Veritasium "How Electricity Actually Works"

[electrodacus]

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

Do you mean a charged cap shorted by a switch ?
If so then all energy contained in the capacitor ends up as heat.

[electrodacus]

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

Yes exactly.

[Naej]

One of the problems with analogies is that you have to have a deep understanding of model's (in this case hydraulic) system. Derek seems to have done his homework. He points out that unlike a hydraulic system (where the water from the pump to the load has high pressure and low velocity and the water in the return pipe has lower pressure and higher velocity), with an electric circuit, there's no difference in current density or drift speed for the  electrons going in and coming out of the battery.

Well in the case you described the "load" (a reducer/nozzle) is analogous to a transformer, and there's no difference in mass flow going in and coming out of the pump. The analogy is tight, here 

What is making the electrons drift is not some kind of pressure. It is just a portion of the electric field generated by the battery that does not contribute to the transfer of energy from the battery to the load. So electrons in the wire are not being pushed by each other like in a fluid. They're just parading in response to an external cause (the electric field) exactly like in the load.

What is making the electrons drift is a kind of pressure, called electric potential. They are being pushed by each other, but unlike in fluids, it happens at long range and can easily go through some solid matter.

[rfeecs]

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

Yes exactly.

I have a suggestion.  Forget the two capacitor question.  Veritasium's experiment uses a battery.  Maybe start a new thread about two capacitors?  It's off topic.

We don't have to understand batteries beyond they are ideal voltage sources.  They maintain a constant voltage across their terminals.  Not complicated.  Go from there.

Here's a nice video to get you started, 1940's style Veritasium:

[electrodacus]

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

Yes exactly.

I have a suggestion.  Forget the two capacitor question.  Veritasium's experiment uses a battery.  Maybe start a new thread about two capacitors?  It's off topic.

We don't have to understand batteries beyond they are ideal voltage sources.  They maintain a constant voltage across their terminals.  Not complicated.  Go from there.

You are wrong about the fact that it is of topic.
There is no difference in this context between a battery and a capacitor as they are both the source of energy in the experiment.
Understanding what energy is and how capacitors work (energy storage devices) is critical to understand the problem.

[timenutgoblin]

If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * V_i

One other thing I want to point out is that if the final voltage on both capacitors is 70.7% of the initial voltage then you will NOT be conserving charge, but only conserving energy instead. Initially, the charge is Q = CV before the switch is closed. After the switch is closed, if the final voltage is 0.707*Vi on both capacitors then the final charge will be C*V*0.707 + C*V*0.707 = 1.414*Q where Q is the initial charge. Where does the extra 41.4% charge come from if it wasn't there initially?

[rfeecs]

You are wrong about the fact that it is of topic.
There is no difference in this context between a battery and a capacitor as they are both the source of energy in the experiment.
Understanding what energy is and how capacitors work (energy storage devices) is critical to understand the problem.

There is a difference between a battery and a capacitor.
I understand how capacitors work.
Now, energy is another issue.    But I can always look up the textbook definition.
What is your definition of energy?

What did you think of the video?

[hamster_nz]

The unasked question so far is:

Those pipes are hollow. What if you put a thick wire inside those pipes, electrically connected only at both ends of the pipe.

You balance the cross-sectional area such that the wire and pipe have the same mm^2.  After a constant current power supply is connected and the system reaches a steady state. How much current would flows through the inner wire, and how much through the outer pipe?

Three different schools of thought:

1) The pipe is a (oddly shaped) Faraday cage. There will be minimal potential difference across the inner wire, so minimal current will flow in it. The bulk of the current flows in the pipe. If connected in parallel outside of the pipe then more current would flow.

2) The current flow is based on the cross-sectional area of the conductor and its bulk resistance, so half the power will be in the inner wire, and half in the outer pipe. The same current would flow the wire was connected in parallel outside the pipe.

3) Well, it's not that simple, as the field from the pipe will still be interacting with the wire and the other way around. You would need to know about a lot of other things about the materials and geometries before you could answer this.

I'm firmly in camp 2 for steady state. But during the transient when current source is connected I'm quickly jumping into "camp 3".

[T3sl4co1l]

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.

Don't let him distract you with added switches, or odd ratios of capacitor: again, it adds nothing to the problem, and only deflects from the central point.

Tim

[electrodacus]

If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * V_i

One other thing I want to point out is that if the final voltage on both capacitors is 70.7% of the initial voltage then you will NOT be conserving charge, but only conserving energy instead. Initially, the charge is Q = CV before the switch is closed. After the switch is closed, if the final voltage is 0.707*Vi on both capacitors then the final charge will be C*V*0.707 + C*V*0.707 = 1.414*Q where Q is the initial charge. Where does the extra 41.4% charge come from if it wasn't there initially?

