How long must a 0.50-mm-diameter aluminum wire be to have a 0.58 a current when connected to the terminals of a 1.5 v flashlight battery? express your answer to two significant figures and include the appropriate units?

In the concept of electric circuits, the most famous and basic equation to be used is the Ohm's Law. Its equation is: I = V/R, where I is the current, V is the voltage and R is the resistance. Since, current and voltage are known, we could determine the resistance of the wire. Rearranging the equation,

R = V/I = 1,5/5.8 = 0.259 Ω

Now, the resistance of the wire depends on many factors. Among these are the type of material used, the cross sectional area, the length and the temperature. For the first three factors mentioned, there is an equation relating all of these to the resistance. The equation is

R = ρL/A, where ρ is the resistivity. This is an inherent property of a material. For aluminum, ρ = 3.2×10⁻⁸ Ω-m. The variable L is for the length and A is for the cross sectional area. For a circular wire, A = πr².

A = π(0.0005 m/2)² = 1.963×10⁻⁷ m²

Then, rearranging the equation,

L = RA/ρ = (0.259 Ω)(1.963×10⁻⁷ m²)/(3.2×10⁻⁸ Ω-m)
L = 1.59 m