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| PlainName:
--- Quote from: hamster_nz on July 01, 2022, 10:38:15 pm ---That may all be true, but is 1x or 2x the energy being converted to infrared by a resistor flowing in the top wire in each circuit? Everybody agrees that both circuits have the same currents and voltages and powers and energy flows. I would have though it is an easy qn. If energy flows only in wires, how much energy is flowing in the upper wire for both circuits? Because the currents and voltages are consistent, so should the energy... Isn't it as simple as remover the top wire and see how much the power in the resistors changes? --- End quote --- That's an interesting setup and question. |
| electrodacus:
--- Quote from: hamster_nz on July 01, 2022, 10:38:15 pm ---That may all be true, but is 1x or 2x the energy being converted to infrared by a resistor flowing in the top wire in each circuit? Everybody agrees that both circuits have the same currents and voltages and powers and energy flows. I would have though it is an easy qn. If energy flows only in wires, how much energy is flowing in the upper wire for both circuits? Because the currents and voltages are consistent, so should the energy... Isn't it as simple as remover the top wire and see how much the power in the resistors changes? --- End quote --- Question is simple you just seems to do not know what you want to ask. Or do not understand what energy is. If you remove the top wire in first diagram then there will be no current through the top resistor just the bottom resistor will be powered by the bottom battery. If you remove the top wire in the second diagram then there is zero current thus zero power and so zero energy which is power integrated over time. |
| hamster_nz:
--- Quote from: electrodacus on July 01, 2022, 10:51:35 pm --- --- Quote from: hamster_nz on July 01, 2022, 10:38:15 pm ---That may all be true, but is 1x or 2x the energy being converted to infrared by a resistor flowing in the top wire in each circuit? Everybody agrees that both circuits have the same currents and voltages and powers and energy flows. I would have though it is an easy qn. If energy flows only in wires, how much energy is flowing in the upper wire for both circuits? Because the currents and voltages are consistent, so should the energy... Isn't it as simple as remover the top wire and see how much the power in the resistors changes? --- End quote --- Question is simple you just seems to do not know what you want to ask. Or do not understand what energy is. If you remove the top wire in first diagram then there will be no current through the top resistor just the bottom resistor will be powered by the bottom battery. If you remove the top wire in the second diagram then there is zero current thus zero power and so zero energy which is power integrated over time. --- End quote --- So you firmly believe that a different quantity of energy is flowing in the top wire, even though the currents and voltages are the same through each element? And removing a wire that carries no energy is resposible for this difference? |
| aetherist:
:popcorn: --- Quote from: Naej on June 29, 2022, 11:45:38 pm --- --- Quote from: aetherist on June 27, 2022, 11:09:05 pm ---Sredni Naej electrodacus & Everybody & Co. What is a wire that has zero resistance? What is a wire that is a perfect conductor? What is a wire that is a superconductor? What are the differences? --- End quote --- I'm not sure what you'll do with the answer but since you asked: - a conductor is usually (not always) modeled with Ohm's law, J=sigma*E - a wire that has zero resistance/a perfect conductor is a conductor in the limit of sigma -> 0. - a superconductor is something which obeys London's equations, i.e. Jsc proportional to -A and div A=0. For more precision (macroscopic quantum effects for example), you have to take other models like Ginzburg-Landau ( https://en.wikipedia.org/wiki/Ginzburg%E2%80%93Landau_theory ). --- End quote --- I notice that wikileaks says that Ohm was crucifyd on the electric forums of the day, but in the end they named "Ohm" (resistance) after him, plus they called his Law "Ohm's Law" – i wonder what unit they will end up naming using my name -- we have kinda run outa new units that need naming -- & what Law they will name using my name (i will havta invent one)(it will need an equation)(damn—my math is weak)…. "…. Ohm's law was probably the most important of the early quantitative descriptions of the physics of electricity. We consider it almost obvious today. When Ohm first published his work, this was not the case; critics reacted to his treatment of the subject with hostility. They called his work a "web of naked fancies"[11] and the Minister of Education proclaimed that "a professor who preached such heresies was unworthy to teach science."[12] The prevailing scientific philosophy in Germany at the time asserted that experiments need not be performed to develop an understanding of nature because nature is so well ordered, and that scientific truths may be deduced through reasoning alone.