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Tungsten wires of all electric light bulbs

  1. Sep 7, 2003 #1
    I'm looking for a jump start with this one. I'm having trouble getting started.

    The tungsten wires of all electric light bulbs are designed to glow at about the same temperature. This requires, as a first approximation, that the power per unit surface area of the filament be the same for all.

    a) Show that this leads to the requirement, at constant voltage, that r/l^2 is constant, where r is the radius and l is the length of the filament.

    b) If P2/P1 = n is the ratio of the power consumption of two different light bulbs, show that r2/r1 = n^(2/3) and that l2/l1 = n^(1/3).

    I've got lots of formulas, but I'm having trouble putting them together to show what a) and b) ask for. I'm sure once I get a), b) will follow easily.

    P = I^2*R => I is current; R is resistance
    R = pl/A => p is resistivity; A is surface area; l is length
    A = [pi]r^2

    P = I^2(pl/A) => P = I^2(pl/([pi]r^2))

    P = V^2/R where V is potential
    V^2 = (I^2)(R^2) => V = IR

    So I'm going around in circles and getting nowhere.

    Any clues would be greatly appreciated. I have a feeling the answer is staring me straight in the face, but I'm not seeing it.

    Thanks much.
    Last edited by a moderator: Feb 5, 2013
  2. jcsd
  3. Sep 7, 2003 #2
    Re: Power

    I think this is talking about exterior surface so:
    Then use P=U2/R=U2*π*r2 / ρl
  4. Sep 7, 2003 #3
    I need to learn to read the problem statement more carefully.

    Thanks for your help Sonty.
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