Author Topic: Veritasium "How Electricity Actually Works"  (Read 64974 times)

0 Members and 1 Guest are viewing this topic.

Offline electrodacus

  • Super Contributor
  • ***
  • Posts: 1858
  • Country: ca
    • electrodacus
Re: Veritasium "How Electricity Actually Works"
« Reply #775 on: May 21, 2022, 08:09:03 pm »
Quote
Charging a capacitor means storing energy in to it.
No energy went around the capacitor but in to capacitor.

So what caused the resistor to heat up? If it had been an LED it would emit light. If it were a motor it would turn. You cannot charge that capacitor without those things happening and yet, according to you, there was no energy available to them. They would be perpetual motion machines running on nothing!

I do not get where did I say anything that looked like a perpetual motion machine ? It is a supper inefficient circuit only 50% efficient.
Supply needs to deliver 144mJ just to charge the capacitor with 72mJ so the other 50% 72mJ are lost on the wires or if you add a lamp or motor they do some work.

Just wires and capacitor 144mJ = 72mJ + 72mJ
Adding a 9Ohm incandescent lamp 144mJ = 64.8mJ + 7.2mJ + 72mJ

Where do you see a perpetual motion machine ?  144mJ are provided and some of that is lost as heat the rest is doing some work and half of it in this case remains stored in the capacitor.
I need to know where do you see the perpetuum mobile ?

Offline snarkysparky

  • Frequent Contributor
  • **
  • Posts: 414
  • Country: us
Re: Veritasium "How Electricity Actually Works"
« Reply #776 on: May 21, 2022, 08:18:50 pm »
Lets do a simple question.

the circuit here

https://en.wikipedia.org/wiki/Capacitive_power_supply

If a DC voltage ( 20 V) is switched into X1 and X1 and left connected indefinitely will there be any current through R4 and R5 temporarily?

Simple yes /no question,

 

Offline electrodacus

  • Super Contributor
  • ***
  • Posts: 1858
  • Country: ca
    • electrodacus
Re: Veritasium "How Electricity Actually Works"
« Reply #777 on: May 21, 2022, 08:31:48 pm »
Lets do a simple question.

the circuit here

https://en.wikipedia.org/wiki/Capacitive_power_supply

If a DC voltage ( 20 V) is switched into X1 and X1 and left connected indefinitely will there be any current through R4 and R5 temporarily?

Simple yes /no question,

Yes there will be a short current spike trough R4 and R5 as the C1 is charged but current will be very small maybe almost impossible to measure because there is a much larger capacitor C2 that will charge while being discharged across R3,R4 and R5
And there is that R2 that will continue to let some small current pass even after this initial spike.
If you remove R2 then it will be just that initial spike.

Why use that much more complex example when you have the super simple capacitor + wires + voltage source that we are discussing ?

If energy was to go through an ideal capacitor instead of in to it when charging then voltage across the capacitor will remain zero.  We know voltage across capacitor increases while it is being charged as all energy goes into the capacitor and remains there as stored energy.

Offline snarkysparky

  • Frequent Contributor
  • **
  • Posts: 414
  • Country: us
Re: Veritasium "How Electricity Actually Works"
« Reply #778 on: May 21, 2022, 08:42:18 pm »
Well I used it cause the simple example seems to be mixed up with a lot of phoohey.

So we know ...  and you agreed ..  that some energy made it through the circuit to R4 and R5 with the application of a dc source to the circuit input.

Do you dispute this ?

This energy seems to have "gotten by" the capacitor.  It is a series element in the circuit. 

So what do you say about this energy.   We know that if an AC source is connected the circuit functions as a power supply allowing energy to be extracted from X3 and X4.   A suddenly applied DC signal has some AC in it. 

If you still say that NO energy is transferred by the capacitor where did the energy for R4 and R5 come from?
Where does it come from if AC is applied to X3 and X4.




