Hi E. Davies; Re: "True, but since I=V/R Power dissipation is also V^2/R. Does that mean it's independent of current too?"
Now you're being silly, and you dam well know it. V^2/R is a perfectly good formula for power, but the V in this formula is not the voltage delivered over the wire (and therefore part of the power delivered through the wire), it's the voltage drop across the wire. In order to compute the percentage loss, you have to divide the total power by the power loss, and that is something that doesn't depend on the voltage transmitted, until you start ionizing stuff.
Re: "Personally I don't believe in the concept of zero resistance. That goes beyond reason. Nothing physical is *ever* zero (or infinite)."
Interesting comment. In a way, I agree, though the physicists would likely argue that what we call the non zero resistance of the wire is not a part of the superconductor, but instead is due to defects .
Re: "If a part of the superconductor sees a greater voltage drop than the rest, as a result of an inhomogeneity, this part will heat up even faster" I didn't write this, and I don't remember seeing it, so I don't know what the context is. But it is possible for a superconducting wire to have a voltage drop across it and still have zero resistance. Just put an AC signal into any transmission line and you'll see the same thing.
-- Carl |
