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

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SUMMARY

The power dissipated by a 100 W (120 V) lightbulb with corroded socket contacts exhibiting 5.0 Ohms of resistance is calculated using the formula P = V^2/R. The total resistance in the circuit is 144 Ohms (lightbulb resistance) plus 5 Ohms (corroded contacts), resulting in a total resistance of 149 Ohms. The current is determined to be 0.8054 A, leading to a power dissipation of 93.4 W for the lightbulb. This calculation confirms the impact of corroded contacts on power efficiency.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Familiarity with power formulas (P = V^2/R, P = RI^2)
  • Basic knowledge of electric circuit analysis
  • Ability to perform calculations involving resistance and power
NEXT STEPS
  • Study the effects of resistance on power dissipation in circuits
  • Learn about the impact of corrosion on electrical connections
  • Explore advanced circuit analysis techniques
  • Investigate methods for improving electrical contact reliability
USEFUL FOR

Students in physics, electrical engineers, and anyone interested in understanding the effects of resistance in electrical circuits.

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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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