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Homework Help: How to treat BULBS in circuits?

  1. Dec 31, 2007 #1
    Two questions

    1) I always get confused when I see bulbs in circuits.. I can handle resistors... I just don't know what to do when I see bulbs

    especially questions asking, "Which bulb shines brighter... the 50 W or the 100 W bulb"

    and different scenarios... parallel vs. series... any help?

    What is the determining factor that determines how bright a bulb shines? Current.. voltage?
    Last edited: Dec 31, 2007
  2. jcsd
  3. Dec 31, 2007 #2

    It does. If you alter anything like this in the circuit you are going to change things in some way.

    If you want to prove this just sub in some numbers to the equation in the link below.


    It would help if you could be more specific about what you are asking here. As I am sure you are aware a 100W bulb will always produce a brighter light than a 50W bulb if they have the same conditions. However I'm going to guess that you are being asked to work out the equivilent resistance (using the formula's in the above wiki link) and say which will shine brighter in these conditions. I'd guess that you would use a formula like P=RI^2 to work out which had the greatest current and compare these values.

    If im on the wrong track tell me what type of question you are asking for.
    Last edited: Dec 31, 2007
  4. Dec 31, 2007 #3
    Well a 100 W light bulb on it's own will shine brighter than a 50 W lightbulb

    But when placed in series, the 50 W lightbulb will be dimmer... and the 100 W lightbulb will be brighter

    I never understood the concept behind this..
  5. Dec 31, 2007 #4
    Put simply its becasue the bulb with the greater power rating incurs a greater resistance.
  6. Dec 31, 2007 #5


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    Power = E^2/R = I^2*R. When they are connected to the same source of EMF, 100 W light bulb on it's own will shine brighter than a 50 W lightbulb. Hence resistance of 100 W bulb is less than 50 W bulb. When they are connected in series, the same current flowns through them. Since P = I^2*R, 50 W bulb will shine brighter than the 100 W bulb.
  7. Dec 31, 2007 #6
    If a 50 W bulb is placed in series with some resistor.... and the resistance of the resistor is increased, the bulb's brightness decreases..

    is this because less current travels through... but then if less current is travelling through it... won't its Power decrease... I don't understand how this can happen if a bulb is already at a preset 50 W state.
  8. Dec 31, 2007 #7


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    the brightness of the bulb (or luminosity) is proportional to the power drop at bulb. now power depends on both Voltage drop and current through bulb, namely P=VI. It depends on your circuit structure, either V or I or both can vary.

    The bulb is not preset at 50W... not in the sense you think. If you look at the specification sheet of a bulb, it always come with a "rated voltage" along with the power in watt. So under correct usage, this bulb can give up a maximum of 50W of power without breaking down. I mean, think about it, if the bulb is not connected to anything, how can it be giving out 50J of energy per second? and if a source can only provide energy of less than this rate, how can the bulb "create" more energy? the bulb is a passive device, it can consume/transfer energy but not create more via eg. chemical reactions.
  9. Jan 2, 2008 #8
    So what is the factor that determines "brightness"?
    Voltage drop? ... greater the voltage drop, the brighter the bulb is?

    Am I correct?
  10. Jan 3, 2008 #9


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    For a lot of commercial incandescent bulbs, this is also marked on the cartons and on the bulbs themselves...
  11. Jan 3, 2008 #10
    Physically you can think of a bulb as basically a resistor you shoot current through, and the current deposits energy in the form of heating up the resistor until it glows. But it's not really a resistor per se

    You don't just have a 100watt bulb, you normally have like a 100 watt bulb at 120V

    So if you hook it up to a 120 V voltage source, a 100 watts of power will be delivered. You can do P=IV and find the current flowing through, the voltage drop is forced courtesy of your wall outlet supplying 120 volts, the only question is what's the "resistance" of the bulb, since P=I^2*R, when comparing two bulbs at the same voltage, the difference in power is accounted for by that resistance

    Problem is that resistance isn't constant by a long shot, so at different voltages, temperatures, etc. it won't be determinable by Ohm's Law or anything

    Next up is the bulb's efficiency, some of the energy deposited becomes wasted heat, and that depends on how the bulb's made. But for two identically constructed bulbs hooked up to an identical power source, the higher power one will be brighter.

    I think, been a while since electronics
  12. Jan 7, 2008 #11
    Going back to why 50W shines brighter than a 100 W bulb...

    I thought I understood this... but I am totally confused about the concept of "power"

    How do we know the 50 W resistor has more resistance than the 100 W resistor?

    If we establish this fact, I understand how we can determine that the 50 W bulb shines brighter in series with the 100 W bulb

    I'm sorry if I seem a little slow... you guys are way above my level physics-wise.. and some of your explanations assume I know certain things.. that I actually don't
    Last edited: Jan 7, 2008
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