Veritasium "How Electricity Actually Works"

[electrodacus]

It is a distributed capacitance and there is inductance involved so charging of each capacitor (seens a series of lumped components) will be delayed.

Doesn't matter if it's distributed - the entire thing has a very very short time constant. All you're saying is that if you imagine it comprised 100 smaller capacitors, each will be 2.1ff (i.e. 2.1pf/100).

And... if there's a 18V pulse going in, how come there's only 5V coming out the other side?

You are looking at the capacitance of a single conductor.  The way that Derek's experiment is setup the capacity is between two wires with 1m distance between them and 10m long in each direction.
So total capacitance will be around 42pf for the two 10m long parallel copper pipes and I have that split in to 100 so 0.42pf for each element.
Voltage on the resistor is proportional with the current passing through the resistor while this distributed capacitance charges so it forms a divider with the line impedance. And there are two lines on each side of the resistor.

[hamster_nz]

It is a distributed capacitance and there is inductance involved so charging of each capacitor (seens a series of lumped components) will be delayed.

Doesn't matter if it's distributed - the entire thing has a very very short time constant. All you're saying is that if you imagine it comprised 100 smaller capacitors, each will be 2.1ff (i.e. 2.1pf/100).

And... if there's a 18V pulse going in, how come there's only 5V coming out the other side?

Just for fun I divided capacitors like this, and the prompts thoughts that energy is stored in the dielectric between the plates. 

[SiliconWizard]

Is the electric field really between the plates?

[electrodacus]

Is the electric field really between the plates?

For a capacitor yes. The plates have large surface area and distance between plates is super small in comparison thus field lines are all between the plates.
With wires separated by 1m of air it still between the wires but a bit more "loose" if this is the correct word.

[electrodacus]

Doesn't matter if it's distributed - the entire thing has a very very short time constant. All you're saying is that if you imagine it comprised 100 smaller capacitors, each will be 2.1ff (i.e. 2.1pf/100).

Ues this calculator https://www.emisoftware.com/calculator/wire-pair-capacitance/
You can use 12mm wire radius 1m separation and 10m length and you will get 62pF total
I also checked with 1mm radius for fire and got about 40pf so I used 42pf in my simulation just to be conservative.

[PlainName]

For the sake of argument I went with 60pF. Derek's result I've attached along with my conservative (that is, relatively slow rise time) simulation. They look completely different, and I'm not convinced that splitting the 1 cap into however many your simulator can put up with will change things significantly. Specifically, the simulation shows a peak of close to the input level and then dies off very quickly, even on this much shorter timescale than the real thing.

[SandyCox]

Doesn't matter if it's distributed - the entire thing has a very very short time constant. All you're saying is that if you imagine it comprised 100 smaller capacitors, each will be 2.1ff (i.e. 2.1pf/100).

Ues this calculator https://www.emisoftware.com/calculator/wire-pair-capacitance/
You can use 12mm wire radius 1m separation and 10m length and you will get 62pF total
I also checked with 1mm radius for fire and got about 40pf so I used 42pf in my simulation just to be conservative.

A lossless transmission line looks like a resistor at its input, not a capacitor. Please read Haus and Melcher or any other good book.

[electrodacus]

For the sake of argument I went with 60pF. Derek's result I've attached along with my conservative (that is, relatively slow rise time) simulation. They look completely different, and I'm not convinced that splitting the 1 cap into however many your simulator can put up with will change things significantly. Specifically, the simulation shows a peak of close to the input level and then dies off very quickly, even on this much shorter timescale than the real thing.

It makes a big difference when you split the capacitance and also inductance.
I guess you did not see my spice simulation that I posted earlier in the thread.
So here it is again.  Keep in mind I simulated only half the circuit the other side of the resistor is directly connected to source negative not through another transmission line and I did not add the influence of oscilloscope probes.
The important part is that shape is exactly the same with a slow rise and then a flat response until about 65ns



 

[electrodacus]

A lossless transmission line looks like a resistor at its input, not a capacitor. Please read Haus and Melcher or any other good book.

This is a real line and during transient the inductance and capacitance of the line play an important role as see from both experimental result and simulation.
During DC after the transient effect is gone it will be just a resistive line.

[SandyCox]

Please read the book.

The experimental line will be close to a lossless line. The theory for lossy lines has also been worked out. Just read and you will learn a lot!

You are openly displaying your own ignorance.

[SandyCox]

Or look at this note I posted a few months ago.

[electrodacus]

Please read the book.

The experimental line will be close to a lossless line. The theory for lossy lines has also been worked out. Just read and you will learn a lot!

You are openly displaying your own ignorance.

I took a looked at the pdf you linked and it seems he only considered the impedance and resistance of the circuit and not the energy storage capacity.
I do not think the ignorance is on my part. The results of my simulation using proper lumped elements perfectly matches the waveform shape that  Derek obtained with his real world test.
The results show that the initial current through the lamp/resistor is due to transmission line capacitance charging so energy flowing through the lamp/resistor is due to resistor being in series with line capacitance (an energy storage device) being charged.

It is the equivalent of charging a capacitor from a voltage source through a lamp/resistor. So during the initial connection the current flows into capacitor (not through it) and that is what is seen in those first 65ns in Derek's experiment as well as in my spice simulation.
After that initial transient phase when at steady state the line will act as a purely resistive line so energy is transferred from the source to the lamp/resistor through the wires and current going in to the lamp/resistor equals current going out of the source.
Voltage across the resistor will be slightly lower due to energy loss on the transmission wires.

The main point Derek tried to make with both videos is that energy flows from source to load outside the wire is completely incorrect not only for DC but for AC/transient as well.

You can consider the lamp and transmission line one and the same so you can imagine a long filament 10m in both directions then you can understand that all energy provided by the source is dissipated in that filament in an uniform way when in DC steady state and a bit uniform during the initial transient due to capacitive and inductive characteristics of the line which is now also the lamp/resistor.   

[PlainName]

The main point Derek tried to make with both videos is that energy flows from source to load outside the wire is completely incorrect

There is no wire between the, uh, two wires. There is energy passing from one side to the other, and you say that's due to the capacitance of the wires. So, let's go with that - the inside of the capacitor is empty, so has not the energy transferred across inside of the capacitor but outside of the wires?

[electrodacus]

The main point Derek tried to make with both videos is that energy flows from source to load outside the wire is completely incorrect

There is no wire between the, uh, two wires. There is energy passing from one side to the other, and you say that's due to the capacitance of the wires. So, let's go with that - the inside of the capacitor is empty, so has not the energy transferred across inside of the capacitor but outside of the wires?

The energy is not passing from one side to the other. Energy is being stored in the capacitor made by the parallel wires.
In order to transfer electrical energy you need a current and that 1m of air between wires is an almost perfect electrical insulator so no current flow no power and if there is no power there is no energy transfer but there is energy storage that everyone seems to want to ignore. 

[IanB]

The results show that the initial current through the lamp/resistor is due to transmission line capacitance charging so energy flowing through the lamp/resistor is due to resistor being in series with line capacitance (an energy storage device) being charged.

If current flows through the lamp/resistor while the capacitance is charging, then the lamp/resistor is consuming power and radiating it as heat and light. (Power = I²R) Where did that power come from?

It is the equivalent of charging a capacitor from a voltage source through a lamp/resistor. So during the initial connection the current flows into capacitor (not through it) and that is what is seen in those first 65ns in Derek's experiment as well as in my spice simulation.

Yes, and during that initial charging period the lamp lights up and consumes power. How did the power reach the lamp?

The main point Derek tried to make with both videos is that energy flows from source to load outside the wire is completely incorrect not only for DC but for AC/transient as well.

But in the transient the load receives power and therefore converts energy to heat and light. Where did that power come from during the transient?

[Naej]

You could say the potential energy is qV so that there'll be two wires at different potentials connected to the resistor-lamp. So energy will flow from the wire into the lamp.

