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 500 A/cm2. What diameter of cylindrical wire should be used to make a fuse that will limit the current to 0.64 A?

Answer 1

Answer:

The diameter of wire should be $4.04 \times 10^{-4}$ m

Explanation:

Given:

Current density $J = 500 \times 10^{4}$ $\frac{A}{m^{2} }$

Current $I = 0.64$ A

From the formula of current density,

  $J = \frac{I}{A}$

Where $A =$ area of cylindrical wire = $\pi r^{2}$

  $\pi r^{2} = \frac{I}{J}$

  $r^{2} = \frac{I}{\pi J }$

   $r = \sqrt{\frac{0.64}{3.14 \times 500 \times 10^{4} } }$

   $r = 2.02 \times 10^{-4}$m

For finding the diameter of wire,

   $d = 2r$

   $d = 4.04 \times 10^{-4}$m

Therefore, the diameter of wire should be $4.04 \times 10^{-4}$ m