Veritasium "How Electricity Actually Works"

[bsfeechannel]

Don't let the pretty pictures distract you.

Like this one by Poynting himself, for instance, showing the energy going from a battery to a resistive wire loop through the empty space.



Don't look at that. It'll shatter your faith in the wires.

[hamster_nz]

You are a hard person to goad something out of, but eventually it works.

Is that your replay ? How about you understand now that if you transfer energy more efficiently from one capacitor to the other that is identical you get close to 0.707 the voltage of the charged capacitor in both.
It should not be me that is doing the work.

You keep on bringing up DC-DC converters

Reply 118:

Just test with a DC-DC converter and you will see very close to ideal is possible.  If the DC-DC converter was 100% efficient (so ideal) then you get 0.707 V_i   but you do not need a DC-DC converter if there is no resistance to get the same result.

Everybody (even I) agree that it is possible:

Replay 119:

(And of course if you have an ultra efficiency DC-DC convertor to make the transfer then you might get close to 71%, but we don't. We have two ideal caps, some ideal wire and an ideal switch).

But you keep on tuning it back to "oh, if I use a DC/DC converter it proves something". So I called your bluff.

So finally we have the proof that everybody else missed. That if you transfer energy to and from a magnetic field, using an inductor, (thus taking the energy "out of the wire")  that energy only flows in the wires. 

[electrodacus]

You keep on bringing up DC-DC converters
...
So finally we have the proof that everybody else missed. That if you transfer energy to and from a magnetic field, using an inductor, (thus taking the energy "out of the wire")  that energy only flows in the wires. 

At first you and others mentioned that you can not have 0.707 * V_i because that will violate the conservation of charge witch likely you confused with energy.
Then after pointing out the formula for energy stored in a capacitor it was still argued that 0.707 * V_i is not possible in an circuit made with superconductors so zero resistance.
Then I mentioned something that is way easier to test (a DC-DC converter) and claim remained the same that voltage at the end of the experiment can not be higher than 0.5 V_i.

The energy seen through the bulb in the first 65ns in Derek's experiment is related to line capacity so no energy travels through air.
Any transmission line has both inductance and capacitance (both are energy storage devices) so the transmission line model that any electrical engineer uses is a realistic and accurate way to predict what happens.

I simplified the two capacitors spice model even more and below is the schematic and the graph
In the schematic you have a 3V supply to charge one of the 1000uF capacitors then the supply is disconnected by the switch S1 after 20ms and then starting at 40ms the S2 switch is closed for just 6ms then open.
The charged capacitor C1 will discharged from 3V down to around 2V while the C2 that started fully discharged is charged using that energy at 2V so you start with C1 at 3V and you end up with both identical capacitors charged at 2V using only energy available in C1
The only other two components are the 47mH inductor and a diode. The inductor is an intermediary energy storage helping reduce the losses due to circuit resistance. There is no outside energy coming in to the circuit and some is still lost as heat but much less than just paralleling the two capacitors directly.
The red graph is the control for the S2 switch showing the period switch was closed. The blue is the voltage across C1 and green voltage across C2.


   

[SandyCox]

Here's another one:

Let's connect and ideal DC voltage source V_d to a fully discharged capacitor C at t=0s:
1. How much energy is delivered by the source?
2. How much energy is stored in the capacitor at t=1s?
3. How much charge is delivered by the source?
4. How much charge ends up in the capacitor?

You can try the same with a current source and an inductor.

[SandyCox]

You keep on bringing up DC-DC converters
...
So finally we have the proof that everybody else missed. That if you transfer energy to and from a magnetic field, using an inductor, (thus taking the energy "out of the wire")  that energy only flows in the wires. 

At first you and others mentioned that you can not have 0.707 * V_i because that will violate the conservation of charge witch likely you confused with energy.
Then after pointing out the formula for energy stored in a capacitor it was still argued that 0.707 * V_i is not possible in an circuit made with superconductors so zero resistance.
Then I mentioned something that is way easier to test (a DC-DC converter) and claim remained the same that voltage at the end of the experiment can not be higher than 0.5 V_i.

