Veritasium "How Electricity Actually Works"

[hamster_nz]

Black cable is held against the negative of the 9V battery.

I see so you where charging and discharging the capacitors. LED's can glow at a few uA so it will take quite some time for them to dim as capacitors charge and those capacitors may have leakage current in that region.
I just tested with a large 4700uF 35Vdc new capacitor and my small white LED had a small glow even after 30 to 50 seconds (did not count) the same level of glow it has if it conducts through my finger so I suspect a few nA
I never used electrolytic capacitors in any of my projects and did a quick search and it seems they do have some leakage current due to the type of electrolyte they use.  Water based ones have the largest leakage but all have even a few uA much more than nA modern LED's can glow at.
So I learned something new and that is that electrolytics are not just capacitors but also a parallel resistor.
There are good reason I never consider electrolytics for my projects (I do not like things with finite life).


Edit: This is the exact one that I just used https://www.mouser.ca/ProductDetail/United-Chemi-Con/EGPA350ELL472MM40S?qs=beQ1fBGcmj2M%2Fui1vdONPg%3D%3D
Looking at the spec it has 4uA of leakage current so fairly significant and this is not a non brand and purchased from mouser.
Is just there on first page and that 4uA is best case can bemore https://www.mouser.ca/datasheet/2/420/GPALL_e-2509122.pdf
In my particular case should be about 705uA based on use case 8V battery 3V on LED so 5V on the capacitor 5V * 4700uF * 0.03 but mine was no where near close to that probably is worse case and was more like 10uA max based on almost invisible glow.

It is a good glow - I put a meter in line with the 10k resistor - current starts out at well over half a mA, falling off over time (as expected). I just left it to settle down to steady-state, and there is no discernable glow, at least during the day, so it isn't leakage.

Actually... I'll just go measure the leakage then update this post... leakage is less than a microamp.

[electrodacus]

Actually... I'll just go measure the leakage then update this post.

Please do as there is sure a leakage. Those red LED's may not be as efficient as the modern white ones.
The white one I have is visible at 1uA and that is about where it settles with that 4700uF cap after about one minute.
You have two capacitors in series and seems like lower capacity not sure what voltage rating but there will be a leakage with any electrolytic.
In any case they are perfectly fine for testing things like charge one cap with the other directly or with a small DC-DC with constant current control.

[T3sl4co1l]

Yea nah, that's a mA or so.  Clearly not leakage.

It also clearly fades over time, as the capacitor(s) charge; exactly as conventional theory would predict.  How odd...

Tim

[SandyCox]

Circuit theory isn't sufficient to solve the two-capacitor problem. We need to take the electrodynamic behavior of the system into account.

So let's do the thought experiment of connecting a charged and a discharged capacitor in parallel at t=0s. We assume that all the conductors are perfect and that the two capacitors are of the parallel plate type with vacuum between the plates. The two capacitors can be modelled as two lossless transmission lines (see for instance House and Melcher example 14.2.1). After the two capacitors are connected in parallel, an electromagnetic wave will bounce back and forth between the two capacitors. The amount of energy stored in each capacitor will fluctuate, but the total amount of stored energy in the two capacitors will remain the same.

If we go to the next level and take electromagnetic radiation into account, then al the energy will be radiated into space over time.

 

[timenutgoblin]

So let's do the thought experiment of connecting a charged and a discharged capacitor in parallel at t=0s.

May I suggest an alternative thought experiment?

Assume lossless capacitors and conductors. Assume two capacitors of equal capacitance, C. Assume one of the capacitors is discharged. Assume that the other capacitor is charged to the exact value 1.602176634×10⁻¹⁹ Coulombs (the charge on an electron). Assume that the charged capacitor has a voltage of 1V, equivalent to 1eV Joules of energy. When the switch is closed, the electron migrates from the initially-charged capacitor to the initially-discharged capacitor. The electron then migrates back to the initially-charged capacitor. Oscillation occurs. Is my alternative thought experiment valid?

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

I was thinking further with this idea of 70.7% of Vi per capacitor after the switch is closed and noted a disparity regarding the issue of time relating to current flow between the two capacitors and the voltages across the capacitors. If both capacitors are 1F and the charged capacitor is charged to 3V then the initial charge is 3 Coulombs.

