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?
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].
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.
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.
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?
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?
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.
Did you even looked closely at the equations you posted from wikipedia to see how absurd and stupid they are.
If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * Vi
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.
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 * Vi
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.
The paradox exists if only the DC steady state is considered and not the AC transient 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.
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\$
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.
Right so there are plenty of problems with S=JV but you are forbidden to discuss them.
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?
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.
You are ignoring the capacitors formed by the transmission line wires one on each side of the lamp.
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.
QuoteYou 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?
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.
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
A good thing Derek is there to stop it. Youtubers shall prevail over mainstream scientists!
[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?
It is a distributed capacitance and there is inductance involved so charging of each capacitor (seens a series of lumped components) will be delayed.