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| Veritasium "How Electricity Actually Works" |
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| electrodacus:
--- Quote from: IanB on May 15, 2022, 04:33:09 am --- 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? --- End quote --- 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:
--- 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. --- End quote --- The paradox exists if only the DC steady state is considered and not the AC transient state. --- Quote from: electrodacus on May 13, 2022, 04:47:24 pm --- Did you even looked closely at the equations you posted from wikipedia to see how absurd and stupid they are. --- End quote --- Did you notice my accidental typo? --- Quote from: electrodacus on May 03, 2022, 12:40:07 am --- If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * Vi --- End quote --- 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: --- Quote ---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. --- End quote --- Here is a post from a few pages back in this thread: --- Quote from: T3sl4co1l on May 04, 2022, 01:07:52 am --- From the left, the voltage steps down corresponding to the switch turning on. Subsequently, the capacitor voltages (Ch1, Ch3) oscillate as predicted. --- End quote --- --- Quote from: electrodacus on May 03, 2022, 12:40:07 am --- If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * Vi --- End quote --- 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:
--- Quote from: electrodacus on May 15, 2022, 02:55:36 am ---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. --- End quote --- 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:
--- Quote from: timenutgoblin on May 15, 2022, 01:42:38 pm --- The paradox exists if only the DC steady state is considered and not the AC transient state. --- End quote --- What is the paradox at DC steady state ? --- Quote from: timenutgoblin on May 15, 2022, 01:42:38 pm ---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. --- End quote --- 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. --- Quote from: timenutgoblin on May 15, 2022, 01:42:38 pm --- 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\$ --- End quote --- 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:
--- Quote from: vad on May 15, 2022, 01:59:45 pm --- 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? --- End quote --- 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. |
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