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 500 A/cm2. What diameter of cylindrical wire should be used to make a fuse that will limit the current to 0.64 A?

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1 day 2021-07-22T07:05:07+00:00 1 Answers 0 views 0

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    2021-07-22T07:06:12+00:00

    Answer:

    The diameter of wire should be 4.04 \times 10^{-4} m

    Explanation:

    Given:

    Current density J = 500 \times 10^{4} \frac{A}{m^{2} }

    Current I = 0.64 A

    From the formula of current density,

      J = \frac{I}{A}

    Where A = area of cylindrical wire = \pi r^{2}

      \pi r^{2} = \frac{I}{J}

      r^{2} = \frac{I}{\pi J }

       r = \sqrt{\frac{0.64}{3.14 \times 500 \times 10^{4} } }

       r = 2.02 \times 10^{-4}m

    For finding the diameter of wire,

       d = 2r

       d = 4.04 \times 10^{-4}m

    Therefore, the diameter of wire should be 4.04 \times 10^{-4} m

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