General > General Technical Chat
Veritasium "How Electricity Actually Works"
electrodacus:
--- Quote from: dunkemhigh on May 18, 2022, 06:52:43 pm ---
Hey, please try not to introduce dead cats. Just stick with the ultra-simple circuit we were discussing: PSU->cap->res->return.
So, if no energy is passed from one side to the other, what is the resistor burning?
Hang on... O. M. G.... Maybe you are converting to aetherwind?
Joking aside, explain how the resistor can show a voltage across it and sink some current. Where is that coming from?
--- End quote ---
No energy passes from one side of the capacitor to the other. Energy is being stored in the capacitor.
Current will from from the source in to capacitor through wires and obviously through the resistor that is also a wire.
Current will decrease as the capacitor is charged until no current flows as capacitor is full.
Say capacitor when discharged start with 1000 free electrons on one plate and 1000 electrons on the other plate.
When battery is connected to capacitor if you connect just the positive nothing will happen same if you connect just the negative.
When battery is connected to capacitor on both sides electrons will flow from battery in to the wire and when that happen there will be an electron wave that will push out electrons on the other side of the wire in to the capacitor plate while at the same time electrons from the other plate will get out of the capacitor plate in to the wire and that will push electrons in to battery.
At the end when capacitor potential equal battery potential current flow stops.
At this point you may have 1100 electrons on one plate and 900 electrons on the opposite plate.
The 100 extra electrons are due to battery pushing those through the wire and the deficit of electrons on the other plate when trough wire in to battery.
No electrons have traveled through the capacitor dielectric.
Now the capacitor contains energy that can be used to do work. So you will say energy flowed in the capacitor where it is stored not through capacitor.
electrodacus:
--- Quote from: snarkysparky on May 18, 2022, 07:09:32 pm ---""Electric current will always require some sort of charged particle either electrons or ions""
Yes but the charged particles need not "transfer" to transmit energy.
--- End quote ---
What do you mean by "transfer"
Charged particles as electrons repel each other thus if you have a region with higher electron density it will want to migrate to a region with lower density.
Since there is not enough energy for them to travel through air even for a few mm distance between switch contacts the contacts need to be close enough so that 20V in this case has enough energy to push an electron from the negatively charged region in to the region with deficit of electrons.
So what you likely call "transfer" has to do with repelling forces that negative charged particles feel relative to one another. That is not high enough at 20V to have electrons fly from the wire trough 1m of air to the other wire and for that you will need way higher voltage about 3400000V
IanB:
--- Quote from: electrodacus on May 18, 2022, 07:10:27 pm ---No energy passes from one side of the capacitor to the other. Energy is being stored in the capacitor.
--- End quote ---
Let's stop bringing up the energy stored in the capacitor. Everyone knows energy is stored in capacitors. Everyone agrees energy is stored in capacitors. Nobody has ever denied that energy is stored in capacitors. Stop acting like this a big mystery.
If we look at an energy balance on the system, we have:
1. The energy leaving the source (battery) is ES
2. The energy stored in the capacitor is EC
3. The energy consumed (dissipated) by the load resistor/LED is EL
Since you tell us that energy is always conserved, we know that ES = EC + EL
Clearly some energy from ES ended up as EL, which means some energy was transferred from the battery to the load, and some energy was stored in the capacitor.
snarkysparky:
Lets take the issue to the instantaneous case.
So the usual battery , switch, capacitor, resistor all in series.
Lets say batter V = 12V
Capacitor is 1000 uf.
Resistor is 10 ohms.
Initially capacitor is discharged.
At the instant the switch is closed we have.
I = 120 ma
Resistor is dissipating power of 12*12/ 10 = 14.4 watts.
capacitor energy = ce = 0.5C * Vc^2
diff that by time
Dce/Dt = Vc
Vc is zero at the beginning. Capacitor is storing No energy.
In this instantaneous point would you say that the 14.4 watts the resistor is dissipating is traveling "through" the capacitor??
If not how is it getting there?
electrodacus:
--- Quote from: snarkysparky on May 18, 2022, 07:31:20 pm ---Lets take the issue to the instantaneous case.
So the usual battery , switch, capacitor, resistor all in series.
Lets say batter V = 12V
Capacitor is 1000 uf.
Resistor is 10 ohms.
Initially capacitor is discharged.
At the instant the switch is closed we have.
I = 120 ma
Resistor is dissipating power of 12*12/ 10 = 14.4 watts.
capacitor energy = ce = 0.5C * Vc^2
diff that by time
Dce/Dt = Vc
Vc is zero at the beginning. Capacitor is storing No energy.
In this instantaneous point would you say that the 14.4 watts the resistor is dissipating is traveling "through" the capacitor??
If not how is it getting there?
--- End quote ---
As soon as an electron passes through the resistor an electron will also end up in the capacitor so voltage across the capacitor is no longer zero (it will be super close to zero but not zero).
There are capacitors everywhere not just that huge by comparison 1000uF capacitor. The wires and even resistor will for a capacitor with the return wire.
So if 1 million electrons (absolute random number but can be calculated exactly) end up stored in the capacitor then that number has passed through wires and the resistor witch is also a wire.
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