(III) The filament of an incandescent lightbulb has a resistan...

Question

(III) The filament of an incandescent lightbulb has a resistance of 12 Ω at 20°C and 140 Ω when hot. In this temperature range, what is the percentage change in resistance due to thermal expansion, and what is the percentage change in resistance due solely to the change in ρ? Use Eq. 25–5.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with resistance. So in this problem, a copper wire has an initial resistance of 8.0 ohms at a temperature of 15 °C upon heating to 2100 °C, its resistance increases to 70 ohms. Given that the thermal coefficient of expansion for copper is alpha L is equal to 16.5 multiplied by 10 to the negative six inverse degrees Celsius. And the temperature coefficient of resistivity. Alpha R is equal to 0.0039 inverse degrees Celsius determine the percent change in the resistance. And we're asked to calculate for two different quantities here for part one, solely due to the thermal expansion and for part two, solely due to the change in resistivity. We're also given four possible choices as our answers. For choice. A for part one, it increases by 1.5%. And for part two, it increases by 810%. For choice B. For part one, it increases by 1.5% and for part two, it decreases by 250%. For choice C for part one, it decreases by 3.3%. And for part two, it increases by 810%. And for choice D, for part one, it decreases by 3.3%. And for part two, it decreases by 250%. Now, for part one, we need to calculate the percent change in resistance solely due to thermal expansion to recall your formula four resistance. That's gonna be R is equal to row, L divided by A where are here is gonna be our resistance row is our resistivity, L is our length and A is our area. So we'll be using this in both case cases. Now for thermal expansion, both our length and area is going to change. So let's rewrite our area in terms of the diameter. So we'll have R is equal to row L divided by. And since I have a wire here, its cross sectional area is a circle. So I write that as pi multiplied by the quantity of D divided by two squared, where D here is going to be our diameter. Now, both the length and the diameter are going to change due to thermal expansion. And so we can use our thermal expansion equation to write out those expressions. So we'll have L is equal to L not multiplied by the quantity of one plus alpha, L multiplied by delta T. And we'll do the same thing for our di diameter. Our diameter D is going to be equal to D not multiplied by the quantity of one plus alpha L multiplied by delta T. So L knot and D not are the original length. And diameters alpha L is our coefficient of thermal expansion. And delta T is our change in temperature. So we can replace these quantities in our formula for the resistance. And so we're gonna write our resistance now as our sub L. So this is gonna be our resistance change due to just the thermal expansion or a change in length. So that's why we're using the subscript L and that's gonna be equal to row multiplied in the place of L. I'm gonna have L not multiplied by the quantity of one plus alpha, L multiplied by delta T. And that's gonna be divided by the quantity of pi multiplying the quantity of D not divided by two squared, multiplying the quantity of one plus alpha L multiplied by delta T and that quantity is squared. Now, the quantity of row L knot divided by pi multiplying the quantity of D not divided by two in quantity squared. That's just our original resistance. So I could rewrite this as RL is equal to or not divided by and one factor of the one plus alpha L delta T cancels out with one of the factors in the denominator and left with just in the denominator, one plus alpha L multiplied by delta T. So now I have my resistance written in terms of my original resistance, I can calculate our percent difference. So our percent difference here, our percent change, we'll write that as percent delta RL and that's gonna be equal to. And here we're just going to use our formula for percent change. So that's gonna be our new resistance RL minus our original resistance, which is R not divided by our original resistance. R not. And we'll take that quantity and multiply it by 100%. I can simplify this a little bit. So I have R percent delta RL is equal to the quantity of RL divided by R not minus one, multiplied by 100 et so we'll plug in a formula for RL. And so we'll have percent delta RL. It's gonna be equal to. And here we have, why not divided by, are not multiplying the quantity of one plus alpha L delta T and that fraction has um one being subtracted from it. And then that quantity is multiplied by 100%. You are not to cancel. And I can plug in for the alpha L and delta T. And so we'll have our percent. Delta RL is equal to the quantity of one divided by one plus. And for alpha L that was given to the problem that is the 16.5 multiplied by 10 to the negative six inverse degrees Celsius, that's gonna be multiplied by our change in temperature. Our initial, our final temperature is the 2100 °C minus our initial temperature of 15 °C don't have minus one. And that's gonna be m that quantity is gonna be multiplied by 100%. So when I plug that value those values into my calculator, I'll get percent delta RL. It's going to be equal to minus 3.3 set. That could be your answer for part one. And the minus sign here just indicates a decrease. So our resistance has decreased by 3.3%. We're gonna do something similar for part two. But here we're going to look at how our resistivity changes. So recall that for your resistivity, we have row is equal to row, not multiplying the quantity of one plus alpha R multiplied by delta T. So this formula here is how our resistance changes with temperature. So we'll plug that into our formula for the resistance. So we'll have RR. So that's gonna be our resistance due to the change in the resistivity. That's gonna be equal to for row, we're gonna replace that with row knot multiplied by the quantity of one plus alpha R delta T in quantity. Now, that's gonna be multiplied by the length and divided by the area. But row knot multiplied by L divided by A is just my original resistance. R not, so we'll have RR is equal to are not multiplying the quantity of one plus alpha R delta T. So again, we can um calculate our percent difference. And so we'll have the percent of delta RR, it's gonna be equal to and we'll use the same formula that we had before, except we'll be replacing RL with RR. So we'll have the quantity of RR minus R not divided by or not multiplied by 100%. Again, simplifying we have percent delta RR is equal to the quantity of RR divided by R not minus one multiplied by 100 per cent. And so we'll have the percent delta RR equal to. And here we'll plug in our formula four RR and that's gonna be, are not multiplying the quantity of one plus alpha R delta T in quantity. And that's gonna be divided by or not. And then I'll have minus one and that whole quantity is multiplied by 100%. The arenas will cancel and then the ones will also cancel as well. And so I can plug in values for the alpha R and the delta T will have the percent delta RR gonna be equal to for alpha R. You're giving that in the problem. That's gonna be the 0.00 39 inverse degrees Celsius. And this could be multiplied by our change in temperature, which is the 2100 °C minus the 15 °C. That's gonna be multiplied by 100%. And so plugging all that into my calculator, I get 5% delta RR is going to be equal to 810 percent. And here, I've just kept two significant figures. And so since this is a positive quantity, this is actually an increase. So now I have both answers for part one and two, I can look at my answer choices and the correct answers to choice in this case is gonna be answer choice. C so just a quick recap of what we've done here for part one, we used our formula for the resistance and we had to expand both our length and the diameter, which meant that both our length and area were changing, which caused our resistance to change as well. We then calculated our new resistance in terms of our original resistance and our coefficient of thermal expansion and the change in temperature. And then use that to calculate the percent change. For part two, we did something similar. We wrote our new resistance in terms of our old resistance using our formula four, our change in resistivity due to a change in temperature. And once we wrote our new resistance in terms of our old resistance, we could use that to calculate our percent change as well. So I hope that this has been useful. And I'll see you in the next video.

(III) The filament of an incandescent lightbulb has a resistance of 12 Ω at 20°C and 140 Ω when hot. In this temperature range, what is the percentage change in resistance due to thermal expansion, and what is the percentage change in resistance due solely to the change in ρ? Use Eq. 25–5.

Key Concepts

Resistance and Temperature Relationship

Percentage Change Calculation

Resistivity and Its Temperature Dependence

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