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

0.34148 m

Step-by-step explanation:

`\rho` = Resistivity of tungsten =
`5.6* 10^(-8)\ \Omega m`

d = Diameter = 0.0018 inch

r = Radius =
`(r)/(2)=(0.0018)/(2)=0.0009\ in`

`r=0.0009* 0.0254=0.00002286\ m`

`\alpha` = Temperature coefficient of tungsten =
`0.0045 /^(\circ)C`

Power is given by

`P=(V^2)/(R)\\\Rightarrow R=(V^2)/(P)\\\Rightarrow R=(120^2)/(100)\\\Rightarrow R=144\ \Omega`

We have the equation

`R_2=R_1[1+\alpha(T_2-T_1)]\\\Rightarrow R_1=(R_2)/(1+\alpha(T_2-T_1))\\\Rightarrow R_1=(144)/(1+0.0045(2550-25))\\\Rightarrow R_1=11.64812\ \Omega`

Resistance is given by

`R=\rho(l)/(A)\\\Rightarrow l=(RA)/(\rho)\\\Rightarrow l=(11.64812* \pi (0.00002286)^2)/(5.6* 10^(-8))\\\Rightarrow l=0.34148\ m`

The length of the filament is 0.34148 m