Question

A standard 100 W (120 V) lightbulb contains a 9.10 cm -long tungsten filament. The high-temperature resistivity of tungsten is 9.0×10−7Ωm. What is the diameter of the filament?

Answer 1

We know power of a bulb

$P=\frac{V^2}{R}$

where is voltage and is the resistance of the filament.

Here watt, $V=120$ volt ,so

$R=\frac{V^2}{P}=\frac{(120)^2}{100}=144$ ohm

Again for a filament of diameter, length and high-temperature resistivity $\rho$, the resistance of the filament is

$R=\rho\cdot\frac{l}{A}=\rho\cdot\frac{l}{\pi (\frac{D}{2})^2}$

$\Rightarrow D=2\times\sqrt{\frac{\rho l}{\pi R}}$

Here $l=9\cdot10$ cm, $l=9\cdot10\times10^-2$ meter ,$\rho=9\cdot0\times10^-7\Omega-m$

So $D=2\times\sqrt{\frac{9\cdot0\times10^-7\times9\cdot10\times10^-2}{\pi\times144}}=2\cdot69\times10^-5$ meter