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.

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= …

To determine the resistance of the light bulb, we can use the formula for power:

Power (P) = (Voltage (V))^2 / Resistance (R)

We are given that the power delivered to the light bulb is 20.7 W and the voltage across the light bulb and the resistor is 120.0 V.

Substituting the given values into the formula, we can rearrange it to solve for resistance:

Resistance (R) = (Voltage (V))^2 / Power (P)

Resistance (R) = (120.0 V)^2 / 20.7 W

Resistance (R) = 14400 V^2 / 20.7 W

Resistance (R) ≈ 694.92 Ω

Therefore, the resistance of the light bulb is approximately 694.92 Ω.

Since there are two possible answers, we need to consider the larger of the two options.