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

The diameter of wire should be
`4.04 * 10^(-4)` m

Step-by-step explanation:

Given:

Current density
`J = 500 * 10^(4)`
`(A)/(m^(2) )`

Current
`I = 0.64` A

From the formula of current density,

`J = (I)/(A)`

Where
`A =` area of cylindrical wire =
`\pi r^(2)`

`\pi r^(2) = (I)/(J)`

`r^(2) = (I)/(\pi J )`

`r = \sqrt{(0.64)/(3.14 * 500 * 10^(4) ) }`

`r = 2.02 * 10^(-4)`m

For finding the diameter of wire,

`d = 2r`

`d = 4.04 * 10^(-4)`m

Therefore, the diameter of wire should be
`4.04 * 10^(-4)` m