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 = V^{2}/R
rearranging for R
R = V^{2}/P R = 120^{2}/100
R = 144 ohms while a 60-watt bulb rated for 120 volts would have an ideal resistance of
R = V^{2}/P R = 120^{2}/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. |