How Much Power is Dissipated by a Lightbulb with Corroded Socket Contacts?

AI Thread Summary
The discussion centers on calculating the power dissipated by a 100 W lightbulb with corroded socket contacts that have a resistance of 5.0 Ohms. The total resistance of the circuit is determined by adding the lightbulb's resistance of 144 Ohms to the socket's resistance, resulting in 149 Ohms. The current in the circuit is calculated to be approximately 0.8054 A. Using this current, the power dissipated by the lightbulb is found to be 93.4 W. The calculations are confirmed as correct, providing clarity on the topic of electric circuit analysis.
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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.
 
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