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| Veritasium "How Electricity Actually Works" |
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| IanB:
--- Quote from: hamster_nz on May 14, 2022, 08:25:14 pm ---I would love to hear electrodacus's analysis of the behavior of this circuit when the switch is closed, with Vi of 3 V and C of 3 F, with idea capacitor, wires and switches, of course, (so no dissipative elements). --- End quote --- It will cause a rift in the fabric of space-time and open up a wormhole to another dimension ;D |
| hamster_nz:
--- Quote from: IanB on May 14, 2022, 08:22:10 pm ---A question (for Tim, ejeffrey, or whoever): How do we precisely define the circumstances in which conservation of charge is applicable when analyzing a system, and when it is not? --- End quote --- Charge conservation, considered as a physical conservation law, implies that the change in the amount of electric charge in any volume of space is exactly equal to the amount of charge flowing into the volume minus the amount of charge flowing out of the volume. If we agree that there is no transfer to charge through the capacitors, and there is no charge flowing in from outside of the schematic, then hopefully we can agree that that the charge in the upper half (A) and the charge in the lower half (B) will remain the same before and after the switch is closed. |
| electrodacus:
--- Quote from: hamster_nz on May 14, 2022, 08:25:14 pm ---I would love to hear electrodacus's analysis of the behavior of this circuit when the switch is closed, with Vi of 3 V and C of 3 F, with idea capacitor, wires and switches, of course, (so no dissipative elements). --- End quote --- You will have an ideal LC circuit thus current through the inductor will increase gradually until capacitor will be fully discharged so a finite current that will create a finite magnetic field that will then collapse and charge the capacitor all they way back up to 3V. At any point after you close the switch the energy will be there in an isolated circuit as no energy is lost through resistive loss as there is no resistance and no energy will be dissipated despite the variable magnetic field around the circuit. But if the circuit is not isolated meaning you have some other loop in range of this and that has resistance then energy will be lost to that eventually all the initial energy will be lost in that magnetically coupled loop. But this will no longer be considered an isolated circuit. The point is that none of the energy will be radiated away but always be there and if you open the switch at the correct time when all energy is in the capacitor then you will be in the same state as the initial state. That is how the charging of a capacitor from another using an inductor as intermediary storage works in order to reduce the amount of lost energy during transfer. Adding the inductor adds no energy to the circuit yet it helps getting very close to ideal where most of the energy is still stored after transfer and not lost as heat. So no energy is "radiated away" by in this case magnetic field as the electric field exists only between the capacitor plates but that also is not "radiated away" While in Derek's example there is of course energy stored in magnetic field around the wires it is not what transfers the energy from one wire to the other. The current in those first 65ns is due to capacitor being charged as current cannot flow through 1m of air at 20V potential (leakage so low that it will not register). Since current flow is not present trough air no energy travels outside the wire let alone all energy as Derek claims. |
| electrodacus:
--- Quote from: IanB on May 14, 2022, 08:22:10 pm ---A question (for Tim, ejeffrey, or whoever): --- End quote --- Just noticed that your profile photo is an illustration of energy traveling through air as there electrical potential is large enough to make the air a usable conductor. Not the case with the 20V in Derek's experiment. |
| hamster_nz:
--- Quote from: electrodacus on May 14, 2022, 09:17:08 pm ---But if the circuit is not isolated meaning you have some other loop in range of this and that has resistance then energy will be lost to that eventually all the initial energy will be lost in that magnetically coupled loop. --- End quote --- Are you saying if there was a suitable loop in range (say with a resistor in it), then all the energy would be lost from this circuit? That all the energy would be transferred through the air to this loop, and it would end up heating the resistor? |
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