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31 August, 08:48

A gold wire is 2.40 m long and 0.760 mm in diameter, and it carries a current of 1.44 A. What is the potential difference between the ends of the wire? Let p gold = 2.44 x 10^-8 Ω * m.

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  1. 31 August, 10:30
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    0.186 v

    Explanation:

    I = V/R

    R = pL/A = 4pL / (3.14159*d^2)

    R = (4 * 2.44 * 10^-8 * 2.40) / 3.14159 * (0.760*10^-3) ^2

    R = 0.129 ohm

    V = I * R

    V = 1.44 * 0.129

    V = 0.186 v
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