If the initially-charged capacitor discharges from 3V to 2.121V then the change in voltage is -29.3% of Vi. If the initially-discharged capacitor charges from 0V to 2.121V then the change in voltage is +70.7% of Vi.

If both capacitors have 70.7% of Vi after the same time period has past then how did the initially-discharged capacitor charge up faster from 0V to 70.7% of Vi than the initially-charged up capacitor discharge from 100% to 70.7% of Vi? This would violate Kirchhoff's Current Law where the current flowing into the node is not equal to the current flowing out of the node. In this case the node is located between the capacitors.

The current flowing out of the initially-charged capacitor must be equal in magnitude to the current flowing into the initially-discharged capacitor. The initially-discharged capacitor would need to charge from 29.3% of Vi to 70.7% of Vi in a time period of zero seconds. This is impossible.

I have seen a similiar problem in a YouTube video.
https://youtu.be/q398AqtTEL8

\$I = C \frac{dV}{dt}\$

\$I.dt = C.dV\$

Time taken for capacitor to charge/dischage to target voltage:
\$dt = C\frac{dV}{I}\$

For reference:

+70.7% \$V_i \approx \frac{V_i}{\sqrt{2}}\$

-29.3% \$V_i \approx \frac{-V_i{(\sqrt{2}} - 1)}{\sqrt{2}}\$

[electrodacus]

Circuit theory isn't sufficient to solve the two-capacitor problem. We need to take the electrodynamic behavior of the system into account.

So let's do the thought experiment of connecting a charged and a discharged capacitor in parallel at t=0s. We assume that all the conductors are perfect and that the two capacitors are of the parallel plate type with vacuum between the plates. The two capacitors can be modelled as two lossless transmission lines (see for instance House and Melcher example 14.2.1). After the two capacitors are connected in parallel, an electromagnetic wave will bounce back and forth between the two capacitors. The amount of energy stored in each capacitor will fluctuate, but the total amount of stored energy in the two capacitors will remain the same.

If we go to the next level and take electromagnetic radiation into account, then al the energy will be radiated into space over time.

The circuit theory is perfectly capable in solving the two parallel capacitor problem.
With ideal conductors so no resistance energy stored in the two capacitors at the end of the experiment will be the same as at the beginning just split between the two capacitors.
With the example I used where capacitors are 1F and charged capacitor starts with 3V so 4.5Ws of stored energy the end result will be that each capacitor will contain 2.25Ws meaning 2.121V across each capacitor.
That is in contrast to normal wires where you have resistance and in that case energy at the end of the experiment in the two capacitors will be just half of the energy at the beginning so just 1.125Ws in each capacitor the other half of the energy 2.25Ws will be lost in the wires as heat because energy travels through wires and wires have resistance.
 

While yes there will be some bounce both in the experiment with resistance and the one with superconductors the speed of electron wave is finite and there will be reflexions at the open ends so voltage will stabilize and not forever slush around as you imagine as reflected waves will interact and in time cancel each other.
You also probably imagine that inductance and capacitance will match but that is not the case in a capacitor where capacitance is much higher than inductance same ways as for an inductor inductance is much higher than capacitance. 

[electrodacus]

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

I was thinking further with this idea of 70.7% of Vi per capacitor after the switch is closed and noted a disparity regarding the issue of time relating to current flow between the two capacitors and the voltages across the capacitors. If both capacitors are 1F and the charged capacitor is charged to 3V then the initial charge is 3 Coulombs.

If the initially-charged capacitor discharges from 3V to 2.121V then the change in voltage is -29.3% of Vi. If the initially-discharged capacitor charges from 0V to 2.121V then the change in voltage is +70.7% of Vi.

If both capacitors have 70.7% of Vi after the same time period has past then how did the initially-discharged capacitor charge up faster from 0V to 70.7% of Vi than the initially-charged up capacitor discharge from 100% to 70.7% of Vi? This would violate Kirchhoff's Current Law where the current flowing into the node is not equal to the current flowing out of the node. In this case the node is located between the capacitors.

The current flowing out of the initially-charged capacitor must be equal in magnitude to the current flowing into the initially-discharged capacitor. The initially-discharged capacitor would need to charge from 29.3% of Vi to 70.7% of Vi in a time period of zero seconds. This is impossible.

