joule heating - transmitting power at higher voltages reduces resistive loss?
Wire 1:

V = 1000V

P = 500W

I = P/V = 500W / 1000V = 0.5A

R_wire = 2 Ohm

Loss in wire 1 = I^2*R = 0.5A^2 * 2 Ohm = 0.5W  

Wire 2:

V = 10000V

P = 500W

I = P/V = 500W / 10000V = 0.05A

R_wire = 2 Ohm

Loss in wire 2 = I^2*R = 0.05A^2 * 2 Ohm = 0.005W 

UPDATE:

To work out the exact loss, assuming voltage is measured at start of wire:







simulate this circuit &ndash; Schematic created using CircuitLab

Total current, I = V1 / (R1 + R2)

Load power, P2 = 500W = I^2 * R2 = (V1^2 R2) / (R1 + R2)^2

This equation can be rearranged into a quadratic form and R2 solved for. Once R2 is found, the loss in the wire can be found by I^2 * R1

To find it using the P = VI formula, calculate V2 as I * R2, the loss is given by (V1 - V2) * I.