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Chapter 18: Problem 109

Consider a \(60.0-\mathrm{W}\) lightbulb and a \(100.0-\mathrm{W}\) lightbulb designed for use in a household lamp socket at \(120 \mathrm{V}\) (a) What are the resistances of these two bulbs? (b) If they are wired together in a series circuit, which bulb shines brighter (dissipates more power)? Explain. (c) If they are connected in parallel in a circuit, which bulb shines brighter? Explain.

Short Answer

Expert verified
Answer: In a series circuit, the 60W (240Ω) lightbulb shines brighter, dissipating approximately 45W of power. In a parallel circuit, the 100W (144Ω) lightbulb shines brighter, dissipating approximately 100W of power.

Step by step solution

01

Calculate the resistance of each bulb

We are given the power and voltage for both bulbs. We can use the formula P = IV to solve for the current (I) in each bulb and then find the resistance using R = V/I. For the 60W bulb: P = 60W, V = 120V Current, I = P/V I = 60/120 = 0.5A Resistance, R = V/I R = 120/0.5 = 240Ω For the 100W bulb: P = 100W, V = 120V Current, I = P/V I = 100/120 ≈ 0.833A Resistance, R = V/I R ≈ 120/0.833 ≈ 144Ω
02

Analyze the series circuit

In a series circuit, the current (I) remains constant through all components. Resistance in series circuit, R(total) = R1 + R2 R(total) = 240Ω (60W bulb) + 144Ω (100W bulb) = 384Ω Since 'current' is the same (I = V/R(total)) in a series circuit, we can now find the power dissipated by each bulb using P = I²R. For the 60W bulb: P = I²R = (120/384)² * 240 ≈ 45W For the 100W bulb: P = I²R = (120/384)² * 144 ≈ 27W Thus, in a series circuit, the 60W bulb (with 240Ω resistance) shines brighter as it dissipates more power (45W) compared to the 100W bulb (27W).
03

Analyze the parallel circuit

In a parallel circuit, the voltage (V) remains constant across all components. Resistance in a parallel circuit: 1/R(total) = 1/R1 + 1/R2 1/R(total) = 1/240 + 1/144 R(total) ≈ 86.15Ω Since 'voltage' is the same (V = 120V) in a parallel circuit, we can now find the power dissipated by each bulb using P = IV. For the 60W bulb: I = V/R = 120V/240Ω = 0.5A P = IV = (0.5A)(120V) = 60W For the 100W bulb: I = V/R = 120V/144Ω ≈ 0.833A P = IV = (0.833A)(120V) ≈ 100W Thus, in a parallel circuit, the 100W bulb (with 144Ω resistance) shines brighter as it dissipates more power (100W) compared to the 60W bulb (60W).

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