I have seen a similiar problem in a YouTube video.
https://youtu.be/q398AqtTEL8

\$I = C \frac{dV}{dt}\$

\$I.dt = C.dV\$

Time taken for capacitor to charge/dischage to target voltage:
\$dt = C\frac{dV}{I}\$

For reference:

+70.7% \$V_i \approx \frac{V_i}{\sqrt{2}}\$

-29.3% \$V_i \approx \frac{-V_i{(\sqrt{2}} - 1)}{\sqrt{2}}\$

You are confusing charge with energy and charge is not energy.
A capacitor is the same as a transmission line and vice versa so there is both inductance and capacitance just a different rates. You can consider each electron and hole pair a capacitor so that first electron that will move from the charged capacitor forms another small capacitor on the other side of you perfect middle point where you have the ideal meter. And while that is fast is not infinitely fast and it is just the beginning as the electron wave needs to travel down to the end of the capacitor.

If you better understand rechargeable batteries then better convert Coulombs to mAh by dividing to 3600 seconds.

Now if you say that the capacitor contains 3C/3600 = 0.833mAh the you will not call this energy instead you will need to multiply this with the average voltage or integrate if voltage will not drop linear like on a capacitor.
So in this case 3V/2 = 1.5V * 0.833mAh = 1.25mWh and this is the energy contained in the capacitor.
1.25mWh * 3600 = 4.5Ws

[IanB]

Circuit theory isn't sufficient to solve the two-capacitor problem. We need to take the electrodynamic behavior of the system into account.

Actually, we don't. It is only necessary to look at the starting and ending states and apply basic principles of physics (no circuit theory required).

At the start we have one charged capacitor with a charge of \$Q = C_a V_0\$ and a second capacitor \$C_b\$ with a charge of zero. The system is fully insulated and no charge can leak, meaning we can apply conservation of charge between any two states.

After we connect the two capacitors in parallel the charge is distributed between them until in the end both have equal voltage. At this point we have \$Q = C_a V + C_b V\$

If both capacitors are identical (\$C_a = C_b\$) we can write \$CV_0 = 2CV\$ and the \$C\$ cancels giving
$$V = \frac{V_0}{2}$$
Thus the final voltage is half the initial voltage.

It is not necessary to consider what path is taken between the initial state and the final state if it is given that any path followed will satisfy the conservation of charge.

[electrodacus]

Circuit theory isn't sufficient to solve the two-capacitor problem. We need to take the electrodynamic behavior of the system into account.

Actually, we don't. It is only necessary to look at the starting and ending states and apply basic principles of physics (no circuit theory required).

At the start we have one charged capacitor with a charge of \$Q = C_a V_0\$ and a second capacitor \$C_b\$ with a charge of zero. The system is fully insulated and no charge can leak, meaning we can apply conservation of charge between any two states.

After we connect the two capacitors in parallel the charge is distributed between them until in the end both have equal voltage. At this point we have \$Q = C_a V + C_b V\$

If both capacitors are identical (\$C_a = C_b\$) we can write \$CV_0 = 2CV\$ and the \$C\$ cancels giving
$$V = \frac{V_0}{2}$$
Thus the final voltage is half the initial voltage.

It is not necessary to consider what path is taken between the initial state and the final state if it is given that any path followed will satisfy the conservation of charge.

You also seems not to understand conservation of energy.
Charge is not conserved energy is.
This is just a special case where you selected two identical capacitors and you have resistance that is why you get exactly half the voltage at the end but that is just a quarter of initial energy in each capacitor so just half the energy you started with. The other half of the energy ended up as heat.
If you add an inductor as an intermediary energy storage you can move the energy from one capacitor to another at much higher efficiency than 50% maybe as high as 90% then the final voltage on those two capacitors will be much closer to 0.707 of the charged capacitor and so much less energy will be wasted as heat.
 

[SandyCox]

When applying circuit theory to the problem, either the law of conservation of charge or the law of conservation of energy is violated.

[electrodacus]

When applying circuit theory to the problem, either the law of conservation of charge or the law of conservation of energy is violated.

There is no such thing as conservation of charge. The conservation of energy is a law.

[SandyCox]

When applying circuit theory to the problem, either the law of conservation of charge or the law of conservation of energy is violated.

There is no such thing as conservation of charge. The conservation of energy is a law.