[electrodacus]

If current flows through the lamp/resistor while the capacitance is charging, then the lamp/resistor is consuming power and radiating it as heat and light. (Power = I²R) Where did that power come from?


Yes, and during that initial charging period the lamp lights up and consumes power. How did the power reach the lamp?


But in the transient the load receives power and therefore converts energy to heat and light. Where did that power come from during the transient?

You have a source (say battery) and you are charging a capacitor from that.
How if the energy from the source transferred to the capacitor ? Through wires.
There is some energy loss on the wires that ends up as heat in the wire and then is radiated to outside world.
Now in series on one of this wires you add a lamp. All you did was making one of those wires a higher resistance as lamp is also a wire.
You are asking how the energy is delivered from source to lamp and answer is through the wires.
All energy used by the lamp flowed through wires and when capacitor is fully charged so same voltage as the source lamp can no longer glow as no energy flows through it (basically a wire).

I guess the confusion is mostly related to what a capacitor is and the fact that energy flowing in capacitor is stored/conserved not used to do any work.

[IanB]

Yes, OK. So let's put a capacitor on both sides, and put the lamp between the two capacitors. Now there is an air gap completely separating the lamp from the battery. But still the lamp lights up?

[hamster_nz]

Yes, OK. So let's put a capacitor on both sides, and put the lamp between the two capacitors. Now there is an air gap completely separating the lamp from the battery. But still the lamp lights up?

We are going back to where we were in reply #266

electrodacus will say that the lamp (or in this case LEDs) are lit, but power doesn't flow through the capacitors, only in and out at the same time 

[electrodacus]

Yes, OK. So let's put a capacitor on both sides, and put the lamp between the two capacitors. Now there is an air gap completely separating the lamp from the battery. But still the lamp lights up?

There is no difference between one capacitor on one side of the lamp or capacitors on both sides.
An electron moved on one plate of a capacitor will push of an electron from the other plate and back in to battery with a single capacitor but if there are two capacitors in series then current will flow on the wire connecting the two capacitors as that one electron pushed off from the plate of first capacitor will on up on the plate of the other capacitor traveling through wire (is not the same electron that it will show up on the plate of the second capacitor but one displaced by that) then that electron on the second capacitor will push an electron in to the battery.
Again this current flow same as with just a capacitor will flow through wires only until capacitor/capacitors are charged. That energy stored in capacitor can be used latter to do work.
Like if you remove the battery and then connect a wire instead of the battery all energy that flowed into capacitors will now flow back out.

[electrodacus]

We are going back to where we were in reply #266

electrodacus will say that the lamp (or in this case LEDs) are lit, but power doesn't flow through the capacitors, only in and out at the same time 

Not "in and out" just in and when capacitors are charged no current will flow other than for the small leakage depending on the quality of those capacitors.

[PlainName]

Er,  hate to point this out but in the simplified circuit it goes: supply -> capacitor -> resistor -> return.

Now, the cap is not charged. It is 0V both sides. Connect the supply and now there is a voltage across the resistor. Where did that come from if not through the capacitor?

[electrodacus]

Er,  hate to point this out but in the simplified circuit it goes: supply -> capacitor -> resistor -> return.

Now, the cap is not charged. It is 0V both sides. Connect the supply and now there is a voltage across the resistor. Where did that come from if not through the capacitor?

In to capacitor not through capacitor.
You can not have a current "through" the dielectric.

You understand that when the capacitor is full may be 30ns or 30s but once the capacitor is fully charged (and I mean charged so it stores energy) current flow will stop. And current flows through wires but in to capacitor not through it.

[hamster_nz]

Er,  hate to point this out but in the simplified circuit it goes: supply -> capacitor -> resistor -> return.

Now, the cap is not charged. It is 0V both sides. Connect the supply and now there is a voltage across the resistor. Where did that come from if not through the capacitor?

You can not have a current "through" the dielectric.

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.

[PlainName]

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?