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^{-8}m

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} m^{2}

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 = V^{2}/R

= 12^{2} / (3.975 * 10^{-5})

= 3.62 * 10^{6} W

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