Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

EXAMPLE 17.3 The Resistance of Nichrome Wire GOAL Combine the concept of resisti

ID: 585770 • Letter: E

Question

EXAMPLE 17.3 The Resistance of Nichrome Wire GOAL Combine the concept of resistivity with Ohm's law PROBLEM (a) Calculate the resistance per unit length of a 22-gauge Nichrome wire of radius 0.321 mm. (b) If a potential difference of 10.0 V is maintained across a 1.00-m length of the Nichrome wire what is the current in the wire? (c) The wire is melted down and recast with twice its original length Find the new resistance Rv as a multiple of the old resistance Ro STRATEGY Part (a) requires substitution into Equation 17.5 after calculating the cross-sectional area whereas part (b) is a matter of substitution into Ohm's law. Part (c) requires some algebra. The idea is to take the expression for the new resistance and substitute expressions for eyand Av, the new length and cross-sectional area, in terms of the old length and cross-section. For the area substitution, remember that the volumes of the old and new wires are the same SOLUTION (A) Calculate the resistance per unit length Find the cross-sectional area of the d = ,2 = (3.210 x 10-4 m)2 = 3.24 x 10-7 m2 wire: Obtain the resistivity of Nichrome, solve for R/, and substitute: R 1.5×10-6 -m 1 d 3.24×10-/ m2 10-6 n-m= 4.6 /m (B) Find the current in a 1.00-m segment of the wire if the potential difference across it is 10.0 V: 1, 10.0V R 4.6 Substitute given values into Ohm's law: = 2.2 A , (C) If the wire is melted down and recast with twice its original length, find the new resistance as a multiple of the old Find the new area Ayin terms of the old area Ao, using the fact the volume doesn't change and IN-210 Substitute to find the new resistance: -= (AO/2) = 440

Explanation / Answer

copper will have higher current.

a) R/L = rho/A = 1.5 x 10-6/(3.14 x (6.42 x 10-4)2) = 1.158 ohm/m

b) current = 11.9/1.158 = 10.27 A

c) LN = 3LO => AN = 1/ 3 x AO

RN = rho x 3LO/(AO/3) = 9RO

resistance of 5.2 m nichrome of radius 0.321 mm = 5.2 x 4.6 = 24 ohm

I = 120/ 24 = 5A