Posted by **Drue** on Monday, April 1, 2013 at 7:35pm.

Multiple-Concept Example 9 reviews the concepts that are important to this problem. A light bulb is wired in series with a 116-Ω resistor, and they are connected across a 120.0-V 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)²-4 R₂²}}/2= …

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