The important part is conservation of energy not charge.
You start with 4.5Ws of stored energy in the charged capacitor 0.5 * 1F * 3V². That is what you need to justify not charge unless you are confusing charge with energy and that seems to be the case.

Initial charge 1F * 3V = 3 Coulomb  = 0.833mAh
4.5Ws / 3600s = 1.25mWh
3V * 0.833mAh = 2.5mWh if the voltage will have been constant but voltage drops linearly so is half that 1.25mWh as calculated with the formula for Energy in a capacitor.

At the end of the experiment you can have for ideal case
2.121V on each capacitor
So 1.25mWh initial energy divided between the two capacitors so 0.625mWh in each
If you care about charge on each capacitor (not relevant) 2.121 Coulomb / 3600 = 0.589mAh
(2.121V /2) * 0.589mAh = 0.625mWh

As you see both energy and charge checks out in my calculation as I did the correct assumption that energy is conserved not charge.   

[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

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

I mean, you're welcome to scrutinize whatever models I use.  But my oscilloscope speaks for itself:



From the left, the voltage steps down corresponding to the switch turning on.  Subsequently, the capacitor voltages (Ch1, Ch3) oscillate as predicted.  I'm not seeing any trend towards 0.71 Vi here.

Tim

[rfeecs]

The unasked question so far is:

Those pipes are hollow. What if you put a thick wire inside those pipes, electrically connected only at both ends of the pipe.

You balance the cross-sectional area such that the wire and pipe have the same mm^2.  After a constant current power supply is connected and the system reaches a steady state. How much current would flows through the inner wire, and how much through the outer pipe?

As with everything it seems, there is a video for that:



He says the currents starts out on the outside of the conductor, then moves inside with time.

A constant current source would be high impedance.  Not sure that's the best choice for a high speed signal.

So who's going to test it?

[electrodacus]

I mean, you're welcome to scrutinize whatever models I use.  But my oscilloscope speaks for itself:



From the left, the voltage steps down corresponding to the switch turning on.  Subsequently, the capacitor voltages (Ch1, Ch3) oscillate as predicted.  I'm not seeing any trend towards 0.71 Vi here.

Tim

What do you have on channel 1 and 3 and also what are you triggering on channel 4?
Also what is the exact circuit ?

[T3sl4co1l]

Channels 1 and 3 are the voltages at the two capacitors in the circuit:



The scope is grounded to one side of the switch, hence the voltage step at turn-on.  This also indicates the initial voltage.  Ch4 is a sync signal triggering the switch.

Digital averaging and bandwidth limiting (100MHz) have also been used to improve signal quality.

The cursor at 4.2µs shows the half wave time.

Tim

[electrodacus]

Channels 1 and 3 are the voltages at the two capacitors in the circuit:



The scope is grounded to one side of the switch, hence the voltage step at turn-on.  This also indicates the initial voltage.  Ch4 is a sync signal triggering the switch.

Digital averaging and bandwidth limiting (100MHz) have also been used to improve signal quality.

The cursor at 4.2µs shows the half wave time.

Tim

A photo of the setup will be super useful.
I do not get how your setup is made and there are to many unknowns.
What capacitors are you using value and type.  How are the capacitors connected (wires if so how long).
How isolated is the system ? Like any power supply connected (even if just one of the wires from the power supply is connected).
What sort of switch do you use ? Is that maybe a mosfet (what model) and how is that controlled ? Is is a photo diode optocoupler ? It may be supper important and likely the reason for osculations if your circuit is not properly isolated.

[T3sl4co1l]

Capacitors: about 20nF each.

Wire: I could measure it, but actually, you have enough information now to calculate this.

The system is isolated on battery power.  Anyway, it's only powered before the switch turns on.

Will my setup produce the fabled 0.71 Vi you have predicted?

Tim

[electrodacus]

Capacitors: about 20nF each.

Wire: I could measure it, but actually, you have enough information now to calculate this.

The system is isolated on battery power.  Anyway, it's only powered before the switch turns on.

Will my setup produce the fabled 0.71 Vi you have predicted?

Tim

20nF so you have build some LC oscillator.
I was for good reason using 1F as an example but at least hundreds or thousands of uF will be useful for such a test.
Someone else here had a video with a test using some larger Electrolytic capacitors. I was expecting you use some similar setup and maybe you had some strange isolation problem due to non isolated switch driver. But matched inductance and capacitance also make sense and that is most likely you case as 20nF is basically nothing.

