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Suppose that the material to be used in a fuse melts when the current density rises to 459 A/cm2. What diameter of cylindrical wire should be used to limit the current to 0.53 A?

User Ganjim
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2 Answers

4 votes

Answer:

diameter is 1 cm

Step-by-step explanation:

diameter is 1 cm

User Nastya Kholodova
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4.2k points
2 votes

Answer:

The required diameter of the fuse wire should be 0.0383 cm to limit the current to 0.53 A with current density of 459 A/cm² .

Explanation:

We are given current density of 459 A/cm² and we want to limit the current to 0.53 A in a fuse wire. We are asked to find the corresponding diameter of the fuse wire.

Recall that current density is given by

j = I/A

where I is the current flowing through the wire and A is the area of the wire

A = πr²

but r = d/2 so

A = π(d/2)²

A = πd²/4

so the equation of current density becomes

j = I/πd²/4

j = 4I/πd²

Re-arrange the equation for d

d² = 4I/jπ

d = √4I/jπ

d = √(4*0.53)/(459π)

d = 0.0383 cm

Therefore, the required diameter of the fuse wire should be 0.0383 cm to limit the current to 0.53 A with current density of 459 A/cm² .

User Eliott
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