Example 20.5 from College Physics, 20.3 Resistance and Resistivity

Calculating Resistor Diameter: A Headlight Filament

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?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
We can rearrange the equation $R = \frac{ρL}{A}$ 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.
Solution
The cross-sectional area, found by rearranging the expression for the resistance of a cylinder given in $R = \frac{ρL}{A}$ , is

$A = \frac{ρL}{R} .$
20.19

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

$\begin{array}{l} A & = & \frac{(5.6 \times 10^{–8} Ω \cdot m) (4.00 \times 10^{–2} m)}{0.350 Ω} \\ = & 6.40 \times 10^{–9} m^{2} . \end{array}$
20.20

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

$A = \frac{{πD}^{2}}{4} .$
20.21

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

$\begin{array}{l} D & = & 2 {(\frac{A}{p})}^{\frac{1}{2}} = 2 {(\frac{6.40 \times 10^{–9} m^{2}}{3.14})}^{\frac{1}{2}} \\ = & 9.0 \times 10^{–5} m . \end{array}$
20.22
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.

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How did it go?

Strategy

Solution

Discussion

Strategy

Solution

Discussion