The diameter of an aluminum wire is 10 millimeters. Find the resistance of an 0.56- kilometer length of such wire used for power transmission. Use a value of 2.65 times 10 to the negative eighth ohm meters for the resistivity of aluminum.
We can start by drawing a sketch of the wire being described. While the sketch isn’t drawn to a scale, it gives us a sense for the dimensions of the aluminum wire. Along with its dimensions, we’re told that this wire has a resistivity, symbolized by the greek letter 𝜚, of 2.65 times 10 to the negative eighth ohm meters. And given all this, we want to solve for the resistance of this wire.
To do that, there is a mathematical connection between resistance and resistivity that we’ll want to recall. The resistance 𝑟 of a wire is equal to its resistivity 𝜚 multiplied by its length divided by its cross-sectional area. And in our case, since our cross- sectional area is circular, we can also recall the area of a circle in terms of its diameter, that area is equal to 𝜋 divided by four times its diameter squared.
Our task then is to use this relationship and the information given to solve for 𝑟, where 𝑟 is the resistance of the aluminum wire. We can start by plugging into this equation the given value for resistivity 𝜚 as well as the length of the wire in units of meters. Note that we’ve converted from kilometers. Once those values for 𝜚 and 𝐿 are in our equation, our last task is to enter in the expression for the cross-sectional area of this wire.
Looking to our equation for the area of a circle, we see that it’s in terms of diameter, which is what we’re given in the problem statement. But our diameter currently is in units of millimeters, and we want to convert it to meters to be consistent with the units in the rest of this expression for 𝑟. So instead of 10 millimeters, we’ll write the diameter as 10 to the negative third meters. And that quantity is squared and multiplied by 𝜋 over four to find the cross-sectional area.
Before we calculate the value for 𝑟, notice what happens to the units in this expression. In particular, the units of meters cancel from numerator and denominator, and we’re left with the units of ohms, the units of resistance. This fraction calculates out to 0.19 ohms. That’s the resistance of this long narrow wire.