Transposition R = RoeαT questions

[JimCrown]

The resistance (R) of a copper wire conducting electricity at a given temperature has the formula:
R = RoeαT

where α is 3.9 x 10-3

Ro is the original resistance = 7.36 x 103 ohms

T is the temperature (ºC) = ?

R is the resistance after an exponential rate of growth under a specific

temperature = 7.95 x 103 ohms.

Calculate the temperature of the copper, (T) which would produce this increase in resistance across the wire.


I am completely stumped with this question. Would I use logs (logarithms) to figure these problems out?

 

[mmm4444bot]

I'm sure that some of your typed expressions need to display an exponent.

For example, ten cubed is typed like this: 10^3

Is the following equation what you're trying to solve?

7950 = 7360 * e^(0.0039T)

If so, then, yes, you need to use logarithms. Please show us what you try.:cool:

 

[JimCrown]

Would I do

7.36 x 10^3 ohms = 7360

e 3.9 x 10^-3 = 0.049402449

0.49402449 x 7360 = 363.6020254

To get the temperature

 

[mmm4444bot]

Would I do

7.36 x 10^3 ohms = 7360 …

That's a good start.

You could also multiply out the other values given in scientific notation (α and R) the same way, as I did in my previous post.

By the way, did you notice the typographical error in your first post?

… temperature = 7.95 x 103 ohms

We do not measure temperature in ohms, heh.

Instead, that's a given value for R, yes?

Substituting for the scientific notation in all three places, yields the equation in my previous post.

R = R₀ * e^(α*T)

7950 = 7360 * e^(0.0039T)

This is the equation that I had asked you to confirm.

… e 3.9 x 10^-3 = 0.049402449

0.49402449 x 7360 = 363.6020254

To get the temperature

These steps do not look correct.

I cannot tell what you did, to obtain 0.049402449, but I can say that you may not separate 0.0039 from T yet (in what I have assumed is the exponent). We will use a property of logarithms (later) to get the expression 0.0039T out of the exponent position, and then a division will solve for T.

The next step is to isolate the power of e. Do that, and then take the natural logarithm of each side.

If you think that you need to review the algebra involved, check out some lessons. Google keywords: solve exponential growth and decay equations

Also, be sure to review the forum guidelines. Thank you! :cool:

PS: There's a link (in the guidelines) titled, "Formatting Math as Text". It takes you to a page showing how to express math using a keyboard.

 

[JimCrown]

That's a good start.

You could also multiply out the other values given in scientific notation (α and R) the same way, as I did in my previous post.

By the way, did you notice the typographical error in your first post?

We do not measure temperature in ohms, heh.

Instead, that's a given value for R, yes?

Substituting for the scientific notation in all three places, yields the equation in my previous post.

R = R₀ * e^(α*T)

7950 = 7360 * e^(0.0039T)

This is the equation that I had asked you to confirm.


These steps do not look correct.

I cannot tell what you did, to obtain 0.049402449, but I can say that you may not separate 0.0039 from T yet (in what I have assumed is the exponent). We will use a property of logarithms (later) to get the expression 0.0039T out of the exponent position, and then a division will solve for T.

The next step is to isolate the power of e. Do that, and then take the natural logarithm of each side.

If you think that you need to review the algebra involved, check out some lessons. Google keywords: solve exponential growth and decay equations

Also, be sure to review the forum guidelines. Thank you! :cool:

PS: There's a link (in the guidelines) titled, "Formatting Math as Text". It takes you to a page showing how to express math using a keyboard.

Yes the equation is:

7950 = 7360 * e^(0.0039T)

Then you divide both sides by 7360

This gets you
1.080163043 = e^(0.0039t)
Then you do logs
Ln (1.080163043) = 0.0039t
Ln( 1.080163043)/0.0039=t
T = 19.77230653
Or 19.8 degrees to 1.d.p.
I feel like this is the correct? Thanks for the advice forsearching on google what I needed to find out

 

[mmm4444bot]

T = 19.77230653

Or 19.8 degrees to 1.d.p.

I feel like this is the correct?

It seems reasonable. Good work. Did you check it? :cool:

Substitute the value 19.7723 for symbol T in the following expression, and then evaluate, to see whether you get 7950:

7360 * e^(0.0039T)

 

[mmm4444bot]

7950 = 7360 * e^(0.0039T)

1.080163043 = e^(0.0039t)

Ln (1.080163043) = 0.0039t

Ln( 1.080163043)/0.0039=t

T = 19.77230653

TIP: Don't switch back and forth between upper- and lower-case symbols. Pick one, and use it consistently throughout.