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A Three-Way Light Bulb. A three- way light bulb has three brightness settings (low, medium, and high) but only two filaments. (a) A particular three-way light bulb connected across a \(120-\mathrm{V}\) line can dissipate \(60 \mathrm{W}, 120 \mathrm{W},\) or 180 \(\mathrm{W}\) . Describe how the two filaments are arranged in the bulb, and calculate the resistance of each filament. (b) Suppose the filament with the higher resistance bums out. How much power will the bulb dissipate on each of the three brightness settings? What will be the brightness (low, medium, or high) on each setting?

Short Answer

Expert verified
(a) Filaments are in series for high; resistances: \( R_1 = 240 \) ohms, \( R_2 = 120 \) ohms. (b) Burned out: low=60W, medium=0W, high=60W.

Step by step solution

01

Analyze Power Distribution

The bulb can dissipate power in three modes: - Low (60 W)- Medium (120 W)- High (180 W). This suggests that the filaments are used in combinations to produce the distinct power outputs. Let's denote filament resistances as \( R_1 \) and \( R_2 \). In low setting, only one filament (say with resistance \( R_1 \)) is used to dissipate 60 W. In medium setting, only \( R_2 \) is used to dissipate 120 W. In high setting, both \( R_1 \) and \( R_2 \) are used together to dissipate 180 W.
02

Apply Electrical Power Formula for Each Setting

Using the power formula \( P = \frac{V^2}{R} \). For low setting, we have: \[ 60 = \frac{120^2}{R_1} \] Solve for \( R_1 \): \[ R_1 = \frac{120^2}{60} = 240 \text{ ohms} \]For medium setting: \[ 120 = \frac{120^2}{R_2} \] Solve for \( R_2 \): \[ R_2 = \frac{120^2}{120} = 120 \text{ ohms} \]
03

Determine Resistance in Series for High Setting

For high setting, both filaments are used together in series:\[ 180 = \frac{120^2}{R} \] where \( R = R_1 + R_2 \). Solve for \( R \): \[ R = \frac{120^2}{180} = 80 \text{ ohms} \]Verify with sum: \[ 240 + 120 = 360 \text{ ohms} \text{ (this is incorrect for parallel)} \]Thus, we confirm they are in parallel for high mode:\[ \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R} \] Verifying:\[ \frac{1}{240} + \frac{1}{120} = \frac{1}{80} \] Both calculations are consistent.
04

Determine Power if Higher Resistance Burns Out

With \( R_2 \) (120 ohms) burned out:- Low setting (\( R_1 \) only with 240 ohms): \[ P = \frac{120^2}{240} = 60 \text{ W} \] So it remains as low brightness. - Medium setting (\( R_2 \) burned out): No power, so no brightness (perhaps defaults to low). - High setting (involves both filaments): \( R_1 \) only remaining, dissipates 60 W, thus low brightness.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Filament Resistance
In an electric circuit, resistance plays a crucial role in controlling the flow of current. Filament resistance specifically refers to the resistance offered by the thin wire inside a light bulb, often made of tungsten. In our three-way light bulb problem, we have two distinct filaments, each with its own resistance: \( R_1 \) and \( R_2 \). These resistances determine how much power the bulb can dissipate under various settings.

For this particular light bulb, when using one filament alone, we calculate the resistance using the formula \( P = \frac{V^2}{R} \). Here, "\( P \)" is the power output you desire, and "\( V \)" is the voltage applied, which in this case, is 120 V.

If set for low power, one filament, \( R_1 \), dissipates 60 W, calculated as \( R_1 = \frac{120^2}{60} = 240 \text{ ohms} \). Meanwhile, in medium settings, \( R_2 \) dissipates 120 W, calculated as \( R_2 = \frac{120^2}{120} = 120 \text{ ohms} \). Understanding these calculations shows us the direct relationship between resistance, power, and the functionality of electric circuits.
Power Dissipation
Power dissipation in an electric circuit is a critical concept that refers to the conversion of electric energy into another form, typically heat. This occurs as current flows through a circuit's resistive components, like the filaments in a light bulb. In our exercise with a three-way light bulb, each setting corresponds to a different power dissipation level: 60 W, 120 W, and 180 W.

The formula used to determine power dissipation is \( P = \frac{V^2}{R} \). For a bulb connected to a 120 V line, each mode utilizes different combinations of filament resistances, resulting in varying power outputs. To understand how power dissipates:
  • **Low brightness (60 W):** Power dissipates through one filament \( (R_1) \).
  • **Medium brightness (120 W):** Power dissipates through another filament \( (R_2) \).
  • **High brightness (180 W):** Both filaments work together in parallel to dissipate the combined power.
Recognizing how each setting impacts power dissipation helps clarify why brightness varies in three-way bulbs.
Parallel and Series Circuits
Electric circuits can configure components either in series or parallel arrangements, significantly affecting the current's flow and the circuit's overall resistance. In a three-way bulb, we encounter both configurations to achieve varying brightness levels.

When components are in **series**, the same current flows through each, and total resistance is the sum of individual resistances. However, this does not apply in our high-setting three-way bulb configuration. In our problem, 'series' referred to calculations that revealed inconsistency, which implies another arrangement.

In the **parallel** configuration, used in the high-brightness setting, the circuits divide the current across two or more paths, leading to different characteristics:
  • Total resistance decreases (less than either component's resistance).
  • Total current is distributed, allowing more power dissipation.
For our three-way bulb, combining \( R_1 \) and \( R_2 \) in parallel effectively allows for maximum light output (180 W), ensuring a higher brightness setting.
Three-Way Light Bulb
A three-way light bulb is a type of light bulb that offers three distinct light levels using two separate filaments. Its unique design is ideal for adjustable lighting conditions in homes and workplaces. The available settings facilitate low, medium, and high light outputs through different filament combinations.

When operating:
  • On **low brightness**, current runs solely through the higher-resistance filament, producing minimal light output.
  • For **medium brightness**, another filament with lower resistance provides more brightness while consuming more power.
  • At **high brightness**, both filaments engage in parallel, offering the maximum light output possible.
This bulb's ability to alter illumination levels beyond a simple on-off switch showcases the sophisticated interplay of electrical components arranged to suit varying lighting needs.

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Most popular questions from this chapter

Three resistors having resistances of \(1.60 \Omega, 2.40 \Omega,\) and 4.80\(\Omega\) are con0nected in parallel to a \(28.0-\mathrm{V}\) battery that has negligible internal resistance. Find (a) the equivalent resistance of the combination; (b) the current in each

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(a) Prove that when two resistors are connected in parallel, the equivalent resistance of the combination is always smaller than that of the smaller resistor. (b) Generalize your result from part (a) for \(N\) resistors.

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