Residential building codes typically require the use of 12-gauge copper wire (diameter 0.2053 cm) for wiring receptacles. such circuits carry currents as large as 20 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.(a) calculate the rate at which internal energy is produced in 2.10 m of 12-gauge copper wire carrying a current of 20.0 a. your response differs significantly from the correct answer. rework your solution from the beginning and check each step carefully. w(b) repeat the calculation for an aluminum wire. your response differs from the correct answer by more than 10%. double check your calculations. wwould a 12-gauge aluminum wire be as safe as a copper wire?

a) E = 4.26 W

b) E' = 6.724 W

c) copper wire is the safer option to use.

Explanation:

Given:

Diameter of the 12 gauge copper wire, d = 0.2053 cm

Thus, Radius of the 12 gauge copper wire, r = 0.2053 cm

/ 2 = 0.10265 cm  = 0.10265 × 10⁻² m

Now.

the area (A) comes out as

A = π × (0.10265 × 10⁻²)²

A = 3.3103 × 10⁻⁶ m²

Length of the copper wire, L = 2.10 m

a) The resisitivity (ρ) of copper = 1.68 × 10⁻⁸ ohm m

Now,

the resistance of the copper , R = ρL/A

or

R = (1.68 × 10⁻⁸ × 2.1) / ( 3.3103 × 10⁻⁶)

or

R = 0.01065 ohm

The Energy (E) is given as,

E = I²R

where, I is the current

I = 20.0 A

on substituting the values, we get

E = 20.0² × 0.01065

E = 4.26 W

(b) For the aluminium

Resisitivity, ρ' = 2.65 × 10⁻⁸  ohm m

Now, the resistance of the aluminium wire, R' = (ρ' × L) / A

Since the cross-section of the aluminium wire is same as the copper wire

thus,

R = (2.65 × 10⁻⁸ × 2.1) / ( 3.3103 × 10⁻⁶)

or

R = 0.0168 ohm

Therefore,

The Rate of energy produced by the aluminium wire, E' = I²R'

or

E' = 20.0² ×  0.0168

or

E' = 6.724 W

(c) From the above results, we can conclude that the power consumed or the rate of energy produced by the aluminium wire is more.

Hence, copper wire is the safer option to use.