did you notice my accidental typo?

If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * V_i

How did you calculate 0.707*Vi for the voltage across the capacitors? Either you were refering to the Vrms of the sinusoidal voltages due to the oscillation (AC conditions) or you were refering to the DC steady state as 0.707*Vi. Vrms is 0.707*Vpeak or 0.707*Vi in this case. If Vi was 3Vdc then the Vrms would be 2.121Vrms with a steady state 1.5Vdc (or 0.5*Vi) across both capacitors. Under these conditions, I think that both the Law of Conservation of Charge and Energy would be satisfied without any violation.

Quoting the Wikipedia article:

If the wires are assumed to have inductance but no resistance, the current will not be infinite, but the circuit still does not have any energy dissipating components, so it will not settle to a steady state, as assumed in the description. It will constitute an LC circuit with no damping, so the charge will oscillate perpetually back and forth between the two capacitors; the voltage on the two capacitors and the current will vary sinusoidally. None of the initial energy will be lost, at any point the sum of the energy in the two capacitors and the energy stored in the magnetic field around the wires will equal the initial energy.

Here is a post from a few pages back in this thread:

From the left, the voltage steps down corresponding to the switch turning on.
Subsequently, the capacitor voltages (Ch1, Ch3) oscillate as predicted.

If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * V_i

If you were refering only to the DC steady state voltage of 0.707*Vi then I think only the Law of Conservation of Charge would be satisfied.

For DC steady state only, \$E_i\$ is the initial energy (before the switch is closed) and \$E_f\$ is the final energy (after the switch is closed).

\$E_i = \frac{1}{2}CV_i^2\$

\$E_f = \frac{1}{2}CV_f^2 + \frac{1}{2}CV_f^2 = CV_f^2\$

\$E_f = E_i\$

\$CV_f^2 = \frac{1}{2}CV_i^2\$

\$V_f^2 = \frac{1}{2}V_i^2\$

\$V_f = \frac{1}{\sqrt{2}}V_i \approx 0.707V_i\$

[vad]

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.

Yes, exactly. Changing EM field carries away energy. Marconi and Popov were the first to demonstrate that in early 20th century.

In 1970s Russians had to build Chernobyl power plant just to power a single Duga radar (huge step up from Popov’s transmitter).

Today I charge my phone and watch wirelessly.

Need more examples?

[electrodacus]

The paradox exists if only the DC steady state is considered and not the AC transient state.

What is the paradox at DC steady state ?

How did you calculate 0.707*Vi for the voltage across the capacitors? Either you were refering to the Vrms of the sinusoidal voltages due to the oscillation (AC conditions) or you were refering to the DC steady state as 0.707*Vi. Vrms is 0.707*Vpeak or 0.707*Vi in this case. If Vi was 3Vdc then the Vrms would be 2.121Vrms with a steady state 1.5Vdc (or 0.5*Vi) across both capacitors. Under these conditions, I think that both the Law of Conservation of Charge and Energy would be satisfied without any violation.

Energy since that is what we are discussing here stored in a capacitor is  0.5 * C * V²
So initial energy before closing the switch
0.5 * 1 * 3² = 4.5Ws

If none of the energy was lost as heat then you will expect the same 4.5Ws split between the two capacitors so a 2F capacitor
0.5 * 2 * 2.121² = 4.5Ws
With just 1.5V at the end of the test half of the energy was lost as heat in the wires due to wire/conductor resistance and the capacitor plates are also wires/conductors.
0.5 * 2 * 1.5² = 2.24Ws just half of the energy you started with.

If you were refering only to the DC steady state voltage of 0.707*Vi then I think only the Law of Conservation of Charge would be satisfied.

For DC steady state only, \$E_i\$ is the initial energy (before the switch is closed) and \$E_f\$ is the final energy (after the switch is closed).

