The resistance of a bulb filament is $100\Omega$ at a temperature of $100^\circ \text{C}$. If its temperature coefficient of resistance be $0.005 \space \text{per} ^\circ \text{C}$, its resistance will become $200\Omega$ at a temperature of

  1. $300^\circ\text{C}$
  2. $400^\circ\text{C}$
  3. $500^\circ\text{C}$
  4. $200^\circ\text{C}$

Now the linear approximation is $R_t=R_0(1+\alpha T)$

Therefore \begin{align*} 100\Omega &= R_0(1+0.005\times 100) \\ \therefore R_0 &= \frac{100}{1.5} \\ \text{Now } 200 &= \frac{100}{1.5}(1+0.005\times t_2) \\ 0.005\times t_2 &= 2 \\ t_2 &=400^\circ\text{C} \end{align*}

But there is also this formula $\alpha=\frac{R_2-R_1}{R_1(t_2-t_1)}$

\begin{align*} 0.005 &=\frac{200-100}{100(t_2-100)} \\ t_2 &= 1/0.005 + 100 \\ t_2 &= 300^\circ\text{C} \end{align*}

Why are these answers differing?

  • 1
    $\begingroup$ I have reopened this question since I do not think that asking why the result of two prima facie applicable formulae is different falls under our homework-like policy in any shape or form. If anyone really disagrees, let me know (or directly post to Physics Meta). $\endgroup$ – ACuriousMind Jan 2 '19 at 23:59

When I use the linear equation, I have always written it as $R_{2}=R_{1}(1+\alpha \ \Delta T)$.

For this problem, $R_{1}=100\Omega$ and $R_{2}=200\Omega$. Also substituting the value for $\alpha$ will give you the change in temperature from 100 degrees. So remember to add the 100 to the $\Delta T$ for the complete answer.

Regarding why you got different answers, it is because linear approximations are linearized about a point and do not always hold. The question indicates that you are to linearize about $100^{\circ}$. Keep in mind that a slope of $0.005\ /^{\circ}C$ is 50% per $100\ ^{\circ}C$, which equates to $50\ \Omega\ per\ 100\ ^{\circ}C$. If you start at $100\Omega$ at $100^{\circ}$, if you get $100^{\circ}$ colder you will lose $50\Omega$. This gives a different result at 0 than what you had.

  • $\begingroup$ So the correct answer is 300C? Also how did you arrive at "a slope of 0.005 Ω/∘C means 50 Ω per 100 ∘C." Why isn't it 0.5 ohm per 100C? $\endgroup$ – user80551 Sep 13 '13 at 6:46
  • $\begingroup$ Yes. Sorry, I had a typo. I edited to fix it. It should have read "a slope of 0.005 /∘C" which is 0.5 (or 50%) per 100C. And with a reference of 100 Ohms 50% = 50 Ohms. The answer is the same 50 Ohms per 100C. $\endgroup$ – jdj081 Sep 13 '13 at 17:53

In your first equation $R_t=R_0(1+\alpha T)$, what does the variable $T$ represent? Investigating this may lead you to your answer.

First, when I look at the equation $R_t=R_0(1+\alpha T)$, I immediately see that the resistance $R_t$ is equal to $R_0$ when $T=0$. How does this make sense? One way to make sense of this is that the $T$ in your expression is really a difference in temperature $\Delta T = T_t - T_0$. If this is the case, then $R_0$ is the resistance at temperature $T_0$ and $R_t$ is the resistance at temperature $T_t$.

Now your "other" equation for $\alpha$ looks like using the original equation (with $\Delta T$) but renaming $T_t$ and $T_0$.

You should be able to fill in the steps I left out. Try it out.

  • $\begingroup$ $R_0$ in the first equation is resistance at a standard temperature. $T$ is difference in final temperature and this standard temperature. The second equation is infact derived from the first one. Here, I've taken standard temp to be 0C so $R_0$ is resistance at 0C $\endgroup$ – user80551 Sep 13 '13 at 4:34

The coefficient $\alpha$ is the fractional change in the resistance per unit temperature:

$$ \frac{dR}{dT} = \alpha R $$

As you note, the equation you (and everyone else in the business) use is a linear approximation to the actual exponential behavior. As such, it is most accurate for small temperature deltas. The larger the temperature delta, the greater the error in the linear approximation.

If you consider your two approaches in that light, I think you can see which is more reliable.

  • $\begingroup$ "I think you can see which is more reliable."--I can't, that's why I asked. $\endgroup$ – user80551 Sep 13 '13 at 6:04

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