(a) Calculate the resistance per unit length.
Find the cross-sectional area of the wire:
A=πr2=π(0.321×10−3m)2=3.24×10−7m2
Obtain the resistivity of nichrome from Table 17.1, solve Equation 17.5 for R/l, and substitute:
lR=Aρ=3.24×10−7m21.5×10−6Ω⋅m=4.6Ω/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=RΔV=4.6Ω10.0V=2.2A
(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 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:
RN=ANρlN=(AO/2)ρ(2lO)=4AOρlO=4RO
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) |
| Material Resistivity Temperature Coefficient of Resistivity [(Ω⋅C)−1] |
| Silver Copper Gold Aluminum Tungsten Iron Platinum Lead Nichrome a Carbon Germanium Silicon Glass Hard rubber Sulfur Quartz (fused) 1.59×10−81.7×10−82.44×10−82.82×10−85.6×10−810.0×10−811×10−822×10−8150×10−83.5×1050.466401010−1014≈1013101575×10163.8×10−33.9×10−33.4×10−33.9×10−34.5×10−35.0×10−33.92×10−33.9×10−30.4×10−3−0.5×10−3−48×10−3−75×10−3 |
| a A nickel-chromium alloy commonly used in heating elements. |