 

Offline electrodacus

  • Super Contributor
  • ***
  • Posts: 1858
  • Country: ca
    • electrodacus
Re: Veritasium "How Electricity Actually Works"
« Reply #779 on: May 21, 2022, 09:04:33 pm »
Well I used it cause the simple example seems to be mixed up with a lot of phoohey.

So we know ...  and you agreed ..  that some energy made it through the circuit to R4 and R5 with the application of a dc source to the circuit input.

Do you dispute this ?

This energy seems to have "gotten by" the capacitor.  It is a series element in the circuit. 

So what do you say about this energy.   We know that if an AC source is connected the circuit functions as a power supply allowing energy to be extracted from X3 and X4.   A suddenly applied DC signal has some AC in it. 

If you still say that NO energy is transferred by the capacitor where did the energy for R4 and R5 come from?
Where does it come from if AC is applied to X3 and X4.

Those resistors are part of the return for the capacitor that is being charged.
You can not charge that capacitor by just connecting one of the terminals from a battery or ideal voltage supply.
You can not have just electrons leaving the battery without same number returning and for that to happen you need a circuit.

How come that after capacitor is charged no more current is flowing ?  Why no energy can go "through" the capacitor after the first few ms ?
The answer should be obvious and that is energy went in to capacitor not trough.
All evidence points to this correct conclusion but for some reason you chose to ignore all evidence.
Once you have 72mJ stored in the 1000uF capacitor there is no more current flow and if you take the capacitor out of the circuit you can still measure 12V across and can calculate that 72mJ are contained as stored energy and you can also discharge that energy and measure that is all there all 72mJ. 

Offline PlainName

  • Super Contributor
  • ***
  • Posts: 6796
  • Country: va
Re: Veritasium "How Electricity Actually Works"
« Reply #780 on: May 21, 2022, 09:14:48 pm »
Quote
How come that after capacitor is charged no more current is flowing ?

The cap is fully charged so cannot be charged more.

Quote
The answer should be obvious and that is energy went in to capacitor not trough.

The CHARGE went into the capacitor. The energy went around the circuit to allow charging. Once the capacitor is charged there can be no more circulating power, so no more energy. But if there is charge going INTO the capacitor then there is power going AROUND the circuit. If that wasn't so you wouldn't NEED a circuit!
 

Offline snarkysparky

  • Frequent Contributor
  • **
  • Posts: 414
  • Country: us
Re: Veritasium "How Electricity Actually Works"
« Reply #781 on: May 21, 2022, 09:24:50 pm »
What difference does it make if the resistors are part of the return circuit of the capacitor.  Nobody would argue that.

What we know is that being in that return circuit for the "Capacitor"  they get to absorb some of the charging energy.

This charging energy is a result of electrons going into one terminal of the capacitor and out the other.

So we have electrons entering and leaving the terminals of the capacitor that go on to impart some energy in other things in the "return path"

And if we keep switching the polarity of the input we can keep getting power out in the return path.

This power will not flow without a route into and out of the capacitor terminals.   the capacitor in integral to the transfer of power to the load.   All this is very simple.

 

Offline electrodacus

  • Super Contributor
  • ***
  • Posts: 1858
  • Country: ca
    • electrodacus
Re: Veritasium "How Electricity Actually Works"
« Reply #782 on: May 21, 2022, 09:26:28 pm »
The cap is fully charged so cannot be charged more.


The CHARGE went into the capacitor. The energy went around the circuit to allow charging. Once the capacitor is charged there can be no more circulating power, so no more energy. But if there is charge going INTO the capacitor then there is power going AROUND the circuit. If that wasn't so you wouldn't NEED a circuit!

Yes capacitor is charged and can not charge more.

Energy and more specifically electrical energy went into capacitor not around it. If it will have been going around it then less energy will have been needed from the supply.
144mJ worth of energy left the supply.
72mJ where dissipated as heat in the wires
72mJ where stored in the capacitor.

