It's a cost vs. energy carried trade-off. Both systems have resistive loses proportional to the square of the current. However: 1. Total power transferred in

It's a cost vs. energy carried trade-off. Both systems have resistive loses proportional to the square of the current. However: 1. Total power transferred in a DC system is proportional to the voltage, whereas power transferred in an AC system is proportional to the RMS voltage (which is roughly 0.7 of nominal for a sine wave), so more energy is transmitted at the same current level in HVDC. 2. AC systems manifest impedance which has a resistive (aka DC) component as above as well as a reactive (aka AC) component, i.e. Z = R + jX. In DC systems X = 0. In a theoretical transmission line no energy is absorbed or supplied from line reactance, but in practice we have to transmit a certain amount of reactive power (VARs) to charge the line capacitance/inductance each AC cycle. This reduces the amount of our current capacity (limited by thermal constraints) that actually carries current that can be delivered to the load as active power (watts). This effect is somewhat although not directly proportional to distance (characteristic impedance has no dependence on line length, but voltage drops along the line due to resistive effects meaning the variation from the optimal reactive power-minimizing voltage level increases). The effect of (1) and (2) is that for any given conductor, at a given voltage level, more usable energy can be transmitted with DC than AC, and that differential increases with distance. That being said, building DC converter and switching stations is much more expensive than AC. So for a shorter line, or one that has many switching stations, I could counter the above by simply generating 5-8% more power at the generating station and still come out ahead (because in real engineering everything is about $). Therefore, DC is only more cost-effective ($/MVA of energy delivered) at long distances.