Energy is amp x volt. Same energy faster is more energy in same time, be it amps or volt. Dunno if your grid can bear it multiple times in each city but still better buffer it. And more volts needs more gum or you get the volts.
That second formula is for how much power gets dissipated in a resistance (hence the R in it) , not how much power travels through a line.
That said the previous poster was indeed incorrect - the required thickness of a cable through which a certain amount of power passes depends only on current, not voltage: make it too thin and it can literally melt with a high enough current and the formula of the power it is dissipating as heat that can cause it to melt is that second formula of yours and the R in that formula is inverselly proportional to the cross-cut area of the cable, which for a round cable is the good old area of a circle formula which depends on the square of the radius - in other words the thicker the cable the less current it can take without heating up too much or, putting it the other way around, the more current you want to safely pass through a cable the thicker it needs to be.
In summary, thinner cables heat up more with higher currents (and if they heat up enough they melt) because even pure copper has some resistance and the thinner the cable the higher the resistance. If you need to move Power, not current specifically (such as to charge something), you can chose more current or to have a higher voltage (because P = V x I), and chosing a higher current means you need thicker cables (because as explained above the cables would overheat and even melt otherwise) but a higher voltage doesn’t require a thicker cable.
Energy is amp x volt. Same energy faster is more energy in same time, be it amps or volt. Dunno if your grid can bear it multiple times in each city but still better buffer it. And more volts needs more gum or you get the volts.
P=i x v
But also
P=i^2 x r
Power goes up with linearly with voltage but to thw square or I.
That second formula is for how much power gets dissipated in a resistance (hence the R in it) , not how much power travels through a line.
That said the previous poster was indeed incorrect - the required thickness of a cable through which a certain amount of power passes depends only on current, not voltage: make it too thin and it can literally melt with a high enough current and the formula of the power it is dissipating as heat that can cause it to melt is that second formula of yours and the R in that formula is inverselly proportional to the cross-cut area of the cable, which for a round cable is the good old area of a circle formula which depends on the square of the radius - in other words the thicker the cable the less current it can take without heating up too much or, putting it the other way around, the more current you want to safely pass through a cable the thicker it needs to be.
In summary, thinner cables heat up more with higher currents (and if they heat up enough they melt) because even pure copper has some resistance and the thinner the cable the higher the resistance. If you need to move Power, not current specifically (such as to charge something), you can chose more current or to have a higher voltage (because P = V x I), and chosing a higher current means you need thicker cables (because as explained above the cables would overheat and even melt otherwise) but a higher voltage doesn’t require a thicker cable.