# Changing Rho, Finding Resistance

1. Oct 5, 2014

### Bluestribute

1. The problem statement, all variables and given/known data
A wire with cross-sectional radius 0.91 mm lying along the x axis from x=0 to x=0[PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmmi10/alpha/100/char3A.png90 [Broken] m is made of an alloy that varies with its length in such a way that the resistivity is given by [PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmmi10/alpha/100/char1A.png=6x^1 [Broken] [PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmr10/alpha/100/char0A.png[PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmsy10/alpha/100/char01.png [Broken] m, for x in meters. What is the resistance of this wire?

2. Relevant equations
dR=(rho)dL/A

3. The attempt at a solution
(I can't find the symbol window so bare with my explanation). Resistance is the integral of rho times length, divided by cross sectional area. I figured since L and A are constants, I could pull them out and simply do: rho, integrated from 0 to 0.9, times L/A, with everything converted to meters. Didn't work.

Then I tried not integrated and just plugging and chugging. Didn't work.

Then I came here.

Last edited by a moderator: May 7, 2017
2. Oct 5, 2014

### Staff: Mentor

What didn't work? Please show what you got and how you got it.

Also, is there a typo in $\rho$? It seems odd to me that it would be 6x1 ohms/m. I don't know why anyone would write x1 instead of just x.

Last edited by a moderator: May 7, 2017
3. Oct 5, 2014

### Bluestribute

So I'm on my phone, but I integrated the rho function and multipled by L/A, L being the difference in the bounds and A being the given mm amount (converted to meters).

It's to the first power cause the numbers are random, including the exponents

4. Oct 5, 2014

### Staff: Mentor

So what value did you get?

5. Oct 5, 2014

### Dick

The notation rho*dL/A doesn't mean integrate rho and then multiply by L/A. It means integrate rho WITH RESPECT to L and then divide by A. By all means, show your work. I think you are multiplying by an extra factor L. Check the units. Remember dL has units of m. Are they ohms or something else?

6. Oct 5, 2014

### RUber

$\int dR = \int_a^b \frac{\rho}{A} dL \implies R=\frac1A\int_a^b \rho dL$ You should not have to multiply by length, it is included in the integral.

7. Oct 6, 2014

### Bluestribute

Ok, I'm back and found out where I was going wrong. Also, the parameters change each time, so don't freak out with me inputting new numbers in my explanation.

So yes, I don't know why I was multiplying by length afterwards . . . Good to catch it now though. Anyways, I integrated the resistivity across the bounds: the integral of 4x from 0 to 0.5. Then I divided by the AREA of the given radius. That's where I was going wrong multiple times. They gave me a radius of 0.46 MM. For some reason I was using this as "A" instead of solving for A . . . So in short, integrated my resistivity function across my wire length and divided by the cross sectional area.

Got it instantly when I started writing out my variables here and realized it was radius, not area . . .