For a capacitor:
Charge is  C * V
Energy is  C * V * V * 0.5
Can you understand the relation and also the difference ?

Offline electrodacus

  • Super Contributor
  • ***
  • Posts: 1858
  • Country: ca
    • electrodacus
Re: Veritasium "How Electricity Actually Works"
« Reply #783 on: May 21, 2022, 09:40:29 pm »
This charging energy is a result of electrons going into one terminal of the capacitor and out the other.

So we have electrons entering and leaving the terminals of the capacitor that go on to impart some energy in other things in the "return path"

And if we keep switching the polarity of the input we can keep getting power out in the return path.

This power will not flow without a route into and out of the capacitor terminals.   the capacitor in integral to the transfer of power to the load.   All this is very simple.

The electrons going in to one plate will never get on the other side. Those electrons that exit the other side where already there in that plate.
This excess of electrons on one plate and deficit on the other plate is the stored energy.


After 144mJ left the source and you get to steady state. You remove the terminal from the supply and connect them with reverse polarity.
What will happen then ? How much energy will be dissipated on the 1Ohm wires ?
Initially it was 72mJ but how much will it be when you change the polarity ?

Offline PlainName

  • Super Contributor
  • ***
  • Posts: 6796
  • Country: va
Re: Veritasium "How Electricity Actually Works"
« Reply #784 on: May 21, 2022, 09:55:34 pm »
Quote
Can you understand the relation and also the difference ?

Can you not understand that to charge the capacitor you need a completed circuit for SOMETHING to go around. If you leave off the return path then nothing will happen. Therefore something must be going down that return path.

Are you saying that this is wrong? That you don't need a completed circuit for the capacitor to charge?
 

Offline electrodacus

  • Super Contributor
  • ***
  • Posts: 1858
  • Country: ca
    • electrodacus
Re: Veritasium "How Electricity Actually Works"
« Reply #785 on: May 21, 2022, 09:59:20 pm »
Quote
Can you understand the relation and also the difference ?

Can you not understand that to charge the capacitor you need a completed circuit for SOMETHING to go around. If you leave off the return path then nothing will happen. Therefore something must be going down that return path.

Are you saying that this is wrong? That you don't need a completed circuit for the capacitor to charge?

Yes that is right. There is no complete circuit that is why energy stops flowing once the capacitor is fully charged.
The circuit is one source supplying a energy storage with limited capacity so current will flow from the source through the wires in the energy storage device (capacitor). No energy flows through the capacitor at any point in time.   

Offline PlainName

  • Super Contributor
  • ***
  • Posts: 6796
  • Country: va
Re: Veritasium "How Electricity Actually Works"
« Reply #786 on: May 21, 2022, 10:19:28 pm »
Quote
so current will flow from the source through the wires in the energy storage device (capacitor)

And the same current will flow back to the source from the other side on your energy storage device. If not, explain WTF is going through that resistor, and why you need a return path.
 

Offline electrodacus

  • Super Contributor
  • ***
  • Posts: 1858
  • Country: ca
    • electrodacus
Re: Veritasium "How Electricity Actually Works"
« Reply #787 on: May 21, 2022, 10:33:04 pm »
Quote
so current will flow from the source through the wires in the energy storage device (capacitor)

And the same current will flow back to the source from the other side on your energy storage device. If not, explain WTF is going through that resistor, and why you need a return path.

See the problem I asked snarkysparky to solve in post 784. Result form that will blow your minds.

I asked to reverse the polarity of the voltage source after the capacitor has charged so circuit is in steady state.

I can provide the answer but likely you won't believe if you do not do that yourself.

Offline PlainName

  • Super Contributor
  • ***
  • Posts: 6796
  • Country: va
Re: Veritasium "How Electricity Actually Works"
« Reply #788 on: May 21, 2022, 10:42:14 pm »
Quote
so current will flow from the source through the wires in the energy storage device (capacitor)

And the same current will flow back to the source from the other side on your energy storage device. If not, explain WTF is going through that resistor, and why you need a return path.

