You appear to be surrounded by confused people. What's the common factor? It's you!
If someone can not understand the difference between stored energy and energy that did work then I can not call him or her anything other than confused.
You seem to lack basic high-school knowledge of physics and electricity. That's why you don't understand what these people are talking about. Then you project your confusion on them, saying they are wrong.
Maybe it's time to revisit the basic concepts and try to understand what they're saying.
You seem to lack basic high-school knowledge of physics and electricity. That's why you don't understand what these people are talking about. Then you project your confusion on them, saying they are wrong.
Maybe it's time to revisit the basic concepts and try to understand what they're saying.
Or maybe high-school physics and even university level is badly done.
School in general seems to be tailored to those that can memorize a lot of facts with no effort put into understanding them.
Here is an example to illustrate the difference between energy used to do work and stored energy.
So say you are the source of energy and you are pushing an electric vehicle on a flat road.
Vehicle is in neutral but there is of course friction loss so you put in 200Wh worth of energy and push the vehicle for 1km.
Now this 200Wh of energy did work.
Then at the same time I also push an identical vehicle and also expend 200Wh on the same flat road but I moved the vehicle just 100m = 0.1km because the regenerative brakes on the vehicle where engaged and 180Wh ended as energy stored in the battery.
When you look at this two systems the input energy was exactly the same 200Wh but in first case all energy input in the system did work while in the second case only 10% was used to do work the other 90% was stored as electricity and that stored energy can be used to do work but at this point in time it is stored energy and anything can be done with that not necessarily move the vehicle the other 900m but maybe just use the energy to listen to music on the car radio.
In this example there was an energy conversion from mechanical to electrical before storing unlike the capacitor problem where no conversion was done.
The 144Ws of energy delivered by the source was used to charge the capacitor with an fairly bad efficiency of just 50% so you end up with 72Ws stored in the capacitor but those 72Ws in the capacitor are still there in the original form of electrical energy ready to be used to do work.
Capacitor is an energy storage device that will not let current pass through thus no energy is passing through (the displacement current is just a mathematical concept not a physical current).
Something like displacement current or displacement energy is a fictional math concept for calculation purposes as maybe energy storage was harder to explain.
You do realize that you will be pushing 10x harder in the second case? 0.1 times the distance means 10x the force...
The person pushing the car has done the same amount of work. Just in the first case more got 'lost' to the environment.
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?
What part exactly is not clear
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.
QuoteWhat part exactly is not clear
How the resistor manages to consume energy without there being any transfer from one side to the other.
This diversion to how capacitors charge and at what cost, etc, is simply your smokescreen that you hope will obscure your inability to deal with the real issue.
I'm not quite clear how you reconcile that with the actual formal definition of work for an electric field:
https://en.wikipedia.org/wiki/Work_(electric_field)
But then again, since you reject the existence of electric fields, that would hardly be surprising.
QuoteWhat part exactly is not clear
How the resistor manages to consume energy without there being any transfer from one side to the other.
This diversion to how capacitors charge and at what cost, etc, is simply your smokescreen that you hope will obscure your inability to deal with the real issue.
You are charging an energy storage device.
Actually, I am heating up a resistor. Where did that energy come from, and how did it get there? The only energy source available is the power supply upwind of the capacitor. So, from what I can gather from your 'explanation' it goes something like this:
PSU: "Hey Cap+, gonna shove you some lovely joules."
Cap+: "Eh? Oh, hang on a mo..."
Cap+: Oy, Cap-, you wanna dump your old electrons and get some new ones?"
Cap-: "Sounds fab to me."
Cap+: "OK, let's have them then"
PSU: "Here you go. Enjoy!"
Cap-: "Hey, resitor, 50gazillion elect(r)ons coming your way!"
Resistor: "WTF? Now?!?! Shit, have to burn them up."
I don't think energy storage has meaning in this context. The cap is series with the resister and you talk as if it's in parallel and providing the juice. It isn't - the PSU is. If anything, the cap is acting as a blockage.
Bringing in a battery complicates things, just as inserting a diode would. Let's keep it simple with the minimal parts necessary to show the problem.
"half that ending as heat due to wire resistance"
It took power to heat the wire. That power came from the flow of electrons into and out of the capacitor wires.
We consider the heat in the wires to be useful work.
And that energy that heated the wires flowed because of charge moving into and out of the capacitor.
Do you agree that a capacitor can not only store energy but can also pass through energy?
If not how does the capacitor power supply above work?
A capacitive dropper works because capacitor is always charged and discharged.
No energy passed through capacitor
QuoteA capacitive dropper works because capacitor is always charged and discharged.
You are confusing charge with energy.QuoteNo energy passed through capacitor
On the contrary, you mean no charge passed through the capacitor. But energy went around the the whole circuit for the duration that the capacitor charged.
""So with DC you have just one pulse as the capacitor chargers and with AC you constantly charge and discharge the capacitor..""
Yes and in that one charging pulse there was a finite amount of energy available to drive a load on the other side of the capacitor.
Charging a capacitor means storing energy in to it.
No energy went around the capacitor but in to capacitor.