Earlier quoted context omitted.
> Any imbalance I watched de-Bauer's analysis this morning, and you've seemingly hit the nail on the head. Even on his test bench it looks like only two of the wires are carrying all of the power (instead of all of them, I think 4 would be nominal?) - using a thermal camera as a measuring tool. The melted specimen also has a melted wire. Maybe 24V or 48V should be considered, and higher gauge wires - yes.
It would be _lovely_ if instead of the 12V only spec we went to 48V for internal distribution. Though that would require an ecosystem shift. USB-PD 2.0~3.0 would also be better supported https://en.wikipedia.org/wiki/USB_hardware#USB_Power_Deliver... As others no doubt mention Power (loss, Watts) = I (amsp) * V (volts (delta~change on the wire)). dV = I*R ==> dV = I * I / R -- That is, other things being equal, amps…
It's one-sixteenth (6.25%) actually. You correctly note that resistive losses scale with the square of the current (and current goes with reciprocal voltage), so at 4 times the voltage, you have 1/4th the current and (1/4)^2 = 1/16th the resistive losses.