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| Veritasium "How Electricity Actually Works" |
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| electrodacus:
--- Quote from: IanB on May 15, 2022, 02:48:57 am --- 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? --- End quote --- 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:
--- Quote from: Sredni on May 15, 2022, 03:13:50 am --- 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]. --- End quote --- 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:
--- Quote from: electrodacus on May 15, 2022, 03:13:59 am --- --- Quote from: IanB on May 15, 2022, 02:48:57 am --- 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? --- End quote --- You will need to clarify what you mean it transmits 10W and 1000W. To where and in what form ? --- End quote --- Electrical power, from a source, to a destination. --- Quote ---I'm assuming you mean this are two isolated systems each with its own source but then where is the return wire ? --- End quote --- The return path is far away and out of sight. --- Quote ---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. --- End quote --- 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:
--- Quote from: electrodacus on May 15, 2022, 03:13:59 am --- --- Quote from: IanB on May 15, 2022, 02:48:57 am --- 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? --- End quote --- 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. --- End quote --- 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:
--- Quote from: IanB on May 15, 2022, 02:48:57 am ---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? --- End quote --- 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]. ;D Tim |
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