posted by Spencer .
The corroded contacts in a lightbulb socket have 8.20 Omega resistance.
How much actual power is dissipated by a 105 W (120 V) lightbulb screwed into this socket?
The resistance of the bulb at operating temperature is R = V^2/P = 137.1 ohms. If it is in series with 8.2 ohm contact resistance, the voltage applied to the bulb filament is reduced to 120*(137.1/145.3) = 113.2 V, and the power dissipated INE THE BULB is reduced to V^2/R = (113.2)^2/R = 93.5 W. There will be additional power dissipated at the corroded contacts, equal to 6.8^2/8.2 = 5.6 W