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Resistance stays the same, loss due to resistance goes down. I’m not sure it halves either, it might be better than halving but I’m not sure myself.

Edit: Basic power loss formula is P=I^2R, so yes power loss is divided by 4 for a 2x increase in voltage assuming the target power delivered is held constant.

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Usually the resistance does not stay the same, because it is preferred to use a thinner cable, to reduce its cost.

At a given power, double voltage means half current. If the resistance is kept the same, that means 4 times lower losses. If the resistance is doubled by using a thinner cable, that still results in two times lower losses.

Yeah I agree, I was just pointing out that a wire won’t change resistance due to voltage going up. Of course notwithstanding the wire heating up or something.

Thank you for the correction.

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