Aeronaut wrote: Long distance transmission lines double the voltage to reduce the current by 50%. Using P=(I^2) R, doubling the transmission voltage results in 25% of the losses. Not sure how this relates to raising voltage to raise current…

There are 2 things here. The wasted energy in collisions and the free electrons. The wasted energy will go like P=(I^2)*R anyway, like in a wire. This is a fraction of the electrons that effectively behave like they were in a resistive medium.The increase of I of an input just makes things go even more wasted.

But note that “P” is not the total energy of the capacitor banks, just the part where it is wasted. So, by increasing the total voltage AND keeping BOTH total resistance (which would happen naturally anyway) and total current CONSTANT, is the same as diminishing the current that goes through the resistive process. So, not only R goes down, as also I (resistive) go down so that conservation of energy is respected. So, more current will be available to be absorbed by the pinch.