The resistivity of gold is $2.44 \times 10 ^ { - 8 } \Omega \cdot \mathrm { m }$ at a temperature of 20°C. A gold wire, 1.8 mm in diameter and 14 cm long, carries a current of 480 mA. What power is dissipated in this wire at 20°C? A)0.31 mW B)0.077mW C)0.14 mW D)0.19 mW E)0.25 mW

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Quizplus · Accepted Answer

The power dissipated in the wire can be calculated using the formula $ P = I^2 R $, where $ R = \rho \frac{L}{A} $. First, calculate the resistance $ R $ using the resistivity $ \rho = 2.44 \times 10^{-8} \Omega \cdot \mathrm{m} $, length $ L = 0.14 \, \mathrm{m} $, and cross-sectional area $ A = \pi \left(\frac{1.8 \times 10^{-3}}{2}\right)^2 $. Then, use the current $ I = 0.48 \, \mathrm{A} $ to find the power. The correct power dissipation is approximately 0.31 mW.

The Resistivity of Gold Is 2.44×10−8Ω⋅m2.44 \times 10 ^ { - 8 } \Omega \cdot \mathrm { m }2.44×10−8Ω⋅m

The Resistivity of Gold Is $2.44 \times 10 ^ { - 8 } \Omega \cdot \mathrm { m }$