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| Veritasium "How Electricity Actually Works" |
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| T3sl4co1l:
--- Quote from: timenutgoblin on May 13, 2022, 12:50:02 pm ---If the area of the floor of the swimming pool is increased from A to 2A (analogous to adding the initially-discharged capacitor) then the height of the water (analogous to the voltage across the capacitors) must be halved from h to h/2. The gravitational potential energy of the water (analogous to the electrical potential energy of the capacitors) must be halved, too. The weight of the water (analogous to the charge on the capacitors) does not change when the floor area (analogous to the total capacitance) is doubled. --- End quote --- Indeed if we perform this experiment, using a pool with one wall able to move freely yet sealed water tight, we get a piston, which as it's moved back and forth, in the steady state, bears a force of F = ρ g l h^2 / 2 (for water height h, gravitational acceleration g, water density ρ, and wall length l). Thus we do work on it when pushing (makes pool smaller, taller), or vice versa. Going between the x and 2x cases (for pool width x, 2x), this provides precisely the energy required to make up for the apparent halving of energy while conserving charge. Energy of the system (pool AND actuator) is conserved, exactly as we should expect; but considering too limited of a subset (i.e. the pool by itself), energy isn't conserved, it's been moved in/out of that boundary somehow. We also have the dynamic case, though an overly messy example of it (Navier-Stokes fluid equations are a bitch!). Suppose we partition a pool in half, filling only one half; then suddenly remove the partition. The water sloshes into the formerly-unoccupied side, and continues to slosh back and forth until friction has absorbed the "AC" energy. Indeed superfluids exist, so we could perform this experiment with such, and demonstrate an apparent perpetual motion machine, where the mean free surface level gives the charge-conserved and half-energy figure, while the AC component (gravity waves on the free surface) continues, in motion, with exactly half the initial energy of the system. I think, because of the nonlinearity of gravity waves, the (potentially?) biphasic nature of superfluids, the free surface moving in gas rather than pure vacuum (the only known superfluids exist at very low temperatures and fairly low pressures, indeed being cooled by evaporation), there will be too many kinds of dispersion (i.e. one long, blocky wave breaks up into numerous higher order waves), friction (due to the gas in the chamber itself, and the liquid's normal phase component) and other effects, dissipating the wave energy in a real superfluid experiment; but even so, again we have the argument: where is the energy BEFORE it's been dissipated to heat, or equivalently, exchanged into other forms of energy besides where it was? And again the answer is clear, it's stored in both static (or average/DC) and dynamic (AC) modes. Tim |
| electrodacus:
--- Quote from: timenutgoblin on May 13, 2022, 12:50:02 pm ---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. --- End quote --- 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. As I clearly mentioned a few times all energy is accounted for. Due to resistance in the two identical capacitors circuit (circuit perfectly symmetrical) after closing the switch half of the energy is wasted as heat in the conductors (that includes wires, switch and capacitor plates) the other half remains in the two capacitors quarter of the energy in each capacitor. There is absolutely no mystery in the two parallel capacitor problems unless you do not understand what capacitors are. Adding an inductor (another energy storage device that people do not understand) and a diode will not bring any extra energy as system is still isolated and the initial total energy at the beginning of the test is the same and contained all in the charged capacitor. The end result with the inductor and diode added is closer to ideal case where circuit will have had no resistance at all case where no energy will be wasted as heat. Did you even looked closely at the equations you posted from wikipedia to see how absurd and stupid they are. |
| SandyCox:
--- Quote from: electrodacus on May 13, 2022, 04:47:24 pm --- --- Quote from: timenutgoblin on May 13, 2022, 12:50:02 pm ---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. --- End quote --- 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. As I clearly mentioned a few times all energy is accounted for. Due to resistance in the two identical capacitors circuit (circuit perfectly symmetrical) after closing the switch half of the energy is wasted as heat in the conductors (that includes wires, switch and capacitor plates) the other half remains in the two capacitors quarter of the energy in each capacitor. There is absolutely no mystery in the two parallel capacitor problems unless you do not understand what capacitors are. Adding an inductor (another energy storage device that people do not understand) and a diode will not bring any extra energy as system is still isolated and the initial total energy at the beginning of the test is the same and contained all in the charged capacitor. The end result with the inductor and diode added is closer to ideal case where circuit will have had no resistance at all case where no energy will be wasted as heat. Did you even looked closely at the equations you posted from wikipedia to see how absurd and stupid they are. --- End quote --- Your point is moot. The whole point is that all the conductors are assumed to be perfect. There is no resistance. |
| electrodacus:
--- Quote from: SandyCox on May 13, 2022, 05:51:56 pm ---Your point is moot. The whole point is that all the conductors are assumed to be perfect. There is no resistance. --- End quote --- Where is that assumption made ? Are you referring to that article on wikipedia ? Take this example two 1F identical capacitors one charged at 3V and one discharged 0V with no resistance anywhere in the circuit including the wires, switch and capacitor plates. Initial energy in the system 0.5 * 1F * 3V2 = 4.5Ws After the switch is closed half of the energy from the charged capacitor is transferred to the discharged capacitor as there are no resistive losses when transfer is done thus final energy in the system will be the same 4.5Ws just that now it is stored in 2F instead of 1F Voltage on both capacitors after switch is closed and energy was transferred is 2.121V 0.5 * 2F * 2.1212 = 4.5Ws no energy is lost to outside as the circuit has zero resistance to current flow. It is as simple as that not the wrong equations used in the Wikipedia page. |
| SandyCox:
Check if your solution satisfies the law of preservation of charge. |
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