Why do these two ways to calculate the resistance at a given temperature give different answers?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.