Please help. Residential building codes typically require the use of 12-gauge co

Question

Please help.

Residential building codes typically require the use of 12-gauge copper wire (diameter 0.205 cm) for wiring receptacles. Such circuits carry currents as large as 20.0 A. If a wire of smaller diameter (with a higher gauge number) carried that much current, the wire could rise to a high temperature and cause a fire. (Table of resistivities) Calculate the rate at which internal energy is produced in 2.00 m of 12-gauge copper wire carrying a current of 20.0 A. W Repeat the calculation for a 12-gauge aluminum wire. W Explain whether a 12-gauge aluminum wire would be as safe as a copper wire.

Explanation / Answer

(a)

12 gauge = 0.205 cm

radius = 0.205 / 2 = 0.1025 cm

area = pi * (0.1025/100)^2 = 3.3 * 10^-6 m^2

resisitivity of copper = 1.68 * 10^-8 ohm m

So R = pho * length / area = 1.68 * 10^-8 * 2 / ( 3.3 * 10^-6)

= 1.018 * 10^-2 ohm

Rate of energy produced = current^2 * R = 20^2 * 1.018 * 10^-2

= 4.072 Watts (answer)

(b)

Resisitivity of Al = 2.65 * 10^-8

R = pho * length / area = 2.65 * 10^-8 * 2 / ( 3.3 * 10^-6)

= 1.606 * 10^-2 ohm

Rate of energy produced = current^2 * R = 20^2 * 1.606 * 10^-2

= 6.424 Watts (answer)

(c)

Since Cu wire produces energy at a lower rate than the allumunium wire, hence the cu wire would be safer to use.

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