Answer:
The diameter of the cylindrical wire is 0.0446 cm or 44.6 mm
Explanation:
Given;
current density of the fuse, J = 480 A/cm²
current in the fuse, I = 0.75 A
Current density, J = Current (I) / Area (A)
[tex]Area \ of \ the \ cylindrical \ wire = \frac{Current \ (I)}{Current \ density \ (J)} = \frac{0.75}{480} = 1.5625*10^{-3} \ cm^2[/tex]
[tex]Area \ of \ the \ cylindrical \ wire, A = \frac{\pi d^2}{4} \\\\d^2 =\frac{4*A}{\pi } \\\\d = \sqrt{\frac{4*A}{\pi } } = \sqrt{\frac{4*1.5625*10^{-3}}{\pi } }= 0.0446 \ cm =44.6 \ mm[/tex]
diameter of the wire = 0.0446 cm or 44.6 mm
Therefore, the diameter of cylindrical wire that should be used to make a fuse that will limit the current to 0.75 A, when the current density rises to 480 A/cm² is 44.6 mm
