Calculating Resistance in a Light Bulb

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SUMMARY

The resistance of a "90-watt" light bulb connected to an rms voltage of 120V is calculated using the formula R = V²/P. Substituting the values, R = 120²/90, which results in a resistance of 160Ω. An initial incorrect calculation of 0.00625Ω was identified as an algebra error. The correct approach confirms that the light bulb's resistance is 160Ω, providing a clear understanding of the relationship between power, voltage, and resistance.

PREREQUISITES
  • Understanding of Ohm's Law
  • Familiarity with electrical power calculations
  • Knowledge of root mean square (rms) voltage
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Study the derivation of Ohm's Law and its applications
  • Explore electrical power formulas in AC circuits
  • Learn about the significance of rms voltage in electrical engineering
  • Practice solving for resistance using various power and voltage values
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Students in electrical engineering, physics enthusiasts, and anyone interested in understanding the principles of electrical resistance and power calculations.

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Homework Statement


A "90-watt" light bulb uses an average power of 90W when connected to an rms voltage of 120V .

What is the resistance of the lightbulb?

Homework Equations



Pav=(V2rms/R)

The Attempt at a Solution



90W = 120V2/R
90W/120V2 = R
R = .00625\Omega
 
Last edited:
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You have an algebra problem with your solving for R.
 
Hrmm...

P = V2/R
PR = V2
R = V2/P
R = 160\Omega

That works better huh? Thank you!
 

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