Veritasium "How Electricity Actually Works"

[electrodacus]

But that's an active circuit.  You didn't say anything about control before.  Did you forget to mention control before?  Then who's controlling the superconductor?

Tim

You understand that a DC-DC converter has no stored energy. So he will be supplied from the existing energy in that stored capacitor.
Even if this active circuit uses some of the energy that we want to transfer it is way more efficient than just paralleling the two capacitors directly.
I do not get your confusion.

[T3sl4co1l]

Shall I quote your numerous claims that capacitors can just be "paralleled directly"?

[electrodacus]

Shall I quote your numerous claims that capacitors can just be "paralleled directly"?

Of course they can be paralleled directly.
What I was saying (main point of the entire discussion) is that energy is transferred trough the wire and not outside the wires.
Since normal wires have resistance you end up with half the initial energy as the other half was lost as heat in the wire thus you end up with just half the V_i in both capacitors.
If you had an ideal setup with no resistance so all conductors are superconductors then you have no loss and end up with 0.707 * V_i in both capacitors.
Then some of you claimed that is not true that with ideal circuit you end up with that high voltage as you confuse voltage with energy.

Since is hard and expensive to setup an experiment with superconductors I offered a much easier solution and that is a DC-DC converter.
Using a 80% efficient DC-DC converter CC-CV working in CC mode will result in almost 90% of the energy still being present in the two capacitors after the energy is split equally between them so close enough to 0.707 * V_i to prove my point.

If you agree that moving the energy from charged capacitor to discharged capacitor will result in a higher final voltage in both capacitors than using wires then my point is made. If you do not then all I can say is you need to do the experiment.
By adding a DC-DC converter to the circuit you do not bring any external energy. You just more efficiently move the existing energy from one place to another.

[T3sl4co1l]

OK so the DC-DC is just an alternative implementation?

I want to do it without a DC-DC.  Can you show me an experimental setup (using superconductors if necessary) to prove the effect?

Tim

[electrodacus]

OK so the DC-DC is just an alternative implementation?

I want to do it without a DC-DC.  Can you show me an experimental setup (using superconductors if necessary) to prove the effect?

Tim

Yes DC-DC is an alternative way to transport energy from the charged capacitor to the discharged one.

The proof is already there with the normal capacitors.  After you parallel the capacitors you end up with only half the energy the other half escaped as heat due to circuit resistance.
There is no electric field in the discharged capacitor and there will only be a field as soon as an electron gets to one of the plates and simultaneously an electron will leave the opposite plate.
So energy transfer is done trough wires.

[T3sl4co1l]

Do you have a setup I can test?

[electrodacus]

Do you have a setup I can test?

What is exactly that you want to test ?
Is that moving energy from one charged capacitor to an identical discharged capacitor using a DC-DC converter will result in final voltage being significantly more than 1/2 and close to 0.707 ?

If you think that is not true then you are basically saying that capacitor energy equation is wrong.
E = 0.5 * C * V²

I already made the example with V_i = 3V and C = 1F
Initial energy in charged capacitor is
0.5 * 1F * 3V² = 4.5Ws

If energy is conserved after the switch was closed and you basically have a 2F capacitor (two 1F in parallel).
0.5 * 2F * (3V * 0.707)² = 4.5Ws

[aetherist]

Two capacitors cant exist like this. This circuit needs 2 switches, not 1 switch.

[electrodacus]

Two capacitors cant exist like this. This circuit needs 2 switches, not 1 switch.

what?
Why will you need a second switch ? redundancy

[SiliconWizard]

If you assume the bottom 'wire' between the two capacitors has zero impedance, then it's essentially just one capacitor. Isn't it?

[electrodacus]

If you assume the bottom 'wire' between the two capacitors has zero impedance, then it's essentially just one capacitor. Isn't it?

One capacitor is charged at V_i  so if you consider that the switch is a capacitor (with it is in real world) the capacitance is so small and negligible that all voltage will be on the (switch/capacitor) and the charge transferred to the discharged capacitor is extremely negligible.
So you can see a real switch as an ideal switch in parallel with a very small value capacitor.
When you close the switch you are basically shorting the small capacity capacitor (switch capacity).