\$E_i = \frac{1}{2}CV_i^2\$

\$E_f = \frac{1}{2}CV_f^2 + \frac{1}{2}CV_f^2 = CV_f^2\$

\$E_f = E_i\$

\$CV_f^2 = \frac{1}{2}CV_i^2\$

\$V_f^2 = \frac{1}{2}V_i^2\$

\$V_f = \frac{1}{\sqrt{2}}V_i \approx 0.707V_i\$

Final energy includes the energy stored in the two capacitors in this case 2.25Ws and the energy dissipated as heat in the wire resistance in this case the remaining 2.25Ws so there is no paradox and all is accounted for.

If you look at few posts back you will see that I showed close to 90% energy transfer efficiency by just adding an inductor to the circuit as intermediary storage device so final voltage was around 2V in both capacitors and just a smaller part of the energy ended as heat in the wires. 

[electrodacus]

Yes, exactly. Changing EM field carries away energy. Marconi and Popov were the first to demonstrate that in early 20th century.

In 1970s Russians had to build Chernobyl power plant just to power a single Duga radar (huge step up from Popov’s transmitter).

Today I charge my phone and watch wirelessly.

Need more examples?

You do not understand how both of those examples work that is why you think the energy is lost.
They are not isolated systems.
The phone is an inductive charger basically a transformer so it works for very short distance a few mm and very inefficiently compared to wires directly.

If you checked the spice simulation I have done with the extra inductor and diode you will see that energy was stored in the magnetic field around the inductor and all of that was retrieved back to charge the empty capacitor and the only losses about 10 or 11% of the total where due to wire resistance and that diode and none of it lost due to that magnetic field.
You can do the real test and you will get the same results.
If half of the energy was lost as radiated energy in the two capacitor test then how come I reduced that loss from 50% to just 11% by adding the inductor and diode (they do not contain any energy at the start of the test and do not collect energy from outside).
With just the capacitors starting with 3V on the charged one and ending with 1.5V in both half the energy is lost as heat in wires (can be both measured and calculated so no mystery) then you add two extra components the inductor and diode and you end up with about 2V in each capacitor so close to 90% of the initial energy and just 10% was lost again all in the wires/conductors and the diode which is a semiconductor.

[vad]

You do not understand how both of those examples work that is why you think the energy is lost.

They are not isolated systems.

The three examples are not isolated systems - they draw power from outside (Duga from Chernobyl station, wireless  charge from power grid)

Now, are you saying that a radio transmitter does not radiate EM energy, unless there is a receiver?

[bsfeechannel]

Right so there are plenty of problems with S=JV but you are forbidden to discuss them.

Of course there's a conspiracy of mainstream scientists whose aim is to silence any dissenters.

[electrodacus]

You do not understand how both of those examples work that is why you think the energy is lost.

They are not isolated systems.

The three examples are not isolated systems - they draw power from outside (Duga from Chernobyl station, wireless  charge from power grid)

Now, are you saying that a radio transmitter does not radiate EM energy, unless there is a receiver?

We are discussing if energy travels through wires or outside the wires.

For example the phone charger where you have two inductors.  Say you put 1Ws of energy into transmitter coil from a source then energy will travel through wire/inductor and say 10% is lost as heat due to IR and half creates a magnetic field around it (all conserved) then the receiver inductor can supply part of that stored magnetic field to a load so say 50% of that magnetic field energy is collected by this inductor and again about 10% of that is lost as heat IR in this inductor the rest is delivered to load. Then the rest of the magnetic field can be retrieved back to the sender coil and put back in to the source or just wasted as heat.
It is basically an inefficient air transformer as the magnetic coupling is not as great as if the inductors share a common high magnetic field conductor core.
In a transformer you also put some energy in with each half wave in the primary and if the secondary is open circuit all that magnetic field will be returned to primary thus the loss will only be in the wire and a bit in the iron core.

But again main question on this thread is if energy is delivered through wires to the lamp/resistor or outside the wire.
Since electrical energy is electrical power integrated over time and electrical power is the product of electrical potential and electrical current and the current can only travel through the wire it means energy is delivered through the wire.
There is absolutely zero evidence that "energy doesn't travel in wire" main claim Derek makes based only on the small current observed at the lamp before the electron wave had the time to travel through wire. That small current is just due to line capacitance being charged so the equivalent of two capacitors on each side of the lamp that are being charged by the battery so since current can not travel through 1m of air that is the dielectric of this capacitors (not at 20V) then no energy travels outside the wire. Not during transient and not after during steady state.

