Temperature dependence of resistance

[tellmesomething]

Homework Statement

Differential form of $R(t)=R_{0}(1+\alpha(t-t_{0})$

Relevant Equations

none

I wanted to find the differential form of the above equation and i get $$\frac{dR(t)}{dt}=R_{0}\alpha$$ ($t_{0}$=0 degree celsius)
So $$\alpha=\frac{dR(t)}{dt} \frac{1}{R_{0}}$$ ($\alpha$= temperature coefficient of resistance $R_{0}$=Resistance at temperature 0 degree celsius)

This idea arises from a question

Here if i use this differential formula i derived i get $$\frac{dR(t)}{dt}\frac{1}{R_{0}}=a+2bt$$ which should be the value of $\alpha$ ?

But that doesnt give me the correct answer, i would get the correct answer if $$\frac{dR(t)}{dt}\frac{1}{R(t)}=\frac{a+2bt}{1+at+bt^2}$$

I dont get why the denominator changes to $R(t)$

 

[kuruman]

The differential form that you are seeking is $$\frac{dR(t)}{dt}=\alpha ~R(t).$$Find the solution and then note that the expression $R(t)=R_{0}[1+\alpha(t-t_{0})]$ shows the first two terms of the series expansion of the solution about $t_0$. If you add one more term to the series, you get the quadratic term.

 

[TSny]

So $$\alpha=\frac{dR(t)}{dt} \frac{1}{R_{0}}$$ ($\alpha$= temperature coefficient of resistance $R_{0}$=Resistance at temperature 0 degree celsius)

Write this a little more explicitly. If $\alpha(t_0)$ denotes the temperature coefficient at temperature $t_0$, then $$\alpha(t_0)= \frac{dR(t)}{dt}\bigg{\rvert}_{t=t_0} \cdot \frac 1 {R(t_0)}$$ If you want the temperature coefficient $\alpha(t_1)$ at some other temperature $t_1$, $$\alpha(t_1) = \frac{dR(t)}{dt}\bigg{\rvert}_{t=t_1} \cdot \frac 1 {R(t_1)}$$ Since $t_1$ could be any temperature, you can drop the subscript and write $$\alpha(t) = \frac{dR}{dt}\bigg{\rvert}_t \cdot \frac 1 {R(t)}$$ Apply this to the question that you quoted in the OP.

 

[tellmesomething]

The differential form that you are seeking is $$\frac{dR(t)}{dt}=\alpha ~R(t).$$Find the solution and then note that the expression $R(t)=R_{0}[1+\alpha(t-t_{0})]$ shows the first two terms of the series expansion of the solution about $t_0$. If you add one more term to the series, you get the quadratic term.

Write this a little more explicitly. If $\alpha(t_0)$ denotes the temperature coefficient at temperature $t_0$, then $$\alpha(t_0)= \frac{dR(t)}{dt}\bigg{\rvert}_{t=t_0} \cdot \frac 1 {R(t_0)}$$ If you want the temperature coefficient $\alpha(t_1)$ at some other temperature $t_1$, $$\alpha(t_1) = \frac{dR(t)}{dt}\bigg{\rvert}_{t=t_1} \cdot \frac 1 {R(t_1)}$$ Since $t_1$ could be any temperature, you can drop the subscript and write $$\alpha(t) = \frac{dR}{dt}\bigg{\rvert}_t \cdot \frac 1 {R(t)}$$ Apply this to the question that you quoted in the OP.

With this equation $R(t)=R_{0}[1+\alpha(t-t_{0})]$ when we say at temperature t1, we mean $R(t1)=R_{0}[1+\alpha(t1-t_{0})]$ ?? Accordingly it follows $\alpha=\frac{dR(t)}{dt} \frac{1}{R_{0}}$ isnt $R_{0}$ acting as a constant here??

 

[TSny]

With this equation $R(t)=R_{0}[1+\alpha(t-t_{0})]$ when we say at temperature t1, we mean $R(t1)=R_{0}[1+\alpha(t1-t_{0})]$ ?? Accordingly it follows $\alpha=\frac{dR(t)}{dt} \frac{1}{R_{0}}$ isnt $R_{0}$ acting as a constant here??

The question you quoted in the OP asks you to find the temperature coefficient $\alpha$ as a function of temperature: $\alpha(t)$.

The coefficient at temperature $t_1$, $\alpha(t_1)$, is used to calculate the resistance at temperatures "near" $t_1$ to a good approximation: $$R(t) \approx R(t_1)\left[1+\alpha(t_1)(t - t_1)\right].$$ Here, $R(t_1)$ is the resistance at temperature $t_1$ as you can see by letting $t = t_1$ in the above formula. The closer $t$ is to $t_1$, the better the approximation.

If you solve the formula for $\alpha(t_1)$, you get $$\alpha(t_1) \approx \frac{R(t) - R(t_1)}{t - t_1} \cdot \frac1 {R(t_1)}.$$ In the limit of letting $t \rightarrow t_1$, the fraction becomes a derivative and the expression is exact: $$\alpha(t_1)= R'(t_1) \cdot \frac1 {R(t_1)} = \frac{R'(t_1)}{R(t_1)}.$$ You can take this equation to be the definition of $\alpha(t_1)$.

In the question in the OP, they give you an example where the resistance varies with temperature as $$R(t) = R_0(1 + at + bt^2).$$ For this example, what do you get for $\alpha(t_1)$ if you use $\alpha(t_1)= \dfrac{R'(t_1)}{R(t_1)}$?

 

[tellmesomething]

The coefficient at temperature $t_1$, $\alpha(t_1)$, is used to calculate the resistance at temperatures "near" $t_1$ to a good approximation: $$R(t) \approx R(t_1)\left[1+\alpha(t_1)(t - t_1)\right].$$ Here, $R(t_1)$ is the resistance at temperature $t_1$ as you can see by letting $t = t_1$ in the above formula. The closer $t$ is to $t_1$, the better the approximation.

This made it all click. Thankyou so much.

In the question in the OP, they give you an example where the resistance varies with temperature as $$R(t) = R_0(1 + at + bt^2).$$ For this example, what do you get for $\alpha(t_1)$ if you use $\alpha(t_1)= \dfrac{R'(t_1)}{R(t_1)}$?

$$\frac{a+2bt_{1}}{1+at_{1}+bt_{1}^2}$$

 

[TSny]

$$\frac{a+2bt_{1}}{1+at_{1}+bt_{1}^2}$$

Yes.