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Homework Help: HELP! Light bulbs and circuits!

  1. Nov 28, 2007 #1
    Two light bulbs, one 60 watt bulb (higher resistance) and one 100 watt bulb (lower resistance) are placed in series. A current is run through them. Which bulb glows brighter?

    The answer detailed that: The bulb with more resistance would draw more current, thus get more voltage, so the 60 watt bulb glows brighter, has more power.
    I thought all resistors in series had the same current...so how is it possible to draw more current (especially with more resistance)?

  2. jcsd
  3. Nov 28, 2007 #2
    yeah, something sounds fishy.. but I agree most power is dissipated by the 60w bulb.
  4. Nov 28, 2007 #3
    Which ever bulb dissipates more power glows brighter.
    Basically, the two bulbs are like two resistors. One higher (60W) and the other lower (100W).
    The higher resistor drops more voltage than the lower resistor.
    The current flowing thru both of them will be the same.

    Power dissipated across the 60W bulb will be higher, hence it will glow brighter.

    If V=120V,
    The resistance of the 60w bulb will be (from P=V*V/R) 240 ohms.
    For 100W, R=144 ohms.
    The total series resistance of the ckt is 384 ohms.

    (From ohm's law) At 120V, current flowing thru the series ckt will be I=120/384= 0.3125Amps.

    Power dissipation at 60W bulb is P=I*I*R = 23.4 W

    Power dissipation at 100w bulb is P= 14.06 W

    So you have a 23W bulb and a 14W bulb. The 23W obviously glows brighter.
    Last edited: Nov 28, 2007
  5. Nov 29, 2007 #4
    Thanks for that additional example. Definitely helped. Just one quick question: What do you mean by "the higher resistor drops more voltage than the lower resistor"... by "drops more" what do you mean?

    Thanks again,

  6. Nov 29, 2007 #5
    "drops more" meaning you see more voltage across the resistor.
  7. Nov 29, 2007 #6
    Incandescent light bulbs have negative resistance. That circuit might even have two metastable states. So its not quite as simple as the two-resistor analogy.
  8. Dec 1, 2007 #7
    Can you give some reference for the negative resistance? I don't know about that.
  9. Dec 2, 2007 #8
    The "temperature coefficient of electrical resistivity" for tungsten is .0046 per degree C at room temperature. This means that if the temperature of the incandescent element of the bulb increases by, say, 3000 degrees C, then the resistance will increase by a factor of something like 3000*.0046 = 13.8. (Actually the coefficient varies a LOT with temperature, but this gives one an idea of the magnitude of the effect).
  10. Dec 2, 2007 #9
    Thanks. I am familiar with the increase in resistance; I just didn't know the name.

    One of the easy ways to answer this question (other than just try it) is to see that the question is general (i.e., higher and lower power bulbs in series) reather than specific (i.e., 100W and 60W). Then you just imagine a 10000W bulb in series with a 10W one and it can be seen that the 10000W bulb is equivalent to a piece of copper wire and thus not bright at all at the current permitted by the 10W bulb, but the 10W is at full brightness since it is not in any way limited.
  11. Jun 22, 2011 #10
    May i ask why " P= V^2/R " was used to calculate the Resistance of the individual light bulbs? In doing so, the voltage in each case was taken constant. (120 V in second reply) How is this possible?. The voltage across the bulbs in series is not constant!!!

    If " P= (I^2)R " is used to calculate the resistances of the bulbs, and considering that "I" for all the bulbs is constant, (being in series) ; we conclude that the 100W bulb has most resistance. You may try it.....

    I dont know, i may be wrong, if im wrong , then im wrong in selecting which formula to use, so do help me out.

  12. Jun 22, 2011 #11


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    Gold Member

    You're not exactly wrong, you have just not understood what was being calculated.

    This was NOT done implying that the bulbs each drop 120volts in the circuit described, it was an individual calculation of the resistance of each bulb based on the fact that they dissipate the stated amount of power WHEN DROPPING 120volts. This resistance was THEN used in the circuit described.
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