Here is the question:
The corroded contacts in a lightbulb socket have 7.6 Omega total resistance. How much actual power is dissipated by a 110 W (120 V) lightbulb screwed into this socket?
Answer is in P=____W
resistance of bulb:
110 = (120)² /R
R = 130.9 ohms
current is 120/(R+7.6) = 0.866A
P = I²R = 98.3 W
This assumes the bulb resistance does not change when operated at lower voltage/current.