# Homework Help: Resistance of hollow copper cylinder

1. Mar 7, 2013

### Siune

1. The problem statement, all variables and given/known data
Hey!

We have a copper pipe, which has a outer diameter $y = 4.1 mm$ and inner diameter $x = 4.0 mm$
Length of the pipe is $L = 120m$ and resistivity of copper is $ρ = 1.7 * 10^{-8}\Omega$

2. Relevant equations
Resistance is $R = ρ * \frac{L}{A}$

3. The attempt at a solution
Now, I calculated the resistances for cylinders of diameter y and diameter x.
Then I subtracted the bigger from lower one and got:

$R = 8*10^{-3} \Omega$

But does the shape of the object matter? Don't we need just the area of the bottom of hollow cylinder?

$A = \pi * ( r_1^{2} - r_2^{2} )$

Using the area above, I get around $R = 3.2 \Omega$

I converted diameters to radius and mm to metres.

P.S. In back of my book the answer is $R=0,8 \Omega$

2. Mar 7, 2013

### voko

The answer in the book is correct. Most likely you are not using the correct units. Show the details of your calculation.

3. Mar 7, 2013

### SteamKing

Staff Emeritus
I get 3.2 ohm when I calculate the resistance.

voko: did you use 4.1 mm and 4 mm as diameters or radii?

4. Mar 7, 2013

5. Mar 7, 2013

### Siune

First option:

$R = 1.7 * 10 ^{-8}\Omega m * \Big(\frac{120 m }{\pi * \big( \frac{0.004 m }{2} \big)^{2}} - \frac{120 m }{\pi * \big( \frac{0.0041 m }{2} \big)^{2}} \Big) = 7,822.. * 10^{-3} \Omega$

Second option:
$R = 1.7 * 10 ^{-8}\Omega m * \Big(\frac{120 m }{ \pi \big( (\frac{0.0041m}{2})^{2} - (\frac{0.0040m}{2})^{2}} \Big) = 3.20667 \Omega$

Last edited: Mar 7, 2013
6. Mar 7, 2013

7. Mar 7, 2013

### Siune

Ye, that's why option 1 seemed pretty weird to me.

I dont see tho how option 2 is about additivity of resistance. As I just calculated the (bottom) surface area of the hollow sylinder and then proceed to calculate the resistance?

8. Mar 7, 2013

### voko

Option 1 is the one one which assumes additivity of resistance. If you use resistance of parallel conductors instead, you should get the same as per option 2. I suggest that you do that symbolically.

9. Mar 7, 2013

### Siune

Okey,

so we can divide the the bottom surface area to n amount of same size segments. Then they all have the same length L.

Now as they are parallel and each has resistance $R_1$ we get

$R^{-1} = \big( \frac{n}{R_{1}}\big)$

Now each segment has surface area

$A_1 = \frac{\phi}{360}\Big(\pi \big( (\frac{y}{2})^{2} - ( \frac{x}{2} ) ^{2} \big ) \Big)$

and then

$R_1 = \frac{ρL}{\frac{\phi}{360}\Big(\pi \big( (\frac{y}{2})^{2} - ( \frac{x}{2} ) ^{2} \big ) \Big)}$

Then as $\frac{\phi}{360} = \frac{1}{n}$

we get the

$R = \frac{ρL}{\Big(\pi \big( (\frac{y}{2})^{2} - ( \frac{x}{2} ) ^{2} \big ) \Big)}$

?

10. Mar 7, 2013

### voko

What you did is puzzling. But the result is correct.

11. Sep 14, 2016