Iniciado por Username is valid
(First, regardless of the temperature that the measurement shows), if a cable fires this temp, all other cables on this string must have a comparable temperature because they are thermally coupled to the mainboard and the board of the power supply.
Assuming the cable harness has a temperature of approx. 65 ° C, then 8 cables have this temperature.
According to the review, each 2x AWG 22, 20, 18, 16 <-> (0.322 | 0.518 | 0.823 | 1.31) mm²
The length is 68cm. In this way, the entire copper volume is calculated in the form of cords on the strand.
Thus (h = 680mm) * (0.322 | 0.518 | 0.823 | 1.31) mm² = (221.68 | 352.24 | 559.64 | 890.8) mm3
The sum is 2024.36mm³. The strand therefore consists of approximately 2024.36 mm³ of copper. The current density is, of course, inhomogeneous due to the different cable cross sections and the resulting current divider, but this is of no interest at the moment as it is about the total power that is passed through the strand. The transport is not loss-free due to the resistance of the copper, so a voltage drops across the strand which is proportional to the total resistance RL1 (AWG 16) || RL2 (AWG 18) || RL3 (AWG 20) || RL4 (AWG22).
The density of copper is 8.92 g / cm 3
The mass of the copper would thus be 2024.36 mm 3 * 8.92 g / cm 3/10 ^ -3 = 18.05gr of copper.
One tends to say now 18.05gr Copper would be everything that dangles at the plug, but one must not forget that we have each cable 2 times. Therefore, the calculated value is multiplied by 2 to obtain 36.1gr of copper
C = delta Q / m * delta T -> delta Q = c * m * delta T with delta T = 45 ° C; C = 0.382; M = 36.1gr.
Delta Q = 620.55 Ws -> 621 Ws
P = U * I -> I = P / U = 621Ws / 12V = 51.75As
That would be proper, it is clear that bullshit was measured. Now you look at the label of the power supply and then you know exactly
Incidentally, cable insulation is permitted up to 105 ° C