chapter 23chapter 23chapter organizer · which bulb burns brighter? one is a 60-watt bulb and the...

530A

7 class sessions

23.1 Simple Circuits, pp. 532–541(3 class sessions)

Pocket Lab: Series Resistance,p. 534

Using a Calculator: TheInverse Key, p. 535

Pocket Lab: ParallelResistance, p. 539

Help Wanted: Electrician, p. 543

Design Your Own Physics Lab:Circuits, p. 545

Pocket Lab: AmmeterResistance, p. 547

How It Works: Electric Switch,p. 549

Transparency 39: SeriesCircuit/Parallel Circuit

Transparency 40: CombinedSeries Parallel Circuits

Lesson Plans: p. 50ELL

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Transparency 41: Fuses

Lesson Plans: p. 51Problems and Solutions

Manual: Chapter 23 L1

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Chapter Assessment:pp. 107–110

Supplemental Problems:Chapter 23

Spanish Resources: Chapter 23ELL

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Study Guide: pp. 133–136 Transparencies 39, 40 Masters

and Worksheets: pp. 83–85,87–88

Physics Lab and Pocket Lab Worksheets: pp.125–126

Enrichment: pp. 45–46 Laboratory Manual: pp. 167–176 L1

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Study Guide: pp. 137–138 Transparency 41 Master and

Worksheet: pp. 89–91

Physics Lab and Pocket Lab Worksheets: pp. 123–124, 127

Reteaching: p. 29 Critical Thinking: p. 30 Tech Prep Applications: Security

System, pp. 35–38 Physics Skills: Skill 19 Using

Ammeters and Voltmeters, pp. 39–40 L1

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ELLL1

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TestCheck SoftwareMindJogger Videoquizzes

ELL

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23.2 Applications of Circuits,pp. 542–549(3 class sessions)

Chapter Review, pp. 550–553(1 class session)

SummaryKey TermsReviewing ConceptsApplying ConceptsProblemsCritical Thinking ProblemsGoing Further

Chapter 23 Chapter OrganizerChapter 23 Chapter Organizer

Contents Text Features Teaching Aids Student Masters

The following Glencoe resources provide opportunities forintegrating science and technology.Student Edition: Help Wanted, p. 543; How It Works, p. 549Teacher Wraparound Edition: Tech Prep, pp. 534, 543;Activity, p. 541; Guest Speaker, p. 542; Applying Physics,pp. 543, 546; Physics Journal, p. 544Teacher Classroom Resources: Tech Prep Applications,pp. 35–38; Physics Skills, pp. 39–40

TECH PREP

KEY TO TEACHING STRATEGIESThe following designations will help you decide which activities are appropriate for your students.

Level 1 activities should be within the ability range of all students including those with learning difficulties.Level 2 activities should be within the ability range of the average to above-average student.Level 3 activities are designed for the ability range of above-average students.ELL activities should be within the ability range of English Language Learners.Cooperative Learning activities are designed for small group work.These strategies represent student products that can be placed into a best-work portfolio.These strategies are useful in a block scheduling format.

P

COOP LEARN

ELL

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Texas Lesson Plans

Reviewing Physics:Mastering the TEKS

530B

Series and Parallel CircuitsSeries and Parallel Circuits

23.1 Simple Circuits1. Describe both a series connection and a parallel

connection and state the important characteristics of each.

2. Calculate current, voltage drops, and equivalentresistance for devices connected in series and in parallel.

3. Describe a voltage divider and solve problemsinvolving one.

23.2 Applications of Circuits4. Explain how fuses, circuit breakers, and ground-

fault interrupters protect household wiring.5. Analyze combined series-parallel circuits and

calculate the equivalent resistance of such circuits.6. State the important characteristics of voltmeters and

ammeters, and explain how each is used in circuits.

Page 545lamp and socket (3)power supply, DC, variable,

0–15 Vwire leads with alligator clips (6)

Page 534, 539ammeterpower supply, DC, variable,

0–15 Vresistor, same value (4)wire leads with alligator

clipsPage 547

ammeterpower supply, DC, variable,

0–15 Vresistor, 10-� 1%voltmeterwire leads with alligator

clips (6)

23–1, Page 538120-V line cord with

pluglightbulb, 100-wattlightbulb, 25-wattlightbulb socket (2)wire

23–2, Page 546battery, 6-Vknife switchmilliammeter, 0–1 mAresistance

substitution boxresistor, 1000-� (2)resistor, unknown

valuewire leads with

alligator clips (9)

DemonstrationsPocket LabPhysics Lab

Science and Technology Videodisc Series (STVS)Physics: Synchrotron; Making Integrated CircuitsChemistry: Battery SciencePhysics for the Computer Age CD-ROM

ELECTRICITY: Electricity Introduction and Series CircuitsThe Mechanical Universe VideotapeQuad 5: Electricity Simple DC CircuitsMindJogger Videoquizzes Chapter 23

The following multimedia resources are available from Glencoe.

1(A), 2(A), 2(C), 2(D), 3(B), 6(E) UCP.1, UCP.2, UCP.3, UCP.5, A.1,A.2, E.2, G.3

Objectives State/Local StandardsNational ScienceContent Standards

1(A), 1(B), 2(A), 2(B), 2(C), 2(D),2(F), 3(B), 3(D), 6(E)

UCP.1, UCP.2, UCP.3, UCP.5, A.1,A.2, E.2

Activity and Demonstration Materials

WhichBulbBurnsBrighter?One is a 60-watt

bulb and the other a

100-watt bulb, and they

are connected in an

electric circuit.

➥ Look at the text on page 541 for the answer.

Series and Parallel Circuits

Chapter Overview

Students calculate equivalent resis-tance in series and parallel circuitsand determine current and voltagedrops across resistors.

In the second section, the designand operation of voltmeters andammeters is described, as well as safety devices such as fuses,circuit breakers, and ground-faultinterrupters.

Key Terms

ammetercircuit breakercombination series-parallel

circuitequivalent resistancefuseground-fault interrupterparallel circuitseries circuitshort circuitvoltage dividervoltmeter

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Learning Styles

Look for the following logo for strategies that emphasize different learning modalities.

Kinesthetic Pocket Labs, pp. 534, 539, 547; Reinforcement, p. 538;Design Your Own Physics Lab, p. 545; Activity, p. 548

Visual-Spatial Activity, p. 532; Tech Prep, p. 534; Assessment, p. 543;Applying Physics, p. 546; Visual Learning, p. 547; How It Works, p. 549

Interpersonal Meeting Individual Needs, p. 535; Reinforcement, p. 546

Intrapersonal Activity, p. 541; Tech Prep, p. 543

Linguistic Physics Journal, pp. 533, 537, 544

Logical- Uncovering Misconceptions, p. 538; Connections To Mathematical Biology, p. 540

LS

23CHAPTER

PHYSICSBe sure to check the GlencoeScience Web site for links to chapter material: science.glencoe.com

Texas TEKS

Pages 530–531: 6(E)

http://science.glencoe.com

Your desk needs to be well lit. You have a 100-watt lightbulband a 60-watt lightbulb available for your desk lamp.Which of the two lightbulbs would you choose? You know

a watt is a unit of power, so you might expect that a 100-wattlightbulb would always produce more light than a 60-watt bulb.But would it?

You’re shopping for a string of holiday lights for a party. Youfind a string of pumpkin lights with a tag that reads “If one goesout, the rest stay lit.” Why is this important? How are the lightsconnected to each other and the power source to make sure thatthe rest stay lit?

Does the type of electric circuit that connects the bulb make adifference? Suppose the lightbulbs are in a circuit unlike the par-allel circuit that normally connects lightbulbs in your home. Forexample, in the photo at the left, the bulbs are connected inseries. Does that make a difference? Can you figure out why? Toanswer these questions, you need to know how a series circuit dif-fers from a parallel circuit.

In Chapter 22, you studied circuits that had one source of elec-tric energy, for example, a battery and one device such as a motoror a lamp that converted the electric energy to another form. Butcircuits can be much more versatile than that. In this chapter,you’ll explore several ways in which devices may be connected inelectric circuits. By the end of this chapter, you’ll have no troubledeciding which of the two bulbs at the left is brighter and why therest of the bulbs stay lit.

Series and ParallelCircuits

WHAT YOU’LL LEARN• You will distinguish between

parallel and series circuitsand series-parallel combina-tions and solve problemsdealing with them.

• You will explain the functionof fuses, circuit breakers,and ground fault interrupters,and describe ammeters and voltmeters.

WHY IT’S IMPORTANT• Electrical circuits are the

basis of every electricaldevice, from electric lights tomicrowave ovens to comput-ers. Understanding circuitshelps you to use them, andto use them safely.

23CHAPTER

531

PHYSICSTo find out more about electrical circuits, visit the Glencoe ScienceWeb site at science.glencoe.com

Introducing the Chapter

USING THE PHOTO

Although students use electricityevery day, they often have miscon-ceptions about how it works.Student comments about the chap-ter opener photo should revealsome of these misconceptions. Thephoto of a 60-W and a 100-W light-bulb connected in a series circuitinvites the comparison of series andparallel circuits and the use of therelationship I � V/R, which wasintroduced in Chapter 22.

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CAUTION: Wear goggles.This demonstration introducesseries and parallel circuits. Selecta battery and three identicallamps and sockets. Wire twolamps in series to the battery.Have students observe theirbrightness. Place the third lampin parallel with the other two, orsimply connect the third lampto the battery. Ask students tocompare the brightness of thelamps. The two lamps in series areequally bright, but not as bright asthe single lamp.

