Question

Aluminum has a resistivity of 2.65 x10^-8 ohm−m. If a 3.0 meter long rectangular aluminum bar, 4.0 cm wide and 5.0 cm tall is hooked to a 12.0V car battery so that current runs along its length, how much power is dissipated (until the battery runs down)?

Answer 1

The resistivity of the aluminum bar, = 2.65 * 10^-8m
Length of the aluminum bar, L = 3.0 m
Area of the aluminum bar, A = width * height
A = (4 * 10^-2) * (5 * 10^-2)
= 2 * 10^-3 m²
The resistance of the aluminum bar, R = * L / A
= [(2.65 * 10^-8) * 3] / (2 * 10^-3)
= 3.975 * 10^-5

The power dissipated, P = V²/R
= 12² / (3.975 * 10^-5)
= 3.62 * 10⁶ W