# Aluminum wire stretched four times longer, resistance?

1. Sep 5, 2007

### redsealelectron

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

A piece of aluminum wire has a resistance of 20C, 15.5 $$\Omega$$. If this wire is melted down and used to produce a second wire having a length four times the original length, what will the resistance of the new wire be at 20C? (Hint: The volume remains the same)

2. Relevant equations

R = $$\rho$$$$\stackrel{l}{a}$$

or

R2 = R1(1 + $$\alpha$$1[t2 - t1])

3. The attempt at a solution

R2 = R1(1 + 0)

R2 = R1 * 4

R2 = 62.0

I have plugged in the information into equation number two. As the delta t = 0, I didn't use the temperature coefficient. My answer was to multiply the initial resistance by 4, which gave me 62.0 $$\Omega$$.

Some guys in my class are saying that you have to multiply by 16, and the correct answer is 248 $$\Omega$$.

I don't see why they would multiply by 16, sure its four times longer, but does that mean the diameter of the wire went down? The volume of the wire remained constant, so I assume yes, it did get longer increasing the resistance by four, but the volume remained the same so that is all that is needed.

Can anyone see if I have missed something?

2. Sep 5, 2007

### Dick

The volume of the wire is A*L, where A is the cross-sectional area and L is the length. So if volume is constant and you multiply the length by four, what happens to the cross-sectional area and how do A and L relate to total resistance.

3. Sep 5, 2007

### redsealelectron

Increasing the length increases the resistance of the wire by a factor of four. If the cross sectional area increases by a factor of four then the resistance should decrease by a factor of 16... this would be r2 = 1/4R1?

4. Sep 5, 2007

### Dick

Noooo. If L->4*L then A->A/4 so the product (which is volume) stays constant. The area DECREASES.

5. Sep 5, 2007

### redsealelectron

So the wire is now four times longer, and is now thinner then it was orginally. The cross sectional area is smaller therefore the resistance is higher then it was. So i must use R = p(L/A)? will I have to solve for the diameter of the wire?

6. Sep 5, 2007

### Dick

Yes, that's exactly what you need. L is four times bigger, A is four times smaller. You don't need the diameter. What happens to R?

7. Sep 5, 2007

### redsealelectron

Its multiplied by 16, because the length has increased by four and the area has decreased by four, initial resistance * the decreased area of four * the increased lenght of four?

8. Sep 5, 2007

### Dick

Yes. Your friends were right. For practice, what happens to the diameter? If you say it decreases by a factor 2, you would be right.

9. Sep 5, 2007

### redsealelectron

Dick, thank you for your help and letting me work through the problem. I have a better understanding now.

10. Sep 5, 2007

### Dick

No problem. Always fun if the victim of my advice finally 'gets it'.