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 390 A/cm2. What diameter of cylindrical wire should be used to make a fuse that will limit the current to 0.67 A

Answer 1

0.0468 cm or 4.68×10⁻⁴ m

Step-by-step explanation:

Current Density = Current/Cross sectional area.

J = I/A.................... Equation 1

Where J = current density, I = current, A = Cross sectional area of the wire.

Since the wire is cylindrical in shape,

A = πd²/4................ Equation 2

Where d = diameter of the cylindrical wire, π = pie

Substitute equation 2 into equation 1

J = 4I/πd²

Make d the subject of the equation

d = √(4I/Jπ)................ Equation 3

Given: I = 0.67 A, J = 390 A/cm², π = 3.14

Substitute into equation 3

d = √[(4×0.67)/(390×3.14)]

d = √(0.002188)

d = 0.0468 cm.

Hence the diameter of the wire = 0.0468 cm or 4.68×10⁻⁴ m

Answer 2

0.047 cm

Step-by-step explanation:

Parameters given:

Current density, J = 390 A/cm²

Current, I = 0.67A

Current Density, J, is given as:

J = I / A

Area, A, is given as:

A = pi * r² = pi * d²/4

Where d = diameter of loop

J = I / (pi * d²/4)

J = (4 * I) / (pi * d²)

Making d² subject of formula:

d² = (4 * I) / (pi * J)

d² = (4 * 0.67) / (3.142 * 390)

d² = 0.0021871

=> d = 0.047 cm