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 540 A/cm2. What diameter of cylindrical wire should be used to make a fuse that will limit the current to 0.57 A

0.0366 m

Explanation:

We are given;

Current density; J = 540 A/cm² = 540 × 10⁴ m

Current; I = 0.57 A

Now, formula for current density is;

J = I/A

Where A is area = πr²

Thus;

J = I/(πr²)

r = √(I/(Jπ))

r = √(0.57/(540π))

r = 0.0183 m

Diameter = 2 × radius

Diameter = 2 × 0.0183

Diameter = 0.0366 m

it takes about 30 days for it to break down

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explanation: