Power dissipation in lightbulb

[Snazzy]

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?

Homework Equations

P = V^2/R
V = IR
P = RI^2

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.

 

[kamerling]

This is completely right, and clearly explained!

 

[Moetasim]

Your approach is correct. By using the formula P = V^2/R, you can calculate the power dissipated by the lightbulb when it is connected to a circuit with a certain resistance. In this case, the total resistance of the circuit is 5 ohms (from the corroded contacts) plus 144 ohms (from the lightbulb), giving a total resistance of 149 ohms. Plugging this into the formula, you get P = (120 V)^2 / 149 ohms = 93.4 watts. This means that 93.4 watts of power is dissipated in the lightbulb due to the resistance in the circuit.