Question 17.3: The Resistance of Nichrome Wire Goal Combine the concept of ...
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 R_{N} as a multiple of the old resistance R_{O} .
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 l_{N} \text { and } A_{N} , the new length and cross-sectional 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 wires are the same.
R=\rho \frac{l}{A}
(a) Calculate the resistance per unit length.
Find the cross-sectional area of the wire:
A=\pi r^{2}=\pi\left(0.321 \times 10^{-3} m \right)^{2}=3.24 \times 10^{-7} m ^{2}
Obtain the resistivity of nichrome from Table 17.1, solve Equation 17.5 for R/l, and substitute:
\frac{R}{l}=\frac{\rho}{A}=\frac{1.5 \times 10^{-6} \Omega \cdot m }{3.24 \times 10^{-7} m ^{2}}=4.6 \Omega / m
(b) Find the current in a 1.00-m segment of the wire if the potential difference across it is 10.0 V.
Substitute given values into Ohm’s law:
I=\frac{\Delta V}{R}=\frac{10.0 V }{4.6 \Omega}=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 A_{N} \text { in terms of the old area } A_{O} , using the fact the volume doesn’t change and l_{N}=2 l_{O} :
\begin{aligned}&V_{N}=V_{O} \rightarrow A_{N} l_{N}=A_{O} l_{O} \rightarrow A_{N}=A_{O}\left(l_{O} / l_{N}\right) \\&A_{N}=A_{O}\left(l_{O} / 2 l_{O}\right)=A_{O} / 2\end{aligned}
Substitute into Equation 17.5:
R_{N}=\frac{\rho l_{N}}{A_{N}}=\frac{\rho\left(2 l_{O}\right)}{\left(A_{O} / 2\right)}=4 \frac{\rho l_{O}}{A_{O}}=4 R_{O}
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 Ω/m, and a 1.00-m length of copper wire of the same radius would carry the same current (2.2 A) 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.
TABLE 17.1
| Resistivities and Temperature Coefficients of Resistivity for Various Materials (at 20°C) |
| \begin{array}{ccc} & &\begin{array}{c}\text { Temperature Coefficient } \\\text { of Resistivity }\end{array} \\\text { Material } & \text { Resistivity } & {\left[(\Omega \cdot C )^{-1}\right]} \\\end{array} |
| \begin{array}{lcc} \text { Silver } & 1.59 \times 10^{-8} & 3.8 \times 10^{-3} \\\text { Copper } & 1.7 \times 10^{-8} & 3.9 \times 10^{-3} \\\text { Gold } & 2.44 \times 10^{-8} & 3.4 \times 10^{-3} \\\text { Aluminum } & 2.82 \times 10^{-8} & 3.9 \times 10^{-3} \\\text { Tungsten } & 5.6 \times 10^{-8} & 4.5 \times 10^{-3} \\\text { Iron } & 10.0 \times 10^{-8} & 5.0 \times 10^{-3} \\\text { Platinum } & 11 \times 10^{-8} & 3.92 \times 10^{-3} \\\text { Lead } & 22 \times 10^{-8} & 3.9 \times 10^{-3} \\\text { Nichrome }^{a} & 150 \times 10^{-8} & 0.4 \times 10^{-3} \\\text { Carbon } & 3.5 \times 10^{5} & -0.5 \times 10^{-3} \\\text { Germanium } & 0.46 & -48 \times 10^{-3} \\\text { Silicon } & 640 & -75 \times 10^{-3} \\\text { Glass } & 10^{10}-10^{14} & \\\text { Hard rubber } & \approx 10^{13} & \\\text { Sulfur } & 10^{15} & \\\text { Quartz (fused) } & 75 \times 10^{16} & \\\end{array} |
| { }^{a} \text { A nickel-chromium alloy commonly used in heating elements. } |