What determines the brightness of bulbs?

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SUMMARY

The brightness of identical bulbs is determined primarily by the filament's temperature and surface area, which are influenced by ohmic heating. The relationship between power, current, and voltage is defined by the equation Power = R I² = V²/R. To increase brightness while maintaining a constant temperature, the filament must be designed with reduced resistance, achieved by making it thicker and longer to enhance surface area. The power utilized is proportional to voltage² multiplied by the radius² divided by the length of the filament, leading to the conclusion that brightness is proportional to the radius raised to the power of 3/2 under constant voltage and temperature conditions.

PREREQUISITES
  • Understanding of electrical concepts such as voltage, current, and resistance
  • Familiarity with the principles of ohmic heating
  • Knowledge of the relationship between power, resistance, and temperature
  • Basic grasp of geometric properties related to filament design
NEXT STEPS
  • Research the effects of filament material on brightness and longevity
  • Explore the principles of thermal radiation and its impact on bulb design
  • Learn about the electrical characteristics of different filament geometries
  • Investigate the role of voltage regulation in maintaining consistent bulb brightness
USEFUL FOR

Students studying physics, electrical engineers, and anyone interested in the design and functionality of incandescent bulbs.

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Homework Statement


This is not a specific problem, but I am wondering, what exactly determines the brightness of identical bulbs? The current going through it? The voltage? Something else? Thanks.


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The Attempt at a Solution


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The brightness will be directly a function of temperature of the filament and how much surface area the filament has. But the cause of the temperature is ohmic heating which is proportional to the current times the voltage drop across the filament (amps times volts = watts). This voltage in turn is proportional to the resistance times the current and so you get Power = R I^2 = V^2/R.

Assuming your voltage source is a constant then making a brighter bulb involves reducing the resistance. You make a thicker filament. But you don't want the temperature to change so you make it thicker and longer so there's more surface area to radiate the energy. (Too hot and the filament will melt and the bulb burns out, too cool and you are making a heat lamp instead of a light).

In general the resistance will be proportional to the length over the cross sectional area of the filament or length/radius^2.

Thus the power utilized will be proportional to voltage^2 * radius^2/length.
Let's say P = C r^2/L

The surface area will likewise be proportional to length * radius and that will be proportional to the power radiated. Let's say P = DrL.
Multiply the two and you'll see that P^2 = CDr^3, or P is proportional to r^3/2 given the conditions of constant voltage and fixed temperature.
 

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