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Light bulb uses current or resistance?

  1. Apr 8, 2012 #1
    Whoosh! Just read electricity.

    Resistance results in heat. The filament of bulb uses heat to light up.

    Keeping this in mind and the equations of power, current, voltage, resistance:
    °Light bulb uses current or resistance to light up and illuminate?

    °What's the difference between a 120 watt and a 60 watt bulb in this context?

    Can someone help me with this terminology.
    Thanks.
     
  2. jcsd
  3. Apr 8, 2012 #2

    tiny-tim

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    hi hasankamal007! :smile:
    hmm … the terminology of the first question is somewhat obscure :confused:

    i suggest you answer the second question first …

    then i think the answer that they want for the first question will become obvious :wink:
     
  4. Apr 8, 2012 #3

    Dale

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    I would say the answer to the first question is "yes". But I agree with tiny-tim's strategy suggestion.
     
  5. Apr 8, 2012 #4

    phinds

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    That question seems to imply that you think either one or the other is responsible but not both.

    Do you think there would be light if there were no resistance?

    Do you think there would be light if there were no current?
     
  6. Apr 8, 2012 #5

    NascentOxygen

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    I'd also accept "the room" here. :wink:
     
  7. Apr 8, 2012 #6

    Dale

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    :rofl: Oh, I like that even better!
     
  8. Apr 8, 2012 #7
    In terms of electromagnetism, one watt is the rate at which work is done when one ampere (A) of current flows through an electrical potential difference of one volt (V). The formula (simplest) is W=A*V where A is ampere and V is volt. When first learning about electricity I found it easiest to think of the electricity as water in pipes. The amperage is the size of the pipe carrying the water,(a small pipe can carry less than a big one), voltage is the "pressure" of the water in the pipe. Power is the rate at which work is being performed, in this case the power is refered to as a watt. Current just refers to the fact that electrons are being moved through some medium(conductive material). Resistance is the ease that those electrons move through the medium. In the case of the light bulb, the filament that glows and produces the light provides the resistance to the electrical current. Take some time to study the definitions, it will go along way in understanding electricity.
     
  9. Apr 8, 2012 #8
    But guys, take a look at this.

    For 2 bulbs A of 120 watt and B of 60 watt have a current of
    A = 1 amp and B = 0.5 amp at constant 120 volts.
    A = 120Ω and B = 240Ω resistance.

    But, we know that 120 watt bulb A will shine brighter. But, bulb A has less resistance as shown before and to shine bulbs use resistance. Aren't these two facts contradicting?

    In short,
    More resisting filament = more brightness.
    But,
    High watt = less resistance.
    What's going on?
     
  10. Apr 8, 2012 #9

    Dale

    Staff: Mentor

    What makes you think that?
     
    Last edited: Apr 8, 2012
  11. Apr 8, 2012 #10
    Because resistance produces heating effect which ultimately produces light. Isn't this right?
     
  12. Apr 8, 2012 #11

    Dale

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    No, power dissipation produces heating. When you open a switch the resistance becomes very high. So according to your logic, turning things off would make them very bright and hot.

    Imagine you had 10000 filaments, would there be any difference if you connect them in series or in parallel? If so, which would be brighter, which would have more resistance, and which would draw more power?
     
  13. Apr 8, 2012 #12

    NascentOxygen

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    The heating is produced by power, and power = V x I

    So for a fixed "V", the more "I" you allow, the brighter the light. ⚡⚡⚡⚡
     
  14. Apr 8, 2012 #13
    I=V/R (I) is the current measured in amperes, (V) is the voltage, and (R) is your resistance. Notice the change in amperage. When you plug your numbers into the above equation you can see that when solved for R, you get the listed resistances.
    R=V/I Bulb A(120W) R=120/1 R=120 and Bulb B(60W) R=120/.5 R=240
    Now consider your W=A*V formula. For light bulb B, you are given half the amperes of the light bulb A. Setting both equations to 0, you get:120watts - (1ampere*120volts)=0 and 60watts - (.5ampere*120volts)=0, so both are true and in accordance with ohm's law, I=V/R the decrease in amperage required an increase in resistance. Hopes this helps.
     
  15. Apr 8, 2012 #14
    I would caution against spending to much time on the filament, a filament can be made of many different materials with individual properties. The important thing is to understand the basic mechanics of the equations and the definitions. The assumption of more light is taken because more work is being done. Lumens measure luminosity "brightness"
     
  16. Apr 8, 2012 #15

    tiny-tim

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    i'll add: P = V2/R, and P = I2R …

    which of them is (probably) fixed, V or I ? :wink:
     
  17. Apr 8, 2012 #16
    One uses [dissipates] twice the power of the other..how much light is produced is a related issue if both bulbs are the same incandescent type.


    As noted, P = i2R = v2/R so how do you adjust R to get the meximum power dissipation......
     
  18. Apr 9, 2012 #17
    Light bulbs in parallel are sometimes used in car indicators. If one fails, it changes the current/resistance, and the indicator speeds up alerting the driver.
     
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