# Electric pipe

1. May 17, 2007

### gamesandmore

A cylindrical aluminum pipe of length 1.36 m has an inner radius of 1.80 10-3 m and an outer radius of 3.00 10-3 m. The interior of the pipe is completely filled with copper. What is the resistance of this unit? (Hint: Imagine that the pipe is connected between the terminals of a battery and decide whether the aluminum and copper parts of the pipe are in series or in parallel.)

Here is what I did:
L = 1.36 m
ri = 1.80e-3m
ro = 3.00e-3 m
R = ? ohms
R = pL/A
pcopper = 1.72e-8 ohm*m
palum = 2.82e-8 ohm*m

In = copper
Out = aluminum

Ai = pi*ri^2 = 1.01788e-5 m^2
Ao= pi*ro^2 = 2.827e-5 m^2

Ri = pcopper * L/Ai = .002298 ohms
Ro = palum * L/Ao = .0013566 ohms

Then I did
1 / ( (1/.002298ohms) + (1/.0013566 ohms) )
and got 8.53e-4 Ohms, but it was wrong...

2. May 17, 2007

### nrqed

The cross sectional area for the aluminum cylinder is not ]$\pi r_0 ^2$, you must use only the area of the outside cyliner which wil be $\pi(r_0^2 - r_i^2)$

3. May 17, 2007

### gamesandmore

So Ao = pi*(ro^2 - ri^2) ?

4. May 17, 2007

### nrqed

Yes, that's what I wrote.

5. May 17, 2007

### gamesandmore

Thanks, worked perfect :)

6. May 17, 2007

### nrqed

Great. Am glad I could help. I hope you see why this is the correct equation to use for the cross sectional area of the aluminum pipe.

good luck