[IanB]

But again main question on this thread is if energy is delivered through wires to the lamp/resistor or outside the wire.
Since electrical energy is electrical power integrated over time and electrical power is the product of electrical potential and electrical current and the current can only travel through the wire it means energy is delivered through the wire.
There is absolutely zero evidence that "energy doesn't travel in wire" main claim Derek makes based only on the small current observed at the lamp before the electron wave had the time to travel through wire. That small current is just due to line capacitance being charged so the equivalent of two capacitors on each side of the lamp that are being charged by the battery so since current can not travel through 1m of air that is the dielectric of this capacitors (not at 20V) then no energy travels outside the wire. Not during transient and not after during steady state.

The small current observed at the lamp before the electron wave had the time to travel through the wire was energy emitted by the lamp. That energy had to come from the battery (there is no other power source), and it didn't travel through the wire (because there wasn't time), so it traveled through the air.

[electrodacus]

The small current observed at the lamp before the electron wave had the time to travel through the wire was energy emitted by the lamp. That energy had to come from the battery (there is no other power source), and it didn't travel through the wire (because there wasn't time), so it traveled through the air.

You are ignoring the capacitors formed by the transmission line wires one on each side of the lamp.
When you are charging two capacitors in series from a source you will not say that energy passed through capacitors but energy went in the capacitor and was stored. While energy was pushed into capacitors it traveled through the wires.  You can can not have electric current through a dielectric in this case 1m of air and with no current you have no energy.

This graph I made to show power at the supply and power at the lamp if understood will explain everything.
The switch is only closed for 30ns then open and stays open
With green is power in mW delivered by the supply and with magenta is power used by the lamp.
The area under the graphs represents the energy.
Notice that energy from supply was delivered only during those 30ns that switch was closed circuit and as soon as circuit was broken no more energy is supplied by the source. The total energy delivered by the source in those initial 30ns is 9.16nJ as you can see in the grey window on the left last value.
The lamp was still getting energy well after the switch was closed. Most of the energy arrives at the lamp after the switch was closed and first large chunk exactly after the energy had the time to travel the entire distance through wire.
In the end 7.68nJ of energy where delivered to the lamp and the difference is all accounted by energy lost as heat in the wire's   
There is no energy radiated away all energy is accounted for and all will eventually be radiated as photons form the lamp and infrared photons from the wire as heat is lost to environment.

[PlainName]

You are ignoring the capacitors formed by the transmission line wires one on each side of the lamp.

While in theory there would be some pumping via line capacitance, are you really saying that the amount shown in the experiment is solely due to the capacitance of two wires 1m apart? Seriously?

[Naej]

Right so there are plenty of problems with S=JV but you are forbidden to discuss them.

Of course there's a conspiracy of mainstream scientists whose aim is to silence any dissenters.

A good thing Derek is there to stop it. Youtubers shall prevail over mainstream scientists!

[electrodacus]

You are ignoring the capacitors formed by the transmission line wires one on each side of the lamp.

While in theory there would be some pumping via line capacitance, are you really saying that the amount shown in the experiment is solely due to the capacitance of two wires 1m apart? Seriously?

Yes that is exactly what I'm saying and the simulation for that showed the exact same result. I did rough estimate of the line capacitance based on two pipes 2.5cm diameter 1 meter apart and so values I used in the simulation should be fairly equivalent to real test Derek made and there result where also very similar.
You need to remember that we are talking about 1Kohm resistor as the lamp/load and about 65ns

[PlainName]

Wouldn't such a small capacitance be quickly discharged, so the scope should show a downward trend right after the leading edge? The time constant would be really really small, after all.

Edit: let's say the wire is 5mm across (actually a tube) and 5m long. The capacitance would be 0.21pf and the time constant 2.10e-12 aka 2.1femtoseconds.

[electrodacus]

Wouldn't such a small capacitance be quickly discharged, so the scope should show a downward trend right after the leading edge? The time constant would be really really small, after all.

