Calculating Resistance in a Light Bulb

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To calculate the resistance of a 90-watt light bulb connected to a 120V rms voltage, the formula used is R = V²/P. Substituting the values, R equals 160 ohms. An initial calculation incorrectly resulted in 0.00625 ohms due to algebraic errors. The correct resistance value is confirmed as 160 ohms. Accurate calculations are essential for understanding electrical properties.
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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!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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