See the problem I asked snarkysparky to solve in post 784. Result form that will blow your minds.

I asked to reverse the polarity of the voltage source after the capacitor has charged so circuit is in steady state.

I can provide the answer but likely you won't believe if you do not do that yourself.

I am wrong there - got carried away. But there is A current going through that resistor isn't there? Go on, you have to admit to that. Let me remind you with the simulation:

 

Offline electrodacus

  • Super Contributor
  • ***
  • Posts: 1858
  • Country: ca
    • electrodacus
Re: Veritasium "How Electricity Actually Works"
« Reply #789 on: May 21, 2022, 10:47:53 pm »
I am wrong there - got carried away. But there is A current going through that resistor isn't there? Go on, you have to admit to that. Let me remind you with the simulation:

Of course there will be a current through the resistor as you are charging the capacitor.  Energy needs to travel through wire and the resistor is a wire.
But the new question is what happens if after you charged the capacitor so no current but capacitor contains those 72mJ and you reverse the polarity on the source.
How much energy will be delivered by the source and how much of that will end up in the resistor ?
I'm fairly certain you will be surprised by the result as it will be very different from initial result because the capacitor is already charged.

The problem data is a 12V ideal voltage source 0.5Ohm resistance for each wire so 1Ohm in total and 1000uF capacitor.
Initial connection when capacitor was discharged is 144mJ supplied by the ideal voltage source = 72mJ dissipated on wires as heat  and 72mJ stored in the 1000uF capacitor.
Then after this is done just remove the supply and connect it again with reverse polarity.
« Last Edit: May 21, 2022, 10:51:14 pm by electrodacus »
 

Offline PlainName

  • Super Contributor
  • ***
  • Posts: 6796
  • Country: va
Re: Veritasium "How Electricity Actually Works"
« Reply #790 on: May 21, 2022, 11:50:09 pm »
Quote
But the new question is what happens if after you charged the capacitor so no current but capacitor contains those 72mJ and you reverse the polarity on the source.

No, you're jumping ahead and going off on a diversion again. We have yet to determine exactly what is going through that resistor since you insist it can't be energy.
 

Offline electrodacus

  • Super Contributor
  • ***
  • Posts: 1858
  • Country: ca
    • electrodacus
Re: Veritasium "How Electricity Actually Works"
« Reply #791 on: May 21, 2022, 11:59:48 pm »
No, you're jumping ahead and going off on a diversion again. We have yet to determine exactly what is going through that resistor since you insist it can't be energy.

Of course energy goes through resistor. When have I ever mentioned that is not ?
It is not going through capacitor but in to capacitor.

The entire point I want to make is that energy travels through wires and wires and resistors are the same thing.

If you reverse the polarity after capacitor was charged you will see that all energy now delivered by the supply ends up as heat on the resistor. All of it not just half as at initial connection when capacitor was discharged.

Offline hamster_nz

  • Super Contributor
  • ***
  • Posts: 2803
  • Country: nz
Re: Veritasium "How Electricity Actually Works"
« Reply #792 on: May 22, 2022, 12:57:58 am »
Are you really arguing that work is not required to store energy in capacitor (or in this case a cars battery)?

Are we in Humpty Dumpty land again, where words only mean what you say the do?

If transfer could be done at 100% efficiency then yes no work will be done to transfer energy from source to capacitor.
In the particular example efficiency was just 50% so half of the energy ended as heat during transportation.
You can use a DC-DC converter and transport the energy with about 90% efficiency so that only 10% is lost for moving the charge from a source to a capacitor (energy storage device).

What part exactly is not clear.
It is clear that you are completely wrong.

