/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 104 A three-way \(120 \mathrm{~V}\) ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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 bulb filaments are wired in parallel. Smaller resistance is 72 \( \Omega \), larger is 144 \( \Omega \).

Step by step solution

01

Identify the power and voltage relationship

For a given electric circuit, the power dissipated is related to the resistance and voltage by the formula: \[ P = \frac{V^2}{R} \]Where:- \( P \) is the power (in watts)- \( V \) is the voltage (in volts)- \( R \) is the resistance (in ohms)Given the power ratings of the bulb (100 W, 200 W, 300 W) and the voltage of 120 V, this formula will be used to determine the individual resistances of the filaments.
02

Determine resistance of filaments from power ratings

Using the formula \( P = \frac{V^2}{R} \), we calculate the resistance for each power setting. For 100 W: \[ R_1 = \frac{V^2}{100} = \frac{120^2}{100} = 144 \, \Omega \]For 200 W:\[ R_2 = \frac{V^2}{200} = \frac{120^2}{200} = 72 \Omega \]For 300 W:\[ R_{combined} = \frac{V^2}{300} = \frac{120^2}{300} = 48 \Omega \]
03

Analyze switch settings based on problem conditions

Since on the highest and lowest switch settings the bulb operates with one filament burnt out, but not on the middle, examine the switch settings. - The highest setting (300 W setting) requires both filaments working in parallel.- The lowest (100 W) setting works with the remaining filament alone, namely \( R_1 = 144 \Omega \).- The middle setting (200 W) would normally use the single filament that's burned out (\( R_2 = 72 \Omega \)). This analysis reveals the switch settings dependently use either or both filaments for different power outputs.
04

Determine which filament burnt out

Since the middle setting does not work, and the functioning on the lowest and highest settings suggests identical resistances at those positions corresponding to \( R_1 = 144 \Omega \), the burnt-out filament is the one rated at 72 \( \Omega \) (only used for the middle 200 W position distinctly on its own).Therefore, the two filaments are typically wired such that they are parallel for the highest wattage and individually selected for the lower wattage position, with the burnt-out filament being the lower resistance 72 \( \Omega \) filament.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Parallel Circuits
In electrical circuit analysis, understanding parallel circuits is essential. Parallel circuits are configurations where components share the same voltage across them. This is unlike series circuits where components share the same current. Parallel circuits have specific characteristics:
  • If one component fails, the rest may continue to operate because each path operates independently.
  • The total current in the circuit is the sum of the currents through each path.
  • The overall resistance of the circuit decreases as more components are added.
With parallel circuits, the voltage remains unchanged across each component. This property was evident in the lamp bulb problem where the filaments are wired in a particular way to manage different power settings. When filaments are connected in parallel, they can handle higher power levels, like the highest 300 W setting in the exercise. By employing Ohm's Law and understanding parallel circuits, one can analyze and solve problems involving complex bulb configurations.
Resistor Calculations
Calculating resistance is a fundamental aspect of understanding electrical circuits, particularly when analyzing components like resistors in light bulbs. The resistance of a filament in a bulb can often be found using the formula:\[ R = \frac{V^2}{P} \]
  • \( R \) is resistance in ohms (\( \Omega \)),
  • \( V \) is voltage in volts,
  • \( P \) is power in watts.
In our exercise, we calculated:
  • For 100 W, \( R_1 = 144 \, \Omega \)
  • For 200 W, \( R_2 = 72 \, \Omega \)
  • For combined 300 W, \( R_{combined} = 48 \, \Omega \)
These calculations help to determine which filament burned out, illustrating the practical importance of accurate resistor calculations in identifying circuit failures. Each resistance value gives insight into the electrical characteristics of the circuit at various settings.
Ohm's Law
Ohm's Law is a core principle in electrical engineering that helps us understand the relationship between voltage, current, and resistance in an electric circuit. The formula associated with Ohm's Law is:\[ V = IR \]Where:
  • \( V \) is the voltage in volts,
  • \( I \) is the current in amperes,
  • \( R \) is the resistance in ohms (\( \Omega \)).
Ohm's Law allows us to predict how electricity will flow through a circuit under different conditions. For example, knowing the resistance and voltage, you can find the current using:\[ I = \frac{V}{R} \]In the exercise, this law was implicit in understanding how the bulb's filaments behaved at different power levels. By understanding the resistance (calculated from power ratings) and voltage (120 V supply), we can determine how the current flows through the bulb, especially in parallel circuits. This is critical for deducing which filament is functional or burnt out.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(15.0 \mathrm{k} \Omega\) resistor and a capacitor are connected in series, and then a \(12.0 \mathrm{~V}\) potential difference is suddenly applied across them. The potential difference across the capacitor rises to \(5.00 \mathrm{~V}\) in \(1.30 \mu \mathrm{s}\). (a) Calculate the time constant of the circuit. (b) Find the capacitance of the capacitor.

A certain car battery with a \(12.0 \mathrm{~V} \mathrm{emf}\) has an initial charge of \(120 \mathrm{~A} \cdot \mathrm{h}\). Assuming that the potential across the terminals stays constant until the battery is completely discharged, for how many hours can it deliver energy at the rate of \(100 \mathrm{~W} ?\)

When resistors 1 and 2 are connected in series, the equivalent resistance is \(16.0 \Omega .\) When they are connected in parallel, the equivalent resistance is \(3.0 \Omega .\) What are (a) the smaller resistance and (b) the larger resistance of these two resistors?

A \(5.0 \mathrm{~A}\) current is set up in a circuit for \(6.0 \mathrm{~min}\) by a rechargeable battery with a \(6.0 \mathrm{~V}\) emf. By how much is the chemical energy of the battery reduced?

A \(3.00 \mathrm{M} \Omega\) resistor and \(\mathrm{a} 1.00 \mu \mathrm{F}\) 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?
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Turning Points in Physics

Read Explanation

Mechanics and Materials

Read Explanation

Solid State Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Medical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.