The rating on a bulb specifies its "room temperature" resistance. Thus a 100-watt bulb rated for 120 volts would have an ideal resistance of
P = IV P = (V/R)V P = V2/R
rearranging for R
R = V2/P R = 1202/100
R = 144 ohms while a 60-watt bulb rated for 120 volts would have an ideal resistance of
R = V2/P R = 1202/60
R = 240 ohms.
The resistance of a wire, or filament, is proportional to its length and inversely proportional to its cross-sectional area. Longer and/or thinner wires have greater electrical resistance. A wire's resistance is also based on the type of substance out of which it is made: copper, zinc, tungsten, iron. This property is called the wire's resistivity, or ρ which is measured in ohm-meters (Ωm).
Putting these properties together gives us the mathematical relationship:
R = ρ(L/A)
Since electrical power is directly proportional to the square of the current running through a device, a high wattage light bulb will draw a larger current. Therefore it has to have a smaller resistance filament, which implies that it has a filament with a larger cross-sectional area. Conversely, low wattage bulbs, which are dimmer, have thinner filaments that present a higher resistance, and for a common voltage, would draw a smaller current. |
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Summary
- If two bulbs share the same current, the bulb which experiences the greater voltage drop will glow more brightly.
- If two bulbs share the same potential, the bulb which draws the greater amount of current will glow more brightly.
If circuits use combinations of the same identical bulb; that is, a circuit with several 60-watt bulbs all having the same resistance, the bulbs experiencing the greater voltage drop or those carrying the greater amount of current will glow more brightly.
In all cases, the greater amount of power used in a bulb's operation, the brighter it will glow.