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You have a fine wire with a cross sectional area of 9.4×10−10 m2 
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#1
Mar1712, 08:54 PM

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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: 2.7E8 Silver: 1.6E8 Copper: 1.7E8 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(1.1/2.95E9) ρ= 1.9E10 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! 


#2
Mar1712, 09:54 PM

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#3
Mar1812, 07:25 AM

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#4
Mar1812, 11:33 AM

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



#5
Mar1812, 12:08 PM

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#6
Mar1812, 12:23 PM

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#7
Mar1812, 12:36 PM

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Mar1812, 12:49 PM

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#9
Mar1812, 12:58 PM

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#10
Mar1812, 01:02 PM

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When I do the calculations I get 2.67E8 ohm meters. So it's aluminum?



#11
Mar1812, 01:02 PM

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#12
Mar1812, 01:04 PM

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#13
Mar1812, 01:10 PM

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#14
Mar1812, 01:14 PM

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#15
Mar1812, 07:46 PM

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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. 


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