DEMOQUICK

Assessment OptionsPORTFOLIO ASSESSMENT

Tech Prep, TWE, pp. 534, 543

PERFORMANCE ASSESSMENT

Physics Lab, SE, p. 545Pocket Labs, SE, pp. 534, 539, 547Performance Assessment, TWE,

pp. 540, 543

KNOWLEDGE ASSESSMENT

Section Review, SE, pp. 541, 548Chapter Review, SE, pp. 550–553

Practice Problems, SE, pp. 534, 537, 538,540, 547

Demonstration, TWE, p. 539

SKILL ASSESSMENT

Demonstration, TWE, p. 547Physics Lab, TWE, p. 545

STUDENT EDITION

Physics Lab: p. 545Pocket Labs: pp. 534, 539, 547

TEACHER EDITION

Demonstrations: pp. 538, 546Quick Demos: pp. 531, 536,

544Activities: pp. 532, 548

LABORATORY MANUAL

Labs 23.1, 23.2


If you have ever had the chance to explore rivers inthe mountains, such as the river shown below in

Figure 23–1, the following description will be familiar. From their sources high in the mountains, rivers make their wayto the plains below. No matter what path they take, the change in eleva-tion, from the mountaintop to the plain, is the same. Some rivers flow in asingle stream, tumbling through a series of rapids and over waterfalls.Other rivers split into two or more streams as they flow over a waterfall orthrough the rapids. Part of the river follows one path; other parts find dif-ferent routes. But no matter how many paths the water takes, the totalamount of water flowing down the mountain is the same, and the verticaldrop from mountaintop to plain is the same distance.

Mountain rivers can serve as a model for electric circuits. The distancethe water drops is similar to potential difference. The amount of waterflowing through the river each second is similar to current. Narrow rapidsare like a large resistance. But what is similar to a battery or generator?Just as in an electrical circuit, an energy source is needed to raise waterto the top of the mountain. As you learned in Earth science, the source ofenergy is the sun. Solar energy evaporates water from lakes and seas andforms clouds that release rain or snow to fall on the tops of mountains.Think about the mountain river model as you read about the current inelectrical circuits.

OBJ ECTIVES

• Describe both a seriesconnection and a parallelconnection and state theimportant characteristics of each.

• Calculate current, voltagedrops, and equivalent resis-tance for devices connectedin series and in parallel.

• Describe a voltage dividerand solve problems involving one.

23.1 Simple Circuits

532 Series and Parallel Circuits

FIGURE 23–1 No matter whatpath the water of a river takesdown a mountain, the amount ofwater and the drop in elevationare the same.

PREPAREContent Refresher

Recall that the potential differenceis the energy supplied per coulombof charge and that current is theamount of charge per second flowingthrough a wire.

Bridging

This chapter extends the relationshipI � V/R to more complex circuits.This relationship applies to elementsin a series just as it does to elementsin parallel, that is, devices having acommon potential difference.

1 FOCUSActivity

Visual-Spatial Have studentsset up the circuit in Figure 23–2using two identical lightbulbs. Theyshould make predictions regardingthe brightness of each lamp andwhether each will light, then testthe circuit. The lights should beequally bright because they have thesame current and resistance. ELLL1

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532

SECTION 23.1

Program Resources

Study Guide, pp. 133–136 Transparencies 39, 40 and Masters

Laboratory Manual, pp. 167–176 Enrichment, pp. 45–46 Physics Lab and Pocket Lab Worksheets,

pp. 125–126 Physics Skills, pp. 39–40 L1

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Texas TEKS

Pages 532–533: 3(B), 6(E)

Series CircuitsPat, Chris, and Ali were connecting two identical lamps to a battery

as illustrated in Figure 23–2. Before making the final connection to thebattery, their teacher asked them to predict the brightness of the twolamps. They knew that the brightness of a lamp depends on the currentflowing through it. Pat said that only the lamp closer to the + terminalof the battery would light because all the current would be convertedinto light. Chris said that the second lamp would light, but it would bedimmer than the other one because some electrical energy would bechanged into thermal and light energy. Consequently, there would beless electrical energy left for the second lamp. Ali said that both lampswould be equally bright because current is a flow of charge, and becausethe charge leaving the first lamp had nowhere else to go in the circuitexcept through the second lamp, the current would be the same in thetwo lamps. Who do you think is right?

If you consider the mountain river model for this circuit, you’ll seethat Ali is correct. Charge has only one path to follow. Recall from Chap-ter 20 that charge cannot be created or destroyed, so the same amountof charge must leave a circuit as enters the circuit. This means that thecurrent is the same everywhere in the circuit. If you connect threeammeters into a circuit as shown in Figure 23–3, they all have the samevalue. A circuit such as this, in which all current travels through eachdevice, is called a series circuit.

But how could you answer Chris? If the current is the same, whatchanges in the lamp to produce the thermal and light energy? Recall thatpower, the rate at which electrical energy is converted, is represented byP � IV. Thus, if there is a potential difference or voltage drop across thelamp, then electrical energy is being converted into another form. Theresistance of the lamp is defined as R � V/I. Thus, the potential differ-ence, also called the voltage drop, is V � IR.

What is the current in the series circuit? From the river model,you know that the sum of the drops in height at each rapid is equal tothe total drop from the top of the mountain to sea level. In the electri-cal circuit, the increase in voltage provided by the generator or otherenergy source, Vsource, is equal to the sum of voltage drops across thelamps A and B.

Vsource � VA � VB

Because the current, I, through the lamps is the same, VA � IRA andVB � IRB. Therefore, Vsource � IRA � IRB or Vsource � I(RA � RB).The current through the circuit is represented by the following.

I � �R

V

A

so�ur

Rce

B�

This equation applies to any number of resistances in series, not justtwo. The same current would exist with a single resistor, R, that has a

23.1 Simple Circuits 533

+––

RA

RB

Vsource

+

–

FIGURE 23–2 What is your prediction about the brightnessof the two lightbulbs?

FIGURE 23–3 The ammetersshow that in a series circuit thecurrent is the same everywhere.

2 TEACHUsing an Analogy

• In terms of the water analogyused in Chapter 22, a series circuitcorresponds to a single closedloop of pipe that has only onepath for water to flow through.

• Students may find it helpful tothink of a series circuit as onedevice connected right afteranother, just as the World Series is one game right after another.

Concept Development

Various terms are used that all mean“potential drop”: potential difference,voltage drop, and IR drop.

533

Linguistic The usualmethod for turning a circuit onor off is by means of a switch.Ask students to make a list intheir journals of the differenttypes of switches they are famil-iar with and can name. They maymention push button, toggle, rocker,chain, and twist switches. Touchswitches that are not mechanical,but entirely electrical, also may bementioned. L1

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Physics JournalPhysics Journal

VideodiscSTVS: ChemistryDisc 2, Side 2Battery Science (Ch. 8)

!7|Ü"

resistance equal to the sum of the resistances of the two lamps. Such aresistance is called the equivalent resistance of the circuit. For resistorsin series, the equivalent resistance is the sum of all the individualresistances.

Equivalent Resistance for Resistors in Series R � RA � RB � . . .

Notice that the equivalent resistance is larger than any single resis-tance. Therefore, if the battery voltage doesn’t change, adding moredevices in series always decreases the current. To find the current, I,through a series circuit, first calculate the equivalent resistance, R, andthen use the following equation to calculate I.

Current I � �Vso

Rurce�


F.Y.I.Henry Cavendish used thedirect approach to measurethe strength of an electriccurrent. Lacking the appropriate instruments, he instead shocked himselfwith the current and thenestimated the pain.

Pocket LabSeries Resistance

Hook up a power supply, aresistor, and an ammeter in aseries circuit. Predict what willhappen to the current in thecircuit when a second, identicalresistor is added in series to thecircuit. Predict the new currentswhen the circuit contains threeand four resistors in series.Explain your prediction. Try it. Analyze and Conclude Makea data table to show yourresults. Briefly explain yourresults. (Hint: Include the ideaof resistance.)

1. Three 20-� resistors are connected in series across a 120-V generator. What is the equivalent resistance of the circuit? What is the current in the circuit?

2. A 10-� resistor, a 15-� resistor, and a 5-� resistor are connectedin series across a 90-V battery. What is the equivalent resistanceof the circuit? What is the current in the circuit?

3. Consider a 9-V battery in a circuit with three resistors connectedin series.a. If the resistance of one of the resistors increases, how will the

series resistance change?b. What will happen to the current?c. Will there be any change in the battery voltage?

4. A string of holiday lights has ten bulbs with equal resistancesconnected in series. When the string of lights is connected to a120-V outlet, the current through the bulbs is 0.06 A.a. What is the equivalent resistance of the circuit?b. What is the resistance of each bulb?

5. Calculate the voltage drops across the three resistors in problem 2, and check to see that their sum equals the voltage of the battery.

Practice Problems

Voltage drops in a series circuit In any circuit, the net change inpotential as current moves through it must be zero. This is because theelectrical energy source in the circuit, the battery or generator, raises thepotential. As current passes through the resistors, the potential drops anamount equal to the increase, and, therefore, the net change is zero.

The potential drop across each resistor in a series circuit can be cal-culated by rearranging the equation that defines resistance, R � V/I, to

Have students refer to Appendix Cfor complete solutions to PracticeProblems.1. 60 �; 2 A2. 30 �; 3 A3. a. increase; b. decrease; c. No4. a. 2000 �; b. 200 �5. 90 V

534

Practice Problems

Series ResistancePurpose To construct a simple seriescircuit and to monitor the effect ofadding resistors by measuring thecurrent

Materials Power supply, ammeter,wires, 4 identical resistors (330-� or470-�, 1/2 W) Resistors of lowerresistance, such as 150-� or 220-�,may be used if they have higherpower ratings such as 1 W.

