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| Veritasium "How Electricity Actually Works" |
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| Naej:
--- Quote from: EEVblog on May 02, 2022, 12:31:57 pm ---I posted this comment on Derek's video, slighly expanded: Riddle me this, a new thought experiment: Let's go extreme and say you have a 100mm diameter copper conductor at pure steady state DC delivering a small amount of power to a pure resitive load, say 1W, but go as low as you want. No transients, no skin effect, no nothing, just pure steady state DC into a resistive load. Is there NO energy WITHIN this comicly large wire? None? It's all on the OUTSIDE of the wire in the fields at DC? Really? REALLY? The classicial field theory math might work at DC, but I just can't get over the feeling that it doesn't pass the sniff test at DC. I don't get The Vibe I get with AC and transients. Quantum Electrodynamics and probability theory in the electron fields within the wire better passes the sniff test at DC. Can someone please convince me that there is no energy flow within this 100mm diameter wire at all, and that all the energy flows outside the wire at DC. --- End quote --- There is no experimental evidence of it. Same for AC, or RF. So no, no one can convince you. --- Quote from: Sredni on May 02, 2022, 07:59:18 pm --- --- Quote from: rfeecs on May 02, 2022, 05:23:18 pm --- --- Quote from: Sredni on May 02, 2022, 03:27:30 pm ---I have a mass of 1 kg in position P at 0 meters over sea level. I take this mass 1000 meters away to drop it from a cliff into a hole deep 10 meters. The potential energy of the mass is converted into kinetic energy and then this is uses to generate heat. Let's say I 'generated' 1 joule of energy. Has this energy traveled along the 1000 meters path? --- End quote --- If we take the Poynting style conservation of energy argument, we know that energy is created in the volume of space around where you 'generated' it. Energy was 'dissipated' in the volume of space around where you dropped it. So we can say energy flowed through the space between. But we can't say exactly the path that the energy took. Edit: However, if we decide that the potential energy is located where the mass of the rock is located (like we do by saying that the fields have energy), then energy flowed along with the rock. How much potential energy the rock contains is a problem without knowing the baseline zero potential energy of the system. I hate analogies. Water analogy, rock analogy, whatever. :rant: --- End quote --- It was not meant to be an analogy. I am considering the mechanical system only. In more detail: one perfectly flat frictionless path goes from point A (the source) to point B (the load). We can put the weight on the path and with an infinitesimal push we make it travel D meters to point B. Now, at point B we have a cusps that leads to lets' say 20 different holes of different depths, from 1 to 20 meters. Chance determines what the final path will be. But once the weight falls into one of the holes, all of its gravitational potential energy from height 0m to the depth of the hole gets converted into heat (to simplify things). At the bottom of each hole there is a path (horizontal and perfectly frictionless) that leads back to the source. At the source the weight is lifted by a machine to sea level and the cycle repeats. Not an analogy, I repeat. It is a mechanical system. Does the energy travel through the one forward path at sea level? If so: HOW MUCH ENERGY does travel along the path? Remember, I do not know which hole the weight will fall into until the weight falls into it. If we cannot get an agreement on this mechanical system, how can we get an agreement on the electromagnetic system where the depth of the hole is determined by the charges themselves (by creating a surface distribution that obeys the constitutive relation in the wires and resistor)? --- End quote --- The relativistic answer (convention) is that a momentum density p of matter corresponds to an energy flux of pc². If you want general relativity then the answer is, well, complicated. https://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html Classical physics is simple and says you have a potential energy of -GMm/r, you're not moving 0 energy, you're just converting: potential->kinetic->heat. |
| electrodacus:
--- Quote from: SiliconWizard on May 02, 2022, 07:31:16 pm ---Well, as I understand it, a medium is needed to hold charges, and if there are no charges, there's no field? The question then is more about fields making charges move rather than charges moving creating fields. Or something. --- End quote --- In this example (Derek's setup) the charges are needed to create the field as the only electric field is inside the battery There is no moving object in this experiment. |
| EEVblog:
--- Quote from: HuronKing on May 02, 2022, 05:14:26 pm ---Can we lock the other thread or something? This discussion is super confusing to read when we've got entire lines of thought being double-posted in both threads. It's super confusing to follow who is responding to what. :( "Let's see - is this the BIG thread talking about how Veritasium is right but actually wrong or the LITTLE thread talking about how Veritasium is wrong but actually right?" :o --- End quote --- The other thread IMO was essentially hijacked by aetherist with all these alternative theories, it's why I stopped reading it. No reason to lock the other thread, they'll just come here. This thread seems to be a clean slate based on the new video as the OP mentioned. |
| timenutgoblin:
Why is it that, in the two capacitor paradox problem, the two capacitors are connected in parallel with each other and the switch connected in series as shown in the circuit below? Source: Wikipedia en.wikipedia.org/wiki/Two_capacitor_paradox Why not have the capacitors connected in series with each other (back to back, of course) and the switch connected in parallel? After the switch is closed, the net voltage across the two capacitors will be zero since both capacitors would be equally charged and zero current would then be flowing. I don't think it would make any difference to the initial and final charge/energy calculations. Also, are the capacitors regarded as ideal voltage sources or ideal current sources? The fact that the capacitors lose/gain voltage and charge when the switch is closed suggests that they are (ideal) current sources and not (ideal) voltage sources. \$I = C \frac{dV}{dt}\$ suggests that for a constant current the terminal voltage of a capacitor must either increase or decrease depending on the direction of current flow (assuming that capacitance is constant and NOT infinite). In order to prevent the terminal voltage of the capacitor from decreasing the capacitance would need to be decreasing to maintain a constant voltage because voltage is energy per unit charge \$V = \frac{E}{Q}\$ If the charge on the capacitor is decreasing then \$\frac{dQ}{dt} < 0\$ Also, if the terminal voltage is decreasing then \$\frac{d{^2}E}{dQ{^2}} < 0\$ The output power is zero before the switch is closed, then peaks before both capacitors are charged and then returns to zero. If the capacitors had infinite capacitance and were already charged to some voltage then the voltages of the capacitors would be constant across their terminals independent of flowing current. This is the definition of an ideal voltage source. On the subject of mechanical analogies, there is a post in a related discussion thread: https://www.eevblog.com/forum/projects/where-does-12-come-from-in-capacitor-energy-calculation/msg1671035/#msg1671035 Using this mechanical analogy, it's as though the two capacitors fuse together when the switch is closed as opposed to colliding elastically where momentum and energy would be conserved. |
| electrodacus:
--- Quote from: timenutgoblin on May 03, 2022, 12:13:41 am ---Why is it that, in the two capacitor paradox problem, the two capacitors are connected in parallel with each other and the switch connected in series as shown in the circuit below? Source: Wikipedia en.wikipedia.org/wiki/Two_capacitor_paradox --- End quote --- Not sure why that Wiki page even exists as there is no paradox. If you understand what capacitors are and how they work you can know exactly what happens and why. If that is made with superconductors so no resistance in wires or capacitor plates then energy will be conserved. If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * Vi If the circuit is not made with superconductors meaning wires, capacitor plates or both have any resistance the voltage after switch is closed will be 0.5 * Vi The reason final energy is half of the initial is because the other half was dissipated as heat due circuit resistance. Due to resistance you have Vi / 2 in the moment switch is closed even before any energy starts to be transferred as you basically have a 1/2 resistor divider at that middle point on a symmetrical circuit. The value of the resistance is irrelevant and will only influence the speed at witch the energy is transferred from charged capacitor to the discharged one still same half of the energy will be wasted as heat. |
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