91Ó°ÊÓ

We have given that a standard $100W\left(120V\right)$lightbulb contains a$7.0cm$-long tungsten filament. The high-temperature resistivity of tungsten is$9.0×{10}^{-7}\mathrm{Î\copyright }m$

We need to find that the diameter of the filament.

The rate of the dissipated energy when the current flows throw the resistor is called power. It is represented as

$P_R=\frac{{\left(∆V_R\right)}^2}{R}$ (1)

It can also be written as

$R=\frac{{\left(∆V_R\right)}^2}{P_R}$

The bulb has power $100W$and a potential difference of $120V$.We plug the values for $P_R$and $∆V_R$into equation (2) to get $R$

$$\begin{gathered} R=\frac{{\left(∆V_R\right)}^2}{P_R} \\ =\frac{{\left(120V\right)}^2}{100W} \\ =144\mathrm{Î\copyright } \end{gathered}$$

The resistance through the filament depends upon the area and the resistivity $\mathrm{ÒÏ}$of the filament and it has a formula by

$R=\frac{\mathrm{ÒÏ}L}{A}=\frac{\mathrm{ÒÏ}L}{{\mathrm{Ï€°ù}}^2}$

Where $L$is the length of the filament, $r$its radius. We solve equation (3) for $r$and plug the values for $R,L$and $\mathrm{ÒÏ}$to get $r$by

$$\begin{gathered} r=\sqrt{\frac{\mathrm{ÒÏ}L}{\mathrm{Ï€¸é}}} \\ =\sqrt{\frac{(9×{10}^{-7}\mathrm{Î\copyright }(m\left)\right(0.07m)}{\mathrm{Ï€}\left(144\mathrm{Î\copyright }\right)}} \\ =1.2×{10}^{-5}\mathrm{m} \\ =12\mathrm{μ³¾} \end{gathered}$$

The diameter $d$is two times the radius, so we find it by

$d=2r=2\left(12\mathrm{μ}m\right)=24\mathrm{μ}m$