Question

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?

Answer 1

Power in a wire where current is flowing can be calculated from the product of the square of the current and the resistance. Resistance is equal to the product of resistivity and length divided by the area of the wire. We do as follows:

Resistance  =  2.44 × 10-8 ( 0.11) / (π)(0.0009)^2 = 1.055x10^-3 Ω

P = I^2R = .170^2 (1.055x10^-3 ) = 3.048x10^-5 W