When there is 0V across a 1F capacitor it takes a very tiny bit of work to remove a charge from the - plate, and add on to the + one.  It doesn't matter if it is the same electron or not - it's a little bit of work to pull the electron out of one plate, leaving it slightly positively charged, and a little bit of work to push it into the other plate, which was quite happily neutral and now won't  be.

But for the next time there is 1/(6.24 x 10^18) of a V between the plates, so that requires more work. If you do the math it will be 4x that of the first electron.

But for the next time there is 2/(6.24 x 10^18) of a V between the plates, so that requires even more work than the previous one. Once again, if you do the math it will be 9x that of the first electron.

For the (6.24 x 10^18)th electron there is now (6.24 x 10^18)/(6.24 x 10^18) = 1V between the plates. It is now taking a whopping  (6.24 x 10^18)^2 times the energy it took to move the first electron.

In fact if you sum up all the work done on each charge it will equal 0.5 * C * Vfinal^2.

And all work being performed by your source capacitor.

The inductor in the DC-DC configuration when the switch is closed it is taking lower current at the higher voltage to make establish a magnetic field, and then when the switch opens it the collapsing magnetic field moves higher current at a lower voltage, at just slightly higher than the voltage of the uncharged capacitor.
Gaze not into the abyss, lest you become recognized as an abyss domain expert, and they expect you keep gazing into the damn thing.
 

Offline electrodacus

  • Super Contributor
  • ***
  • Posts: 1858
  • Country: ca
    • electrodacus
Re: Veritasium "How Electricity Actually Works"
« Reply #793 on: May 22, 2022, 01:12:47 am »
It is clear that you are completely wrong.

When there is 0V across a 1F capacitor it takes a very tiny bit of work to remove a charge from the - plate, and add on to the + one.  It doesn't matter if it is the same electron or not - it's a little bit of work to pull the electron out of one plate, leaving it slightly positively charged, and a little bit of work to push it into the other plate, which was quite happily neutral and now won't  be.

But for the next time there is 1/(6.24 x 10^18) of a V between the plates, so that requires more work. If you do the math it will be 4x that of the first electron.

But for the next time there is 2/(6.24 x 10^18) of a V between the plates, so that requires even more work than the previous one. Once again, if you do the math it will be 9x that of the first electron.

For the (6.24 x 10^18)th electron there is now (6.24 x 10^18)/(6.24 x 10^18) = 1V between the plates. It is now taking a whopping  (6.24 x 10^18)^2 times the energy it took to move the first electron.

In fact if you sum up all the work done on each charge it will equal 0.5 * C * Vfinal^2.

And all work being performed by your source capacitor.

The inductor in the DC-DC configuration when the switch is closed it is taking lower current at the higher voltage to make establish a magnetic field, and then when the switch opens it the collapsing magnetic field moves higher current at a lower voltage, at just slightly higher than the voltage of the uncharged capacitor.

You forget one super important thing.
That is reversible so energy storage and not work.
The work will be done when you discharge that capacitor.
So all that energy you put in the capacitor 0.5 * C * V2 is still there in the capacitor ready to do work.

In my most recent example with 12V ideal voltage source 1Ohm total circuit resistance 0.5Ohm for each wire and the 1000uF capacitor.
Energy stored in capacitor was 0.5 * 0.001F * 12V2 = 72mWs
But the ideal voltage source needed to supply 2x as much 144mWs in order to cover the losses in transportation of that energy that happens to also be 72mWs drop on that 1Ohm resistance as energy traveled through wires to get from voltage source to capacitor.

Now try to calculate what happens if you after the capacitor is charged you disconnect the voltage source and connect it back but with reverse polarity.
What you will notice is that now all the energy delivered by the voltage source ends up as heat in the wire (all of it not just half like when capacitor was discharged).

Offline hamster_nz

  • Super Contributor
  • ***
  • Posts: 2803
  • Country: nz
Re: Veritasium "How Electricity Actually Works"
« Reply #794 on: May 22, 2022, 04:58:24 am »
It is clear that you are completely wrong.