For all intents and purposes you can consider the empty capacitor on the right as containing no charge.
Also if you where to add another switch on the bottom you are just adding another small capacitor in series and the difference between one switch and two switches is likely not even measurable with normal lab equipment.

[aetherist]

Two capacitors cant exist like this. This circuit needs 2 switches, not 1 switch.

what?
Why will you need a second switch ? redundancy

If u had 2 open switches, with a charged capacitor on leftside & non-charged capacitor on righthandside, then if u closed one switch i think that the 2 capacitors would not end up with the same (say half each) charge. They would have i think a very mixed arrangement of 4 different charges.

[T3sl4co1l]

What is exactly that you want to test ?
Is that moving energy from one charged capacitor to an identical discharged capacitor using a DC-DC converter will result in final voltage being significantly more than 1/2 and close to 0.707 ?

As you said earlier, and as I said earlier: without a DC-DC converter.

Can you do it?

If energy is conserved

Is a very, very big 'if'.

So can you do it?  With just wires, superconducting or otherwise?  As you said to do it earlier?  I'm really curious to see what you propose.  I haven't seen a 'no', no indication that it be impossible.

Granted, it's been a challenge getting anything of substance out of you at all...  I'm beginning to think you don't actually know what you're talking about.

Tim

[electrodacus]

Two capacitors cant exist like this. This circuit needs 2 switches, not 1 switch.

what?
Why will you need a second switch ? redundancy

If u had 2 open switches, with a charged capacitor on leftside & non-charged capacitor on righthandside, then if u closed one switch i think that the 2 capacitors would not end up with the same (say half each) charge. They would have i think a very mixed arrangement of 4 different charges.

If you consider the switch ideal (no capacitance) then if you close just one of two switches absolutely nothing will happen.
Real switches do have some capacitance depending on how they are constructed but that will be so small so many orders of magnitude that they can be ignored.
If you where to go to extreme and say that switch capacitance is the same as the two main capacitors in the test say 1F as in the example then when you charge the capacitor on the left to 3V you are also charging the two series capacitors the one from the switch and that one drawn in diagram as discharged thus you will see 1.5V on the right side capacitor.

To get rid of even the smallest switch capacitance that you can ignore anyway. You can just short the empty capacitor before closing the switch to get rid of the infinitesimal amount of charge it got there when you charged the one on the left.
Adding another switch will not help with anything as it will be just another small capacitor in series.

[electrodacus]

As you said earlier, and as I said earlier: without a DC-DC converter.

Can you do it?

Do you have the equipment to construct two capacitors form super conductive materials to do such a test ?
Also the equations show clearly what the result will be.

If energy is conserved

Is a very, very big 'if'.

So can you do it?  With just wires, superconducting or otherwise?  As you said to do it earlier?  I'm really curious to see what you propose.  I haven't seen a 'no', no indication that it be impossible.

Granted, it's been a challenge getting anything of substance out of you at all...  I'm beginning to think you don't actually know what you're talking about.

Tim

Energy is always conserved it is the law. If you do not agree with that then it is you that will need to prove it.

In the case of identical capacitors with resistance half of the energy is in the capacitors after the switch is closed and half is dissipated as heat from the conductors (that includes wires and capacitor plates).
You are lacking fundamental understanding thus you do not understand my replays.
It is sad to see influential science communicators and university professors not understanding the subject that they should learn others.   

[T3sl4co1l]

I do.  I performed the test.

I observed an oscillating waveform, with the average voltage being 0.5 Vi, and the peak sine amplitude being 0.5 Vi.  Put another way, the capacitors slosh alternately between 0 and 1 Vi.  At no point are they simultaneously 0.71 Vi; and the current in the wire connecting them (inductor, really), when one is at 0.71 Vi, is nonzero, so opening the switch at that instant would result in energy loss.

Can you explain my findings?

Tim

[electrodacus]

I do.  I performed the test.

I observed an oscillating waveform, with the average voltage being 0.5 Vi, and the peak sine amplitude being 0.5 Vi.  Put another way, the capacitors slosh alternately between 0 and 1 Vi.  At no point are they simultaneously 0.71 Vi; and the current in the wire connecting them (inductor, really), when one is at 0.71 Vi, is nonzero, so opening the switch at that instant would result in energy loss.

