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!
 
Thread 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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