Look at section 1.5 of Haus and Melcher.

[IanB]

When applying circuit theory to the problem, either the law of conservation of charge or the law of conservation of energy is violated.

If the two capacitors form a closed system then charge is conserved within that system. The model then indicates that electrical energy is not conserved, but this says nothing about total energy. Energy could be converted to other forms such as heat, or it could cross the system boundaries and leave. Neither of these possibilities violate causality.

If you look at the situation from a physics perspective, the connecting together of the two capacitors is an irreversible process, which therefore causes an increase in entropy. This increase in entropy translates to a decrease in the free energy of the system, which means you can get less useful work out of the two capacitors after you have joined them than you could get out of the one capacitor at the start.

[electrodacus]

Look at section 1.5 of Haus and Melcher.

Never heard of them before now and likely for a good reason.

People seems to confuse charge with energy.
Energy is the one that can not be created or destroyed just converted from one form to another.

[SandyCox]

When applying circuit theory to the problem, either the law of conservation of charge or the law of conservation of energy is violated.

If the two capacitors form a closed system then charge is conserved within that system. The model then indicates that electrical energy is not conserved, but this says nothing about total energy. Energy could be converted to other forms such as heat, or it could cross the system boundaries and leave. Neither of these possibilities violate causality.

If you look at the situation from a physics perspective, the connecting together of the two capacitors is an irreversible process, which therefore causes an increase in entropy. This increase in entropy translates to a decrease in the free energy of the system, which means you can get less useful work out of the two capacitors after you have joined them than you could get out of the one capacitor at the start.

That's very true. The problem is that circuit theory, on its own, cannot explain how this energy is converted from one form to another. It is problems like these that indicate that we need a deeper understanding, i.e. electrodynamics.

[SandyCox]

Look at section 1.5 of Haus and Melcher.

Never heard of them before now and likely for a good reason.

People seems to confuse charge with energy.
Energy is the one that can not be created or destroyed just converted from one form to another.

https://web.mit.edu/6.013_book/www/

One of the best books on Electromagnetics.

[electrodacus]

https://web.mit.edu/6.013_book/www/

One of the best books on Electromagnetics.

Is either that you misinterpreted what you read in the book or the book is just not as good as you say.
For a single capacitor that say it is charged so there is an imbalance of electrons on the two plates. What do you think it will happen with charge if you change the size of the plates?

I guess you realized that Q=C*V and E_c=0.5 * C * V²  are both true and nobody ever disproved the conservation of energy so it seems clear that charge will not be conserved in this example.

Q = 3C at the beginning and for case where there is no resistance so no energy lost as heat end will be Q = 4.242C
It just happens that due to the perfect symmetry of the two identical capacitors you end up with half of the energy lost as heat and so charge is conserved but this is just thanks to sysmentry you can call this a coincidence.
Try to calculate for two capacitors that are not identical and see then what you will get even with resistance and heat loss. You will be surprised. 


I guess those two are professors at MIT and same as the other professor at University of California Alex have a poor understanding of energy storage.
Alex kusenko knew that energy conservation can not be broken yet he lost due to his ability to understand energy storage.
He knew Derek was wrong he was just unable to understand himself how vehicle works thus he lost the bet.
I try to explain this first as it is simpler to explain than the "faster than wind direct down wind vehicle" but there the exact same issue exist and that is understanding what energy is.

[T3sl4co1l]

While yes there will be some bounce both in the experiment with resistance and the one with superconductors the speed of electron wave is finite and there will be reflexions at the open ends so voltage will stabilize and not forever slush around as you imagine as reflected waves will interact and in time cancel each other.

Consequent does not match the antecedent.

Think: if you didn't know a damn thing about waves, you just heard someone say, "this thing is reflective therefore energy is absorbed", would you be not the least bit concerned?

The definition of something being reflective, is that it does not absorb energy, at least not predominantly so.

Indeed, some very excellent mirrors can be constructed, whether using superconductors (at radio frequencies), or stacks of dielectrics (at optical frequencies).

If the mirrors are indeed perfect, where pray tell is the energy going to go?

