OpenStax College Physics, Chapter 20, Problem 38 (Problems & Exercises)

This is College Physics Answers with Shaun Dychko. This question asks us to calculate the resistance of a tungsten filament in the incandescent headlamp of a car taking into account the fact that the filament will expand due to thermal expansion and we are given the coefficient of thermal expansion as 12 times 10 to the minus 6 per Celsius degree and so the filament will both increase its length, which would have a tendency to increase resistance according to this formula, and the filament will also increase its cross-sectional area because its diameter will expand as well and that would tend to decrease resistance because that term is in the denominator. We will see that the net effect of taking into account expansion results in a decrease in resistance compared to if you don't account for expansion just because this denominator has this diameter squared term in it and that exponent 2 means that it has a greater effect than the linear dependency on length in the numerator here. Okay! So we have an initial temperature of 20 degrees Celsius and a final temperature we are told in example [20.6], is 2850 degrees Celsius that's the typical temperature of a tungsten filament when the light is turned on and the thermal coefficient of resistivity we look up in table [20.2] for tungsten and it's 4.5 times 10 to the minus 3. The diameter of this particular filament we are given in example [20.5] and it's 9.0 times 10 to the minus 5 meters and the resistivity of tungsten at 20 degrees Celsius— so its ρ naught—is found in table [20.1] and for tungsten, it's 5.6 times 10 to the minus 8 so we collected that information and for comparison in part (b), we know that at 20 degrees Celsius, the tungsten filament has a resistance of 0.350 ohms. Okay! So in part (a), we are gonna figure out the resistance of the filament after it increases its temperature taking into account thermal expansion. So resistance is resistivity multiplied by the length of the wire divided by its cross-sectional area and we have expressions for each of these factors here. So the resistivity at the higher temperature will be the original resistivity at 20 degrees Celsius times 1 plus the temperature coefficient of resistivity multiplied by the change in temperature. The new length of the wire will be its original length plus the coefficient of expansion multiplied by the original length times the change in temperature and we can factor out the L naught and write it this way if we like. The area of the wire— its cross-sectional area—is π times its radius squared and so that's π times the radius after expansion which is R naught plus coefficient of linear expansion times r naught times ΔT; it's the same expression for length but it's just going radially now instead of along the length of the wire. Factor out the r naught and its squared and then we have 1 plus α EΔT all squared. Then substitute in diameter divided by 2 because that's what we are given in this example [20.5] is the diameter so we divide that diameter by 2 as a substitution for radius and we end up with the cross-sectional area then is π times original diameter squared over 4 times 1 plus α EΔT squared. Then we plug in this expression, this expression and this expression for their respective factors A, L and ρ in our equation for the resistance. and that's what we have here: this is substituting for ρ, substituting for length and substituting for cross-sectional area. We see there is a common factor here 1 plus α EΔT and so this cancels with one of these leaving us with 1 plus α EΔT to the power of 1 in the denominator. And then multiply top and bottom by 4 so the 4 cancels on the bottom leaving us with 4 times ρ naughtL naught on the top times 1 plus α RΔT over π times original diameter squared times 1 plus α EΔT then plug in numbers. So we have 4 times the resistivity of tungsten at 20 degrees Celsius multiplied by the original length of this filament, which is 4.00 centimeters, and then multiply that by 1 plus the temperature coefficient of resistivity multiplied by the change in temperature so this is the final temperature minus the initial temperature and then divide that by π times the diameter squared times 1 plus the coefficient of linear expansion times the change in temperature, this works out to 4.7 ohms. Now part (b), we are asked to compare what our answer is with consideration for thermal expansion to the resistance with no thermal expansion. So if we ignore thermal expansion, we end up with an answer of 4.807 ohms instead of 4.677 ohms. And so we get that by taking this resistance of the filament at 20 degrees Celsius, which we are told is 0.350 ohms, and multiply that by 1 plus the temperature coefficient of resistivity multiplied by the temperature change and we get this 4.807 ohms. So find the difference between that and the answer when you do consider thermal expansion and divide that by the original answer times by a 100 percent and you get negative 2.7 percent. So the resistance when you do consider thermal expansion is 2.7 percent less than the resistance where that was ignored.