Question

Determine the length of the tungsten filament in a 100 watt lightbulb given: (i) the resistivity of tungsten is 5.6 × 10−8 Ω · m, (ii) the filament diameter is 0.0018 inches, and (iii) that the filament temperature is about 2550°C when it is "on". Show your work. How is possible to use this length filament in the bulb? Explain! Hint: Assume a ambient temperature of 25°C and an operating voltage of 120Vrms (more on this later!), and you will have to look up the temperature coefficient of tungsten to complete the calculations.

Answer 1

Answer:

0.34148 m

Explanation:

$\rho$ = Resistivity of tungsten = $5.6\times 10^{-8}\ \Omega m$

d = Diameter = 0.0018 inch

r = Radius = $\dfrac{r}{2}=\dfrac{0.0018}{2}=0.0009\ in$

$r=0.0009\times 0.0254=0.00002286\ m$

$\alpha$ = Temperature coefficient of tungsten = $0.0045 /^{\circ}C$

Power is given by

$P=\dfrac{V^2}{R}\\\Rightarrow R=\dfrac{V^2}{P}\\\Rightarrow R=\dfrac{120^2}{100}\\\Rightarrow R=144\ \Omega$

We have the equation

$R_2=R_1[1+\alpha(T_2-T_1)]\\\Rightarrow R_1=\dfrac{R_2}{1+\alpha(T_2-T_1)}\\\Rightarrow R_1=\dfrac{144}{1+0.0045(2550-25)}\\\Rightarrow R_1=11.64812\ \Omega$

Resistance is given by

$R=\rho\dfrac{l}{A}\\\Rightarrow l=\dfrac{RA}{\rho}\\\Rightarrow l=\dfrac{11.64812\times \pi (0.00002286)^2}{5.6\times 10^{-8}}\\\Rightarrow l=0.34148\ m$

The length of the filament is 0.34148 m