If you use large capacitors Large electrolytic or super-capacitors then you can use a DC-DC converter with constant current control and get very close to 0.7 V_i

The probing was also strange as I have no idea why you will connect the oscilloscope ground to the switch. But is fairly clear that you have build an LC resonator so not at all what is shown in that diagram two parallel capacitors and a switch.

[T3sl4co1l]

20nF so you have build some LC oscillator.
I was for good reason using 1F as an example but at least hundreds or thousands of uF will be useful for such a test.

Sure, but I don't have superconducting capacitors that large.  Are you saying this effect disappears for small values, then?  By what mechanism?  That's quite a peculiar effect if so!

Someone else here had a video with a test using some larger Electrolytic capacitors. I was expecting you use some similar setup and maybe you had some strange isolation problem due to non isolated switch driver. But matched inductance and capacitance also make sense and that is most likely you case as 20nF is basically nothing.

If you use large capacitors Large electrolytic or super-capacitors then you can use a DC-DC converter with constant current control and get very close to 0.7 V_i

The probing was also strange as I have no idea why you will connect the oscilloscope ground to the switch. But is fairly clear that you have build an LC resonator so not at all what is shown in that diagram two parallel capacitors and a switch.

But I was told a DC-DC converter is not required, this effect will be observed for superconducting wires?

I have wired the diagram exactly as shown, I'm not sure what is missing?

Tim

[electrodacus]

Sure, but I don't have superconducting capacitors that large.  Are you saying this effect disappears for small values, then?  By what mechanism?  That's quite a peculiar effect if so!

Are you saying those 20nF capacitors are made of superconductor material ? Even the wires ?
Do you also have the means to cool them down to whatever temperature is needed to become superconductors ? You will not be able to use a mosfet switch in that case anyway.

But I was told a DC-DC converter is not required, this effect will be observed for superconducting wires?

I have wired the diagram exactly as shown, I'm not sure what is missing?

Tim

The effect can be observed for superconducting wires (this need to include the switch and the capacitor plates). Unless you have access to some university lab with this sort of equipment I doubt this is an option for you.

You did not wired as shown in diagram as you added significant inductance connecting the super small 20nF capacitors with wires. There is no inductor drawn in the schematic for a good reason.
And yes you can not get rid of inductance or capacitance but when the diagram shows capacitors only you understand that inductance in that circuit will need to be negligibly small and that is not possible when you connect 20nF capacitors with wires.


So use some few mF electrolytic capacitors and low power and efficient DC-DC converter as it is way easier and less expensive to setup than superconductors.
I also want to insist on the fact that I already proved multiple times with the correct equations that what I say is correct.
Main equation is the one for Energy stored in a capacitor = 0.5 * C * V²

initial condition  3V charged capacitor identical 1F capacitors

0.5 * 1 * 3² = 4.5Ws

end result after switch is closed and steady state is reached ideal case or very close with efficient DC-DC

0.5 * 2 * 2.121² = 4.5Ws

end result with resistance in series

0.5 * 2 * 1.5² = 2.25Ws with the other half of the energy lost as heat due to circuit resistance.

[hamster_nz]

So use some few mF electrolytic capacitors and low power and efficient DC-DC converter as it is way easier and less expensive to setup than superconductors.

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?

You are now worrying about wires (even superconducting ones) having inductance. Why worry? A moving charge induces a magnetic field. That is how the underlying physical reality works - a wire with zero self-inductance cannot exist! (well, is infinitely short... so no longer a wire)

Do you agree that charges moving between the capacitors will induce changes in the magnetic field? If so, there is the source of the L for the LC resonator - it's not a flawed component, it is built into nature.

[T3sl4co1l]

Are you saying those 20nF capacitors are made of superconductor material ? Even the wires ?
Do you also have the means to cool them down to whatever temperature is needed to become superconductors ? You will not be able to use a mosfet switch in that case anyway.

FETs work fine at superconducting temperatures, just need the right ones.

The effect can be observed for superconducting wires (this need to include the switch and the capacitor plates). Unless you have access to some university lab with this sort of equipment I doubt this is an option for you.

You did not wired as shown in diagram as you added significant inductance connecting the super small 20nF capacitors with wires. There is no inductor drawn in the schematic for a good reason.

But you specifically requested superconductors; there is no resistance in the circuit.  And it is, well, wired -- I'm not sure where you suggest I go and get noninductive wires from, I mean, that would be preposterous!  ...Wouldn't it?

So I've done the nearest thing to the diagram I can do, and assumed the wire links mean wires, that, well, sometimes have inductance to them.  I can't exactly help it, okay!

And yes you can not get rid of inductance or capacitance but when the diagram shows capacitors only you understand that inductance in that circuit will need to be negligibly small and that is not possible when you connect 20nF capacitors with wires.

