Question

A fuse in an electric circuit is a wire that is designed to melt, and thereby open the circuit, if the current exceeds a predetermined value. Suppose that the material to be used in a fuse melts when the current density rises to 440 A/cm2. What diameter of cylindrical wire should be used to make a fuse that will limit the current to 0.46 A

Answer 1

Answer:

3.64×10⁻² cm or 3.64×10⁻⁴ m

Explanation:

Current density = Current/cross sectional area.

δ = I/A................. Equation 1

Where δ = current density, I = current, A = cross sectional area of the wire.

make A the subject of the equation

A = I/δ............. Equation 2

Given: δ = 440 A/cm², I = 0.46 A

Substitute into equation 2

A = 0.46/440

A = 1.045×10⁻³ cm²

But,

A = πd²/4................ Equation 2

Where d = diameter of the wire

make d the subject of the equation

d = √(4A/π)................ Equation 3

Given: A = 1.045×10⁻³ cm², π = 3.14

Substitute into equation 3

d = √(4×1.045×10⁻³/3.14)

d = √(1.33×10⁻³)

d = 0.0364 cm

d = 3.64×10⁻² cm or 3.64×10⁻⁴ m