What diameter wire of this material will do the job?

Question

Design a 0.60 A fuse that "blows" if the current exceeds 0.60 A . The fuse material in your stockroom melts at a current density of 450 A/cm^2 .

What diameter wire of this material will do the job?

Answer 1

Solution:

The fuse material strength has a density of 450 Amp/cm². The fuse has an allowable current of 0.6Amp. The area of the wire needed for the fuse would be the allowable current divided by the density of the material strength

Required Area = A = 0.6 Amp / 450 Amp /cm² = 1.333 x10^-3 cm²

The diameter of the wire can be found from the area formula, A = π D²/4 ., rewriting D² = 4 A / π

D² = 4 ( 1.333 x10^-3 cm²) / π = 1.693 x10^-3 cm², and D= √1.639x10^-3 cm² = 0.041 cm = 0.41 mm