When there is 0V across a 1F capacitor it takes a very tiny bit of work to remove a charge from the - plate, and add on to the + one.  It doesn't matter if it is the same electron or not - it's a little bit of work to pull the electron out of one plate, leaving it slightly positively charged, and a little bit of work to push it into the other plate, which was quite happily neutral and now won't  be.

But for the next time there is 1/(6.24 x 10^18) of a V between the plates, so that requires more work. If you do the math it will be 4x that of the first electron.

But for the next time there is 2/(6.24 x 10^18) of a V between the plates, so that requires even more work than the previous one. Once again, if you do the math it will be 9x that of the first electron.

For the (6.24 x 10^18)th electron there is now (6.24 x 10^18)/(6.24 x 10^18) = 1V between the plates. It is now taking a whopping  (6.24 x 10^18)^2 times the energy it took to move the first electron.

In fact if you sum up all the work done on each charge it will equal 0.5 * C * Vfinal^2.

And all work being performed by your source capacitor.

The inductor in the DC-DC configuration when the switch is closed it is taking lower current at the higher voltage to make establish a magnetic field, and then when the switch opens it the collapsing magnetic field moves higher current at a lower voltage, at just slightly higher than the voltage of the uncharged capacitor.

You forget one super important thing.
That is reversible so energy storage and not work.
The work will be done when you discharge that capacitor.
:-// Oh come on, you know the formula: work = force x distance, 

If you or I push down on a lever, lifting a weight, it is still work, even though the weight on the far end can move the lever back to the original position, undoing our efforts.
Gaze not into the abyss, lest you become recognized as an abyss domain expert, and they expect you keep gazing into the damn thing.
 

Offline electrodacus

  • Super Contributor
  • ***
  • Posts: 1858
  • Country: ca
    • electrodacus
Re: Veritasium "How Electricity Actually Works"
« Reply #795 on: May 22, 2022, 05:26:59 am »
:-// Oh come on, you know the formula: work = force x distance, 

If you or I push down on a lever, lifting a weight, it is still work, even though the weight on the far end can move the lever back to the original position, undoing our efforts.

You will be storing potential energy.
So say you put in 144mWs by dropping a weight from some height and it is connected through a pulley to another weight that is lifted (this will be the input).
Say lifting that other weight gained you 72mWs in potential gravitational energy and 72mWs were wasted as heat due to lifting mechanism friction.
Now all you have lost is the 72mWs due to friction the other 72mWs are still available in the same form as the original so you can say you moved potential gravitational energy from one weight to another not very efficiency just 50% efficiency.

I guess the above is a decently accurate analogy to what happens with the capacitor circuit.
Best case ideal case when you have no friction you are able to lift the same weight as the one providing the input to the exact same height.
So will you say that you did some work or did you just transferred the potential kinetic energy from one weight to another ?

Offline hamster_nz

  • Super Contributor
  • ***
  • Posts: 2803
  • Country: nz
Re: Veritasium "How Electricity Actually Works"
« Reply #796 on: May 22, 2022, 09:19:16 am »
Here is the mechanical analogy to the two capacitor problem that actually fits.

Springs can store energy, Stored energy is proportional to the 'stretch' squared - just like a how capacitors store energy. In this analogy, the spring is our capacitor.

Flywheels can store kinetic energy. They also have very good bearings that cause minimal energy loss. This is our deliberately introduced inductance.

Ropes can transfer the forces (admittedly only when it tension, but we will make sure they stay in tension). Here I am making no statement if these are the wires, or the electric field. These are just a mechanical way to transfer the forces.  When the rope is moving it is assumed to have minimal momentum. Consider the momentum of the rope as self-inductance - if the flywheel is very light enough (or things are arranged so no flywheel is needed), this might becomes important in in how the system behaves. But with any sensible flywheel it becomes unimportant.

