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Homework Help: Lightbulbs and power/resistance

  1. May 25, 2013 #1
    My questions are:
    1. are we assuming constant voltage?
    (And I'm going to ask a really basic one: are we supplied constant voltage at our homes?)

    2. brightness = k*(power)?

    3. is it just a fact that higher power output devices have less resistance?
  2. jcsd
  3. May 25, 2013 #2


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    wall power (well, I actually should say voltage) is constant. As for your other questions, what do you think?
  4. May 25, 2013 #3


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    As a general rule, yes.
    I had to look this up. I didn't trust the wiki answer, as it didn't quite look right.

    But they referenced the source, and I found the answer:


    I plugged in 125v & 120v and came up with 115%.
    125/120 = 1.04
    So a 4% increase in voltage increased luminosity by 15%

    Thank you for asking this question. I didn't know that.

    You can plug numbers into the following equations and find out.

    P = I*E = I2*R = E2/R
    P = Watts
    I = Amps
    E = Volts
    R = Ohms
    Last edited: May 25, 2013
  5. May 26, 2013 #4

    I like Serena

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    Wiki is not wrong. It says the same thing.

    Anyway, this formula applies only for a voltage outside of the design specification, which is not something you would normally want to do. (The lamp will break.)

    The wiki article also contains a table, specifying how the perceived brightness (in Lumen) varies with varying types of incandescent light bulbs when used at their design specification.
    The relation in that table is:
    ##\qquad Brightness = k \times Power^{1.2} \quad## up to ##\quad k \times Power^{1.4}##.
    depending on the type of lamp.

    The amount of radiated power is always exactly the power you put into the lamp.
    It's just that by far most of it is in the infrared spectrum, that is, radiated as heat.
    The efficiency of a lamp is the ratio between amount of visible light (in Lumen) and the radiated power (in Watt).

    The reason that you get better efficiency from a lamp with higher power specification is because the filaments burn hotter, meaning a larger percentage of power is radiated in the visible spectrum. But burning hotter also means quicker degradation.
  6. May 26, 2013 #5


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    More or less constant.

    Apply conservation of energy. Where does the incoming power go?

    If you are talking about devices with the same design voltage then in general yes.

    If the design voltage is different then it's not allways true.

    You can design a light bulb to suit a car (12V) or mains (110V).

    V = IR
    P= IV

    Some combinations and the resulting resistance...

    12V, 50W... R = 12^2/50 = 2.88 Ohms
    110V, 50W ... R = 110^2/50 = 200 ohms

    12V, 100W.... R = 12^2/100 = 1.44 Ohms
    110V, 100W ...R = 100^2/100 = 100 Ohms

    So in this case a 12V, 50W bulb has a lower resistance (2.88Ohms) that a 110V, 100W bulb (100 Ohms).
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