No it depends on power, but the power dissipated by the cable, not the power through the cable. The power dissipated is i^2 * r, where r is the resistance of the cable and i, crucially, is the current through the cable which depends on the power it's supplying (which with a resistive load, in this case, is i * 12v).
Yes, but it seems the connectors, not the entire cable are too high resistance.
Using a larger diameter wire would drop resistance in the cable, but if it has to go through the same connector, it will likely still get hot.
Also, NVDA might be telling the truth about poorly seated connectors, that could raise resistance and heat significantly. That could also be handwaving away a business decision to move forward with a design with too little margin.
I’m not sure what you’re trying to suggest here… the PSU is also connected to the wall with a C13 connector, and is able to supply 600W at 12VDC to the card.
Yes, but the voltage on the wall side of the psu is not 12V it's 120-240V. And 600W at 120V requires only 5A of current.
It's easier to handle more power at higher voltages because heat produced depends on current and not on voltage.
If you take that puny C13 connector and move it to the 12V side with its 50A it would just melt.
