Respuesta :
Answer:
0.45 mm
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A = [tex]\frac{\pi d^{2} }{4}[/tex]
where d is the diameter of the wire
1.613 x 10^-7 = [tex]\frac{3.142* d^{2} }{4}[/tex]
6.448 x 10^-7 = 3.142 x [tex]d^{2}[/tex]
[tex]d^{2}[/tex] =[tex]\sqrt{ 2.05*10^-7}[/tex]
d = 4.5 x 10^-4 m = 0.45 mm
The value of the diameter of the wire is 0.45 mm.
Given that:
Current in the fuse = 1 Ampere
Current density = 620 A/cm²
Area of the wire = [tex]\dfrac{\text I}{\rho}[/tex]
Area = [tex]\dfrac{1}{620}[/tex]
Area = 1.613 x 10⁻³ cm²
Also, we know that:
10000 cm² = 1 m²
1.613 x 10⁻³ cm² = 1.613 x 10⁻⁷ m²
The area of a wire can be calculated as:
Area = [tex]\dfrac{\pi \text d^2}{4}[/tex]
where, d = diameter
Substituting the values in the above equation, we get:
1.613 x 10⁻⁷ = [tex]\dfrac{3.14 \times \text d^2}{4}[/tex]
6.448 x 10⁻⁷ = 3.142 x d²
Hence, the value of d will be:
d² = 2.05 x 10⁻⁴
d = [tex]\sqrt{2.05 \times 10^{-7}}[/tex]
d = 4.5 x 10⁻⁷
Thus, the value of the diameter of the wire is 0.45 mm.
To know more about current, refer to the following link:
https://brainly.com/question/2285102
