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Brightness of a Bulb

  1. Dec 11, 2004 #1
    We have two bulbs, X and Y. Y has thinner and longer filament than X. When connected to a power supply which bulb will give out most light and why? I think the answer should be Y as it has heigher resistance and therefore will dissipate more energy.
    Thanks in advance. :approve:
    Last edited: Dec 11, 2004
  2. jcsd
  3. Dec 11, 2004 #2


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    Energy absorbed:W=UIt.Energy emitted (light energy) W_{emitted}=\eta W.
    Assume the 2 bulb are connected at the same tension U.Then,assuming by absurdum that the bulbs have the same effieciency of converting electric energy into light energy,then your discussion should stick with comparing the intensities.And from that respect,i think your conclusion is wrong.

  4. Dec 11, 2004 #3
    Well thanks for your reply but I'm afraid I didnt get a single thing. Can you please explain this in simple 10th grade physics. My teacher said the same thing. He proved it by using the equation P=V^2/R (where, I believe, P is rate of dissipation of energy).
    So does it mean the tungsten filament in the bulb has extremely less resistance? If thats the case then anything with low resistance should convert more electrical energy into other forms of energy? Man, I'm confused!
    I'll be gateful to anybody who explains it and correct me or even tries to do so.
    Last edited: Dec 11, 2004
  5. Dec 11, 2004 #4


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    Let's supersimplify things.I'm assuming that the 2 filaments are connected in a circuit and the difference of potential between the ends of the filaments are kept constant at,let's say,a common value U.Assume your filaments are made from the same material (if u want,tungsten/wolfram),so we shouldn't worry about different internal processes.Assume that the filaments are ideal,that is they convert all the electric energy absorbed from the circuit into electric energy.U know that the resitance is proportional with the length of the conductor/filament and inverse proportional with the area of a transverse section made through the conductor.
    So it means that the bulb with the shorter and thicker filament will have a smaller resistance.
    Since u assumed all the above,u have to compare the powers (energies emitted in each second) for each bulb.Since for both bulbs u have to apply the same formula P=UI=U^2/R.
    So u write for the first bulb (X):P_{X}=U^2/R_{X}.
    For the second (Y):P_{Y}=U^2/R_{Y}.

    From prior assumption,u have found that R_{X}<R_{Y},so that should mean:
    P_{X}>P_{Y}.So the X bulb will radiate more energy.

  6. Dec 11, 2004 #5
    Ummm. Do you mean that the two bulbs are connected one at a time, or both together in series? (If they're connected in parallel, that's the same as one at a time).

    OK, the rest of this post assumes the 'one at a time' or parallel wired case.

    Now think of a filament that's really long - say ten miles. And really thin too, so that it's resistance is like 10 million mega-ohms. Now this filament will conduct hardly any current, so there will be hardly any power flow, and it won't shine at all.

    Take it to the other extreme and have a filament with a resistance of 0.00001 ohms. Now with a 10 volt perfect battery (and perfect wiring to the bulb, like they always have in these physics questions) then you'd get about a million amps flowing, so the power dissipated in the bulb would be ten million watts. I think that would shine more than the long spindly filament job.
  7. Dec 11, 2004 #6


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    Obviously in parallel connected so he can look at both at the time and compare the intensities of light emitted.One at a time,should require some larger gap between the resistances.
    The case in series connection,he would find more useful the other formula:
    P=RI^2,since the intensity is the same through every bulb.And again,series connection would mean putting the bulbs in the same circuit and use the eye to compare the light intensities,and the formula to check whether the sight is wrong. :tongue2:
  8. Dec 11, 2004 #7


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    Questions like this are confusing because your instinct is to think of a steady flow of electricity. Then it is agreed that the higher resistance has more dissipated power. That's called "constant current" case.

    But batteries and house electricity don't work that way. They operate at "constant voltage". It's like a constant force, causing the higher resistance case to pass far less electricity (I mean current).

    Think of a car battery. Short it with a thick copper wire and the wire will melt in an instant. (OK, on second thought, don't try it; just imagine it.) That's a lot of power; because the resistance is very low. Now connect a small light bulb. No melting because the resistance is very high.
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