General > General Technical Chat
Veritasium "How Electricity Actually Works"
electrodacus:
--- Quote from: dunkemhigh on May 18, 2022, 08:08:45 pm ---
OK, you're on.
Here is the waveform and the circuit that created it. You may have seen it before.
Now, the green line is the voltage applied to the circuit. Clearly, it is the cause of red line, which is the voltage across the resistor. (Or, alternatively, the voltage over the resistor - shown by the red line - is caused by the voltage applied to the circuit.)
There's the proof. Now, perhaps you can say just how that happens? How does the one cause the other?
--- End quote ---
You forgot to draw the source that is also connected to GND. Maybe that is what confuses you.
electrons from the source will flow in to resistor and in to the capacitor plate while on the other side that other capacitor plate will send the exact same number of electrons from that plate to the source.
Resistor is nothing else than a wire helping to complete the circuit. You need a closed loop to charge that capacitor and maybe because that is not seen in your schematic you are confused about how the energy is stored in to capacitor.
You also understand that red waveform also represents the current through the resistor and as you can see it will fast trend to zero meaning once the capacitor plate has enough electrons on one side with similar deficit on the other the capacitor is fully charged and at same voltage as the source thus no more energy transfer from the source to capacitor.
electrodacus:
--- Quote from: snarkysparky on May 18, 2022, 08:10:39 pm ---Energy is conserved. At the instant the switch is closed the battery is providing 14.4 watts of power.
That 14.4 watts need to be accounted for everywhere. Assuming zero resistance wires. So for the capacitor we have Joules / second = watts and same for the resistor.
The capacitor is gaining zero energy at the instant the switch is closed.
The capacitor IS passing energy through it. Suppose the capacitor was infinite in capacitance. No energy would ever be stored in it but load current could flow into one plate an out of the other plate with no transfer across the dielectric.
Maybe we are saying the same thing.
What is happening is that the for the plates of the capacitor to "charge" relative to each other current must flow into and out of the plates but NOT across the dielectric barrier.
And instantaneously the energy balance at the start is that no energy is being stored in the capacitor. And somehow energy shows up at the resistor.
--- End quote ---
Exact moment that switch is closed the electron from the tip of the switch will jump to the other side on the wire so there will be no power through the resistor and no energy in the capacitor.
You will also need to consider the energy stored in wire inductance as wire has some inductance that opposes electron flow converting that in to magnetic field around the conductor.
At about the speed of light there will be an electron passing through resistor and arriving in the capacitor at witch point you can no longer say there is no energy delivered to capacitor.
Yes the initial part is super inefficient as say capacitor is at 0.001V and battery 12V that means 11.999V drops across the resistor and that means terrible charging efficiency with a small fraction of a percent efficiency and it will take the capacitor to get to 6V for the efficiency to get to 50% still horrible but not as crazy.
Even if you do not have the capacitor there just the resistor with just one terminal connect to battery the other terminal just not connected to anything and then on the other side of the battery a switch and a piece of wire again that wire not connected to anything due to the capacitance between the resistor and that not connected wire after the switch you will have some capacitance maybe a few pF but still enough so that when you close the switch energy will flow through the resistor for a very short period of time.
electrodacus:
--- Quote from: dunkemhigh on May 18, 2022, 08:08:45 pm ---OK, forget it. It's a lost cause and I am out.
--- End quote ---
It is a lost cause for you as you are the one that has no understanding of the subject.
hamster_nz:
--- Quote from: electrodacus on May 18, 2022, 08:31:44 pm ---
--- Quote from: dunkemhigh on May 18, 2022, 08:08:45 pm ---OK, forget it. It's a lost cause and I am out.
--- End quote ---
It is a lost cause for you as you are the one that has no understanding of the subject.
--- End quote ---
As a rough rule, would you say that capacitors allow voltage & current transients to pass through, and but do not pass steady voltages and currents? (deciding what is a transient and what is a steady voltage can be computed based on the Rs Cs & Ls in the circuit)
So at DC they have infinite resistance, and with signals of high enough frequency they offer very, very little resistance?
TimFox:
The elementary features I was taught in the first week of electronics 101 were that you can't change the voltage across a capacitor instantaneously, and you can't change the current through an inductor instantaneously.
Analysis of most simple AC and pulse circuits can start from this, and then introduce the finite resistances and time constants.
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