Question

A copper wire that is 1 mm thick and 30 cm long is connected to a 1 V battery.

(The resistivity of copper is 1.69 x 10-8 Ω⋅m.) Find the current in the wire in [A] assuming it's at room temperature. What is the power in [W] dissipated in the wire at room temperature?

Answer 1

154.83 W

Step-by-step explanation:

Applyying,

P = V²/R..................... Equation 1

Where P = power dissipated, V = voltage R = resistance

But,

R = Lρ/A................. Equation 2

Where L = length of the wire, A = crosssectional area, ρ = resistivity of the wire.

Substitute equation 2 into equation 1

P = V²/(Lρ/A)

P = AV²/Lρ........................ Equation 3

From the question,

Given: L = 30 cm = 0.3 m, ρ = 1.69 x 10⁻⁸Ω⋅m, V = 1 V.

But,

A = πd²/4, Where d = 1 mm

A = (3.14×0.001²/4) = 7.85×10⁻⁷ m²

Substitute these values into equation 3

P = (7.85×10⁻⁷×1²)/(0.3×1.69 x 10⁻⁸)

P = 154.83 W