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. Calculate the rate at which internal energy is produced in 1.00 m of 12-gauge copper wire carrying 20.0 A.

Given Information:  

Current = I = 20 A

Diameter = d = 0.205 cm = 0.00205 m

Length of wire = L = 1 m

Required Information:  

Energy produced = P = ?

P = 2.03 J/s

Explanation:  

We know that power required in a wire is

P = I²R

and R = ρL/A

Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m

L is the length of the wire and A is the area of the cross-section and is given by

A = πr²

A = π(d/2)²

A = π(0.00205/2)²

A = 3.3x10⁻⁶ m²

R = ρL/A

R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶

R = 5.09x10⁻³ Ω

P = I²R

P = (20)²*5.09x10⁻³

P = 2.03 Watts or P = 2.03 J/s

Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A

and 1and 1