General > General Technical Chat
Veritasium "How Electricity Actually Works"
electrodacus:
--- Quote from: T3sl4co1l on May 04, 2022, 05:37:13 pm ---I mean, how would you know? Given the above information, can you solve for the resistance (if any) in my circuit?
Tim
--- End quote ---
Any resistance value will result in half of the energy lost as heat on the resistance.
A smaller resistance will mean wires will heat faster and a larger resistance value means wires will heat slower but the exact same total amount will be wasted as heat.
With 20nF capacitors energy transferred is small so that half energy ending as heat is to small for you to be able to measure.
But use larger capacitors and a large value resistor in series then you will be able to measure that lost energy.
You know that half the energy was wasted by the end of the experiment
1F cap with 3V charging a discharged 1F capacitor
Start energy 4.5Ws end energy 2.25Ws
Current was limited by resistance so current trough a resistor means wasted heat.
You can add a 1Ohm, 100Ohm and 10kOhm between the two capacitors to increase the resistance and you will see that no matter what resistance you use you end up with the same 0.5 Vi on the two capacitors.
T3sl4co1l:
Actually, power peaks for R = Zo. If it were true that smaller resistance heats faster -- what would superconductors do? Just explode? :D
Tim
electrodacus:
--- Quote from: T3sl4co1l on May 04, 2022, 07:15:28 pm ---Actually, power peaks for R = Zo. If it were true that smaller resistance heats faster -- what would superconductors do? Just explode? :D
Tim
--- End quote ---
Of course the speed of electron wave is not infinite so there is a limit to how fast the energy will be transferred from one capacitor to the other.
Yes a superconductor will explode if the temperature increases and it gets out of the super-conduction zone while high current is passing trough it.
Superconductor has no resistance not a small resistance but absolute zero.
Any resistance value no matter how small will dissipate the same amount of energy as heat in this two capacitor example assuming proper ratio of inductance and capacitance with inductance much, much smaller than capacitance.
If you have this resonant circuit as it was in your example energy will flow in multiple directions multiple times significantly increasing the losses then you no longer end up at 0.5Vi.
That is why industrial users of electricity are charged for apparent power so that they will try and correct the power factor.
If we use this same example with two parallel capacitors but the wires are super long so switch is super far from the two capacitors.
How long to you think it will take for the first unit of energy to get to the discharged capacitor?
Will energy get to the discharged capacitor based on the short distance between the two capacitors or it will be based on the long distance between capacitors and switch ?
This should demonstrate if energy travels inside or outside the wires.
hamster_nz:
--- Quote from: electrodacus on May 04, 2022, 07:37:31 pm ---If we use this same example with two parallel capacitors but the wires are super long so switch is super far from the two capacitors.
How long to you think it will take for the first unit of energy to get to the discharged capacitor?
Will energy get to the discharged capacitor based on the short distance between the two capacitors or it will be based on the long distance between capacitors and switch ?
This should demonstrate if energy travels inside or outside the wires.
--- End quote ---
We already know the answer to that from the video - some energy will travel a shorter distance, via the EM fields, due to the changes in that fields. The bulk of the energy will start once the a stable field has built up around the whole length of the wire.
It doesn't answer if energy is transported inside or outside of the wires, just that energy can travel outside of wires (through the EM field), and that energy can travel along wires.
For me the interesting observation is that only the tiniest bit of EM energy travels the shortest distance, and after that the energy is then being supplied by changes in the EM field that are happening further and further away on the wires.
Sredni:
--- Quote from: rfeecs on May 02, 2022, 08:44:59 pm ---
--- Quote from: Sredni on May 02, 2022, 07:59:18 pm ---If so: HOW MUCH ENERGY does travel along the path?
--- End quote ---
If we assume that potential energy is located in the rock, and arbitrarily say sea level is zero energy, then zero energy travels along the surface path, and negative energy travels back along the underground path.
Between the two paths (passing through a plane perpendicular to the paths), there is a net energy flow from A to B. The amount of energy is the energy converted to heat. So it depends on which hole it goes down.
--- End quote ---
This is equivalent to say that the path the energy follows depends on where we arbitrarily choose to set the zero reference. We have seen it with the electrical system as well: for a given convention we can have all energy traveling along the top wire, for another half in the top and half in the bottom wire, and for another one yet all energy travels along the bottom wire. They can't be all true, can they?
Now, in this particular example you elected the return path because it allows you to remove the chance element, but what if I put a second cusp on the return path? Do I have to wait till the weight has fallen in the return hole to know how much energy has traveled?
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