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Veritasium "How Electricity Actually Works"

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TimFox:
I mean that in the context of mathematical limit calculations.
(I was once criticized here for suggesting "epsilon-delta" (q.v.) calculations of limits, but that's how it works.)
Even if the resistance is 1x10-6 ohms (unphysical), the loss of total energy is still 1/2.
Therefore, for any specified resistance value the loss is unchanged.
The factor (1/2), of course, is only valid for two equal capacitors.
In the calculation, the voltages across the two capacitors must be equal to each other after a long time, with corresponding charges (unequal if the capacitors are unequal).

This is an example of a one-sided limit, since a physical passive resistor (as opposed to an active device with external power applied, such as a tunnel diode or transitron vacuum tube) cannot have a negative value.
A physical inductor with finite resistance will give an "evanescent" (exponentially dying sinusoidal) solution, which should go to the same asymptotic state at long time as the non-inductive resistor.

DC-DC converters are not passive, and will have different behavior.  For example, if you want to charge a capacitor efficiently, ramping the voltage in a controlled manner is better than throwing a knife switch through a resistance from a battery.

electrodacus:

--- Quote from: TimFox on May 13, 2022, 07:18:45 pm ---I mean that in the context of mathematical limit calculations.
(I was once criticized here for suggesting "epsilon-delta" (q.v.) calculations of limits, but that's how it works.)
Even if the resistance is 1x10-6 ohms (unphysical), the loss of total energy is still 1/2.
Therefore, for any specified resistance value the loss is unchanged.
The factor (1/2), of course, is only valid for two equal capacitors.

--- End quote ---

Using an inductor to store that energy that otherwise will end up as heat and then discharging that energy stored in inductor in to the discharged capacitor can get you very close to ideal energy transfer with very little loss.
Even from my simple setup I got slightly above 2V in both capacitors at the end of the energy transfer.

In my example I used 1000uF = 1mF capacitors so
0.5 * 1mF * 32 = 4.5mWs as start energy

At the end with say rounded 2V in both we have
0.5 * 2mF * 22 = 4mWs

So 0.5mWs still lost as heat but it is way less than having to lose half and is close to 90% energy transfer efficiency.

Q = C * V at the start is  1mF * 3V = 3mC = 0.833uAh

Q at the end is 2mF * 2V = 4mC = 1.111uAh   


Edit: What difference does it make if you use active or passive components when we talk about energy ?
No energy is added to the isolated system by adding non charged components in the experiment.
inductor same as capacitor is a passive component and if you want I can eliminate the diode replace that with another switch and show you the same thing.
Not that the diode can add any energy in to the system unless it is a photo diode and is hit by photons coming from outside of this system but then it will no longer be an isolated system.

TimFox:
In the Spice model you posted above (reply 477), there is zero resistance in series with your finite inductor.
The switches have finite resistance, but the second one is not on for a long time.
However, the diode model (an actual Schottky power diode) included in the circuit can bleed charge from the system, assuming it includes losses.
In a resistor-only circuit with two capacitors, only resistors in parallel with the capacitors (e.g., capacitor leakage) will remove charge from the system as the capacitors slowly discharge with the power switch open.

(PS:  when I distinguished active and passive circuits, I mean that an active circuit has an external power source.  So a diode in the dark is passive, but it has losses.) 
What happens if you leave S2 on for a long time in your simulation?

electrodacus:

--- Quote from: TimFox on May 13, 2022, 07:42:41 pm ---In the Spice model you posted above (reply 477), there is zero resistance in series with your finite inductor.
The switches have finite resistance, but the second one is not on for a long time.
However, the diode model included in the circuit can bleed charge from the system, assuming it includes losses.
In a resistor-only circuit with two capacitors, only resistors in parallel with the capacitors (e.g., capacitor leakage) will remove charge from the system as the capacitors slowly discharge with the power switch open.

--- End quote ---

Sorry I did not mentioned but the inductor has a 0.3Ohm resistance is just not visible in the diagram.
Yes the reverse current of the diode will slowly discharge the capacitor but that can be disconnected after the energy was transferred by adding another switch.
And yes capacitors are not perfect and have some leakage thus energy will be lost over time but that is not the point of the experiment.
Main point of this experiment is to show that in the two parallel capacitors half of the energy is lost because energy travels through wires and wires have a resistance to current flow.
If energy was not transferred through wires then you will either have no loss associated with the wire resistance or the losses will be manifested somewhere else other than in wires.
You need to move electrons through wires from one capacitor to another in order to move energy. Unless voltage is so high that air can become a conductor and electrons can be transferred that way from one capacitor to another the energy will travel through wires.

electrodacus:

--- Quote from: TimFox on May 13, 2022, 07:42:41 pm ---(PS:  when I distinguished active and passive circuits, I mean that an active circuit has an external power source.  So a diode in the dark is passive, but it has losses.) 
What happens if you leave S2 on for a long time in your simulation?

--- End quote ---

What do you think it will happen ?
The energy will flow back and forth between the two capacitors and inductor (RLC resonant circuit) until half of the initial energy will be dissipated as heat so you end up with half the voltage.
   

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