Posted by Drue on Monday, April 1, 2013 at 7:35pm.
MultipleConcept Example 9 reviews the concepts that are important to this problem. A light bulb is wired in series with a 116Ω resistor, and they are connected across a 120.0V source. The power delivered to the light bulb is 20.7 W. What is the resistance of the light bulb? There are two possible answers. Give the larger of the two.

Physics  Elena, Tuesday, April 2, 2013 at 5:23am
R₂= 116 Ω,
ℰ=120 V,
P₁=20.7 W,
The resistance of light buble R₁=?
R=R₁+R₂,
I=V/R,
R₁=P₁/I²=P₁R²/V²=P₁(R₁+R₂)²/V².
R₁²+(2R₂  V²/P₁)R₁+R₂²=0
Solve for R₁
R₁ =[ (2R₂  V²/P₁) ±sqrt{(2R₂  V²/P₁)²4 R₂²}}/2= …
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