question

More expert minds will add to this, I’m sure, but I think of fusing as a protection of the cabling more than the devices connected in a properly sized system. You are on the right track about figuring the load, and should size cable to that. I use the nearest size fuse above that highest expected load, which puts me safely under the cable rating, while accounting for the anticipated start/run loads of the circuit.

Best wishes! Rick

Hi

For the section of the cables, this also depends on the outward and return length between the two elements to be connected. You can use the following formula:

Ω = copper 0.0179 (but be careful with aluminum: 0.028!)

S = ρ x A / 0.03 x V or

S = (Ω x (L x 2)) x A / 0.03 x V or

S = (Ω x (L x 2)) x (W / V) / 0.03 x V

Example with 2 meters of cable, outward and return, in 12V:

S = 0.0179 x (2 x 2) x (400/12) / 0.03 x 12 -> S = 6.62 mm2 -> 10 mm2 is the section to be installed because the previous standardized section (6 mm2) is exceeded.

For the battery, with 1 meter of cable, outward and return, in 12V:

S = 0.0179 x (1 x 2) x 405 / 0.03 x 12 -> S = 40.275 mm2 -> 50 mm2

For fuses, the formula is often used:

A x 1.25 or

(W / V) x 1.25

Example with 400W peak and 12V:

(400/12) x 1.25 = 41.667 -> 45A. But it also depends on the connected elements. This is for the W peak. If elements do not have a strong load at start-up, the 250W gives 30A rounding. Everything must be taken into account. But in all cases, the wiring in mm2 is sectioned according to the W peak.

Hope this helps you.

Over the years I have collected some info on sizing.

One is: Fuse size is 1.52 x the amps you use(33.4) gives 50.77A.

Round off to 50Amp.

Althoug 35mm calbe looks oversized, this will be also my choice.