Question

1. (50 points) In household wiring, usually 12-gauge copper wire (diameter 2.05 mm) is used. A light bulb is connected to power supply through such a wire. A 1.5-Ampere current runs through the wire and the light bulb. a. (10 points) What is the resistance of 10-meter length of the wire? b. (15 points) How many electrons pass through the light bulb each second? c. (10 points) What is the current density in the wire? d. (15 points) If we were to use wire of twice the diameter, which of the above answers would change? Would they increase or decrease? You do not need to do any calculations.

Answer 1

Answer:

a. 50.9 mΩ b. 9.32 × 10¹⁸ electrons/s c. 4.55 × 10⁵ A/m² d. i. resistance and current density  ii. They would decrease.

Explanation:

a. The resistance of the copper wire is given by

R = ρl/A where ρ = resistivity of copper wire = 1.68 × 10⁻⁸ Ωm, l = length of copper wire = 10 m and A = cross-sectional area of coper wire = πd²/4 where d = diameter of copper wire = 2.05 mm = 2.05 × 10⁻³ m

A =  πd²/4  

= π(2.05 × 10⁻³ m)²/4

= 13.2025/4  × 10⁻⁶ m²

= 3.3 × 10⁻⁶ m².

So R = ρl/A

=  1.68 × 10⁻⁸ Ωm × 10 m/3.3 × 10⁻⁶ m²

= 0.0509 Ω

= 50.9 mΩ

b. Since a current of 1.5 A flows through the wire, it means that 1.5 C/s, that is 1.5 Coulombs of charge flows through it per second.

Since 1 electron = 1.609 × 10⁻¹⁹ C, the number of electrons in 1.5 A is 1.5 C/s ÷ 1.609 × 10⁻¹⁹ C per electron = 1.5/1.609 × 10⁻¹⁹ C = 9.32 × 10¹⁸ electrons/s

c. The current density J = I/A where I = current = 1.5 A and A = cross-sectional area of copper wire = 3.3 × 10⁻⁶ m²

J = 1.5 A/3.3 × 10⁻⁶ m²

= 4.55 × 10⁵ A/m²

d. i. If the diameter were twice the initial diameter, d' = 2d, then the resistance and current density would change since they are dependent of the cross-sectional area of the wire which is then dependent on the diameter of the wire.

ii. If the diameter were twice the initial diameter, d' = 2d, then since the cross-sectional area A' = πd'²/4 = π(2d)²/4 = 4πd²/4 = 4A. So, the cross-sectional area increases by a factor of four.

The new resistance R' =  ρl/4A = R/4

The new current density J' = I/4A = J/4.

So the resistance and current density would decrease.