# Resistance in a light bulb

## 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?

Pav=(V2rms/R)

## The Attempt at a Solution

90W = 120V2/R
90W/120V2 = R
R = .00625$$\Omega$$

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gneill
Mentor
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!