Can you explain my findings?

Tim

Let me guess. You used some simulation tool that was not setup properly or the simulation tool was wrongly designed.

[hamster_nz]

Discussing the case where you have switches in both the top and bottom wires...

If you consider the switch ideal (no capacitance) then if you close just one of two switches absolutely nothing will happen.

Well, not absolutely nothing. If only one switch is closed, there will always be 10V measured over the other switch, so the act of closing one of the switches changes the voltages on the right hand side.

With both switches open the voltages on the right hand side are not at all clearly defined, as the two circuits are not connected.

(I'm waiting for some "the quantum wave function collapsed because you need to a meter to make the measurements" handwaving... from the person who adds DC-DC converters)

[electrodacus]

Discussing the case where you have switches in both the top and bottom wires...

If you consider the switch ideal (no capacitance) then if you close just one of two switches absolutely nothing will happen.

Well, not absolutely nothing. If only one switch is closed, there will always be 10V measured over the other switch, so the act of closing one of the switches changes the voltages on the right hand side.

With both switches open the voltages on the right hand side are not at all clearly defined, as the two circuits are not connected.

(I'm waiting for some "the quantum wave function collapsed because you need to a meter to make the measurements" handwaving... from the person who adds DC-DC converters)

We are talking about energy transfer.
I think I mentioned before that your multi-meter is basically 1Mohm resistor so it is like closing the switch and measuring the voltage drop on that 1MOhm resistor.
When you have two switches if you had a high resolution oscilloscope you will see a spike when you connect the oscilloscope probe as the probe impedance closes that circuit and there is a circuit because the other open switch (a real one has capacitance) thus you rearrange some of the charges.


You are concentrating on insignificant details that you do not even understand instead of the main subject of how the energy is transferred (trough wire my claim or outside of the wire Derek's claim that you seems to be defending without any basis).
Also your do not provide any feedback on where my math is wrong. Math that shows the amount of energy in a capacitor.
 

[hamster_nz]

Also your do not provide any feedback on where my math is wrong. Math that shows the amount of energy in a capacitor.

I have done that many times.... but one more time...

Can we agree that the total charge on the top half of the system is:

  Q_total = C_c1 V_c1 + C_c2 V_c2

So here is the initial state of the system:

  Q_total =   1F * 3V + 1F * 0V = 3 Coulomb.

I hope both agree that no charge can cross either capacitor regardless of the position of the switch. So the total charges on the top and bottom halves must remain the same before and after the switch is closed. If not, please let me know what mechanism is taking electrons out of the top half and moving them to the bottom half, as I am sure we both agree that no positively charged atoms are moving around either.

The switch is closed, and allowing for the incorrect assumption that everything settles down to a stable state when using ideal components:

  Q_total = C_c1 V_c1 + C_c2 V_c2
  Q_total =   1F * 1.5V + 1F * 1.5 = 3 Coulomb.

This is the same as you would have if you consider the two 1F capacitors now a single 2F capacitor:

  Q_total =   2F * 1.5V = 3 Coulomb.

We have the same amount of charge, but are left with only half the voltage over each capacitor. The incorrect assumption that the system becomes stable has let us down, as half the energy is missing, but ideal capacitors and wires have no resistive element to dissipate it.

A simpler example that shows how flawed your position is

What about a single charged ideal capacitor (no ESR), ideal wires (zero ohms) and a single ideal switch (zero ohms when closed, perfect isolation when open), that shorts the capacitor.  If, as you say, energy must be conserved, what happens if the capacitor has been charged to 3V and the switch is closed?

What happens with "close to ideal wires", that have zero ohms resistance, but a small amount of inductance?

The inductance is important, as it gives a solution where the total energy in the system is conserved.

[electrodacus]

Also your do not provide any feedback on where my math is wrong. Math that shows the amount of energy in a capacitor.

I have done that many times.... but one more time...

Can we agree that the total charge on the top half of the system is:

  Q_total = C_c1 V_c1 + C_c2 V_c2

So here is the initial state of the system:

  Q_total =   1F * 3V + 1F * 0V = 3 Coulomb.

