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23 November, 17:59

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?

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Answers (2)
  1. 23 November, 18:22
    0
    diameter is 1 cm

    Explanation:

    diameter is 1 cm
  2. 23 November, 18:22
    0
    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.
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