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 459 A/cm2. What diameter of cylindrical wire should be used to limit the current to 0.53 A?

Answer 1

Given Information:

Current density = j = 459 A/cm²

Limiting current = I = 0.53 A

Required Information:

Diameter of wire = d = ?

Answer:

Diameter of wire = d = 0.0383 cm

Step-by-step explanation:

We want to limit the current in a fuse wire and for that we want to find the required diameter of the fuse wire.

We know that current density is current divided by area

j = I/A eq. 1

where A is given by

A = πr²

also we know that

r = d/2

A = π(d/2)²

A = πd²/4

so eq. 1 becomes

j = I/πd²/4

d² = 4I/jπ

d = √4I/jπ

d = √4*0.53/459π

d = 0.0383 cm

Therefore, the diameter of the required fuse wire is 0.0383 cm