Question

A number 12 copper wire has a diameter of 2.053 mm. Calculate the resistance of a 40.0 m long piece of such wire. (Use 1.72x 10-8 2m for the resistivity of copper.) Submit Answer Tries 0/99 For safety, the National Electrical Code limits the allowable amount of current which such a wire may carry. When used in indoor wiring, the limit is 20.0 A for rubber insulated wire of that size. How much power would be dissipated in the wire of the above problem when carrying the maximum allowable current? Submit Answer Tries 0/99 What would be the voltage between the ends of the wire in the above problem? Submit Answer Tries 0/99 What is the current density in the wire when it is carrying the maximum allowable current? (Current density is the current in the wire hate is the hurcens ecsilby in the irethehwret is carrying the maximum allowable current? (Current density is the current in the wire ded by the cross sectional area of the wire.) Submit Answer Tries 0/99 What is the drift velocity of the electrons when the wire is carrying the maximum allowable current? The densitv of electrons in copper is 8.47x1028 m3.) Submit Answer Tries 0/99

Answer 1

(A) R = rho L / A = rho L / ( pi D^2 / 4)

R = 4 (1.72 x 10^-8) (40) / (pi (2.053 x 10^-3)^2)

R = 0.208 Ohm

(B) P = I^2 R = (20^2)(0.208)

P = 83 W

(C) V = I R = 4.15 Volt

(D) J = I / ( pi D^2 / 4) = 6.04 x 10^6 A/m^2

(E) v = J / n q = (6.04 x 10^6) / (8.47 x 10^28 x 1.6 x 10^-19)

v = 4.46 x 10^-4 m/s