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

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 440 A/cm2. What diameter of cylindrical wire should be used to make a fuse that will limit the current to 0.46 A

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RI SƠ 1 week 2021-07-21T13:23:40+00:00 1 Answers 2 views 0

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    2021-07-21T13:25:03+00:00

    Answer:

    3.64×10⁻² cm or 3.64×10⁻⁴ m

    Explanation:

    Current density = Current/cross sectional area.

    δ = I/A…………….. Equation 1

    Where δ = current density, I = current, A = cross sectional area of the wire.

    make A the subject of the equation

    A = I/δ…………. Equation 2

    Given: δ = 440 A/cm², I = 0.46 A

    Substitute into equation 2

    A = 0.46/440

    A = 1.045×10⁻³ cm²

    But,

    A = πd²/4……………. Equation 2

    Where d = diameter of the wire

    make d the subject of the equation

    d = √(4A/π)……………. Equation 3

    Given: A = 1.045×10⁻³ cm², π = 3.14

    Substitute into equation 3

    d = √(4×1.045×10⁻³/3.14)

    d = √(1.33×10⁻³)

    d = 0.0364 cm

    d = 3.64×10⁻² cm or 3.64×10⁻⁴ m

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