1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Tungsten wires of all electric light bulbs
  1. Light bulbs (Replies: 3)