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.