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

In summary: Simply using Ohm's Law and selecting the material with the highest resistivity will yield the correct answer. Resistance is given by R = ρL/A using resistivity, length and cross-section area. R = V/I from Ohms lawCombining you have ρL/A = V/Ictica result 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.
  • #1
Snape1830
65
0
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!
 
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  • #2


Snape1830 said:
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.
 
  • #3


PeterO said:
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?
 
Last edited:
  • #4


Snape1830 said:
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.
 
  • #5
gneill said:
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.
 
  • #6


Snape1830 said:
I'm sorry, I meant that I divided the voltage by the current.
Okay, so what resistance does that give you?
 
  • #7


gneill said:
Okay, so what resistance does that give you?

the resistance was .072 ohms.
 
  • #8


Snape1830 said:
the resistance was .072 ohms.

Recheck that value!
 
  • #9


gneill said:
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?
 
  • #10


When I do the calculations I get 2.67E-8 ohm meters. So it's aluminum?
 
  • #11


Snape1830 said:
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?

:confused: You're given the cross sectional area. What other number do you have in mind?
 
  • #12


gneill said:
:confused: 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.
 
  • #13


Snape1830 said:
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 :smile:
 
  • #14


gneill said:
If it's closest to the corresponding value given for aluminum, then yes :smile:

Well, aluminum is Aluminum: 2.7E-8 ohm meters, so I'm going to go with yes. Thanks!
 
  • #15


Snape1830 said:
Well, aluminum is Aluminum: 2.7E-8 ohm meters, so I'm going to 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.
 

1. What is the significance of the cross sectional area of the wire?

The cross sectional area of a wire is an important factor in determining its resistance and current-carrying capacity. It represents the area of the wire that is perpendicular to the direction of current flow.

2. How does the cross sectional area affect the resistance of the wire?

The resistance of a wire is inversely proportional to its cross sectional area. A smaller cross sectional area means that there is less space for electrons to flow, resulting in higher resistance and lower current flow. A larger cross sectional area allows for more space and easier flow of electrons, resulting in lower resistance and higher current flow.

3. What is the unit of measurement for cross sectional area?

The unit of measurement for cross sectional area is square meters (m2).

4. Can the cross sectional area of a wire be changed?

Yes, the cross sectional area of a wire can be changed by altering its physical dimensions. For example, stretching or compressing a wire can change its cross sectional area, which in turn affects its resistance and current-carrying capacity.

5. How does the cross sectional area of a wire relate to its strength?

The cross sectional area of a wire is directly proportional to its strength. A larger cross sectional area means that the wire can withstand more force and tension without breaking. This is why thicker wires are often used for heavy-duty applications or in structures that require high strength and durability.

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