The energy seen through the bulb in the first 65ns in Derek's experiment is related to line capacity so no energy travels through air.
Any transmission line has both inductance and capacitance (both are energy storage devices) so the transmission line model that any electrical engineer uses is a realistic and accurate way to predict what happens.

I simplified the two capacitors spice model even more and below is the schematic and the graph
In the schematic you have a 3V supply to charge one of the 1000uF capacitors then the supply is disconnected by the switch S1 after 20ms and then starting at 40ms the S2 switch is closed for just 6ms then open.
The charged capacitor C1 will discharged from 3V down to around 2V while the C2 that started fully discharged is charged using that energy at 2V so you start with C1 at 3V and you end up with both identical capacitors charged at 2V using only energy available in C1
The only other two components are the 47mH inductor and a diode. The inductor is an intermediary energy storage helping reduce the losses due to circuit resistance. There is no outside energy coming in to the circuit and some is still lost as heat but much less than just paralleling the two capacitors directly.
The red graph is the control for the S2 switch showing the period switch was closed. The blue is the voltage across C1 and green voltage across C2.
(Attachment Link)
(Attachment Link)

You cannot add inductors and diodes or DC-to-DC converters. The problem is about connecting two capacitors in parallel without any other components.

How do you think energy is transferred through a capacitor? Let's take a parallel plate capacitor with vacuum between the plates.

What about the airgap of a transformer?

[hamster_nz]

You cannot add inductors and diodes or DC-to-DC converters. The problem is about connecting two capacitors in parallel without any other components.

How do you think energy is transferred through a capacitor? Let's take a parallel plate capacitor with vacuum between the plates.

What about the airgap of a transformer?

Electrodacus already knows that, from message #169 when a DC-DC convertor was originally suggested:

It is also 'cheating'. I too could get any answer I want if I am free to add to the system. If I said "let me put an inductor in there, a diode, a switch and a trained imp that can push the switch really quickly" would you not agree that that is not the same problem any more?

Even knowing that Electrodacus went there, and put the diode, an inductor, a switch, and even the trained imp to push it. 

[SandyCox]

The problem is that circuit theory doesn't take the physical dimensions and the electrodynamic behavior of the system into account. That is why we get a contradiction.

Let's look at the more realistic example of using two lossless coaxial cables as "capacitors".

Let's see if Electrodacus' modern outlook on electronics can derive the equations that govern this example.

[abebarker]

Is it possible to have an electric field without an associated particle? Is it possible to have an electric field without a charged particle?

I'm sure there are situations where it is much easier to calculate considering only the fields. I guess it is important to remember that numbers can be used to roughly describe reality and roughly predict things but they are not reality itself. It would be preposterous to think otherwise, absolutely posterior backwards. However, it does seem more common than not for people to think that words make reality rather than reality dictating the words (the little angels, little to no connection to reality).

[vad]

Is it possible to have an electric field without an associated particle? Is it possible to have an electric field without a charged particle?

Absolutely. Changing magnetic field creates electric field (Faraday’s law).

And you don’t need charged particles to have a magnetic field. For example, neutron has magnetic dipole moment. Magnetars are neutron stars that create strongest magnetic fields in the Universe.

[timenutgoblin]

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.

https://en.m.wikipedia.org/wiki/Charge_conservation

According to the Wikipedia entry, half of the energy is 'lost' in the initially-charged capacitor after the switch is closed as shown in the mathematical expression below:

\$W_i = \frac{1}{2}CV_i\$

\$W_f = \frac{1}{2}CV_f^2 + \frac{1}{2}CV_f^2 = CV_f^2 = C(\frac{V_i}{2})^2 = \frac{1}{4}CV_i^2 = \frac{1}{2}W_i\$