The flywheel has a clamp on it. When the clamp is on the rope can't move, so no force can be transferred through a clamped flywheel. This is our switch.

The setup
See the lefthand side of the attached image.

We have two springs bolted to the ground, and a very heavy flywheel attached to the roof. We tie a rope to the lefthand spring, wrap it around the flywheel a couple of times so it can't slip, and tie it to the righthand spring, so it is nice and snug, but with minimal stretch on either spring. We check that the springs are both at equal height, and then clamp the flywheel in place.

This is our "zero energy in either capacitor" state.

Charging
With the flywheel clamped, we shorten the righthand rope till we get one unit of stretch in it. We tie the rope off, and trim it off nice and neat.

All of the energy is now in the righthand spring and the apparatus is armed and the experiment can proceed.

The experiment.

We release the clamp. The extra tension of the right hand side causes the flywheel to spin up in the clockwise direction, and the left-hand spring start to stretch.

What I expect to see, in an 'idea' world
When the springs get to equal length, in a perfect world half the energy in the flywheel, so it keeps the rope moving, slowly spinning down, finally coming to a stop when the LH spring now has ALL the energy in it. The cycle reverses, and continues on forever - i.e. the flywheel start turning counterclockwise, and ALL the energy is transferred back to the RH spring.

What I expect to see, in the real world
Energy gets lost due to friction and other effects (like rope slippage and stretch). Regardless of the weight and quality of the flywheel, springs and rope the system will settle down with only half the initial energy stored, shared equally between both springs. The only thing that is different is how long it takes to settle down.

That is it. there is no more to the "Two capacitor problem" than that.
« Last Edit: May 22, 2022, 09:24:23 am by hamster_nz »
Gaze not into the abyss, lest you become recognized as an abyss domain expert, and they expect you keep gazing into the damn thing.
 

Offline snarkysparky

  • Frequent Contributor
  • **
  • Posts: 414
  • Country: us
Re: Veritasium "How Electricity Actually Works"
« Reply #797 on: May 22, 2022, 12:36:41 pm »
Electrodacus has agreed that there is power available to do work in the return circuit. And that this current flowed in and out of  the capacitor terminals.   As to whether this power went "through" the capacitor i guess we will never agree. 

Am I correct ?
 

Offline PlainName

  • Super Contributor
  • ***
  • Posts: 6796
  • Country: va
Re: Veritasium "How Electricity Actually Works"
« Reply #798 on: May 22, 2022, 02:26:16 pm »
That appears to be the gist of it.
 

Offline electrodacus

  • Super Contributor
  • ***
  • Posts: 1858
  • Country: ca
    • electrodacus
Re: Veritasium "How Electricity Actually Works"
« Reply #799 on: May 22, 2022, 02:48:24 pm »
Here is the mechanical analogy to the two capacitor problem that actually fits.

Springs can store energy, Stored energy is proportional to the 'stretch' squared - just like a how capacitors store energy. In this analogy, the spring is our capacitor.

Flywheels can store kinetic energy. They also have very good bearings that cause minimal energy loss. This is our deliberately introduced inductance.

Ropes can transfer the forces (admittedly only when it tension, but we will make sure they stay in tension). Here I am making no statement if these are the wires, or the electric field. These are just a mechanical way to transfer the forces.  When the rope is moving it is assumed to have minimal momentum. Consider the momentum of the rope as self-inductance - if the flywheel is very light enough (or things are arranged so no flywheel is needed), this might becomes important in in how the system behaves. But with any sensible flywheel it becomes unimportant.

The flywheel has a clamp on it. When the clamp is on the rope can't move, so no force can be transferred through a clamped flywheel. This is our switch.


You got to complicated and the analogy is wrong.
I guess you wanted to discuss the two parallel capacitor problem but more recently we are discussing a ideal voltage source and a 1000uF capacitor charged by that.


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf