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**1. Homework Statement**

The corroded contacts in a lightbulb socket have 5.0 Ohms of resistance. How much actual power is dissipated by a 100 W (120 V) lightbulb screwed into this socket?

**2. Homework Equations**

P = V^2/R

V = IR

P = RI^2

**3. The Attempt at a Solution**

I found the resistance of the lightbulb to be 120^2/100 = 144 Ohms, then added 5 Ohms to it to get the resistance of the entire circuit. Then I found the current in the circuit by using I = V/R = 0.8054 A, and then I found the power dissipated by the lightbulb by using P = R(of the lightbulb)I^2 = 144 x 0.8054 = 93.4 W as my final answer, but I don't know if this is right and electric circuit analysis is NOT my specialty in physics.