Source: en.m.wikipedia.org/wiki/Two_capacitor_paradox

In order to satisfy Kirchhoff's Voltage Law after the switch is closed, both capacitors must have equal voltage. In this case, it is half the initial voltage of the initially-charged capacitor. Also, when the initially-discharged capacitor is connected after the switch is closed, this has the effect of doubling the total capacitance. If Q = CV and total capacitance doubles, then the voltage must be halved (V = Q/C) assuming charge is conserved. It then follows that if voltage is energy per unit charge, and charge is conserved after the switch is closed, then the initial energy must be halved if voltage is halved (across both capacitors). If the Law of Conservation of Charge is satisfied after the switch is closed, half of the initial energy must be lost to satisfy Kirchhoff's Voltage Law.

https://en.m.wikipedia.org/wiki/Voltage

Using the swimming pool analogy, assume that a swimming pool is filled with water and the floor of the swimming pool has a surface area of A (metres squared) and a height of h (metres). The weight and the height of the water is analogous to the charge and the voltage of the capacitor. If weight of the water can be expressed as F = mg then the gravitational potential energy of the water can be expressed as E = mgh.

https://en.m.wikipedia.org/wiki/Gravitational_energy

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.

Elaborating further with the swimming pool analogy, if the surface area of the floor is increased to the point where all of the water molecules are in direct contact with the floor then the effective height of the water would be the height of a single water molecule. The effective height of the water in the swimming pool would be approaching zero metres. The gravitational potential energy would be approaching zero, too.



I found a YouTube video showing the magnetic field of a toroidal transformer. You can see that the field exists only within the windings of the electromagnet. It's as though there are discrete magnets stacked with north to south poles arranged in a loop. What if the electromagnet was replaced with batteries connected in a loop with the positive terminal of one battery connecting to the negative terminal of the next battery? Would the electric field only be within the batteries or would it also extend/exist outside of the batteries?

Toroidal Electromagnet video: https://youtu.be/0-KaSGlDQ-8

[T3sl4co1l]

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.

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]

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.

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]

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.

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.

Your point is moot. The whole point is that all the conductors are assumed to be perfect. There is no resistance.

[electrodacus]

Your point is moot. The whole point is that all the conductors are assumed to be perfect. There is no resistance.

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.

[TimFox]

Also, check what the charge and voltage ratios as a function of the series resistance and what the limits as R-->0 are.

[electrodacus]

Check if your solution satisfies the law of preservation of charge.

You do not understand what conservation of charge means and how it is applied.
The clue is in the fact that adding another charged particle to a capacitor that is at 1V and same capacitor that is at 2V requires different amounts of energy.

[electrodacus]

Also, check what the charge and voltage ratios as a function of the series resistance and what the limits as R-->0 are.

Any resistance different from zero will result in the same exact final state for two identical parallel capacitors and that will be that half of the energy will be lost as heat in conductors due to resistance to current flow.
You basically have a resistor divider. The value of the resistance will only influence the time it takes for the energy to be transferred and wasted as heat.
To test this just add a 1 Ohm resistor between the two capacitors and then add a 1KOhm resistor and you will see that same amount of energy will be wasted as heat and you will have the same half of the original voltage in the two identical capacitors.

[TimFox]

Exactly.
Since the final result is independent of the finite resistance in the circuit, it will also be the result in the limit as R-->0.

Although the famous conservation of energy law is true, it is not always relevant to a given situation unless you can include the exact losses due to, for example, resistor heat loss and EM radiation.
In first-year physics, we learned to distinguish between conservation of momentum (true) and conservation of kinetic energy (not always true) in simple mechanical calculations.

[electrodacus]

Exactly.
Since the final result is independent of the finite resistance in the circuit, it will also be the result in the limit as R-->0.

Can you rephrase that ? Id not not understand what you mean.
As long as resistance is higher than absolute zero half of the energy will be lost as heat in that resistance so you end up with just half the energy in this particular systemic case of two identical capacitors.
If capacitors are not identical then percentage of loss as heat will be different than half.
All energy is accounted for in any setup. Q on the other hand that is Q = C*V will not be the same in any other experiment other than the two identical capacitor case with resistance higher than zero. This is due to symmetry so it is just a coincidence and not a rulle.

[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]

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.

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]

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.

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]

(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?

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.