Veritasium "How Electricity Actually Works"

[snarkysparky]

"The transmit and receive antenna are each the plate of a large gap capacitor and the circuit is closed through the ground (earth) as the common conductor."

absolutely incorrect.   You know we receive radio waves from space right ?

[electrodacus]

So you are saying that what we see in that circuit is false? That the input energy from the PSU does not cause the resistor to consume energy?

What I'm saying is that you can not connect just one end of a resistor to one terminal of the battery and expect a current except for the one needed to charge the newly created capacitor that is made up by the other battery terminal and the resistor with air being the dielectric.

It is about how current flows in to a capacitor when being charged and not through a capacitor.

Say you have an ideal voltage source 12V so there is no internal series resistance considered for this ideal voltage source.
Then say you have two wires each 0.5Ohms connected to this 12V supply and then you short the ends of those wires you will get 12V/ (0.5+0.5) =12A of current. This 12V * 12A will result in 144W energy wasted as heat so each second 144Ws (144J same thing) or 40mWh again same thing.
So 144Ws delivered by the source and 144Ws end as heat nothing stored.

Then say you connect a 5Ohm resistor and energy will go through wires and through resistor toal 6Ohms so 2A * 12V = 24W in one second it will be 24Ws delivered by the source and 24Ws all end as heat dissipated by wires and by resistor as energy travels through them.

Now say you leave the wires and instead of a resistor you connect a capacitor say it is 1000uF = 1mF and say internal DC ESR is basically zero to keep things simple.

a) How much energy will power supply deliver in one second after the discharged capacitor was connected ?
b) How much energy will end up as heat dissipated by wires?
c) How much energy is stored in the capacitor after this second?
d) How much energy is delivered by the source in the next second? so after first second ended and another second from that point.

Answer correctly to all this questions and you should understand what happens. Answer wrong and I may be able to help you understand what part you got wrong.

[electrodacus]

absolutely incorrect.   You know we receive radio waves from space right ?

I know but do you know how that works ?

If you think you do give an answer to the above problem the one in post #701

[snarkysparky]

I know but do you know how that works ?

Yes,  Maxwell's equations predict that a time varying field produces a propagating wave.  Just like shaking your hand on the surface of a pool creates a propagating wave on the surface of the water.

So wave a magnet about and you have in fact sent a tiny propagating wave out to the universe.

[snarkysparky]

if current flows into one terminal of a capacitor and out the other terminal I think most of us here call that "through"  the capacitor.

Are you saying that this situation is not correctly described by using the word "through" to describe the currents ?

[electrodacus]

if current flows into one terminal of a capacitor and out the other terminal I think most of us here call that "through"  the capacitor.

Are you saying that this situation is not correctly described by using the word "through" to describe the currents ?

You are calling it through but that is incorrectly. Saying electrical current flows through dielectric is incorrect and you know that.
Try and solve the problem in post #701 and see if going "through" is helpful.
Electric current flows through a resistor but it will not flow through a capacitor but in and out as it is stored there not doing any work.

[IanB]

Then say you connect a 5Ohm resistor and energy will go through wires and through resistor toal 6Ohms so 2A * 12V = 24W in one second it will be 24Ws delivered by the source and 24Ws all end as heat dissipated by wires and by resistor as energy travels through them.

Now say you leave the wires and instead of a resistor you connect a capacitor say it is 1000uF = 1mF and say internal DC ESR is basically zero to keep things simple.

No, this is the wrong question. Keep the resistor there and insert the capacitor as well, so you have both the resistor and the capacitor in series in the circuit. Now answer your questions below:

a) How much energy will power supply deliver in one second after the discharged capacitor was connected ?
b) How much energy will end up as heat dissipated by wires and resistor?
c) How much energy is stored in the capacitor after this second?
d) How much energy is delivered by the source in the next second? so after first second ended and another second from that point.

[snarkysparky]

a) How much energy will power supply deliver in one second after the discharged capacitor was connected ?
b) How much energy will end up as heat dissipated by wires?
c) How much energy is stored in the capacitor after this second?
d) How much energy is delivered by the source in the next second? so after first second ended and another second from that point.


The circuit you describe is basically  12V  zero impedance source,    1 ohm resistor ( both 0.5 ohm wires )  ,  and 1000 u capacitor all in series.

Vc  =  12 * ( 1 - exp(-R*C* t))

Vc  =  12 * (1 - exp(- 1 * 1000e-6* t)

plot of the energy stored in the capacitor vs time

[electrodacus]

a) How much energy will power supply deliver in one second after the discharged capacitor was connected ?
b) How much energy will end up as heat dissipated by wires and resistor?
c) How much energy is stored in the capacitor after this second?
d) How much energy is delivered by the source in the next second? so after first second ended and another second from that point.

