You're writing the instruction manual for a power saw, and you have to specify the maximum permissible length for an extension cord made from 18-gauge copper wire (diameter 1.0 mm). the saw draws 9.0 a and needs a minimum of 115 v across its motor when the outlet supplies 120 v.

12.7 meters You want a maximum voltage drop of 5 volts while carrying a load of 9.0 amperes. So using ohms law I=V/R and solving for R, gives R=V/I. So R = 5/9 = 0.5556 ohms. The formula for the resistance of a conductor is R = pL/A where p = resistivity 1.72x10^-8 L = length A = cross sectional area Since the wire has a diameter of 1.0 mm, its radius is 0.5 mm. So the cross section is pi r^2 = pi 0.5^2 = pi 0.25 = 0.785398163 mm^2 We need to convert that cross section to square meters, so we divide by 1 million, giving a result of 7.85398163x10^-7 m^2 The resistivity of copper is 1.72x10^-8 Now substitute the known values into the formula and solve for L R = pL/A 0.5556 = 1.72x10^-8 * L / 7.85398163 x 10^-7 0.5556 * 7.85398163 x 10^-7 = 1.72x10^-8 * L 0.5556 * 7.85398163 x 10^-7 / 1.72x10^-8 = L 25.4 = L So the longest length of 18 gauge wire you can use is 25.4 meters. But you need to take into account that the current is flowing to the saw and back from the saw through the extension cord, so you need to divide that length by 2, giving 12.7 meters.

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