OpenStax College Physics for AP® Courses, Chapter 20, Problem 25 (Problems & Exercises)

This is College Physics Answers with Shaun Dychko. It's important to begin a question by writing down all the stuff that we're given and also take care of unit conversions in this step when you're just writing things down so that you don't have to think about it during your algebra and the work that follows. So the diameter of our zero gauge copper wire is 8.252 millimeters which we convert into meters by multiplying by ten to the minus three. The length is one kilometer which is one times ten to the three meters. Since we're told this is copper, its resistivity if we look it up in the table given to us in the textbook, is 1.72 times ten to the minus eight ohm meters. So the resistance of the wire is resistivity multiplied by its length divided by its cross-sectional area. The cross-sectional area will be pi times d squared over four, and then if you multiply top and bottom by four you get rid of the four that's nested within this fraction which looks kind of ugly, so we'll put the four in the numerator. So we have four rho l over i d squared. So that's four times 1.72 times ten to the minus eight ohm meters, times one times ten to the three meters divided by pi times 8.252 times ten to the minus three meters squared, giving us 0.322 ohms of resistance.