No, this is the wrong question. Keep the resistor there and insert the capacitor as well, so you have both the resistor and the capacitor in series in the circuit. Now answer your questions below:

Wires already have a resistance 1Ohm in total so it will not make any difference if you keep the 5Ohm resistor in series. If you want keep the resistor in series and solve the problem.

[electrodacus]

a) How much energy will power supply deliver in one second after the discharged capacitor was connected ?
b) How much energy will end up as heat dissipated by wires?
c) How much energy is stored in the capacitor after this second?
d) How much energy is delivered by the source in the next second? so after first second ended and another second from that point.


The circuit you describe is basically  12V  zero impedance source,    1 ohm resistor ( both 0.5 ohm wires )  ,  and 1000 u capacitor all in series.

Vc  =  12 * ( 1 - exp(-R*C* t))

Vc  =  12 * (1 - exp(- 1 * 1000e-6* t)

plot of the energy stored in the capacitor vs time

(Attachment Link)

The graph looks wrong (x axis) just answer the questions. Those are us so 5000us not 5000ms

[snarkysparky]

Yes the xaxis should be us

here are the other questions.





These are the complete answers to your questions. 

[electrodacus]

Yes the xaxis should be us

here are the other questions.

These are the complete answers to your questions.

I want you to provide the numbers not the graphs only. After you provide the numbers you can tell me exactly how much energy passes through capacitor.

[snarkysparky]

""You are calling it through but that is incorrectly. Saying electrical current flows through dielectric is incorrect and you know that.""


This is the crux of the discussion.   Just simply an obscure definition of "through"

Electrodous...   What word would you use to describe the power delivered to the other side of the capacitor.   You do agree that power is found on the other side of the capacitor ?

[electrodacus]

""You are calling it through but that is incorrectly. Saying electrical current flows through dielectric is incorrect and you know that.""


This is the crux of the discussion.   Just simply an obscure definition of "through"

Electrodous...   What word would you use to describe the power delivered to the other side of the capacitor.   You do agree that power is found on the other side of the capacitor ?

There is no energy delivered to the other side. Just answer the questions in that problem and you will see that is the case.

[PlainName]

So you are saying that what we see in that circuit is false? That the input energy from the PSU does not cause the resistor to consume energy?

What I'm saying is that you can not connect just one end of a resistor to one terminal of the battery and expect a current except for the one needed to charge the newly created capacitor that is made up by the other battery terminal and the resistor with air being the dielectric.

It is about how current flows in to a capacitor when being charged and not through a capacitor.

Say you have an ideal voltage source 12V so there is no internal series resistance considered for this ideal voltage source.
Then say you have two wires each 0.5Ohms connected to this 12V supply and then you short the ends of those wires you will get 12V/ (0.5+0.5) =12A of current. This 12V * 12A will result in 144W energy wasted as heat so each second 144Ws (144J same thing) or 40mWh again same thing.
So 144Ws delivered by the source and 144Ws end as heat nothing stored.

Then say you connect a 5Ohm resistor and energy will go through wires and through resistor toal 6Ohms so 2A * 12V = 24W in one second it will be 24Ws delivered by the source and 24Ws all end as heat dissipated by wires and by resistor as energy travels through them.

Now say you leave the wires and instead of a resistor you connect a capacitor say it is 1000uF = 1mF and say internal DC ESR is basically zero to keep things simple.

a) How much energy will power supply deliver in one second after the discharged capacitor was connected ?
b) How much energy will end up as heat dissipated by wires?
c) How much energy is stored in the capacitor after this second?
d) How much energy is delivered by the source in the next second? so after first second ended and another second from that point.

Answer correctly to all this questions and you should understand what happens. Answer wrong and I may be able to help you understand what part you got wrong.

Jesus F Christ!

A simple yes or no question results in waffling that isn't relevant yet again! You're incapable of following a logical process and always jump to the conclusion before working back, then just fill up reams of inappropriate gumpf to support that foregone concluision instead of just answering the damn question as stated.

[electrodacus]

Jesus F Christ!

A simple yes or no question results in waffling that isn't relevant yet again! You're incapable of following a logical process and always jump to the conclusion before working back, then just fill up reams of inappropriate gumpf to support that foregone concluision instead of just answering the damn question as stated.

Not sure what that fictional character has to do with this problem. Other that he is as fictional as you energy going through a capacitor.
Answer the question and you will see the answer for yourself.

