Current in a hollowed nichrome wire

AI Thread Summary
The discussion revolves around calculating the current in a hollow nichrome tube connected to a 2.50 V battery. The relevant equations include Ohm's law (V=IR) and the resistance formula (R = p*(L/A)), where resistivity for nichrome is given as 1.5E-6. One participant attempts to calculate the resistance using the tube's dimensions but encounters issues with their calculations. After corrections, the resistance is determined to be 0.00902 ohms, resulting in a calculated current of 277 A. Accurate calculations and proper use of brackets in formulas are emphasized for correct results.
Minhtran1092
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Homework Statement



A 10.0 cm long hollow nichrome tube of inner diameter 2.20 mm , outer diameter 5.10 mm is connected to a 2.50 V battery. What is the current in the tube?

Homework Equations


V=IR
R = p*(L/A);p = resistivity of nichrome (1.5E-6), L = length of wire, A = Surface area of cross section


The Attempt at a Solution


V = I*R → V/R = I; V = 2.5V
R = p*(.1cm)/(π([5.1/2]/1000))2)-π((1.1/1000)2))

I = 2.5/R; Isn't correct
 
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Looks good to me, you must have done some calculations wrong... If you used wolframalpha or similar, your brackets are wrong so that might have messed it up...

I got R=0.00902 ohms, I=277 A
 

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