Outcome Students will find that, forexample, tripling the number of resis-tors reduces the current to one-thirdits original level.

Analyze and Conclude The current isinversely proportional to the numberof resistors. LS

CD-ROMPhysics for the Computer Age ELECTRICITY: ElectricityIntroduction and Series Circuits

ELEMENTS OF CIRCUITS

Visual-Spatial Have students identifyindividual resistance elements and series cir-cuits within a system. The electrical diagramfrom an automobile owner’s manual providesa convenient example of a fairly complex DC electrical system. A radio, television, orreceiver-amplifier owner’s manual also mayhave a schematic diagram that could be used.

Students could add their marked-up diagramsto their portfolios. PELLL1LS

TECHPREP


solve for V, V � IR. First, find the equivalent resistance, R, in the circuitby calculating the sum of all the individual resistances. Then, to find thecurrent, which is the same everywhere in the circuit, use the equivalentresistance and the equation I � V/R, where V is the potential drop. Hav-ing determined the current in the circuit, multiply I by the resistance ofthe individual resistor to find the potential drop across that resistor.

An important application of series resistors is the voltage divider. Avoltage divider is a series circuit used to produce a voltage source ofdesired magnitude from a higher-voltage battery. Suppose you have a 9-V battery but need a 5-V potential source. A voltage divider can supplythis voltage. Consider the circuit shown in Figure 23–4. Two resistors,RA and RB, are connected in series across a battery of magnitude V. Theequivalent resistance of the circuit is R � RA � RB. The current, I, is rep-resented by the following equation.

I � �V

R� � �

RA �

V

RB�

The desired voltage, 5 V, is the voltage drop, VB, across resistor RB.

VB � IRB.

I is replaced by the preceding equation.

VB � IRB � ��RA �

V

RB�� RB

VB � �RA

V

�

RBRB

�

Voltage dividers are often used with sensors such as photoresistors.The resistance of a photoresistor depends upon the amount of light thatstrikes it. Photoresistors are made of semiconductors such as silicon,selenium, and cadmium sulfide. A typical photoresistor can have a resis-tance of 400 � when light strikes it, but 400 000 � when in the dark.The output voltage of a voltage divider that uses a photoresistordepends upon the amount of light striking the photoresistor sensor.This circuit can be used as a light meter, such as the one in Figure 23–5.In this device, an electronic circuit detects the potential difference andconverts it to a measurement of illuminance that can be read on the digital display.

RA

V I

RB

+

–VB

AmplifiedVoltmeter

Sensitivity adjustment(potentiometer)

Photoresister sensor

Light

Drycells

FIGURE 23–4 The values of RAand RB are chosen such that thevoltage drop across RB is thedesired voltage.

FIGURE 23–5 The output voltageof this voltage divider dependsupon the amount of light strikingthe photoresistor sensor. This isthe basis for the light metershown at left.

The Inverse Key

Because multiplication is theinverse of division, you can usethe inverse key, 1/x, to performcalculation without having toreenter numbers.

�40

(90

V�

)(�

50500�

0)�

�

Keys Display

900

1.11 . . . �03

5

Answer 5 V

�500�9�

�1/x

�500�400

Uncovering

Misconceptions

Students often believe that the cur-rent gets used up or decreases as itgoes around a series circuit and thatthe first circuit element receives allof the applied potential difference.They often forget that the potentialdifference divides up through eachelement in a series circuit.

535

THE ¤MECHANICAL ¤

¤UNIVERSEHIGH SCHOOL ADAPTATION

VideotapeQuad 5: ElectricitySimple DC Circuits

Learning Disabled Even students withgood abstract and conceptual abilities mayhave difficulty juggling the properties of seriesand parallel circuits in their heads. Havesmall groups of students make a summarychart of the properties of series and parallelcircuits. The chart should include the behav-ior of voltage, resistance, and current; thedependence or independence of circuitdevices; and the number of conducting paths.

Have students share their results. Each studentshould transfer the information to his or herjournal or notebook.

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COOP LEARNELLL1

Meeting Individual NeedsMeeting Individual Needs

Texas TEKS

Pages 534–535: 1(A), 2(A), 2(C),2(D), 3(B), 6(E)


Voltage Drops in a Series Circuit

Two resistors, 47.0-� and 82.0-�, are connected in series acrossa 45.0-V battery.

a. What is the current in the circuit?

b. What is the voltage drop across each resistor?

c. The 47.0-� resistor is replaced by a 39.0-� resistor. Will the currentincrease, decrease, or remain the same?

d. What will happen to the voltage drop across the 82.0-� resistor?

Sketch the Problem

• Draw a schematic of the circuit. • Include an ammeter and voltmeters.

Calculate Your Answer

Known: Unknown:

Vsource � 45.0 V I � ?

RA � 47.0 � VA � ?

RB � 82.0 � VB � ?

effects of changing RA

Check Your Answer

• Are the units correct? Current, A � V/�, voltage, V � A�.• Is the magnitude realistic? Numerically, R > V, so I < 1.

R decreases so I increases. VB changes because it depends on I.

Strategy:

a. To determine the current, first find the equivalent resistance.

b. Use V � IR for each resistor.

c. Calculate current again using RA as 39.0 �.

d. Determine new voltage drop.

Calculations:

R � RA � RB

I � �Vso

Rurce� � �

R

V

A

so�ur

Rce

B�

I ��47.0 �

45

�

.0

8

V

2.0 �� 0.349 �

VA � IRA � (0.349)(47.0 �) � 16.4 V

VB � IRB � (0.349)(82.0 �) � 28.6 V

I � �R

V

A

so�ur

Rce

B� ��

39.0 �

45

�

.0

8

V

2.0 �� 0.372 �

The current will increase.

VB � IRB � (0.372)(82.0 �) � 30.5 V

The voltage will increase.

+

–

RA

RB

A

VA

VB

V

Example Problem

536

CAUTION: Wear goggles. Set up a power supply and con-nect three lightbulbs in series.Arrange the wires so that studentscan easily see that the lamps arestrung end to end and that thereis only a single path for current.Turn on the lightbulbs and havestudents observe the brightness.Unscrew one bulb so that stu-dents will see that all the bulbsgo out. Now connect the light-bulbs in parallel, this timearranging the wire so studentscan see that the pattern is like agroup of stacked bridges. Havethem observe the brightness andthen unscrew one bulb. Studentswill observe that the remainingbulbs stay lit. Ask students tocompare the brightness of thebulbs in parallel with those inseries. Discuss the voltage or cur-rent differences in the two typesof circuits.

DEMOQUICK

Content Background

An ammeter is made by placing a low-resistance shunt across the terminals of a galvanometer so that it is parallel with themeter, thus providing a low-resistance paththrough the ammeter. It can, therefore, beplaced in series without affecting the circuit.The multiplier is placed in series with the galvanometer. When it is hooked into the circuit parallel with a component to be measured, the current through it is very

small, although the voltage drop is the same. The resistance of a typical galvanometer is 100–2000 �.

Voltage Divider

A 9.0-V battery and two resistors, 400 � and 500 �, are connectedas a voltage divider. What is the voltage across the 500-� resistor?

Sketch the Problem

• Draw the battery and resistors in a series circuit.


Known:

Vsource � 9.0 V

RA � 400 �

RB � 500 �

Unknown:

VB � ?

Check Your Answer

• Are the units correct? V � V �/�.• Is the magnitude realistic? The voltage drop is less than the battery

voltage. Because 500 � is more than half of the equivalent resistance,the voltage drop is more than half the battery voltage.


+

RA

I

RB VB

–V

Strategy:

Write the expression for thecurrent through the circuit.

Determine equivalentresistance, R.

Use voltage drop equation todetermine VB.

Calculations:

I � �Vso

Rurce�

R� RA � RB

VB � IRB ��(V

Rso

A

ur�ce)

R

(R

B

B)�

VB � � 5 V(9.0 V)(500 �)

��400 � � 500 �

6. A 20.0-� resistor and a 30.0-� resistor are connected in seriesand placed across a 120-V potential difference.a. What is the equivalent resistance of the circuit?b. What is the current in the circuit?c. What is the voltage drop across each resistor?d. What is the voltage drop across the two resistors together?

7. Three resistors of 3.0 k�, 5.0 k�, and 4.0 k� are connected inseries across a 12-V battery.a. What is the equivalent resistance?b. What is the current through the resistors?c. What is the voltage drop across each resistor?d. Find the total voltage drop across the three resistors.

F.Y.I.The largest voltages in yourhome are in your televisionset, where 15 000 V to 20 000 V are common. The largest currents arelikely to be the 40 A in anelectric range.

Example Problem

Practice Problems

Continued on next page

Have students refer to Appendix Cfor complete solutions to PracticeProblems.6. a. 50.0 �; b. 2.4 A; c. 48 V; 72 V;

d. 1.20 � 102 V7. a. 12.0 k�; b. 1.0 � 10�3 A;

c. 3.0 V; 5.0 V; 4.0 V; d. 12.0 V

537

Linguistic The humanbody uses electricity in nerveimpulses, muscle action, brainfunction, and bone growth.Have students research bioelec-tricity and write a journal entrydescribing their findings. L2

LS


Practice Problems

Cultural DiversityCultural Diversity

In 1918, the Edison Pioneers group wasformed to honor some of the people whoworked with Thomas Edison as the “creatorsof the electric industry.” One member, LewisH. Latimer, was born in Massachusetts in1848, the son of a freed slave. As a teenager,he taught himself to be a draftsman and con-vinced a businessman to hire him. In 1876,Latimer prepared the drawings that AlexanderGraham Bell needed to patent the telephone.Three years later, he was hired by the United

States Electric Lighting Company, where heinvented and patented the first carbon-filamentelectric lamp. He also invented an inexpensiveway to produce the filaments, which made itpossible for more people to afford electriclighting. In 1883, Latimer helped Edisonorganize the patents derived from Edison’scompanies, and in 1890, he published thefirst textbook on electrical lighting in Edison’shonor. Latimer died in 1928.