[HuronKing]

""You are calling it through but that is incorrectly. Saying electrical current flows through dielectric is incorrect and you know that.""


This is the crux of the discussion.   Just simply an obscure definition of "through"

Electrodous...   What word would you use to describe the power delivered to the other side of the capacitor.   You do agree that power is found on the other side of the capacitor ?

It's worse than that. electrodacus doesn't understand displacement current.

He keeps saying "electrical current" hoping this smokescreen will distract from his ignorance of there being TWO terms for current in Ampere's Law - the conduction current AND the displacement current. In the region between two capacitors, there is no conduction current, but there IS displacement current.

Asserting that displacement current propagates no energy is anti-Maxwell.

[TimFox]

In profane usage, the middle initial is "H".

[electrodacus]

It's worse than that. electrodacus doesn't understand displacement current.

He keeps saying "electrical current" hoping this smokescreen will distract from his ignorance of there being TWO terms for current in Ampere's Law - the conduction current AND the displacement current. In the region between two capacitors, there is no conduction current, but there IS displacement current.

Asserting that displacement current propagates no energy is anti-Maxwell.

Provide an answer to the problem then let me know the amount of energy that passed "through" capacitor.

[Naej]

You will need to understand Maxwell's equations to understand what they say. They do not say energy flows outside the wires.

They absolutely say that. That's what led Maxwell to unify the electric and magnetic fields into electromagnetism... and it's what led Hertz to try transmitting an electrical signal from one radiator to another without any wires whatsoever.

No they don't. They are equations giving the curl/divergence of two functions, and none of the two are energy.

Also, "displacement current" is a concept made up by Maxwell, grouping various effects into one.
Much like you can use "magnetic monopoles current/density" and it can simplify computations, it does NOT mean magnetic monopoles exist.

Asserting that displacement current propagates no energy is anti-Maxwell.

1) How is it a problem.
2) Asserting it does is anti-Poynting.

[HuronKing]

It's worse than that. electrodacus doesn't understand displacement current.

He keeps saying "electrical current" hoping this smokescreen will distract from his ignorance of there being TWO terms for current in Ampere's Law - the conduction current AND the displacement current. In the region between two capacitors, there is no conduction current, but there IS displacement current.

Asserting that displacement current propagates no energy is anti-Maxwell.

Provide an answer to the problem then let me know the amount of energy that passed "through" capacitor.

It's the same energy you believe only exists when conduction current flows - this is what Ampere's Law is trying to tell you.

@Naej

No they don't. They are equations giving the curl/divergence of two functions, and none of the two are energy.

https://web.mit.edu/6.013_book/www/chapter11/11.2.html

The electric and magnetic fields are confined to the free space regions. Thus, power flow and energy storage pictured in terms of these variables occur entirely in the free space regions.

The Poynting Vector is derived from Maxwell's Eqs. If Maxwell's Eqs are true - then so is Poynting.

[PlainName]

In profane usage, the middle initial is "H".

Ah, but this was crude profanity 

[snarkysparky]

Ahite.  I got your numbers.  And discovered how rusty i am at this stuff.   skool was a long time ago

a) How much energy will power supply deliver in one second after the discharged capacitor was connected ?       ->  0.1439 Joules

b) How much energy will end up as heat dissipated by wires?    -> 0.0719 joules 

c) How much energy is stored in the capacitor after this second?   ->  0.0720  Joules

d) How much energy is delivered by the source in the next second? so after first second ended and another second from that point.     -> 0    ( time constant is 1 ms ) 


What does this mean?

[snarkysparky]

I tell ya what I think it means.

The resistance of the wires got to dissipate as much energy as is stored in the capacitor.

And the current that made this possible went into one terminal of the capacitor and out the other terminal.   I call this "through"   what do you call it ?

[Naej]

No they don't. They are equations giving the curl/divergence of two functions, and none of the two are energy.

https://web.mit.edu/6.013_book/www/chapter11/11.2.html

The electric and magnetic fields are confined to the free space regions. Thus, power flow and energy storage pictured in terms of these variables occur entirely in the free space regions.

The Poynting Vector is derived from Maxwell's Eqs. If Maxwell's Eqs are true - then so is Poynting.

These are 2 completely different things: Maxwell's equations describe an approximation of physics (a very good one when the number of photons is large, and their energy small compared to electrons); Poynting's theorem is well, a theorem (not an approximation), which is true if you assume Maxwell's equations are correct. Maxwell had nothing do to with ExH, and there's no dD/dt in ExH.

The important part of the quote is what I put in bold, it's a picture, and you might as well take S=JV.