Physics
posted by Drue on .
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.

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= …