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Filament failure mode

  1. Jul 7, 2015 #1


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    1. The problem statement, all variables and given/known data

    Not homework but this seems the best place to post this. In a private conversation someone posted this explanation from a book on how a lamp filament fails..

    I'm having some trouble doing the maths to show that the thinner higher resistance part gets hotter...

    2. Relevant equations
    Ohms law

    3. The attempt at a solution

    Imagine the filament is made up of two resistors in series R1 and R2 (See diagram). Let R2 be the part that's getting thinner for some reason...


    The voltage on R2 is given by the potential divider rule..

    VR2 = V * R2/(R1+R2)

    The power dissipated in R2 is

    WR2 = (V * R2/(R1+R2)2/R2

    If you expand and simplify you get..

    WR2 = (V2 * R2) / (R12 + 2R1R2 + R22)

    So if R2 increases the power in R2 decreases due to the R22. That should reduce the temperature in R2 which is the opposite of what the book suggests. Have I made an error?
  2. jcsd
  3. Jul 7, 2015 #2


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    I suspect the problem is to do with the physics of the filament. eg the thinner part has less surface area so although the power dissipated in that bit is lower the surface area is also lower so it's harder to dissipate the power.
  4. Jul 7, 2015 #3
    If R2 is small compared with the total resistance, the current through the combination will be roughly constant. The dissipation in in R2 will be I^2 x R2, so it is proportional to the resistance of R2. In your formula, R2 appears in both top and bottom, so your statement is not quite accurate.
  5. Jul 7, 2015 #4


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    Ah yes I'm forgetting that the square of a number less than 1 is smaller than the original number.
  6. Jul 8, 2015 #5

    rude man

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    If R2 is the part getting thinner then, assuming R2 is same length as the rest of the filament R1, R2 will be bigger than R1, and
    power in R2 = i*R2 = [V/(R1 + R2)]R2 which is > [V/(R1 + R2)]R1.

    Question for you: why is R2 bigger than R1?
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