Texas TEKS

Pages 536–537: 3(B), 6(E) Pages 538–539: 1(A), 2(A), 2(C),2(D), 3(B), 6(E)

Parallel CircuitsLook at the circuit shown in Figure 23–6. How many current paths

are there? The current from the generator can go through any of the threeresistors. A circuit in which there are several current paths is called a parallel circuit. The three resistors are connected in parallel; both endsof the three paths are connected together. In the mountain river modelfor circuits, such a circuit is illustrated by three paths for the water over awaterfall. Some paths may have a large flow of water, others a small flow.The sum of the flows, however, is equal to the total flow of water over thefalls. In addition, it doesn’t matter which channel the water flowsthrough because the drop in height is the same. Similarly, in a parallelelectrical circuit, the total current is the sum of the currents through eachpath, and the potential difference across each path is the same.

What is the current through each resistor? It depends upon the indi-vidual resistance. For example, in Figure 23–7, the potential differenceacross each resistor is 120 V. The current through a resistor is given byI � V/R, so you can calculate the current through the 24-� resistor asI � (120V)/(24 �) � 5 A. Calculate the currents through the other tworesistors. The total current through the generator is the sum of the cur-rents through the three paths, in this case, 38 A.

What would happen if the 6-� resistor were removed from the cir-cuit? Would the current through the 24-� resistor change? That currentdepends only upon the potential difference across it and its resistance,and neither has changed, so the current is unchanged. The same is true


8. A photoresistor is used in a voltage divider as RB. V � 9.0 V andRA � 500 �.a. What is the output voltage, VB, across RB, when a bright light

strikes the photoresistor and RB � 475 �?b. When the light is dim, RB � 4.0 k�. What is VB?c. When the photoresistor is in total darkness, RB � 0.40 M�

(0.40 � 106 �). What is VB?9. A student makes a voltage divider from a 45-V battery, a 475-k�

(475 � 103 �) resistor, and a 235-k� resistor. The output ismeasured across the smaller resistor. What is the voltage?

Generator

RC

RB

RA

120 V

38 A

5 A

24 � 9 �120 V

A

A

V

13 AA

V 6 �120 V

20 A

120 V

A

V

FIGURE 23–6 The parallel pathsfor current in this diagram areanalogous to the paths that a river may take down the mountain.

FIGURE 23–7 In a parallel circuit, the reciprocal of the totalresistance is equal to the sum ofthe reciprocals of the individualresistances.

Have students refer to Appendix Cfor complete solutions to PracticeProblems.8. a. 4 V; b. 8 V; c. 9 V9. 15 V

Uncovering

Misconceptions

Check to see that studentsknow how to manipulate the recip-rocals in the equation for calculatingthe equivalent resistance. Ask themif 1/a � 1/b � 1/(a � b). Have themreview the steps they would take toobtain the equivalent resistance ofthree resistors—60 �, 40 �, and 10 �—connected in parallel. Guide them through the keystrokeson their calculators. R � 7.1 �.Emphasize that the equivalent resis-tance of several resistors in parallelis always less than the smallestresistor in the parallel combination.This fact can be used as a check onthe reasonableness of their results.

Reinforcement

Kinesthetic The abstractnessof circuit diagrams may cause diffi-culty for many students. It may helpto have students construct many ofthe circuits that are discussed in thetext to help them become betterable to visualize diagrams. ELLL1

LS

LS

538

Practice Problems

DEMONSTRATION 23-1

PURPOSE

To increase students’ understanding ofhow voltage divides in a series circuit

MATERIALS

Hookup wire, 120-VAC cord with plug,two 120-VAC incandescent lamp sockets,100-W lightbulb, 15-W or 25-W lightbulb

PROCEDURE

CAUTION: This circuit exposes you to poten-tially dangerous voltage and current. Use a

commercial board or have an electriciandesign the circuit. There should be no barewires. Do not touch any exposed wires whenthe circuit is turned on. 1. Wire one of the lightbulb sockets to the

power cord. Place the 100-W lightbulb inthe socket and plug it in. Have studentsobserve the intensity of the light.

2. Replace the 100-W lightbulb with thesmaller lightbulb. Observe the intensity.

3. Wire two 120-VAC lightbulb sockets in series with the power cord as in thechapter opening photo. Place the twolightbulbs into the sockets. Plug in the circuit. Have students observe theintensity of the two bulbs.

RESULTS

The 100-W lightbulb by itself will bebright while the smaller bulb will be dim-mer. When placed in series, the 100-W

for the current through the 9-� resistor. The branches of a parallel circuit are independent of each other. The total current through the gen-erator, however, would change, and the sum of the currents in thebranches would then be 18 A.

How can you find the equivalent resistance of a parallel circuit? InFigure 23–7, the total current through the generator is 38 A. A singleresistor that would have a 38-A current when 120 V were placed acrossit would be represented by the following equation.

R � �V

I� � �

1

3

2

8

0

A

V� � 3.2 �

Notice that this resistance is smaller than that of any of the three resis-tors in parallel. Placing two or more resistors in parallel always decreasesthe equivalent resistance of a circuit. The resistance decreases becauseeach new resistor provides an additional path for current, increasing thetotal current while the potential difference remains unchanged.

To calculate the equivalent resistance of a parallel circuit, first notethat the total current is the sum of the currents through the branches. IfIA, IB, and IC are the currents through the branches and I is the total cur-rent, then I � IA � IB � IC.

The potential difference across each resistor is the same, so the cur-rent through each resistor, for example, RA, can be found from IA �

V/RA. Therefore, this becomes the equation for the sum of the currents:

�V

R� � �

R

V

A� � �

R

V

B� � �

R

V

C�

Dividing both sides of the equation by V provides an equation for theequivalent resistance of the three parallel resistors.

Equivalent Resistance for Resistors in Parallel �R

1� � �

R

1

A� � �

R

1

B� � �

R

1

C�

This equation can be used for any number of resistors in parallel.


Equivalent Resistance and Current in a Parallel Circuit

Three resistors, 60.0 �, 30.0 �, and 20.0 �, are connected in parallel across a 90.0-V battery.

a. Find the current through each branch of the circuit.

b. Find the equivalent resistance of the circuit.

c. Find the current through the battery.

Sketch the Problem• Draw a schematic of the circuit. • Include ammeters to show the paths of the currents.

90.0 V

60.0 � 30.0 � 20.0 �

IA IB ICA

A A A

Pocket LabParallel Resistance

Hook up a power supply, aresistor, and an ammeter in aseries circuit. Predict what willhappen to the current in thecircuit when a second, identicalresistor is added in parallel tothe first. Predict the new cur-rents when the circuit containsthree and four resistors in parallel. Explain your prediction.Try it. Analyze and Conclude Makea data table to show yourresults. Briefly explain yourresults. (Hint: Include the ideaof resistance.)

Example Problem

Continued on next page

539

Parallel ResistancePurpose To construct a simple paral-lel circuit and to monitor the effect ofadding resistors by measuring thecurrent

Materials Power supply, ammeter,wires, four identical resistors (330-�

or 470-�, 1/2 W) (Resistors of lowerresistance may be used if they havehigher power ratings.)

Outcome Students are often sur-prised to find that the currentincreases as more parallel branchesare added. With three branches, thecurrent will be three times the origi-nal value.

Analyze and Conclude The resistancedecreases with the addition of moreparallel branches. LS

lightbulb will not visibly light or it will bevery dim while the smaller 25-W lightbulbwill be fairly bright, but not as bright aswhen it was operated alone.

KNOWLEDGE

1. When the two lightbulbs were connectedin series, was the 100-W bulb working,

or was it burned out? Ask students to suggest a way of proving their contention. It was working. This can beshown by unscrewing it from the circuit,which will cause the 25-W lightbulb toturn off. This indicates that the bulb is notburned out because it is completing theseries circuit.

2. Propose a possible explanation for thebehavior of the two lightbulbs wired in

series. The combined resistance of the two lightbulbs in series is fairly high so the current through the circuit is lowerthan it would be for either bulb by itself.The current through both bulbs is thesame, but the resistance of the 25-W bulbis higher. Because P=I2R, more power isdissipated by the 25-W bulb and it glowsmore brightly.

Assessment



Known: Unknown:

RA � 60.0 � RC � 20.0 � IA � ? IC � ? R � ?

RB � 30.0 � V � 90.0 V IB � ? I � ?

Check Your Answer• Are your units correct? Currents are in amps, resistances are in ohms.• Do the signs make sense? All are positive, as they should be.• Is the magnitude realistic? IA � IB � IC � I

Calculations:

IA � �R

V

A� � �

6

9

0

0

.

.

0

0

�

V� � 1.50 A

IB � �R

V

B� � �

3

9

0

0

.

.

0

0

�

V� � 3.00 A

IC � �R

V

C� � �

2

9

0

0

.

.

0

0

�

V� � 4.50 A

�R

1��

R

1

A� � �

R

1

B� � �

R

1

C�

�R

1��

60.

