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 520 A/cm2. What diameter of cylindrical wire should be used to make a fuse that will limit the current to 0.62 A

Answer 1

Answer:

0.0389 cm

Explanation:

The current density in a conductive wire is given by

$J=\frac{I}{A}$

where

I is the current

A is the cross-sectional area of the wire

In this problem, we know that:

- The fuse melts when the current density reaches a value of

$J=520 A/cm^2$

- The maximum limit of the current in the wire must be

I = 0.62 A

Therefore, we can find the cross-sectional area that the wire should have:

$A=\frac{I}{J}=\frac{0.62}{520}=1.19\cdot 10^{-3} cm^2$

We know that the cross-sectional area can be written as

$A=\pi \frac{d^2}{4}$

where d is the diameter of the wire.

Re-arranging the equation, we  find the diameter of the wire:

$d=\sqrt{\frac{4A}{\pi}}=\sqrt{\frac{4(1.19\cdot 10^{-3})}{\pi}}=0.0389 cm$