A 10-gauge copper wire diameter = 2.588 mm, ρ = 1.7 × 10^-8 Ω · m) has a total resistance of 0.32 Ω. How long is the wire?

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The resistance of a wire is given by:$$R = ho \frac{L}{A}$$where:* R is the resistance in ohms (Ω)* ρ is the resistivity of the wire in ohm-meters (Ω·m)* L is the length of the wire in meters (m)* A is the cross-sectional area of the wire in square meters (m²)We are given the following information:* R = 0.32 Ω* ρ = 1.7 × 10^-8 Ω·m* d = 2.588 mm = 2.588 × 10^-3 mThe cross-sectional area of the wire is:$$A = \frac{\pi d^2}{4} = \frac{\pi (2.588 	imes 10^{-3} 	ext{ m})^2}{4} = 5.262 	imes 10^{-6} 	ext{ m}^2$$Substituting these values into the equation for resistance, we get:$$0.32 \Omega = 1.7 	imes 10^{-8} \Omega \cdot 	ext{m} \cdot \frac{L}{5.262 	imes 10^{-6} 	ext{ m}^2}$$Solving for L, we get:$$L = \frac{0.32 \Omega \cdot 5.262 	imes 10^{-6} 	ext{ m}^2}{1.7 	imes 10^{-8} \Omega \cdot 	ext{m}} = 99 	ext{ m}$$Therefore, the length of the wire is 99 m.

A 10-Gauge Copper Wire Diameter = 2