The resistivity of gold is 2.44 × 10^-8Ω·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.The power dissipated in the wire is closest to:

Question

Quizplus · Accepted Answer

The power dissipated in the wire is given by P = I^2 * R, where I is the current and R is the resistance. The resistance of the wire is given by R = ρ * L / A, where ρ is the resistivity, L is the length, and A is the cross-sectional area. Substituting the given values, we get:R = (2.44 × 10^-8 Ω·m) * (0.14 m) / (π * (1.8 mm / 2)^2) = 0.011 ΩP = (0.48 A)^2 * 0.011 Ω = 0.0026 W = 2.6 mWThe closest answer choice is A) 0.31 mW.

The Resistivity of Gold Is 2