# You have a fine wire with a cross sectional area of 9.4×10−10 m2

 P: 65 You have a fine wire with a cross sectional area of 9.4×10−10 m2 and a length of 1.1 m. You wish to determine what it is made out of, so you connect it to a standard AA battery (1.5 V) and measure a current of 0.048 A. What material is the wire most likely made out of? The options for the resistivities are as follows: Iron (but iron is wrong, so discard) Aluminum: 2.7E-8 Silver: 1.6E-8 Copper: 1.7E-8 I found the resistance by multiplying the current (.048 A) and voltage (1.5). I got .072 ohms. I'm sort of confused about the area though. Would it be A=∏(9.4×10−10)2, or what? Then when I solve I get .072=ρ(1.1/2.95E-9) ρ= 1.9E-10 ohm meters. I only have one shot of getting the answer. I'm confused, do it right, and in that case is it closest to aluminum or silver? My friend had aluminum, and we might have the same numbers. PLEASE HELP!
HW Helper
P: 2,318
 Quote by Snape1830 You have a fine wire with a cross sectional area of 9.4×10−10 m2 and a length of 1.1 m. You wish to determine what it is made out of, so you connect it to a standard AA battery (1.5 V) and measure a current of 0.048 A. What material is the wire most likely made out of? The options for the resistivities are as follows: Iron (but iron is wrong, so discard) Aluminum: 2.7E-8 Silver: 1.6E-8 Copper: 1.7E-8 I found the resistance by multiplying the current (.048 A) and voltage (1.5). I got .072 ohms. I'm sort of confused about the area though. Would it be A=∏(9.4×10−10)2, or what? Then when I solve I get .072=ρ(1.1/2.95E-9) ρ= 1.9E-10 ohm meters. I only have one shot of getting the answer. I'm confused, do it right, and in that case is it closest to aluminum or silver? My friend had aluminum, and we might have the same numbers. PLEASE HELP!
Read those two bits in red and see if you are still confused.
P: 65
 Quote by PeterO Read those two bits in red and see if you are still confused.
Yes, I am, because my friend did the same thing as me and it's wrong. When I used the area like that I got 6.0E-11 ohm meters. Neiter one of those options is remotely close to that number. Or do I put the area into the Area equation? Divide it by pi to get the radius and then solve?

Mentor
P: 11,676
You have a fine wire with a cross sectional area of 9.4×10−10 m2

 Quote by Snape1830 I found the resistance by multiplying the current (.048 A) and voltage (1.5). I got .072 ohms.
Check what Ohm's Law states. Your resistance calculation is not correct.
P: 65
 Quote by gneill Check what Ohm's Law states. Your resistance calculation is not correct.
I'm sorry, I meant that I divided the voltage by the current.
Mentor
P: 11,676
 Quote by Snape1830 I'm sorry, I meant that I divided the voltage by the current.
Okay, so what resistance does that give you?
P: 65
 Quote by gneill Okay, so what resistance does that give you?
the resistance was .072 ohms.
Mentor
P: 11,676
 Quote by Snape1830 the resistance was .072 ohms.
Recheck that value!
P: 65
 Quote by gneill Recheck that value!
31.25 ohms...oh I see what I did with the resistance. I had the equation as V/I but I still multiplied them. Silly me. But what about the area? Is that actually the number I use?
 P: 65 When I do the calculations I get 2.67E-8 ohm meters. So it's aluminum?
Mentor
P: 11,676
 Quote by Snape1830 31.25 ohms...oh I see what I did with the resistance. I had the equation as V/I but I still multiplied them. Silly me. But what about the area? Is that actually the number I use?
You're given the cross sectional area. What other number do you have in mind?
P: 65
 Quote by gneill You're given the cross sectional area. What other number do you have in mind?
I don't know, my friend said he did it a different way, but he got the right answer. He confused me, because I thought it was the area. I guess it is. Just wondering.
Mentor
P: 11,676
 Quote by Snape1830 When I do the calculations I get 2.67E-8 ohm meters. So it's aluminum?
If it's closest to the corresponding value given for aluminum, then yes
P: 65
 Quote by gneill If it's closest to the corresponding value given for aluminum, then yes
Well, aluminum is Aluminum: 2.7E-8 ohm meters, so I'm gonna go with yes. Thanks!
HW Helper
P: 2,318
 Quote by Snape1830 Well, aluminum is Aluminum: 2.7E-8 ohm meters, so I'm gonna go with yes. Thanks!
Resistance is given by R = ρL/A using resistivity, length and cross-section area

R = V/I from Ohms law

Combining you have ρL/A = V/I

so ρ = VA/LI

since you were given the values of V, A, L and I it should just have been a matter of plugging in the values and see what you get.

Note that the actual Resistance value does not ever have to be calculated.