1

0 ��

30.

1

0 ��

20.

1

0 �� 0.100 ��1

R � 10.0 �

I � �V

R� � �

1

9

0

0

.

.

0

0

�

V� � 9.00 A

Strategy:

a. The voltage across each resistor is the

same, so use I � �V

R� for each branch.

b. Use the equivalent resistance equationfor parallel circuits.

c. Use I � �V

R� to find the total current.

10. Three 15-� resistors are connected in parallel and placed acrossa 30-V battery.a. What is the equivalent resistance of the parallel circuit?b. What is the current through the entire circuit?c. What is the current through each branch of the circuit?

11. A 120.0-� resistor, a 60.0-� resistor, and a 40.0-� resistor areconnected in parallel and placed across a 12.0-V battery.a. What is the equivalent resistance of the parallel circuit?b. What is the current through the entire circuit?c. What is the current through each branch of the circuit?

12.Suppose one of the 15.0-� resistors in problem 10 is replacedby a 10.0-� resistor.a. Does the equivalent resistance change? If so, how?

Practice Problems

Have students refer to Appendix Cfor complete solutions to PracticeProblems.10. a. 5.0 �; b. 6 A; c. 2 A11. a. 20.0 �; b. 0.600 A;

c. 0.100 A; 0.200 A; 0.300 A12. a. yes, smaller

b. yes, gets largerc. No, it remains the same.

540

Practice Problems

Connections To Biology

YOUR BODY AS A RESISTORThe electrical resistance across yourskin is high, around 40 000 �.

However, the resistance from one part ofyour body to another is much smaller,closer to 100 �. If, for example, you touchthe terminals of a 9-V battery with yourhands, the current encounters three resis-tances in series: skin, internal organs, andskin. The total resistance is greater than

80 000 �, so the current is small, lessthan 1/8 mA. In some cases, dry skin mayhave much higher resistance. Water lowersthe resistance of skin by a factor of 10. Ifboth of your hands are wet, current couldbe high enough to be felt. Even low volt-ages can be dangerous. In hospitals,instrument probes are sometimes placeddirectly into the body, thus bypassing thehigh resistance of the skin that tends to

keep currents low. In such cases, currentsas high as 1 mA should be avoided.

Ask students what the largest voltage isthat can be placed across two internalprobes if the body’s resistance between theprobes is 50 �. V � IR � (1 mA)(50 �)� 50 mV or 0.05 V LSL1

PERFORMANCE To furtherassess student’s understandingof parallel circuits, ask if the cur-rent in the first resistor changeswhen you connect a second andthird resistor in parallel. Havestudents predict the answer andthen devise a method for testingit. Ask them to explain how theywould connect an ammeter tomeasure the current through justone resistor. One lead must bedisconnected from the first resistorfrom the point where the secondand third resistors join. Connectone lead of the ammeter to thispoint and the other lead to theresistor lead that was disconnected.In general, it should appear in thelocation of the ammeter that reads5 A in Figure 23–7. ELLL1

Assessment


b. Does the amount of current through the entire circuit change?In what way?

c. Does the amount of current through the other 15.0-� resistors change? In what way?

Now that you have learned about both parallel and series circuits,you can analyze the brightness of the 60-W and 100-W lamps shown inthe photo at the beginning of this chapter. The brightness of a lightbulbis proportional to the power dissipated by it. Used in the normal way,each bulb would be connected across 120 V. Based on what you learnedin Chapter 22, the resistance of the 60-W bulb is higher than that of the100-W bulb. But when the bulbs are connected in series, the currentthrough the two bulbs is the same, so P � I2R. The higher-resistancelamp, the 60-W lamp, now dissipates more power and glows brighterthan the 100-W lamp.

Section Review

1. Are car headlights connected in seriesor parallel? Draw on your experience.

2. Lamp dimmers often contain variableresistors.

a. Would a dimmer be hooked inseries or in parallel with the lampto be controlled? Why?

b. Should the resistance of the dimmer be increased or decreasedto dim the lamp?

3. A switch is connected in series with a75-W bulb to a source of 120 V.

a. What is the potential differenceacross the switch when it is closedor on?

b. What is the potential differenceacross the switch when it is open,or off? Explain.

4. Critical Thinking The circuit inFigure 23–8 has four identical

resistors. Suppose a wire is added toconnect points A and B. Answer thefollowing questions, explaining yourreasoning.

a. What is the current through the wire?

b. What happens to the currentthrough each resistor?

c. What happens to the current drawnfrom the battery?

d. What happens to the potential difference across each resistor?

23.1

R

A

R

R

B

R

FIGURE 23–8

Which Bulb Burns Brighter?

➥ Answers question from page 530.

3 ASSESSChecking for

Understanding

Have students describe the way cur-rent divides in a parallel circuit.Which resistor has the largest cur-rent? the smallest resistor Which resis-tor has the smallest current? thelargest resistor What is the same forall of the resistors in parallel? thepotential difference

Reteaching

Review common elements of a circuit by actually displaying wires,switches, resistors, power supplies,voltmeters, and ammeters. Branchingcan be discussed as a lead into par-allel circuits. Series circuits can bereviewed by asking how many 3-V lightbulbs are needed to attachto a 120-V circuit. 40

Extension

For students who have mastered thelessons, assign problems from theSupplemental Problems booklet.

4 CLOSEActivity

Pass around an appliancethat does not have a labeldescribing its power rating,or cover the label. Ask stu-

dents to write down the measure-ments they would have to make todetermine the power rating. Becauseyou know that the voltage is 120 V, youneed only measure the current or theappliance resistance (unless it is amotor-driven device), then calculate theproduct of current and voltage to getpower or divide the square of the voltageby the resistance. LSL1

L2

L1

541

TECHPREP

1. parallel, because you have seen a car withonly one headlight functioning

2. a. series, so it can control the currentthrough the lamp b. Increase the resistance to reduce the current and dim the lamp.

3. a. 0 V; V � IR with R � 0. b. 120 V, There is no current in the circuit,so the voltage across the bulb is 0. Becausethe sum of the voltages across the switch

and bulb is 120 V, there must be 120 Vacross the switch.

4. The answer to each is “nothing.” Thepotentials of points A and B are the same.Therefore, no current flows through thewire and nothing changes.

Texas TEKS

Pages 540–541: 3(B), 6(E)

Section Review23.1

You have already learned some of the elementsof household wiring circuits. It’s important to

understand the requirements and limitations ofthese systems. Above all, it is important to be awareof the safety measures that must be practiced to prevent accidents.

Safety DevicesIn an electric circuit, fuses and circuit breakers are switches that act as

safety devices. They prevent circuit overloads that can occur when toomany appliances are turned on at the same time or a short circuit occursin one appliance. When appliances are connected in parallel, each addi-tional appliance placed in operation reduces the equivalent resistance inthe circuit and causes more current through the wires. The additionalcurrent may produce enough thermal energy (P � I2R) to melt insula-tion on the wires and cause a short circuit in the wires or even a fire.

A fuse is a short piece of metal that melts if too large a current passesthrough it. The thickness of the metal to be used is determined by theamount of current that can be safely handled by the circuit. Should therebe a larger current through the circuit, the fuse will melt and break thecircuit. A circuit breaker, shown in Figure 23–9, is an automatic switchthat opens when the current reaches some set value. If current greaterthan the set value flows in the circuit, the circuit is overloaded. The cir-cuit breaker will open and thereby stop all current.

A ground-fault interrupter is often required by law in electrical out-lets in bathrooms and kitchens. Current follows a single path from thepower source into the electrical outlet and back to the source. Some-times, when an appliance such as a hair dryer is used, the appliance oruser might touch a cold water pipe or a sink full of water and in this waycreate another current path through the user. If a current as small as5 mA should follow this path through a person, it could result in seri-ous injury. The ground-fault interrupter contains an electronic circuit

OBJ ECTIVES

• Explain how fuses, circuitbreakers, and ground-faultinterrupters protect house-hold wiring.

• Analyze combined series-parallel circuits and calcu-

late the equivalent resis-tance of such circuits.

• State the important charac-teristics of voltmeters andammeters, and explain howeach is used in circuits.

23.2 Applications of Circuits


Current in fromcentral switch

Bimetallic strip

Switchcontacts

LatchCircuit Breaker

Current outto loads

On-offreset switch handle

FIGURE 23–9 When too muchcurrent flows through thebimetallic strip, it will bend downand release the latch. The handlemoves to the off position causingthe switch to open, and thatbreaks the circuit.

PREPAREContent Refresher

Recall that V � IR and that as theequivalent resistance of a circuitbecomes smaller, the current in thecircuit becomes larger. If the currentexceeds specific values, it can melt a fuse or cause a circuit breaker toturn off.

Bridging

This section relates directly to thefirst section of this chapter and toChapter 22.

542

SECTION 23.2

VideodiscSTVS: Physics Disc 1, Side 1 Synchrotron (Ch. 20)

!9,Ä"Disc 1, Side 2Making Integrated Circuits(Ch. 15)

!8^Ñ"

1 FOCUSGuest Speaker

Invite a local electrician orfire marshall to talk to theclass about safety consider-ations in electrical circuits.

Fuses, circuit breakers, grounding,and ground-fault interrupters areonly the beginning of this subject.

TECHPREP

Program Resources

Study Guide, pp. 137–138 Transparency 41 and Master

Critical Thinking, p. 30 Tech Prep Applications, pp. 35–38 Reteaching, p. 29 Physics Lab and Pocket Lab Worksheets,

pp. 123–124, 127 L1

L1

L1

L1

ELLL1

L1

543

that detects small differences in current caused by an extra current pathand opens the circuit, thereby preventing dangerous shocks.

