Resistance and temperature problem; answer does not agree with book

[clairez93]

Homework Statement

What is the fractional change in the resistnace of an iron filament when its temperature changes from 25.0 degrees Celsius to 50 degrees Celsius?

Homework Equations

R = R$$_{0}$$(1 + $$\alpha$$(T - T$$_{0}$$))

The Attempt at a Solution

R = R$$_{}0$$(1+5x10^3x25)
R = 1.125R$$_{}0$$
1.125R$$_{}0$$/R = 1.125


The book's answer is 0.125, and i have 1.125, what did I do wrong?


Sorry; I can't figure out how to make the 0's in the R's go into subscript, not superscript. Those 0's are supposed be subscripted.

 

[alphysicist]

Hi clairez93,

Homework Statement

What is the fractional change in the resistnace of an iron filament when its temperature changes from 25.0 degrees Celsius to 50 degrees Celsius?

Homework Equations

R = R$$_{0}$$(1 + $$\alpha$$(T - T$$_{0}$$))

The Attempt at a Solution

R = R$$_{}0$$(1+5x10^3x25)
R = 1.125R$$_{}0$$
1.125R$$_{}0$$/R = 1.125

In this case, this is not the quantity they are asking for. They want the fractional change. So your fraction has to have the change in resistance in it. Do you see what it needs to be?


About the subscripts: put the entire equation between the tex tags, and don't put the {} brackets you have right before the zero subscript.

 

[clairez93]

So I'm guessing I have to subtract 1.125R_0 - R, so that would be 0.125, and that is the change?

 

[alphysicist]

So I'm guessing I have to subtract 1.125R_0 - R, so that would be 0.125, and that is the change?

No, not exactly. Since R=1.125 R_0, then 1.125R_0 - R would just equal zero.


The change in anything is the final value minus the initial value. So first find the change in resistance.

Then, since they want the fractional change, use the change in resistance that you found in your fraction. The fraction will cause an unknown variable to vanish.