Edit: let's say the wire is 5mm across (actually a tube) and 5m long. The capacitance would be 0.21pf and the time constant 2.10e-12 aka 2.1femtoseconds.

It is a distributed capacitance and there is inductance involved so charging of each capacitor (seens a series of lumped components) will be delayed.
You can see the downward trend when the switch is open after was just 30ns closed but the wave and associated storage both as inductance and capacitance will travel all around the transmission line until it gets to lamp after about 65ns or so.
So the transmission line is not just a single capacitance and inductance plus resistance it is a continuous set of this.
I think I used 100 RLC elements for the simulation of a 10m transmission line so I divided the total line capacity in 100 small pcs then use that as a repeating circuit.
You where probably considering the capacitance of a single wire while that is fairly small in Dereks experiment you have the capacitance between two parallel wires that are 1m apart and I assumed 25mm diameter pipes (looked like 1" copper pipe).

[AndyBeez]

[As a board newbie] there is not much I can add to the original Veritasium debate other than voice my own opinion. Which is, all Veritasium did was to prove the concept of close coupled inductance and mutual capacitance (cross talk) by building the world's longest (hypothetical) dipole antenna /slash speaker cable. Nice thought experiment, but how does this prove the hypothesis in a universe owned by the EMC demons of EMF?

I ponder what would happen if, rather than being stretched out one light second apart, the cables are laid out serpentine fashion - like differential traces on a PCB? It's the same distance for the electron wave to propogate. So is the outcome the same if the round trip time is the same?

A not so hypothetical experiment is to stand under a high tension power line whilst holding a flourescent tube. Spooky stuff. Is this an example of "vampire capacitance?"

On this theme, [thanks to Dave's debunking videos ] I note there are free energy harvesting devices than can run entire hospitals from a single wifi signal. So maybe we don't know how electricity really travels? So then, why does MY electricy have to flow through a meter when, I can just run out a very long wire and close couple with the ether for free?

ERROR

[electrodacus]

On this theme, [thanks to Dave's debunking videos ] I note there are free energy harvesting devices than can run entire hospitals from a single wifi signal. So maybe we don't know how electricity really travels? So then, why does MY electricy have to flow through a meter when, I can just run out a very long wire and close couple with the ether for free?

ERROR

When I was a young child in the late 80's I used to listen to a radio that was made only out of a speeked (the high impedance one from an old rotary dial phone) and a diode in parallel with that.
One end will be connected to any large metallic object that was not grounded like the water gutters on the house acting as the receive antenna and the other end connected to my body (a finger was enough) with me basically being capacitively connected to ground that was the return path.
This way I was able to listen to the closest AM (amplitude modulated) radio station with no batteries using energy from the radio station antenna.
The two antennas are just the plates of a capacitor and the ground is the return path for the loop.
There were stories that people leaving very close to AM transmit antenna will steal energy by capacitively coupling to the radio station antenna.
A bit later I build a similar radio but with an inductor and variable capacitor in order to be able to tune any of the close by AM stations and also then I powered a red LED in a similar way.
But this will not be correctly named wireless energy transfer as I was in a closed loop in series with a capacitor that was charged and discharged at AM frequency of the radio station.

[bsfeechannel]

A good thing Derek is there to stop it. Youtubers shall prevail over mainstream scientists!

Unless he's colluded with them.

[bsfeechannel]

[As a board newbie] there is not much I can add to the original Veritasium debate other than voice my own opinion. Which is, all Veritasium did was to prove the concept of close coupled inductance and mutual capacitance (cross talk) by building the world's longest (hypothetical) dipole antenna /slash speaker cable. Nice thought experiment, but how does this prove the hypothesis in a universe owned by the EMC demons of EMF?

Inductance, capacitance and "antennance" are just aspects of the same phenomenon.

[PlainName]

It is a distributed capacitance and there is inductance involved so charging of each capacitor (seens a series of lumped components) will be delayed.

Doesn't matter if it's distributed - the entire thing has a very very short time constant. All you're saying is that if you imagine it comprised 100 smaller capacitors, each will be 2.1ff (i.e. 2.1pf/100).

And... if there's a 18V pulse going in, how come there's only 5V coming out the other side?