The resistivity of gold is 2.44 × 10^-8 Ω ∙ m at room temperature. A gold wire that is 1.8 mm in diameter and 11 cm long carries a current of 170 mA. How much power is dissipated in the wire?

Question

Quizplus · Accepted Answer

The power dissipated in a wire is given by the equation P = I^2 * R, where P is the power in watts, I is the current in amperes, and R is the resistance in ohms. The resistance of a wire is given by the equation R = ρ * L / A, where ρ is the resistivity of the wire in ohm-meters, L is the length of the wire in meters, and A is the cross-sectional area of the wire in square meters.In this problem, we are given the resistivity of gold, the diameter of the wire, the length of the wire, and the current flowing through the wire. We can use these values to calculate the resistance of the wire and then use the equation P = I^2 * R to calculate the power dissipated in the wire.First, we need to calculate the cross-sectional area of the wire. The cross-sectional area of a wire is given by the equation A = π * r^2, where r is the radius of the wire in meters. In this problem, the diameter of the wire is given as 1.8 mm, which is equal to 0.0018 meters. Therefore, the radius of the wire is 0.0009 meters.Now we can calculate the cross-sectional area of the wire:A = π * r^2A = π * (0.0009 m)^2A = 2.544 × 10^-6 m^2Next, we can calculate the resistance of the wire. The resistance of a wire is given by the equation R = ρ * L / A, where ρ is the resistivity of the wire in ohm-meters, L is the length of the wire in meters, and A is the cross-sectional area of the wire in square meters. In this problem, we are given the resistivity of gold as 2.44 × 10^-8 ohm-meters, the length of the wire as 11 cm, which is equal to 0.11 meters, and the cross-sectional area of the wire as 2.544 × 10^-6 square meters. Therefore, the resistance of the wire is:R = ρ * L / AR = (2.44 × 10^-8 ohm-meters) * (0.11 meters) / (2.544 × 10^-6 square meters)R = 0.0106 ohmsFinally, we can calculate the power dissipated in the wire. The power dissipated in a wire is given by the equation P = I^2 * R, where P is the power in watts, I is the current in amperes, and R is the resistance in ohms. In this problem, we are given the current flowing through the wire as 170 mA, which is equal to 0.17 amperes, and the resistance of the wire as 0.0106 ohms. Therefore, the power dissipated in the wire is:P = I^2 * RP = (0.17 amperes)^2 * (0.0106 ohms)P = 0.0030 wattsP = 3.0 mWTherefore, the power dissipated in the wire is 3.0 mW. The correct answer is A.

The Resistivity of Gold Is 2