Resistance of New Wire: Answer the Question

[Magna1]

I hate to ask for the answer on this one but the arguments are getting louder amongst us.


Here is the ??
A wire has a resistance of [tex]5\Omega[\tex]. It is melted down and drwan into a new wire of [tex]\frac{1}{2}]the original diameter. What is the resistance of the new wire?<br /> <br /> Answer #1 [tex]5\Omega[\tex] because everything is proportional the amount of wire is the same just longer and not as thick.<br /> <br /> Answer #2 This is where we want to get numbers for any kind of wire and plug them into the formula $$R=\frac{\rho*\ell}{A}. then hopefully see something that makes sense.<br /> <br /> So our class is tomorrow and we can't settle on an answer. I hate to ask for the answer but we are out of time.<br /> <br /> I tried to Latex but not sure how it will look.<br /> I will repost if it doesn't make sense.<br /> <br /> Thank You in Advance$$[/tex][/tex][/tex]

 

[Magna1]

tex didn't work

The wire is 5 ohms and we want to get it to 1/2 it's original diameter.
the formula is R=[rho)*l(length)]/Area

 

[jamesrc]

Start with the formula for resistance:

$$R = \frac{\rho L}{A}$$

The resistivity, &rho;, remains constant throughout the problem (assuming you make your two measurements at the same temperature).

The volume of the wire also remains the same, so

$$V = A_1L_1 = A_2L_2$$

Given that the new diameter is half the old diameter, we know that the new area is 1/4 the old area:

$$A_1 = \frac{\pi d_1^2}{4}$$
$$A_2 = \frac{\pi d_2^2}{4} = \frac{\pi\left(\frac{d_1}{2}\right)^2}{4} = \frac{\pi d_1^2}{16} = \frac{A_1}{4}$$

And from the first equation, this means that the length of the new wire is 4 times that of the old one. Now check the resistances:

$$R_1 = \frac{\rho L_1}{A_1}$$
$$R_2 = \frac{\rho L_2}{A_2} = \frac{\rho (4L_1)}{\left(\frac{A_1}{4}\right)} = 16\frac{\rho L_1}{A_1} = 16*R_1$$

You should see that lowering the area raises the resistance. You should also see that raising the length raises the reistance. (Both of these concepts should make some intuitive sense.) In this problem, you do both so it's like a double whammy.

P.S. When you're entering LaTeX, watch the direction of your slashes.

 

[Blu3eyes]

Hello, can somebody please explain to me this point?

And from the first equation, this means that the length of the new wire is 4 times that of the old one. Now check the resistances:

$$R_1 = \frac{\rho L_1}{A_1}$$
$$R_2 = \frac{\rho L_2}{A_2} = \frac{\rho (4L_1)}{\left(\frac{A_1}{4}\right)} = 16\frac{\rho L_1}{A_1} = 16*R_1$$

Thank you in advance!