Electric wiring in homes uses parallel circuits, such as the one dia-grammed in Figure 23–10, so that the current in any one circuit doesnot depend upon the current in the other circuits. The current in adevice that dissipates power, P, when connected to a voltage source, V, isrepresented by I � P/V. Suppose, that in the schematic diagram shownin Figure 23–10, a 240-W television is plugged into a 120-V outlet. Thecurrent that flows is represented by I � (240 W)/(120 V) � 2 A. Then, a720-W curling iron is plugged in. The current through the curling ironis I � (720 W)/(120 V) � 6 A. Finally, a 1440-W hair dryer is pluggedin. The current through the hair dryer is I � (1440 W)/(120 V) � 12 A.

The current through these three appliances can be found by consider-ing them as resistors in a parallel circuit in which the current througheach appliance is independent of the others. The value of the resistanceis found by calculating the current the appliance draws and then usingthe equation R � V/I. The equivalent resistance of the three appliances is

�R

1� � �

10

1

��

20

1

��

60

1

��

6

1

��

R � 6 �.

The 15-A fuse is connected in series with the power source so theentire current passes through it. The current through the fuse is

I � �V

R� � �

1

6

20

�

V� � 20 A.

The 20-A current exceeds the rating of the 15-A fuse, so that the fuse willmelt, or blow, cutting off current to the entire circuit.

A short circuit occurs when a circuit is formed that has a very lowresistance. The low resistance causes the current to be very large. If therewere no fuse or circuit breaker, such a large current could easily start afire. A short circuit can occur if the insulation on a lamp cord becomesold and brittle. The two wires in the cord could accidentally touch. Theresistance of the wire might be only 0.010 �. When placed across 120 V,this resistance would result in the following current.

I � �V

R� � �

0

1

.0

2

1

0

0

V

�� 12 000 A

120 V

15 A fuse

HELP WANTEDELECTRICIANElectrical contractor needselectricians who have suc-cessfully completed a 5-yearapprenticeship program con-ducted by a union or profes-sional builder’s association.You must be a high schoolgrad, be in good physicalcondition, have excellentdexterity and color vision,and be willing to work whenand where there is work. Youwill do all aspects of the job,including reading blueprints,dealing with all types ofwires, conduits, and equip-ment. Safety and qualitywork must be your highestpriorities. For informationcontact:International Brotherhood ofElectrical Workers1125 15th Street, N.W.Washington, DC 20005


FIGURE 23–10 This parallelwiring arrangement permits theuse of more than one appliancesimultaneously, but if all threeappliances are used at once, thefuse could melt.

2 TEACHConcept Development

• Continue to emphasize that one volt is equal to one joule percoulomb. This will help keep theunits more meaningful and makeunit cancellation easier.

• If you have a broken electricalmeter in the lab, remove the backand pass it around for students tosee the circuit elements.

543

HOUSEHOLDPROTECTIONAsk students where themain household fuse orcircuit breaker should

be placed so that it protects allthe circuits in the home, not justone of the parallel circuits. Itshould be placed on one of the mainwires that brings electricity into thehome so that the fuse or circuitbreaker will shut off everything. L1

TECHPREP

Enrichment

Ask students if there are other waysto hook up circuit devices besidesseries and parallel. Let them drawtheir suggestions. Two other ways arethe delta—three devices connected in a triangular arrangement; and thewye—three devices connected at onepoint. These two arrangements areneither strictly parallel nor seriesarrangements. L2

HOUSEHOLD WIRING

Intrapersonal Some students may beinterested in finding out how to wire a lightso that it can be turned off or on from twodifferent locations, such as at the top and bot-tom of a staircase. This type of circuit is calleda three-way switch. Another arrangement inwhich a switch can be turned off and on inthree locations is called a four-way switch.

Have students draw a diagram of one or bothof these circuits and explain to the class howthey work. The diagrams could be placed instudent portfolios. If equipment is available,students could build and demonstrate aswitch circuit. PELLL1

LS

TECHPREP

PERFORMANCE To determinehow well students are able torelate a circuit diagram to anactual electrical circuit, set upsome simple series, parallel, and series-parallel circuits at stations around your classroom.Have students visit each stationand draw a diagram of thecircuit. LSL1

Assessment

Applying Physics

Such a current would cause a fuse or a circuit breaker to open the circuitimmediately, thereby preventing the wires from becoming hot enoughto start a fire.

Combined Series-Parallel CircuitsHave you ever noticed the light in your bathroom dim when you

turned on a hair dryer? The light and the hair dryer were connected inparallel across 120 V. This means that the current through the lampshould not have changed when you plugged in the dryer. Yet the lightdimmed, so the current must have changed. The dimming occurredbecause the house wiring had a small resistance in series with the parallel circuit. This is a combination series-parallel circuit. The fol-lowing is a strategy for analyzing such circuits. Refer to Figure 23–11which illustrates the procedure described in steps 1, 2, and 3 of theProblem Solving Strategy.

Series-Parallel Circuits1. Draw a schematic diagram of the circuit.2. Find any parallel resistors. Resistors in parallel have separate

current paths. They must have the same potential differencesacross them. Calculate the single equivalent resistance thatcan replace them. Draw a new schematic using that resistor.

3. Are any resistors (including the equivalent resistor) now inseries? Resistors in series have one and only one currentpath through them. Calculate a single new equivalent resis-tance that can replace them. Draw a new schematic diagramusing that resistor.

4. Repeat steps 2 and 3 until you can reduce the circuit to a sin-gle resistor. Find the total circuit current. Then go backwardsthrough the circuits to find the currents through and the voltages across individual resistors.

60 V

RA 8 �

30 �

IA

IB

RB

20 �

RC = =IC

RA 8 �

RBC

12 �

VA

VPR

20 �

FIGURE 23–11 Use these diagramsas you study the following ProblemSolving Strategy.


Uncovering

Misconceptions

Propose this scenario to your stu-dents and ask them to comment onit: A fuse burns out in your bedroomcircuit. You find a box of assorted fuses,grab one, and slip it into the fuse socket.If students understand that a fusemust have the appropriate rating for the intended circuit, they willnote that the fuse should have beenselected according to its amp rating.To reinforce this concept, use anautomobile owner’s manual to showthat a number of fuses of differentresistance values are used in themany circuits that keep electricaldevices operating.

544

Gifted There are rules for analyzing electri-cal circuits known as Kirchhoff’s rules orlaws. These rules extend the basic conceptsdeveloped in this chapter to more compli-cated circuits. The rules can be used toderive systems of simultaneous equationsfor calculating unknown circuit voltages andcurrents. Have capable students investigateKirchhoff’s laws and present a summary tothe class. L3

MeetingIndividual Needs

MeetingIndividual Needs Have students make a survey of

circuits in their homes and recordtheir findings in a table in theirjournals. They should considernot only series and parallel circuits,

but more complex circuits, as well. Forexample, ask if they found two switches in series controlling one lamp, or if there is one switch that controls several lamps.

LSL1


TECHPREP

CAUTION: Wear goggles.After the discussion of short cir-cuits, show students a three-holereceptacle and a three-prongedplug properly wired to a piece ofelectric cable. Point out that athird wire connects the (metal)case of an appliance to the thirdprong in the plug. Have studentshypothesize what the third holein the receptacle might be con-nected to. It is attached to theground. Such a system protects theoperator of an appliance fromreceiving a dangerous or fatal shockshould the high voltage in theappliance accidentally come intocontact with the case. The currentfollows a low-resistance path intothe ground.

DEMOQUICK

Texas TEKS

Pages 542–543: 3(D), 6(E) Pages 544–545: 1(A), 1(B), 2(A),2(B), 2(C), 2(D), 2(F), 3(A), 3(B),6(E)

5. Assuming that the power supply is at 6.0 V,the bright lamp will be measured at 6.0 Vand each dim lamp will be at 3.0 V.

6. Measurements should verify the predictions.

Apply

1. One wall switch can control many overheadlights. The lamps are in parallel with eachother, but the switch is in series with the lamps.Opening the switch stops the electrical flowthrough all of the overhead lamps.

SKILL To further assess students’ understand-ing of circuits, ask the following questions.1. Draw the diagram of a circuit in which four

identical lamps have the same brightness,but when one lamp is loosened in its socket,only two lamps remain on.

545

Problem

Suppose that three identical lamps are con-nected to the same power supply. Can a cir-cuit be made such that one lamp is brighterthan the others and stays on if either of theothers is loosened in its socket?

HypothesisOne lamp should be brighter than the othertwo and remain at the same brightness wheneither of the other two lamps is loosened inits socket so that it goes out.

Possible Materialspower supply with variable voltagewires with clips3 identical lamps and sockets

Plan the Experiment1. Sketch a series circuit and predict the

relative brightness of each lamp. Predictwhat would happen to the other lampswhen one is loosened so that it goes out.

2. Sketch a parallel circuit and predict therelative brightness of each lamp. Predictwhat would happen to the other lampswhen one is loosened so that it goes out.

3. Draw a combination circuit. Label thelamps A, B, and C. Would the bulbs havethe same brightness? Predict what wouldhappen to the other two lamps when eachlamp in turn is loosened so that it goes out.

4. Check the Plan Show your circuits andpredictions to your teacher before starting to build the circuits.

5. When you have completed the lab, disposeof or recycle appropriate materials. Put awaymaterials that can be reused.

Analyze and Conclude1. Interpreting Data Did the series circuit

meet the requirements? Explain.

2. Interpreting Data Did the parallel circuitmeet either of the requirements? Explain.

3. Formulating Hypotheses Explain thecircuit that solved the problem in terms of current.

4. Formulating Hypotheses Use the defi-nition of resistance to explain why onelamp was brighter and the other two wereequally dim.

5. Making Predictions Predict how thevoltages would compare when measuredacross each lamp in the correct circuit.

6. Testing Conclusions Use a voltmeter tocheck your prediction.

Apply1. Can one wall switch control several lights

in the same room? Are the lamps in paral-lel or series? Are the switches in parallelor series with the lamps? Explain.

Circuits

23.2 Applications of Circuits 545

Assessment

Process Skills

Predicting, sequencing, recognizingcause and effect, experimenting, formulating hypotheses, interpret-ing data

Safety Precautions

Caution students to handle electricalequipment carefully to avoid beingshocked.

Possible Hypotheses

Students’ hypotheses could varygreatly depending upon their graspof circuits.

Possible Procedures

Students could make predictionsbefore building the circuits, or theycould build the circuits, observe the results, and then explain whathappened and why.

Teaching Suggestions

If students loosen one of the dimlamps and the bright lamp gets dim-mer and the dim lamp gets brighter,they have not built the correct circuit.

Analyze and Conclude

1. No. The three lamps go on andoff together and have the samebrightness.

2. The lamps can be turned on andoff individually, but they all havethe same brightness.

3. The correct circuit has two parallelbranches. One branch containsonly one lamp that is brighterthan the other two. The otherparallel branch has two dimmerlamps in series. Loosening eitherof these lamps will turn off theother dim lamp, but will not affectthe flow through the bright lamp.

4. The lamp in its own branch hadthe full voltage of the power sup-ply. The two dim lamps each hadhalf the voltage of the powersupply. Because the two dimlamps in series with each otherhave more resistance, the currentthrough these lamps is less thanthrough the bright lamp.

LS


Series-Parallel Circuit

A hair dryer with a resistance of 12.0 � and a lamp with a resistanceof 125 � are connected in parallel to a 125-V source through a 1.50-�resistor in series.

a. Find the current through the lamp when the hair dryer is off.

b. Find the current when the hair dryer is on.

c. Explain why the lamp dims when the hair dryer is on.

Sketch the Problem• Draw a diagram of the simple series circuit when the dryer is off. • Draw the series-parallel circuit including the dryer and lamp. • Replace RA and RB with a single equivalent resistance, RP.


Known: Unknown:RA � 125 � IA1 � ?

RB � 12.0 � R � ?

RC � 1.50 � IA2 � ?

Vsource � 125 V I2 � ?V

IA1

RA

RC

IA2

IA

RA

RC RC

RRB

Strategy:

a. When the hair dryer is off, the circuit is asimple series circuit.

b. Find the equivalent resistance for parallel circuit.

Find the equivalent resistance for entire circuit.

Use equivalent resistance to determine thecurrent when the hair dryer is on.

c. The greater current when the hair dryer is on means a greater voltage drop acrossRC, which causes less voltage across RA. A decrease in voltage means currentdecreases, which is less power.

Calculations:

IA1 � �RA �

V

RC� ��

125 �

12

�

5

1

V

.50 �� 0.988 A

�R

1

p� � �

R

1

A� � �

R

1

B�, so Rp � �

RA

RA�

RBRB

�

R � RC � Rp � RC � �RA

RA�

RBRB

�

R �1.50 � ��(

1

1

2

2

5

5

�

�

�

)(1

1

2

2

.0

.0

�

�

)�� 12.5 �

IA2 � �Vso

Rurce� � �

1

1

2

2

.

5

5

V

�� 1.00 � 101 A

VC � IRC � (1.00 � 101 A)(1.50 �) � 15.0 V

VA � Vsource � VC � 125 V � 15.0 V

� 1.10 � 102 V

IA � �V

RA

A� ��1.10

12

�

5

1

�

02 V�� 0.880 A

Example Problem

Current drops from 0.988 A to 0.880 A. Power, P � I2R, is smaller, consequently the light dims.

546

THREE-WAYLAMPS ANDSWITCHESAsk students to considera lamp having a three-

way switch and fitted with a singlelightbulb capable of delivering50 W, 100 W, and 150 W. Havestudents observe such a lamp andexplain how it works. The light-bulb has two filaments. The switchdetermines which filament haspower applied to it or if bothreceive power at the same time. The threaded portion of the base of the lightbulb is common for bothfilaments, and there are two con-nections for power in the center ofthe bulb base, one going to eachfilament. LSL1

Applying Physics

TECHPREP

Reinforcement

Interpersonal If you taughtthe resistor color code, here’s anidea for a quick quiz and review. If students do not know the code,make a chart of the code for studentsto refer to. Obtain a supply ofassorted resistors or remove somefrom a discarded electronic device.Give one resistor to each studentand ask students to identify theirresistor by means of the color code.Then, have pairs of students calculatethe equivalent resistance of theirtwo resistors in series and in parallel.Students could fasten their resistorsto their papers so that you can checktheir work. COOP LEARNELLL1

LS

DEMONSTRATION 23-2

PURPOSE

To demonstrate a series-parallel circuitcalled a bridge circuit and show how it canbe used to measure an unknown resistance

MATERIALS

Resistance substitution box, two 1000-�resistors, 0-to 1-mA meter (center zeroreading, if possible), knife switch,unknown resistor or a covered knownresistor, hookup wire, 6-V or 9-V battery

PROCEDURE

1. Attach the two 1000-� resistors to thepositive terminal of the battery.

2. Attach one side of the resistance substitution box and one side of theunknown resistance to the switch.Attach the other side of the switch tothe negative terminal of the battery.

3. Connect one end of the mA meter tothe free end of the unknown resistor

and to one of the 1000-� resistors.Connect the free end of the resistancesubstitution box to the other end of themeter and also to the free end of theother 1000-� resistor.

4. Adjust the resistance substitution box toapproximately 1000 �. Now close theswitch to apply power to the circuit.

5. Increase or decrease the resistance ofthe resistance substitution box until the

13.Two 60-� resistors are connected in parallel. This parallelarrangement is connected in series with a 30-� resistor. Thecombination is then placed across a 120-V battery.a. Draw a diagram of the circuit.b. What is the equivalent resistance of the parallel portion of

the circuit?c. What single resistance could replace the three original resistors?d. What is the current in the circuit?e. What is the voltage drop across the 30-� resistor?f. What is the voltage drop across the parallel portion of

the circuit?g. What is the current in each branch of the parallel portion of

the circuit?


0.01 �Ammeter

0.01 ��10 ��10 �=20.01 �

20 V

RA10.00 �

RB10.00 �

FIGURE 23–12 An ammetermeasures current and so it isalways placed in series in a circuit.

Check Your Answer

• Are the units correct? Current is in amps, potential drops are in volts.• Is the magnitude realistic? Decreased parallel resistance increases

the current, causing a voltage drop in the series resistor. This leavesless voltage across the combination, so the current is smaller.

Pocket LabAmmeter Resistance

Design an experiment using apower supply, a voltmeter, anammeter, a resistor, and somewires to determine the resis-tance of the ammeter. Make asketch of your setup and includemeasurements and equations. Communicating Results Whatis the resistance of the ammeter?Be prepared to present yourexperiment to the class.

Practice Problems

Ammeters and VoltmetersAn ammeter is used to measure the current in any branch or part of

a circuit. If, for example, you want to measure the current through aresistor, you would place an ammeter in series with the resistor. Thisrequires opening a current path and inserting an ammeter. The use of anammeter should not change the current in the circuit you wish to mea-sure. Because current would decrease if the ammeter increased the resis-tance in the circuit, the resistance of an ammeter should be as low aspossible. Figure 23–12 shows a real ammeter as an ideal, zero-resis-tance meter placed in series with a 0.01-� resistor. The ammeter resis-tance is much smaller than the values of the resistors. The currentdecrease would be from 1.0 A to 0.9995 A, too small to notice.

Another instrument, called a voltmeter, is used to measure the voltagedrop across some part of a circuit. To measure the potential drop across aresistor, connect the voltmeter in parallel with the resistor. A voltmetershould have a very high resistance so that it causes the smallest possiblechange in currents or voltages in the circuit. Consider the circuit shown inFigure 23–13. A typical inexpensive voltmeter consists of an ideal, zero-resistance meter in series with a 10-k� resistor. When it is connected inparallel with RB, the equivalent resistance of the combination is smallerthan RB alone. Thus, the total resistance of the circuit decreases, increasing

Have students refer to Appendix Cfor complete solutions to PracticeProblems.13. a. Refer to Appendix C.;

b. 30 �; c. 60 �; d. 2 A; e. 60 V; f. 60 V; g. 1 A

Visual Learning

Visual-Spatial Open the backor bottom of one of your laboratoryammeters and one of your laboratoryvoltmeters so students can see thecircuitry and draw a diagram ofeach meter. Compare their diagramsto Figures 23–12 and 23–13.

ELL

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547

Ammeter ResistancePurpose To determine if studentsunderstand the concept of resistanceand electrical measurements

Materials Power supply, voltmeter,ammeter, wires, 5-� or 10-�, 1% or0.1% precision resistor

Outcome Students should draw andbuild a series circuit with the powersupply, ammeter, and resistor, andmeasure the voltage drop across theammeter.

Analyze and Conclude The readingon the voltmeter should be very lowbecause of the low resistance of theammeter. LS

Practice Problems

ammeter reads zero current.

