General > General Technical Chat
Veritasium "How Electricity Actually Works"
vad:
--- Quote from: electrodacus on May 15, 2022, 03:19:30 pm ---You do not understand how both of those examples work that is why you think the energy is lost.
They are not isolated systems.
--- End quote ---
The three examples are not isolated systems - they draw power from outside (Duga from Chernobyl station, wireless charge from power grid) :)
Now, are you saying that a radio transmitter does not radiate EM energy, unless there is a receiver?
bsfeechannel:
--- Quote from: Naej on May 13, 2022, 08:26:38 pm ---Right so there are plenty of problems with S=JV but you are forbidden to discuss them.
--- End quote ---
Of course there's a conspiracy of mainstream scientists whose aim is to silence any dissenters.
electrodacus:
--- Quote from: vad on May 15, 2022, 04:02:35 pm ---
--- Quote from: electrodacus on May 15, 2022, 03:19:30 pm ---You do not understand how both of those examples work that is why you think the energy is lost.
They are not isolated systems.
--- End quote ---
The three examples are not isolated systems - they draw power from outside (Duga from Chernobyl station, wireless charge from power grid) :)
Now, are you saying that a radio transmitter does not radiate EM energy, unless there is a receiver?
--- End quote ---
We are discussing if energy travels through wires or outside the wires.
For example the phone charger where you have two inductors. Say you put 1Ws of energy into transmitter coil from a source then energy will travel through wire/inductor and say 10% is lost as heat due to IR and half creates a magnetic field around it (all conserved) then the receiver inductor can supply part of that stored magnetic field to a load so say 50% of that magnetic field energy is collected by this inductor and again about 10% of that is lost as heat IR in this inductor the rest is delivered to load. Then the rest of the magnetic field can be retrieved back to the sender coil and put back in to the source or just wasted as heat.
It is basically an inefficient air transformer as the magnetic coupling is not as great as if the inductors share a common high magnetic field conductor core.
In a transformer you also put some energy in with each half wave in the primary and if the secondary is open circuit all that magnetic field will be returned to primary thus the loss will only be in the wire and a bit in the iron core.
But again main question on this thread is if energy is delivered through wires to the lamp/resistor or outside the wire.
Since electrical energy is electrical power integrated over time and electrical power is the product of electrical potential and electrical current and the current can only travel through the wire it means energy is delivered through the wire.
There is absolutely zero evidence that "energy doesn't travel in wire" main claim Derek makes based only on the small current observed at the lamp before the electron wave had the time to travel through wire. That small current is just due to line capacitance being charged so the equivalent of two capacitors on each side of the lamp that are being charged by the battery so since current can not travel through 1m of air that is the dielectric of this capacitors (not at 20V) then no energy travels outside the wire. Not during transient and not after during steady state.
IanB:
--- Quote from: electrodacus on May 15, 2022, 04:26:44 pm ---But again main question on this thread is if energy is delivered through wires to the lamp/resistor or outside the wire.
Since electrical energy is electrical power integrated over time and electrical power is the product of electrical potential and electrical current and the current can only travel through the wire it means energy is delivered through the wire.
There is absolutely zero evidence that "energy doesn't travel in wire" main claim Derek makes based only on the small current observed at the lamp before the electron wave had the time to travel through wire. That small current is just due to line capacitance being charged so the equivalent of two capacitors on each side of the lamp that are being charged by the battery so since current can not travel through 1m of air that is the dielectric of this capacitors (not at 20V) then no energy travels outside the wire. Not during transient and not after during steady state.
--- End quote ---
The small current observed at the lamp before the electron wave had the time to travel through the wire was energy emitted by the lamp. That energy had to come from the battery (there is no other power source), and it didn't travel through the wire (because there wasn't time), so it traveled through the air.
electrodacus:
--- Quote from: IanB on May 15, 2022, 05:33:12 pm ---
The small current observed at the lamp before the electron wave had the time to travel through the wire was energy emitted by the lamp. That energy had to come from the battery (there is no other power source), and it didn't travel through the wire (because there wasn't time), so it traveled through the air.
--- End quote ---
You are ignoring the capacitors formed by the transmission line wires one on each side of the lamp.
When you are charging two capacitors in series from a source you will not say that energy passed through capacitors but energy went in the capacitor and was stored. While energy was pushed into capacitors it traveled through the wires. You can can not have electric current through a dielectric in this case 1m of air and with no current you have no energy.
This graph I made to show power at the supply and power at the lamp if understood will explain everything.
The switch is only closed for 30ns then open and stays open
With green is power in mW delivered by the supply and with magenta is power used by the lamp.
The area under the graphs represents the energy.
Notice that energy from supply was delivered only during those 30ns that switch was closed circuit and as soon as circuit was broken no more energy is supplied by the source. The total energy delivered by the source in those initial 30ns is 9.16nJ as you can see in the grey window on the left last value.
The lamp was still getting energy well after the switch was closed. Most of the energy arrives at the lamp after the switch was closed and first large chunk exactly after the energy had the time to travel the entire distance through wire.
In the end 7.68nJ of energy where delivered to the lamp and the difference is all accounted by energy lost as heat in the wire's
There is no energy radiated away all energy is accounted for and all will eventually be radiated as photons form the lamp and infrared photons from the wire as heat is lost to environment.
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