If the 0.100mm diameter tungsten filament in a light bulb is to have a resistance of $\mathbf{0}\mathbf{.}\mathbf{200}\mathit{\Omega }$ at$\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{C}$, how long should it be?

In an electrical circuit, resistance is a measure of the resistance to current flow. The Greek letter omega $\left(\Omega \right)$is used to represent resistance in ohms.

The electrical resistance of the value $R$of a resistive circuit element depends on its geometric structure (its length symbolized as$L$ and cross-sectional area symbolized as $A$) and on the resistivity $p$of the material of which it is made:

$R=p\frac{L}{A}$

• Resistance of the filament is:$R=0.200\Omega$.
• The diameter of the wire is:$D=0.100\mathrm{mm}\times \frac{1\mathrm{m}}{1000\mathrm{mm}}=0.100\times {10}^{-3}\mathrm{m}$.
• The radius of the wire is:$r=D/2=0.0500\times {10}^{-3}m$.
• The resistivity of the tungsten is:$p=5.6\times {10}^{-8}\Omega .m$

The resistance of the tungsten filament is found from the above equation as:

$L=\frac{RA}{p}$

The value of which is the cross-sectional area of the filament:

$L=\frac{R{\mathrm{\pi r}}^2}{p}$

Entering the values, we obtain:

$$\begin{gathered} R=\frac{\mathrm{\pi }{\left(0.0500\times {10}^{-3}\mathrm{m}\right)}^2\times \left(0.200\mathrm{\Omega }\right)}{5.{610}^{-8}\Omega .\mathrm{m}} \\ =2.80\times {10}^{-2}\mathrm{m}\times \frac{100\mathrm{cm}}{1\mathrm{m}} \\ =2.80\mathrm{cm} \\ \mathrm{Therefore},\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{tungsten}\mathrm{filament}\mathrm{is}:\mathrm{L}=2.80\mathrm{cm}. \end{gathered}$$