Tuesday
March 28, 2017

Post a New Question

Posted by on .

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 - ,

    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= …

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question