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 a wire is given by the equation:$$P = I^2R$$where:* P is the power in watts* I is the current in amperes* R is the resistance in ohmsThe resistance of a wire is given by the equation:$$R = \rho \frac{L}{A}$$where:* R is the resistance in ohms* ρ is the resistivity of the material in ohm-meters* L is the length of the wire in meters* A is the cross-sectional area of the wire in square metersIn this problem, we are given the following information:* ρ = $2.44 \times 10 ^{- 8} \Omega \cdot \mathrm { m }$* L = 14 cm = 0.14 m* d = 1.8 mm = 0.0018 m* I = 480 mA = 0.48 AFirst, we need to calculate the cross-sectional area of the wire:$$A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.0018 \mathrm { m}}{2}\right)^2 = 2.54 \times 10^{-6} \mathrm { m}^2$$Now we can calculate the resistance of the wire:$$R = \rho \frac{L}{A} = (2.44 \times 10^{-8} \Omega \cdot \mathrm { m}) \frac{0.14 \mathrm { m}}{2.54 \times 10^{-6} \mathrm { m}^2} = 0.14 \Omega$$Finally, we can calculate the power dissipated in the wire:$$P = I^2R = (0.48 \mathrm { A})^2 (0.14 \Omega) = 0.031 \mathrm { W} = 31 \mathrm { mW}$$Therefore, the correct answer is A) 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 $2.44 \times 10 ^{- 8} \Omega \cdot \mathrm { m }$