Asked by Drue
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.
Answers
Answered by
Elena
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= …
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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