Indeed it is exactly the reflection which causes it to oscillate.  If the energy could be absorbed, it would, but there is insufficient resistance in the circuit to absorb it, at least before it's reflected around a few times.  There is a direct equivalence between a lumped-equivalent circuit (like the present CLC network) and a transmission line construction, and further to a full-fields (mirrors and light beams) model, assuming of course the geometry keeps the waves confined.  So it is perfectly consistent to say that energy is reflecting in this network, and that it is precisely that reflection, at each end, that causes the energy to "slosh" back and forth.

Tim

[T3sl4co1l]

Is either that you misinterpreted what you read in the book or the book is just not as good as you say.
For a single capacitor that say it is charged so there is an imbalance of electrons on the two plates. What do you think it will happen with charge if you change the size of the plates?

I guess you realized that Q=C*V and E_c=0.5 * C * V²  are both true and nobody ever disproved the conservation of energy so it seems clear that charge will not be conserved in this example.

Pray tell, how much force is required to "change the size of the plates"?

Also, what meaning does "change size" have?  Surely you aren't changing just one, that accomplishes [almost] nothing.

But I would be inclined to understand this as "separation of plates", which is perfectly meaningful, a standard experiment.

And indeed, if there is a force, and changing the separation of the plates implies a change in distance... then does that not also imply.......?

I guess those two are professors at MIT and same as the other professor at University of California Alex have a poor understanding of energy storage.

Tim

[electrodacus]

Consequent does not match the antecedent.

Think: if you didn't know a damn thing about waves, you just heard someone say, "this thing is reflective therefore energy is absorbed", would you be not the least bit concerned?

The definition of something being reflective, is that it does not absorb energy, at least not predominantly so.

Indeed, some very excellent mirrors can be constructed, whether using superconductors (at radio frequencies), or stacks of dielectrics (at optical frequencies).

If the mirrors are indeed perfect, where pray tell is the energy going to go?

Indeed it is exactly the reflection which causes it to oscillate.  If the energy could be absorbed, it would, but there is insufficient resistance in the circuit to absorb it, at least before it's reflected around a few times.  There is a direct equivalence between a lumped-equivalent circuit (like the present CLC network) and a transmission line construction, and further to a full-fields (mirrors and light beams) model, assuming of course the geometry keeps the waves confined.  So it is perfectly consistent to say that energy is reflecting in this network, and that it is precisely that reflection, at each end, that causes the energy to "slosh" back and forth.

Tim

There is not just a single wave that moves from one side to the other. and they interact with each other.
Not sure why we discus so much the ideal superconductor when you can get very similar result by just adding an inductor as intermediary energy storage to move the energy from one capacitor to another. 

[electrodacus]

Pray tell, how much force is required to "change the size of the plates"?

Also, what meaning does "change size" have?  Surely you aren't changing just one, that accomplishes [almost] nothing.

But I would be inclined to understand this as "separation of plates", which is perfectly meaningful, a standard experiment.

And indeed, if there is a force, and changing the separation of the plates implies a change in distance... then does that not also imply.......?

No energy is needed if there is no friction. Variable capacitors are build this way but they do not care about friction. In any case energy from the capacitor will always be lost as heat due to plate resistance when you move the plates in any direction.
That force * speed * time is energy and it will have nothing to do with electrical energy stored in a capacitor.
There is an attraction force between the two plates of the capacitor so it takes energy to move the plates apart but that will not translate in any increase in electrical energy stored in the capacitor.   

[electrodacus]

I guess those two are professors at MIT and same as the other professor at University of California Alex have a poor understanding of energy storage.

Tim

I will need to correct this as it is unfair since I have no read the book written by them. In case of Alex I have seen his inability to understand energy by being unable to explain how that vehicle works and win the bet with Derek.

[T3sl4co1l]

TIL this machine doesn't exist!

[electrodacus]

TIL this machine doesn't exist!

That is not what you think. It is an electrostatic energy generator. There are also piezoelectric capacitors that can convert mechanical energy to electrical energy.

[PlainName]

I guess those two are professors at MIT and same as the other professor at University of California Alex have a poor understanding of energy storage.

Tim

I will need to correct this as it is unfair since I have no read the book written by them. In case of Alex I have seen his inability to understand energy by being unable to explain how that vehicle works and win the bet with Derek.

As you asked someone else a few messages back, "I need to ask what is your qualification." Seems that literally everyone is wrong except for you, so what is your qualification for having the definitive knowledge about energy?