I hope both agree that no charge can cross either capacitor regardless of the position of the switch. So the total charges on the top and bottom halves must remain the same before and after the switch is closed. If not, please let me know what mechanism is taking electrons out of the top half and moving them to the bottom half, as I am sure we both agree that no positively charged atoms are moving around either.

The switch is closed, and allowing for the incorrect assumption that everything settles down to a stable state when using ideal components:

  Q_total = C_c1 V_c1 + C_c2 V_c2
  Q_total =   1F * 1.5V + 1F * 1.5 = 3 Coulomb.

This is the same as you would have if you consider the two 1F capacitors now a single 2F capacitor:

  Q_total =   2F * 1.5V = 3 Coulomb.

We have the same amount of charge, but are left with only half the voltage over each capacitor. The incorrect assumption that the system becomes stable has let us down.

A simpler example that shows how flawed your position is

What about a single charged ideal capacitor (no ESR), ideal wires (zero ohms) and a single ideal switch (zero ohms when closed, perfect isolation when open), that shorts the capacitor.  If, as you say, energy must be conserved, what happens if the capacitor has been charged to 3V and the switch is closed?

What happens with "close to ideal wires", that have zero ohms resistance, but a small amount of inductance?

The inductance is important, as it gives a solution where the total energy in the system is conserved.

As I mentioned you are confusing charge with energy.
The important law is called Conservation of energy and not conservation of charge.

Please do a energy balance not a charge balance and you will understand what I'm saying.
When you do the transfer from one capacitor to the other you will be losing half of the energy as heat.

Once you understand the difference between charge and energy it should be easier to also understand what I'm saying.

3 Coulomb is like saying 0.833mAh so you say nothing about energy.

[TimFox]

If one ignores the inductance in the circuit, and just treats it as having a finite resistance R in series with the two equal capacitors:
The final distribution of charge (1/2 on each capacitor) is independent of R, but the time required to equilibrate decreases as R decreases.
Since the final distribution is independent of R, it will still be true in the limit as R --> 0.
Also, the total energy after equilibrium in the two-capacitor system will also be 1/2 the original total energy in only one capacitor, independent of R and therefore in the limit as R --> 0.

[hamster_nz]

A simpler example that shows how flawed your position is

What about a single charged ideal capacitor (no ESR), ideal wires (zero ohms) and a single ideal switch (zero ohms when closed, perfect isolation when open), that shorts the capacitor.  If, as you say, energy must be conserved, what happens if the capacitor has been charged to 3V and the switch is closed?

What happens with "close to ideal wires", that have zero ohms resistance, but a small amount of inductance?

The inductance is important, as it gives a solution where the total energy in the system is conserved.

As I mentioned you are confusing charge with energy.
The important law is called Conservation of energy and not conservation of charge.

Please do a energy balance not a charge balance and you will understand what I'm saying.
When you do the transfer from one capacitor to the other you will be losing half of the energy as heat.

Once you understand the difference between charge and energy it should be easier to also understand what I'm saying.

3 Coulomb is like saying 0.833mAh so you say nothing about energy.

But voltage is also clearly defined, through the capacitance formula of Q = C V, and you can calculate energy directly from charge and capacitance U = 1/2 Q^2 / C

And the answer in the "one cap, one switch" situation?

[electrodacus]

If one ignores the inductance in the circuit, and just treats it as having a finite resistance R in series with the two capacitors:
The final distribution of charge (1/2 on each capacitor) is independent of R, but the time required to equilibrate decreases as R decreases.
Since the final distribution is independent of R, it will still be true in the limit as R --> 0.

Yes 1/2 of the charge but 1/4 of energy.  Half of the total energy was dissipate as heat on the series resistance no matter the value of the series resistance.
Please understand that discussion is about transfer of energy.

[TimFox]

If one ignores the inductance in the circuit, and just treats it as having a finite resistance R in series with the two capacitors:
The final distribution of charge (1/2 on each capacitor) is independent of R, but the time required to equilibrate decreases as R decreases.
Since the final distribution is independent of R, it will still be true in the limit as R --> 0.

Yes 1/2 of the charge but 1/4 of energy.  Half of the total energy was dissipate as heat on the series resistance no matter the value of the series resistance.
Please understand that discussion is about transfer of energy.

1/4 of the energy in each capacitor, for a total energy in the two-capacitor system half the original energy.