Question

A number 12 copper wire has a diameter of 2.053 mm. Calculate the resistance of a 37.0 m long piece of such wire.
(Use 1.72×10^-8 Ωm for the resistivity of copper.)

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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?

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What would be the voltage between the ends of the wire in the above problem?

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What is the current density in the wire when it is carrying the maximum allowable current? (Current density is the current in the wire divided by the cross sectional area of the wire.)

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What is the drift velocity of the electrons when the wire is carrying the maximum allowable current?
(The density of electrons in copper is 8.47×10²⁸ m^-3.)

Answer 1

d = diameter of copper wire = 2.053 mm = 2.053 x 10^-3 m

A = area of cross-section of wire = (0.25) $\pi$d² = (0.25) (3.14) (2.053 x 10^-3)² = 3.31 x 10^-6 m²

L = length of the wire = 37 m

$\rho$ = resistivity of copper = 1.72 x 10^-8 Ωm

Resistance of wire is given as

R = $\rho$L/A

R = (1.72 x 10^-8) (37)/(3.31 x 10^-6 )

R = 0.192 ohm

i = current allowed to flow = 20 A

Power dissipated in the wire is given as

P = i² R = (20)² (0.192) = 76.8 Watt

using ohm's law , voltage between the ends of the wire is given as

V = i R

V = 20 x 0.192 = 3.84 Volts

J = current density = i /A = 20/(3.31 x 10^-6) = 6.04 x 10⁶ A/m²

drift velocity is given as

v_d = i/(n e A) = 20/((8.47 x 10²⁸) (1.6 x 10^-19)(3.31 x 10^-6))

v_d = 4.5 x 10^-4 m/s