Re:A Room-Temperature Superconductor? New Developments
by Workaccount2 on Tuesday August 01, 2023 @03:26PM
(#36961272)
>Higher resistance means more of the electrical energy is converted to heat per "time".
This isn't really accurate because increasing the resistance of a resistor in a given circuit will actually decrease the amount of heat dissipated.
W = IV where I = V/R plug that in we get W = (V/R)V = V^2/R
So Watts = Volts^2/Resistance. Increase resistance, decrease watts.
Its better to just say that resistance converts voltage to heat, and leave it at that. Also is why in the orginal paper (and other superconductor work) they measure voltage drop across the conductor. No voltage drop(loss) = no resistance.
Re:A Room-Temperature Superconductor? New Developments
by delusional on Tuesday August 01, 2023 @05:04PM
(#36962897)
Yeah, my explanation is clearly wrong since it also breaks down at a pretty obvious extreme. If your resistor is non-conductive (infinitely resistive) my model would suppose it would convert all the electricity into heat. What would actually happen is that no electricity would flow and we would therefore get no heat.
I presented it as an intuitive "feel" based idea of what a resistor does. It's very much not a numerically useful or physically accurate one.
Magnetic fields moving through conductors induce electrical energy in the conductors. Normally this is quickly dissipated as heat, but in a superconductor this energy can't go anywhere, and the conductor therefore can't move through the magnetic field.
We've already done a lot of experiments with superconductors, since we've found some that work at extreme cold, but room temperature superconductors would allow us to productive some of those cool ideas by making them economically viable.