Ask Question
15 December, 17:09

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

+1
Answers (1)
  1. 15 December, 18:22
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your 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 ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers