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Chapter 27: Problem 104

A three-way \(120 \mathrm{~V}\) lamp bulb that contains two filaments is rated for \(100-200-300 \mathrm{~W}\). One filament burns out. Afterward, the bulb operates at the same intensity (dissipates energy at the same rate) on its lowest as on its highest switch positions but does not operate at all on the middle position. (a) How are the two filaments wired to the three switch positions? What are the (b) smaller and (c) larger values of the filament resistances?

Short Answer

Expert verified
The filaments are wired in a way that 100 W corresponds to filament A, and 200 W corresponds to filament B. The smaller resistance is 72 惟, and the larger is 144 惟.

Step by step solution

01

Understanding the Circuit

The lamp bulb has two filaments and three switch positions with power ratings of 100 W, 200 W, and 300 W. These filaments are wired to allow different combinations for these power settings. When one filament burns out, the bulb still operates at the same power in the highest and lowest positions, but not in the middle.
02

Analyzing Filament Operation

Based on the problem, initially, let's assume that the two filaments are referred to as A and B. When both filaments work, the configurations are as follows: 100 W for filament A, 200 W for filament B, and 300 W for both A and B together. After one filament burns out, the lamp operates at the same power for the lowest and highest positions.
03

Identifying the Burned-Out Filament

Since the bulb does not operate at all at the middle setting, the middle power (200 W, which uses only one filament) corresponds to the burned-out filament. This means that filament B (initially 200 W) must be the one that burned out.
04

Finding Resistance Values

To find resistance, use the power formula: \[ P = \frac{V^2}{R} \]Where \(P\) is the power, \(V\) is the voltage, and \(R\) is the resistance. - For the 100 W setting (active filament A): \[ R_A = \frac{(120)^2}{100} = 144 \Omega \]- For the 200 W setting: \[ R_B = \frac{(120)^2}{200} = 72 \Omega \] (but this filament is burnt out)- When both filaments are functional, and using the highest setting (300 W), both are supposed to be combined: \[ R_{total} = \frac{(120)^2}{300} = 48 \Omega \]Given that as a parallel combination: \[ \frac{1}{R_{total}} = \frac{1}{R_A} + \frac{1}{R_B} \]

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Key Concepts

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

Electrical Circuits
Electrical circuits are pathways that allow electricity to flow. They are composed of various components, such as resistors and conductors, connected by wires. In the case of our lamp bulb, the electrical circuit contains two filaments. When both filaments function properly, they can be arranged to provide different power settings, such as 100 W, 200 W, and 300 W. Each filament acts as a resistor. The power settings correspond to different configurations of resistors in the circuit. When a filament burns out, it alters the circuit's pathways, changing its functionality and power output.
Ohm's Law
Ohm's Law is a fundamental principle in physics relating voltage (V), current (I), and resistance (R) in electrical circuits. It is given by the formula: \( V = I imes R \).This law helps us determine how different elements in a circuit interact with one another. For a lamp with filaments, we can also calculate the resistance of a filament using power (P) and voltage (V), rearranging the formula for power: \( R = \frac{V^2}{P} \).Understanding Ohm's Law is crucial when analyzing problems involving burnt-out filaments or calculating how much resistance a functioning filament offers.
Burned-Out Filament Analysis
When analyzing a lamp with a burned-out filament, it's essential to understand how the internal circuit of the lamp is affected. A burned-out filament means that electricity can no longer pass through it, effectively removing one of the resistors from the circuit. In this scenario, the lamp operates normally at low and high settings but not at the middle one. This suggests that the filament corresponding to the middle power setting (200 W) is the one that is burned out. This analysis uses the fact that the remaining functional configurations are dependent on the filament that is still intact. Such logical steps help determine the dynamics of the circuit post-failure.
Power Ratings in Lamps
Power ratings in lamps indicate the amount of electrical energy the lamp can convert into light and heat per second. These ratings are critical for understanding a lamp's performance and energy consumption.In a multi-setting lamp with different power ratings, each setting corresponds to a different filament or combination of filaments operating at specific resistances. When a filament burns out, these power ratings help us determine which filament is no longer functional based on which settings remain operational.Using the formula \( P = \frac{V^2}{R} \), where \( V \) is the voltage and \( R \) is resistance, students can calculate and understand better the resistance values that relate to each power setting, thus solving complex circuit analysis problems.

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Most popular questions from this chapter

A simple ohmmeter is made by connecting a \(1.50 \mathrm{~V}\) flashlight battery in series with a resistance \(R\) and an ammeter that reads from 0 to \(1.00 \mathrm{~m} A\), as shown in Fig. 27-59. Resistance \(R\) is adjusted so that when the clip leads are shorted together, the meter deflects to its full- scale value of \(1.00 \mathrm{~m} \mathrm{~A}\). What external resistance across the leads results in a deflection of (a) \(10.0 \%,(\mathrm{~b}) 50.0 \%\), and (c) \(90.0 \%\) of full scale? (d) If the ammeter has a resistance of \(20.0 \Omega\) and the internal resistance of the battery is negligible, what is the value of \(R\) ?

A capacitor with an initial potential difference of \(100 \mathrm{~V}\) is discharged through a resistor when a switch between them is closed at \(t=0 .\) At \(t=10.0 \mathrm{~s}\), the potential difference across the capacitor is \(1.00 \mathrm{~V}\). (a) What is the time constant of the circuit? (b) What is the potential difference across the capacitor at \(t=17.0 \mathrm{~s}\) ?

An automobile gasoline gauge is shown schematically in Fig. \(27-74\). The indicator (on the dashboard) has a resistance of \(10 \Omega\). The tank unit is a float connected to a variable resistor whose resistance varies linearly with the volume of gasoline. The resistance is \(140 \Omega\) when the tank is empty and \(20 \Omega\) when the tank is full. Find the current in the circuit when the tank is (a) empty, (b) half-full, and (c) full. Treat the battery as ideal.

A \(3.00 \mathrm{M} \Omega\) resistor and a \(1.00 \mu \mathrm{F}\) capacitor are connected in series with an ideal battery of emf \(\mathscr{E}=4.00 \mathrm{~V}\). At \(1.00 \mathrm{~s}\) after the connection is made, what is the rate at which (a) the charge of the capacitor is increasing, (b) energy is being stored in the capacitor, (c) thermal energy is appearing in the resistor, and (d) energy is being delivered by the battery?

What multiple of the time constant \(\tau\) gives the time taken by an initially uncharged capacitor in an \(R C\) serics circuit to be charged to \(99.0 \%\) of its final charge?
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