### Author Topic: Veritasium "How Electricity Actually Works"  (Read 21812 times)

T3sl4co1l and 3 Guests are viewing this topic.

#### electrodacus

• Super Contributor
• Posts: 1109
• Country:
##### Re: Veritasium "How Electricity Actually Works"
« Reply #600 on: Yesterday at 10:03:59 pm »

Demonstrations such as placing an ammeter on both legs of a capacitor as it charges, showing the same current flowing both terminals, at the same time are complete fabrications...

Unless it has to be the same electron flowing out as flowed in - in which case long wires are out due to the slow drift velocity. On a 1m x 2mm diameter copper wire,  carrying 1A DC , it will take on average about 12 hours for an electron to travel down that wire.

It is absolutely not the same electron not even the same electron after some delay that you find living the other plate of the capacitor.
On wires is not the electron that flows into the wire that gets on the other end at the speed of light but the electron that enters the wire will create a wave that pushes out an electron on the other end after about the time light needs to travel that distance.
Electron drift speed is a different subject.

So when you charge a capacitor from a battery none of the electrons entering one plate will be found on the other side.
The electrons leaving from the other plate where already on that plate and they leave in order for that plate to become positively charged.

#### electrodacus

• Super Contributor
• Posts: 1109
• Country:
##### Re: Veritasium "How Electricity Actually Works"
« Reply #601 on: Yesterday at 10:12:20 pm »
Something goes through to use up energy across that resistor (or lamp, or anything else that consumes power). Where does it come from? It is energy of some form, isn't it?

No energy will not go through but in to capacitor.
Those electrons that travel through the resistor (basically a wire) where never in the battery or the wire connected on the other side of the capacitor.

The analogy I can make is a compressed air cylinder that is split in two by an rubbery elastic membrane.
Discharged cylinder means the membrane is relaxed in the middle then when you push more air molecules in one of the sides same amount of air molecules will leave from the other side but they will not be the same as that elastic membrane (dielectric) will prevent any air molecules passing from one half of the cylinder to the other.
Electrons can not pass through dielectric so no current no energy flows through.
As I mentioned before a switch is also a capacitor where close circuit it means you shorted the two plates. As long as switch is open so just a capacitor no energy will flow through it.

#### dunkemhigh

• Super Contributor
• Posts: 4161
##### Re: Veritasium "How Electricity Actually Works"
« Reply #602 on: Yesterday at 10:17:26 pm »
Something goes through to use up energy across that resistor (or lamp, or anything else that consumes power). Where does it come from? It is energy of some form, isn't it?

No energy will not go through but in to capacitor.

Well, hang on a minute. There is 4mA going through that resistor, enough to light up an LED. Energy must be involved there, no? Where did that come from?

Quote
The analogy I can make is a compressed air cylinder that is split in two by an rubbery elastic membrane.

In that case the air on one side pushes against the membrane and that pushes against whatever is on the other side. You've transferred energy from the pushing side to the pushed side, haven't you?

#### electrodacus

• Super Contributor
• Posts: 1109
• Country:
##### Re: Veritasium "How Electricity Actually Works"
« Reply #603 on: Yesterday at 10:37:17 pm »

Well, hang on a minute. There is 4mA going through that resistor, enough to light up an LED. Energy must be involved there, no? Where did that come from?

In that case the air on one side pushes against the membrane and that pushes against whatever is on the other side. You've transferred energy from the pushing side to the pushed side, haven't you?

Yes those 4mA are due to electrons leaving that plate of the capacitor (the discharged capacitor has equal amount of free electrons on each plate) so when you charge a capacitor you push electrons on to one plate while the same amount of electrons leave the other plate same as in that compressed air tube analogy air molecules enters one side and others are leaving from the other side.

So a discharged compressed air cylinder with two partitions will have air in it say 500 molecules one one side and 500 molecules on the other side.
Say you charged this discharged cylinder from another one that is charged but 10x larger that has 8000 in one side and 2000 on the other side.
Now with two hoses you connect the large one to the small one to charge it
You will get almost 800 on one side and more than 200 on the other side (to lazy to do the exact calculation) then you are left with about 7700 on the large one and about 2300 on the other side. You get the idea the pressure (voltage) will equalize.
After pressure has equalized there is no flow so no energy transfer.  You can add some turbine (lamp) in between the two and it will work during transfer but stop when pressure is equalized.
As any analogy you can not go to far and still be correct but for the limited purpose of explaining why energy flows in or out of capacitors but not through should be good enough.

#### hamster_nz

• Super Contributor
• Posts: 2629
• Country:
##### Re: Veritasium "How Electricity Actually Works"
« Reply #604 on: Yesterday at 10:39:51 pm »
Something goes through to use up energy across that resistor (or lamp, or anything else that consumes power). Where does it come from? It is energy of some form, isn't it?

No energy will not go through but in to capacitor.

Well, hang on a minute. There is 4mA going through that resistor, enough to light up an LED. Energy must be involved there, no? Where did that come from?

Quote
The analogy I can make is a compressed air cylinder that is split in two by an rubbery elastic membrane.

In that case the air on one side pushes against the membrane and that pushes against whatever is on the other side. You've transferred energy from the pushing side to the pushed side, haven't you?

Careful, you may encourage them to go full Humpty Dumpty mode on you:

Quote
'When I use a word,' Humpty Dumpty said, in rather a scornful tone, 'it means just what I choose it to mean — neither more nor less.'
Gaze not into the abyss, lest you become recognized as an abyss domain expert, and they expect you keep gazing into the damn thing.

#### dunkemhigh

• Super Contributor
• Posts: 4161
##### Re: Veritasium "How Electricity Actually Works"
« Reply #605 on: Yesterday at 11:39:32 pm »

Well, hang on a minute. There is 4mA going through that resistor, enough to light up an LED. Energy must be involved there, no? Where did that come from?

In that case the air on one side pushes against the membrane and that pushes against whatever is on the other side. You've transferred energy from the pushing side to the pushed side, haven't you?

Yes those 4mA are due to electrons leaving that plate of the capacitor (the discharged capacitor has equal amount of free electrons on each plate) so when you charge a capacitor you push electrons on to one plate while the same amount of electrons leave the other plate same as in that compressed air tube analogy air molecules enters one side and others are leaving from the other side.

OK, I guess we are on the same page, then. The energy going into the capacitor on the plus side is pushing energy out of the capacitor on the negative side, right? In a simplistic manner of speaking.

So, back to our capacitor with 1m of free air in the middle... there is a net energy transfer from one side to the other. The PSU is pushing some energy into the cap which is consequently pushing the same amount of energy out to the resistor which is consuming it (aka converting to heat/light). The PSU has lost it, the resistor has used it, yes?

#### SandyCox

• Regular Contributor
• Posts: 131
• Country:
##### Re: Veritasium "How Electricity Actually Works"
« Reply #606 on: Today at 11:42:09 am »
Please look at Equation 18. Zo is real. It is a resistance and has no imaginary part. Note that the two transmission lines and the resistor form a resistive divider. Let's assume that the line is perfectly matched, i.e. R = 2Zo. Before the first reflection arrives back, 3/4 of the energy delivered by the battery is sorted in the two transmission lines. 1/4 is dissipated in the resistor. No energy is transferred from the transmission line to the resistor.

Smf