I don't quite understand what your calculations are or what you are trying to say but 2.5 kw doesn't seem right for losses in the conductor. Could you explain in more detail? I have a pretty good understanding of what I'm doing for the conductors because I have a power inverter which has 150 amp rails. If I'm not understanding something I should know before I build these rails.
I'm basically saying you are not going to have 500 amps in a home build construction. It is beyond reasonable limits for an operating current. To give you an idea of what 500 amps means, your bus bars to carry that current would have to be copper rods 0.5 inches in diameter, and switching it would require components and heat sinking of a size far beyond what is available to the home constructor.
What the linked photo was intended to show was that 500 amps is the kind of current you find in electric traction supplies for trains, and the gear to handle that current is enormous and heavy.
Broadly speaking, what my calculations were trying to show is that conductor size is roughly proportional to the square of the current. So in going from 150 amps to 500 amps everything basically has to be ten times bigger. Circuits that electronically switch 500 amps are built with big metal lugs, bolts and copper cables as thick as ropes. (With powerful fans for cooling.)