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?
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