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This was one of the questions for the repetition paper before the test. "A lit light bulb has a resistance of 1320 Ω. It uses tungsten as a filament, and when hot, has a resistivity of 0.92 µΩm. Assume the filament is 2.0 cm long, how thick is then the filament?"

I work it out as follows: Using the formula for resistance in conductors:

  • ρ = Resistivity
  • l = length
  • A = cross section area
  • R = resistance

    1. R = ρl/A
    2. A = ρl/R
    3. A = 0.92 µΩm * 0.02m / 1320 Ω
    4. A = 1.39×10^-11 m^2
    5. A ~= 13.9pm^2

That's just ridiculous. The answer given by my professor is 4.2µm^2. What am I doing wrong here?

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4.2 microns, not 4.2 microns squared. Area is not a unit of thickness. Divide your answer by pi, take the square root, and multiply by 2 to yield diameter

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1000 mm = 1 meter & 1,000,000 mm^2 = 1 m^2

$$ A = 1.39 x10^{-11} m^2 = 1.39 x10^{-5} mm^2 $$

Assume the filament is circular wire...

$$ A = \pi r^2 $$ $$ r = \sqrt{1.39x10^{-5} / \pi} = \sqrt{4.44x10^{-6}} = 0.002106 mm $$

Double the radius to get the diameter (thickness): 0.004213 mm = 4.213 microns.

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