Negligibly small in relation to......what?

It's not like there's an RC time constant in there.  You said no resistance, I went to quite extreme lengths to eliminate it!  But then what can be left?!

So use some few mF electrolytic capacitors and low power and efficient DC-DC converter as it is way easier and less expensive to setup than superconductors.
I also want to insist on the fact that I already proved multiple times with the correct equations that what I say is correct.
Main equation is the one for Energy stored in a capacitor = 0.5 * C * V²

Yes yes yes I get the refrain, if I wanted perfect energy balance I could do that.  But you don't understand, this isn't an application -- this is a curiosity.  Surely you have a curiosity about this conundrum as well? -- Else, why stick in this thread so long?

Your claim that a sqrt(2) should pop up, is quite a curious one, and as a scientist, you should be as interested to disprove it, as I am to prove it!  Yes, to disprove ones' own ideas, such a strange notion to some people, but it is the scientific method; there is none easier to fool than the self, as a famous scientist once said.

If you're curious, I reconstructed the experiment with much shorter, non-superconducting wires, and obtained this waveform:



It again does not show a sqrt(2) ratio, but shows the perfect 0.5 as predicted by charge balance.  Energy is of course conserved because the excess is dissipated in the resistance, and no worries about whether the energy flowed around or through the wires, it got where it needed to go all the same.

Tim

[ejeffrey]

And yes you can not get rid of inductance or capacitance but when the diagram shows capacitors only you understand that inductance in that circuit will need to be negligibly small and that is not possible when you connect 20nF capacitors with wires.

I think this is the core of the problem.  You cannot make a circuit with current flowing where both inductance and resistance are negligible. It is not physical to ignore both at the same time, and when you try to you get incorrect answers.

There are two things that can happen: The circuit will have significant resistance and the circuit will reach a steady state at half the initial voltage with a time constant RC.  Or the circuit will not have significant resistance and the circuit will oscillate at it's resonance frequency given by w= 1/sqrt(LC).  Reducing L will increase the resonant frequency but the overall behavior will stay the same.  The first case does not conserve electrical energy because it has resistance.  The second does conserve electrical energy but never reaches a steady state, instead it (ideally) oscillates forever.  Both conserve charge of course, globally and locally (i.e., the net charge on each metal island remains the same). 

I also want to insist on the fact that I already proved multiple times with the correct equations that what I say is correct.
Main equation is the one for Energy stored in a capacitor = 0.5 * C * V²

Of course the correct answer has to conserve both charge and (total) energy, not just one or the other.

[electrodacus]

Is late and I do not want to answer all the silly questions you (all) have.

You have no understanding on the basics so my answers are done considering you at least understand some basic parts.
Like the fact that a 20nF capacitor has a very small capacity and so even a relatively short wire will have an inductance large enough that you form a resonant circuit.
If you use a few mF then even with fairly long wires the inductance is super negligible and it is not a problem.

The mosfet can work at low temperatures but it will not be a superconductor so even if you have superconductor wires and capacitors adding the mosfet that will have a resistance will cancel everything.
I mentioned no resistance in the circuit and superconductors just to inform you that is not just theoretical but also possible in real life to setup such a test setup but for probably very few physics labs in the world not for any of you to attempt. Not to mention none of you even have the knowledge necessary to attempt this.
The DC-DC converter is fairly easy to do but likely not for last few people that replayed to me.

The correct answer is just energy conservation not charge conservation as I already showed.
If any of you have an engineering degree then you should be ashamed for not knowing what energy is.
Do not think I will have any time in the next few day's. I may check once a day and replay if there is a serious question for someone that understand the subject.

[ejeffrey]

Like the fact that a 20nF capacitor has a very small capacity and so even a relatively short wire will have an inductance large enough that you form a resonant circuit.

Any indicance and capacitance will (at least in an ideal situation) form a resonant circuit. 

If you use a few mF then even with fairly long wires the inductance is super negligible and it is not a problem.

Only to the extent that the parasitic L and R of a real world 1 mF capacitor are likely to be larger.  But an ideal 1 mF capacitor plus a 10 nH inductor will resonate at about 50 kHz. A more realistic 100 nH inductance will lower the frequency to 15 kHz.

The DC-DC converter is fairly easy to do but likely not for last few people that replayed to me.

Everybody here agrees that the DC DC converter will do exactly as you have described.  Nobody is trying because there isn't any disagreement on this topic so no point arguing.  Our contention is that simply shorting a capacitor out will do something different and Tim did experiments to demonstrate that the circuit either oscillates or decays to V/2.

[cj]

Removed as it didn't really answer Daves question.

CJ