A cylindrical cable with 2.38 x 10-8 m resistivity carries a current of 1100 A. There is a potential difference of 1.5 x 10-2 V between two

Question

A cylindrical cable with 2.38 x 10-8 m resistivity carries a current of 1100 A. There is a potential difference of 1.5 x 10-2 V between two points on the cable that are 0.22 m apart. What is the radius of the cable

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Tryphena 2 months 2021-07-30T13:06:06+00:00 1 Answers 3 views 0

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    2021-07-30T13:07:17+00:00

    Answer:

    11.1 mm

    Explanation:

    The resistance of the cable R = ρl/A where ρ = resistivity of cable = 2.38 × 10⁻⁸ Ωm, l = distance between the two points on the cable = 0.22 m and A = cross-sectional area of cable = πr² where r = radius of cable. Also, from Ohms’ law, R = V/I where V = potential difference between the two points = 1.5 × 10⁻² V and I = current in cable = 1100 A

    So,

    ρl/A = V/I

    A = ρlI/V

    πr² = ρlI/V

    r² = ρlI/πV

    taking square root of both sides, we have

    r = √(ρlI/πV)

    substituting the values of the variables into the equation, we have

    r = √(ρlI/πV)

    r = √(2.38 × 10⁻⁸ Ωm × 0.22 m × 1100 A/[π × 1.5 × 10⁻² V)

    r = √(575.96 × 10⁻⁸ Ωm²A/4.712 × 10⁻² V)

    r = √(122.2225 × 10⁻⁶ m²)

    r = 11.06 × 10⁻³ m

    r = 11.06 mm

    r ≅ 11.1 mm

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