Find study content
Learning Materials

Discover learning materials by subject, university or textbook.

Explanations

Textbooks

Exams
All Subjects

Anthropology

Archaeology

Architecture

Art and Design

Bengali

Biology

Business Studies

Greek

Chemistry

Chinese

Combined Science

Computer Science

Economics

Engineering

English

English Literature

Environmental Science

French

Geography

German

History

Hospitality and Tourism

Human Geography

Italian

Japanese

Law

Macroeconomics

Marketing

Math

Media Studies

Medicine

Microeconomics

Music

Nursing

Nutrition and Food Science

Physics

Polish

Politics

Psychology

Religious Studies

Sociology

Spanish

Sports Sciences

Translation

All Subjects

Biology

Business Studies

Chemistry

Combined Science

Computer Science

Economics

English

Environmental Science

Geography

History

Math

Physics

Psychology

Sociology

EXAM TYPES

GCSE

IGCSE

AS

A Level

International A Level

University Admissions Tests

GCSE SUBJECTS

GCSE 1

GCSE 2

GCSE 3

GCSE 4

GCSE 5

SOME-TEXT

IGCSE SUBJECTS

IGCSE 1

AS SUBJECTS

AS 1

A Level SUBJECTS

A Level 1

International A Level SUBJECTS

International A Level 1

University Admissions Tests SUBJECTS

University Admissions Tests 1
Features
Features

Discover all of these amazing features with a free account.

Flashcards

Vaia AI

Notes

Study Plans

Study Sets
What’s new?

Flashcards
Study your flashcards with three learning modes.

Study Sets
All of your learning materials stored in one place.

Notes
Create and edit notes or documents.

Study Plans
Organise your studies and prepare for exams.
Resources
Discover

All the hacks around your studies and career - in one place.

Find a job

Find a degree

Student Deals

Magazine

Mobile App
Featured

Magazine
Trusted advice for anyone who wants to ace their studies & career.

Job Board
The largest student job board with the most exciting opportunities.

Vaia Deals
Verified student deals from top brands.

Our App
Discover our mobile app to take your studies anywhere.

Go to App

Learning Materials

Features

Discover

University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition · 1596 pages

ISBN: 9780321973610

Chapter 4: Q2E (page 873)

A machine part has a resistor X protruding from an opening in the side. This resistor is connected to three other resistors, as shown in Fig. E26.2. An ohmmeter connected across a and b reads 2.00 Ω. What is the resistance of X?

Short Answer

Expert verified

The resistance of X is 7.5 $Ω$ .

Step by step solution

01

Definition of resistance

The resistance that is measured by the ohmmeter represents the equivalent resistance of the total resistors because these four resistors are connected in parallel.

Therefore, use an equation that represents the connection of the resistors in parallel as:

$\frac{1}{R_{e q}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \frac{1}{R_{x}} \frac{1}{R_{x}} = \frac{1}{R_{e q}} - \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$

02

Resistance of X

Substitute to get R_x as:

$\frac{1}{R_{x}} = \frac{1}{R_{e q}} - \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} = \frac{1}{2 Ω} - \frac{1}{15 Ω} - \frac{1}{5 Ω} - \frac{1}{10 Ω} = \frac{2}{15} R_{x} = \frac{15}{2} Ω$

The resistance of X is 7.5 $Ω$ .

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

In the circuit in Fig. E25.47, find (a) the rate of conversion of internal (chemical) energy to electrical energy within the battery; (b) the rate of dissipation of electrical energy in the battery; (c) the rate of dissipation of electrical energy in the external resistor.

Section 27.2 describes a procedure for finding the direction of the magnetic force using your right hand. If you use the same procedure, but with your left hand, will you get the correct direction for the force? Explain.

The text states that good thermal conductors are also good electrical conductors. If so, why don’t the cords used to connect toasters, irons, and similar heat-producing appliances get hot by conduction of heat from the heating element?

A particle with charge $- 5.60 nC$ is moving in a uniform magnetic fieldrole="math" localid="1655717557369" $B = - (1.25 T) \hat{k}$

The magnetic force on the particle is measured to berole="math" localid="1655717706597" $\overset{⇀}{F} = - (3.40 \times 10^{- 7} N) \hat{i} - (7.40 \times 10^{- 7} N) \hat{j}$ (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there
components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product $\overset{⇀}{v} ֏ \overset{⇀}{F}$ . What is the angle between velocity and force?

A particle with mass $1.81 \times 10^{- 3} k g$ and a charge of has $1.22 \times 10^{- 8} C$ , at a given instant, a velocity $\overset{⇀}{V} = (3.00 \times 10^{4} m / s)$ .What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field $B = (1.63 T) i + (0.980 T) \hat{j}$ ?

See all solutions

Recommended explanations on Physics Textbooks

Famous Physicists

Read Explanation

Waves Physics

Read Explanation

Fields in Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Force

Read Explanation

Fluids

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.

Sign-up for free