Question

The resistivity of tungsten increases approximately linearly with temperature from 56.0 nΩ · m at 293 K to 1.10 µΩ · m at 3500 K. A lightbulb is powered by a 120-V dc power supply. Under those operating conditions the temperature of the tungsten filament is 2500 K, the length of the filament is equal to 5.30 cm, and the power delivered to the filament is 36.4 W. Estimate the following.

(a) the resistance of the filament
Ω

(b) the diameter of the filament
µm

Answer 1

Given data:
resistivity = 56 n-ohm at Temparure T= 293K
resistivity = 1.10 micro-ohm at Temparure T= 3500K
Volage V= 120 V

Resistance of a wire in Ω
R = ρL/A
ρ is resistivity of the material in Ω-m
L is length in meters
A is cross-sectional area in m²
A = πr², r is radius of wire in m

Temperature Change in Resistance
Rt = Ro(1 + α∆T)
α is temperature coef in /ºC

∆T = 3500–293 = 3207
resistance is proportional to resistivity.
ρt/ρo = 1.1*10^-6 / 56*10^-9 = 19.6
1 + α∆T = 19.6
α∆T = 18.6
α = 0.00581 /ºC, which is fairly close to the published value of 0.0045

Now we need the resistivity at 2500
∆T = 2500–293 = 2207
again, substituting resistivity for resistance
ρt = ρo(1 + α∆T) = 56 nΩm(1+(0.00581)(2207))
ρt = 774 nΩm

Power = E/R²
R² = 120/36.4
R = 1.815 ohms

R = ρL/A
A = ρL/R = (774 nΩm)(0.0530) / (1.815) = 2.260*10^-8 m²
A = πr² = 2.260*10^-8 m²
r = 84.8µm
diameter of the filament is =169.6µm