[13] Also, Ohm's brother Martin, a mathematician, was battling the German educational system. These factors hindered the acceptance of Ohm's work, and his work did not become widely accepted until the 1840s. However, Ohm received recognition for his contributions to science well before he died ….. " Old (electron) electricity i think says that resistivity is due to energy transfer (loss) tween jostled electrons (ie the drifting electrons) & the Cu atoms of the wire, thusly giving a temperature rise, & heat loss. As a starting point for my new (electon) electricity i use the same definition for resistivity. Skoolkids might be satisfyd with that definition/explanation, but, if u look at it, it is not an explanation at all. That kind of explanation works for a car. The max speed depends on the power pushing & the resistance retarding. In/on a wire it is the voltage pushing & the ohms resisting. For a car we can theoretically calculate the power, & the resistance, & the top speed. For a wire we can calculate the power, but we cant calculate the amps, koz we cant calculate the resistance. We don’t know what electric resistance iz or iznt. Hence there is no possible calculation from first principles. My electons i suppose fill up the available surface of the wire. They saturate the surface. The saturation is i suppoze limited by the available area & the available supply (eg lead acid battery). Re the area, electons probly jostle each other, shoulder to shoulder, like pedestrians. Or, praps they (photons) are like helical coil springs, in which case they can occupy the same space/place (if the coils are not tight), in which case they can pass through each other (eg as they cross, or as they meet head-on when going in opposite directions). Photons have a helical central main body, with photaenos radiating out to infinity (ie the em radiation). The supply/source will involve a pressure (voltage), & an amperage (the nett number of electons per second)(electons going the wrong way will rob amps). The resistance of the wire is a necessary factor that helps electons to hug the wire, else they would fly away & be lost as ordinary free photons. Or praps they would form a loop & bekum free-ish electrons (& would either stick to the wire or fly away). Metals that are poor conductors (eg Fe) would have a lower saturation level, ie Fe might accommodate say a half of the number of electons compared to the same sized Cu wire, for a given voltage. At saturation the backpressure would be such that the numbers of electons leaving the negative terminal of the battery (into the wire) would equal the number leaving the wire (into that terminal). If the wire is made of say plastic then no electons would leave the negative terminal (koz in effect the surface of plastic is saturated when there are zero electons on the surface)(what i mean is that an electon can't hug plastic). Anyhow, even if electons can pass through each other, & can in a way occupy the same place (or nearly)(temporarily), there must be some sort of limit of congestion (ie u can't have an infinite number of electons on a wire). Photaenos fight for the use of the available aether (koz everything that we see & feel is a process of the aether)(eg an excitation or vibration or pulsation or spin etc)(possibly including an annihilation of the aether). Re annihilation, a photon is largely a hole in the aether, the hole propagating at the speed of light. Photons make everything that we see & feel, hence matter iz made of holes in the aether (not hard little nuts). So, where woz i – the surface of the Pb of the negative terminal of the lead acid battery is saturated with electons going in every direction. But there is no heat loss/gain in/on the terminal, even tho we know that the Pb has some Ohms, ie the Pb has some resistance. How is it that there is no heat gain/loss in the Pb? The reason is that when 2 electons propagate in opposite directions their magnetic fields cancel, leaving only a doubled charge field. It is the magnetic field that jostles the free-ish conduction electrons in the Cu. Zero nett field equals zero jostling equals zero heat gain/loss. Anyhow, i need to fathom resistivity & conductivity & saturation etc. Still thinking. |
| electrodacus:
--- Quote from: hamster_nz on July 01, 2022, 11:43:19 pm --- So you firmly believe that a different quantity of energy is flowing in the top wire, even though the currents and voltages are the same through each element? And removing a wire that carries no energy is resposible for this difference? --- End quote --- What are you talking about ? By top wire I hope you mean the wire between the positive of top battery and top resistor ? If you remove that in first circuit then obviously can not be any current trough that wire as it is missing and so also no current from the top battery and no current through the top resistor. But since there is a wire in the middle the lower resistor and battery will see a current 1A if battery is 10V on a 10Ohm resistor so 10W total as only half the circuit is closed circuit. In second diagram since the middle wire is not present removing the top wire means no current at all and so no energy is being used. |
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