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Veritasium "How Electricity Actually Works"

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electrodacus:

--- Quote from: SandyCox on May 08, 2022, 09:49:28 am ---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.

--- End quote ---

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:

--- Quote from: timenutgoblin on May 08, 2022, 01:58:29 pm ---
--- Quote from: electrodacus on May 03, 2022, 12:40:07 am ---If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * Vi
--- End quote ---

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}}\$

--- End quote ---

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:

--- Quote from: SandyCox on May 08, 2022, 09:49:28 am ---Circuit theory isn't sufficient to solve the two-capacitor problem. We need to take the electrodynamic behavior of the system into account.
--- End quote ---

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:

--- Quote from: IanB on May 08, 2022, 05:14:22 pm ---
--- Quote from: SandyCox on May 08, 2022, 09:49:28 am ---Circuit theory isn't sufficient to solve the two-capacitor problem. We need to take the electrodynamic behavior of the system into account.
--- End quote ---

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.

--- End quote ---

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.

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