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 350 A/cm2. What diameter of cylindrical wire should be used to make a fuse that will limit the current to 0.58 A?

Answer 1

Answer:

0.04594 cm

Explanation:

So, we need the wire melt when the max current density its $350 \frac{A}{cm^2}$. Now, we got our limit current, 0.58 A.

The current density its $\frac{current}{area}$, so, with our data, we can obtain the cross area of the wire.

For a cylinder, the area its given by:

$area =\pi r^2$

We can put all this in the equation for the max current density:

$density_{max} =\frac{current_{limit}}{area}$

$density_{max} =\frac{current_{limit}}{\pi r^2}$

And now, we can work it a little:

$r^2 =\frac{current_{limit}}{\pi * density_{max}}$

$r =\sqrt{ \frac{current_{limit}}{\pi * density_{max}}$

using our values, max current density = $350 \frac{A}{cm^2}$ ,   and limit current = 0.58 A,

$r =\sqrt{ \frac{0.58 A}{\pi * 350 \frac{A}{cm^2} }}$

And, of course, the diameter its two times the radius:

$d = 2 * r = 2 * 0.02297 cm$

$d = 0.04594 cm$