Example 17.4 The Resistance of Nichrome Wire Goal Combine the concept of resisti

Question

Example 17.4 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 10.2 V is maintained across a 1.00 m length of wire, what is the current in the wire?

(c) The wire is melted down and recast with twice it's original length. Find the new resistance RN as a multiple of the old resistance RO. Strategy Part(a) requires substitution into Equation 17.5, after calculating the cross-sectional area, while 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 lN and AN, the new length and cross section area, in terms of the old length and cross-section. For the area substitution, use the fact that the volumes of the old and new wire are the same.
Solution (a) Calculate the resistance per unit length. Obtain the cross-sectional area of the wire. A = ?r2 = ?(0.321 10-3 m)2 = 3.24 10-7 m2 Obtain the resistivity of nichrome from Table 17.1, solve Equation 17.5 for R/L, and substitute.

= 1/m
(b) Find the current in a 1.00 m segment of wire if the potential difference across it is 10.2 V. Substitute given values into Ohm's law. = 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 AN in terms of the old area AO using the fact that the volume doesn't change and lN = 2lO. VN = VO ANlN = AOlO AN = AO(lO/lN)
AN = AO(lO/2lO) = AO/2 Substitute into Equation 17.5
= 3RO

Remarks From Table 17.1, the resistivity of nichrome is about 100 times that of copper, a typical good conductor. Therefore, a copper wire of the same radius would have a resistance per unit length of only 0.052 /m, and a 1.00 m length of copper wire of the same radius would carry the same current (2.2A) with an applied voltage of only 0.115 V. Because of its resistance to oxidation, nichrome is often used for heating elements in toasters, irons, and electric heaters.

Explanation / Answer

resistivity of nichrome wire = 1.5*10^(-6) ohm-m

resistance R = l/a

a = r^2 = *(0.321*10^(-3))^2 = 3.24*10^(-7) m^2

a) resistance per unit length = R/l = /a

R/l = 1.5*10^(-6) / [3.24*10^(-7)] = 4.634 ohm/m^2

b) here length = 1m

so R = 4.634 ohm

potential difference = 10.2 V

from ohm's law, V=IR

so I = V/R = 10.2 / 4.634 = 2.201 A

c) as wire is melted down an recasted, its volume remains constant.

initial volume = a*l

if l1 and a1 are final length and area, then final volume = a1*l1

we have a*l = a1*l1

also l1 = 2l

so a1 = a/2.

now final resistance RN= l1 / a1 = *2l / (a/2) = 4l/a

initial resistance R0 = l/a

so RN = 4*R0

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