Question

An incandescent light bulb uses a coiled filament of tungsten that is 580 mmmm long with a diameter of 46.0 μm. At 20.0∘C tungsten has a resistivity of 5.25×10^−8Ω⋅m. Its temperature coefficient of resistivity is 0.0045 (C∘)−1(C∘)−1, and this remains accurate even at high temperatures. The temperature of the filament increases linearly with current, from 20∘C when no current flows to 2520∘C at 1.00 A of current.

What is the resistance of the light bulb at 20∘C?

Answer 1

The resistance of the light bulb at 20∘C is 22.36 Ω.

Step-by-step explanation:

To calculate the resistance of the light bulb at 20∘C, we can use the formula R = ρL/A, where R is the resistance, ρ is the resistivity, L is the length of the filament, and A is the cross-sectional area of the filament.

Given that the resistivity of tungsten is 5.25×10^-8 Ω⋅m and the temperature coefficient of resistivity is 0.0045 (C∘)^-1, we can use the formula ρ = ρ₀(1 + αΔT) to calculate the resistivity at 2520∘C. By rearranging the formula, we can find the resistivity at 20∘C: ρ₀ = ρ/(1 + αΔT).

Substituting the values into the formula, we get ρ₀ = 5.25×10^-8 Ω⋅m/(1 + 0.0045(2520-20)), which gives us ρ₀ = 1.547×10^-7 Ω⋅m. Now, we can calculate the resistance at 20∘C using the formula R = ρL/A. The cross-sectional area A can be calculated using the formula A = πr², where r is the radius of the filament.

Given that the diameter of the filament is 46.0 μm, we can calculate the radius as r = (46.0 μm)/2 = 23.0 μm. Converting μm to meters, we get r = 23.0×10^-6 m. Substituting the values into the formula, we finally get R = (1.547×10^-7 Ω⋅m)(580×10^-3 m)/[(π(23.0×10^-6 m)²)], which gives us R = 22.36 Ω.

Answer 2

The resistance of the light bulb at 20∘C is 18.65 Ω.

Step-by-step explanation:

To calculate the resistance of the light bulb at 20∘C, we need to use the formula for resistance, which is R = ρL/A, where ρ is the resistivity, L is the length of the filament, and A is the cross-sectional area of the filament.

Given that the filament is 580 mmmm long and has a diameter of 46.0 μm (which we can convert to meters by dividing by 10^6), we can calculate the cross-sectional area as follows:

A = πr^2 = π(d/2)^2 = π(46.0 μm/2)^2 = π(23.0 μm)^2 = π(23e-6 m)^2 = 1.66e-11 m^2

Now we can calculate the resistance:

R = ρL/A = (5.25e-8 Ω⋅m)(580e-3 m)/(1.66e-11 m^2) = 18.65 Ω

Therefore, the resistance of the light bulb at 20∘C is 18.65 Ω.