1. Limited time only! Sign up for a free 30min personal 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!

Question on bulb and brightness

  1. Jun 16, 2010 #1
    EDIT: NOT A HOMEWORK PROBLEM

    in a bulb increase in resistance means that there is a increase in brightness.


    if there is a decrease in resistance means that decrease in brightness.

    is the statement correct ?

    one more question does the brightness depend on current
    say we have same R , if i increase the current does the brightness change?

    say you have the same resistance if you increase the current does the brightness change?
     
    Last edited: Jun 16, 2010
  2. jcsd
  3. Jun 16, 2010 #2

    russ_watters

    User Avatar

    Staff: Mentor

    Not sure if this is homework so a small hint:

    Voltage is constant, right? Keep that in mind and examine the two equations that relate voltage, amperage, wattage and resistance.
     
  4. Jun 16, 2010 #3
    that helps but this is not a home work question this is something that bothered me as i was reading the text book
     
  5. Jun 16, 2010 #4
    so i think since resistance increases the thermal energy as the resistance increase the thermal energy increases so as thermal energy increases the brightness increase.

    but not sure if the current affects it..?
     
  6. Jun 16, 2010 #5

    K^2

    User Avatar
    Science Advisor

    The power output of a resistive element is V²/R. If you keep applied voltage constant, increasing resistance would reduce power output.

    This is even more critical to light bulbs, because reducing power output reduces temperature, and that reduces amount of visible light generated as a fraction of total output.

    In other words, given a fixed voltage supply, you want as low a resistance as possible to maximize brightness.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook