A car headlight filament is made of tungsten and has a cold resistance of 0.350 Ω . If the filament is a cylinder 4.00 cm long (it may be coiled to save space), what is its diameter?
Strategy
We can rearrange the equation $R=\frac{\rho L}{A}$ to find the cross-sectional area A of the filament from the given information. Then its diameter can be found by assuming it has a circular cross-section.

Question

A car headlight filament is made of tungsten and has a cold resistance of 0.350 Ω . If the filament is a cylinder 4.00 cm long (it may be coiled to save space), what is its diameter?
Strategy
We can rearrange the equation [latex]R=\frac{\rho L}{A}[/latex] to find the cross-sectional area A of the filament from the given information. Then its diameter can be found by assuming it has a circular cross-section.

Accepted Answer

The cross-sectional area, found by rearranging the expression for the resistance of a cylinder given in [latex]R=\frac{ ho L}{A}[/latex], is

[latex]A=\frac{ ho L}{R}[/latex]. (20.19)

Substituting the given values, and taking ρ from Table 20.1,

Material	Resistivity ρ ( Ω ⋅ m )
Conductors
Silver	[latex]1.59 imes 10^{-8}[/latex]
Copper	[latex]1.72 imes 10^{-8}[/latex]
Gold	[latex]2.44 imes 10^{-8}[/latex]
Aluminum	[latex]2.65 imes 10^{-8}[/latex]
Tungsten	[latex]5.6 imes 10^{-8}[/latex]
Iron	[latex]9.71 imes 10^{-8}[/latex]
Platinum	[latex]10.6 imes 10^{-8}[/latex]
Steel	[latex]20 imes 10^{-8}[/latex]
Lead	[latex]22 imes 10^{-8}[/latex]
Manganin (Cu, Mn, Ni alloy)	[latex]44 imes 10^{-8}[/latex]
Constantan (Cu, Ni alloy)	[latex]49 imes 10^{-8}[/latex]
Mercury	[latex]96 imes 10^{-8}[/latex]
Nichrome (Ni, Fe, Cr alloy)	[latex]100 imes 10^{-8}[/latex]
[latex] ext { Semiconductors }^{[1]}[/latex]
Carbon (pure)	[latex]3.5 imes 10^{5}[/latex]
Carbon	[latex](3.5-60) imes 10^{5}[/latex]
Germanium (pure)	[latex]600 imes 10^{-3}[/latex]
Germanium	[latex](1-600) imes 10^{-3}[/latex]
Silicon (pure)	2300
Silicon	0.1–2300
Insulators
Amber	[latex]5 imes 10^{14}[/latex]
Glass	[latex]10^{9}-10^{14}[/latex]
Lucite	[latex]>10^{13}[/latex]
Mica	[latex]10^{11}-10^{15}[/latex]
Quartz (fused)	[latex]75 imes 10^{16}[/latex]
Rubber (hard)	[latex]10^{13}-10^{16}[/latex]
Sulfur	[latex]10^{15}[/latex]
Teflon	[latex]>10^{13}[/latex]
Wood	[latex]10^{8}-10^{11}[/latex]

yields

[latex]A=\frac{\left(5.6 imes 10^{-8} \Omega \cdot m ight)\left(4.00 imes 10^{-2} m ight)}{1.350 \Omega}[/latex] (20.20)

[latex]=6.40 imes 10^{-9} m ^{2}[/latex].

The area of a circle is related to its diameter D by

[latex]A=\frac{\pi D^{2}}{4}[/latex]. (20.21)

Solving for the diameter D , and substituting the value found for A , gives

[latex]D=2\left(\frac{A}{p} ight)^{\frac{1}{2}}=2\left(\frac{6.40 imes 10^{-9} m ^{2}}{3.14} ight)^{\frac{1}{2}}[/latex] (20.22)

[latex]=9.0 imes 10^{-5} m[/latex].

Discussion
The diameter is just under a tenth of a millimeter. It is quoted to only two digits, because ρ is known to only two digits.

Question 20.5: A car headlight filament is made of tungsten and has a cold ...

Related Answered Questions

Material	Resistivity ρ ( Ω ⋅ m )
Conductors
Silver	$1.59 \times 10^{-8}$
Copper	$1.72 \times 10^{-8}$
Gold	$2.44 \times 10^{-8}$
Aluminum	$2.65 \times 10^{-8}$
Tungsten	$5.6 \times 10^{-8}$
Iron	$9.71 \times 10^{-8}$
Platinum	$10.6 \times 10^{-8}$
Steel	$20 \times 10^{-8}$
Lead	$22 \times 10^{-8}$
Manganin (Cu, Mn, Ni alloy)	$44 \times 10^{-8}$
Constantan (Cu, Ni alloy)	$49 \times 10^{-8}$
Mercury	$96 \times 10^{-8}$
Nichrome (Ni, Fe, Cr alloy)	$100 \times 10^{-8}$
$\text { Semiconductors }^{[1]}$
Carbon (pure)	$3.5 \times 10^{5}$
Carbon	$(3.5-60) \times 10^{5}$
Germanium (pure)	$600 \times 10^{-3}$
Germanium	$(1-600) \times 10^{-3}$
Silicon (pure)	2300
Silicon	0.1–2300
Insulators
Amber	$5 \times 10^{14}$
Glass	$10^{9}-10^{14}$
Lucite	$>10^{13}$
Mica	$10^{11}-10^{15}$
Quartz (fused)	$75 \times 10^{16}$
Rubber (hard)	$10^{13}-10^{16}$
Sulfur	$10^{15}$
Teflon	$>10^{13}$
Wood	$10^{8}-10^{11}$