RESULTS

When the meter reads zero, the value ofthe resistance substitution box equals theunknown resistance.

SKILL

1. Have students draw the circuit.

A

AssessmentTexas TEKS

Pages 546–547: 1(A), 2(A), 2(B),2(C), 2(D), 2(F), 3(B), 6(E)

the current. RA has not changed, but the current through it has increased,increasing the potential drop across it. The battery, however, holds thepotential drop across RA and RB constant. Thus, the potential drop acrossRB must decrease. The result of connecting a voltmeter across a resistor isto lower the potential drop across it. The higher the resistance of the volt-meter, the smaller the voltage change. Using a voltmeter with a 10 000-�resistance changes the voltage across RB from 10 V to 9.9975 V, too smalla change to detect. Modern electronic multimeters have even higher resis-tances, 107 �, and so produce even smaller changes.


Section Review

1. Consider the circuit in Figure 23–14made with identical bulbs.

a. Compare the brightness of the three bulbs.

b. What happens to the brightness ofeach bulb when bulb 1 is unscrewedfrom its socket? What happens to thethree currents?

c. Bulb 1 is screwed in again and bulb 3 isunscrewed. What happens to the bright-ness of each bulb? What happens to thethree currents?

d. What happens to the brightness of eachbulb if a wire is connected betweenpoints B and C?

e. A fourth bulb is connected in parallelwith bulb 3 alone. What happens tothe brightness of each bulb?

2. Research and describe the connectionbetween the physics of circuits and future careers.

3. Critical Thinking In the circuit inFigure 23–14, the wire at point C is broken and a small resistor is inserted in series with bulbs 2 and 3. What happens to the brightness of the two bulbs? Explain.

23.2

V

A

B

l2

l3

l1

1 2

3

C

A

A

A

20 V

10.00 �

10 k�

Voltmeter

RA

RB10.00 �

FIGURE 23–13 A laboratoryvoltmeter such as this one mea-sures potential difference. Volt-meters are placed in parallel.

FIGURE 23–14

3 ASSESSChecking for

Understanding

Ask students to explain why anammeter is hooked up in series andwhy a voltmeter is hooked up inparallel. Alternatively, ask studentswhy a voltmeter does not workproperly if it is placed in a circuitand attached in series. Voltmetershave a very high internal resistance;thus, the current will be small.

Reteaching

Have students work with a partnerto summarize properties of a fuse,circuit breaker, ground-fault inter-rupter, ammeter, and voltmeter.

Extension

For students who have mastered thelesson, assign problems from theSupplemental Problems booklet.

4 CLOSEActivity

Kinesthetic Have studentgroups hook up two lightbulbs inparallel using a battery and a switchbut no meters. Check the circuits.Now have them place an ammeterin the circuit so that it will measurethe total current. Have them add athird lightbulb and observe the cur-rent increase. Students should thenuse a voltmeter to measure the volt-age across the lightbulbs and seethat the voltage drops are all thesame. In order to measure the cur-rent through each branch, studentsmust open the circuit between theresistor and the junction and insertthe ammeter, making sure that thepolarity is correct. Students willobserve that the current through eachbranch is less than the circuit’s totalcurrent. When the current in allbranches is added together, it equalsthe total current. Have students reducethe circuit to one lightbulb so they cansee that the voltage is still the same,and that the total current is lower thanwhen there were several lightbulbs inthe circuit. COOP LEARNELLL1

LS

L2

COOP LEARN

L1

L1

548

1. a. Bulbs 2 and 3 are equal in brightnessbut dimmer than bulb1.

b. Bulb 1 goes out. The other two are notchanged. I1 goes to zero, I2 remains thesame, and I3 is reduced.

c. Bulb 1 is its original brightness; the othertwo are out. I1 returns to its originalvalue, I2 is zero, and I3 is reduced.

d. Bulb 2 goes out (short circuited). Bulb 3is now as bright as bulb 1; the brightness

of bulb 1 is unchanged. e. Bulb 1 is unchanged. The current in

the end of the path increases, so bulb 2brightens and bulb 3 dims.

2. Student answers may vary. As circuitsand technology improve, as we have witnessed in the computer industry, this drives the need for people in careerscapable of producing and repairingthese devices.

3. Both dim equally. The current in each isreduced by the same amount.

Section Review23.2

549

Content Background

Faulty electrical devices can deliverlethal shocks. An electric heater canbe a potential hazard if the highvoltage becomes connected to thecase because of frayed insulation on

the wires. If a person who is grounded touchesthe case, he or she will complete a path forthe current to the ground. This kind of hazardcan be avoided by the use of a three-way system.

The third prong of a plug is connected by awire to the case of the appliance. The thirdhole in the receptacle is connected to theground. Therefore, if there is faulty wiring inthe appliance, the current is routed throughthe ground and not through the person becausethe case and the ground have essentially noresistance between them.

TECHPREP

Purpose

Students learn the physicsinvolved in the operationof a simple snap-actiontoggle switch.

Background

Electric switches are classified by theway in which their metal contactsare arranged. The switch shown is asingle-pole, single-throw switch thateither starts or stops the currentalong a single path. A pair of single-pole, double-throw switches is usedto operate a light from differentlocations by allowing electrons toflow when the metal contacts inboth switches are connected to thesame wire.

Visual Learning

Have students use a do-it-yourselfbook on household electricity tofind out about the different types ofswitches commonly found in homes.Ask them to generate sketches of atleast three different types of switchesshowing the path of current througheach device.

Teaching Strategies

• Ask students to explain how alight can be turned off and onfrom different locations in a room.Students should be able to deducethat both switches have metalcontacts that can move betweenthe wires of the switch. When themetal contacts are connected to thesame wire, the circuit is completeand the light turns on.

• Have students do a quick inven-tory of the switches they use daily.Examples might include computerswitches, hair dryers, lights, soundequipment, and flashlights.

Thinking Critically

1. A circuit breaker automaticallyopens a circuit when too large acurrent threatens to overheat a system.

2. Metals are good conductors of electricity. Plastic is a goodinsulator.

LS

TECHPREP


Thinking Critically

1. Circuit breakers areautomatic switches.Explain why this is so.

2. Why must the contactsin a switch be metal?Why are switch handlesoften plastic?

Electric SwitchAn electric switch is a device that is used to interrupt,complete, or divert an electrical current in a circuit.Switches are found on everything from hair dryers andtoaster ovens, to calculators and video games, to com-puters and airplane instrument panels. Some switchesare simple mechanical switches. In certain devices,such as computers, however, mechanical switches aretoo slow and are replaced by electronic switches madefrom semiconducting materials.

Probably the most common type of switch is themechanical switch you use to operate the small appli-ances and lights in your home or school. The switchshown below is a snap-action toggle switch typicallyused to turn lights off and on.

OFFOFF

Contacts

Handle

Wall plate

Grounding wire

Incoming current

3

12

4

Insulated casing

5

3

The insulated handle of a snap-action toggle switch can be flippedin either of two positions–off or on.

2. When the handle is in the offposition, as it is in this diagram,metal contacts within the switchare separated, interrupting thepath of the current.

When the handle is in the on posi-tion, the metal contacts, which arelinked by wires, come together tocomplete the circuit.

1

2

5

One wire leads to a grounding loca-tion. Then if a problem occurs withthe wires or switch contacts, the cur-rent is routed to the ground, not tothe person touching the switch.

The switch casing is insulated, soif a wire becomes loose or frayedand touches the casing, currentwill not flow.

4

Review Summary statements andKey Terms with your students.

Reviewing ConceptsSection 23.1

1. When one bulb burns out, thecircuit is incomplete and all ofthe bulbs go out.

2. Each new resistor provides anadditional path for the current.

3. The equivalent resistance willbe less than that of any of theresistors.

4. Appliances in parallel can berun independently of oneanother.

5. In a series circuit, the current isopposed by each resistance inturn. The total resistance is thesum of the resistors. In a parallelcircuit, each resistance providesan additional path for current.The result is a decrease in totalresistance.

6. The amount of current enteringa junction is equal to theamount of current leaving.

Section 23.2

7. The purpose of a fuse is to pre-vent conductors from beingoverloaded with current, caus-ing fires due to overheating. Afuse is simply a short length ofwire that will melt from theheating effect if the currentexceeds a certain maximum.

8. A short circuit is a formed cir-cuit that has extremely lowresistance. A short circuit isdangerous because a low resis-tance across any potential dif-ference will draw a large cur-rent. The heating effect of thecurrent can cause a fire.

9. An ammeter must have lowresistance because it is placeddirectly in the circuit. If itsresistance were high, it wouldchange the total resistance ofthe circuit and thus serve toreduce the current in the circuit, thereby defeating itsown purpose.

550

23 CHAPTER REVIEW

10. A voltmeter is placed in parallel with theportion of the circuit whose difference inpotential is to be measured. A voltmetermust have very high resistance for the samereason that an ammeter has low resistance.If the voltmeter had low resistance, itwould lower the resistance of the portionof the circuit it is across and increase thecurrent in the circuit. This would produce ahigher voltage drop across the part of thecircuit where the voltmeter is located,defeating its purpose.

11. An ammeter is connected in series; a voltmeter is connected in parallel.

VideotapeMindJogger VideoquizzesChapter 23: Series and Parallel CircuitsHave students work in groups as they playthe videoquiz game to review key chapterconcepts.

CHAPTER 23 REVIEW


Key Terms

23.1

• series circuit• equivalent

resistance• voltage divider• parallel circuit

23.2

• fuse• circuit breaker• ground-fault

interrupter• short circuit• combination

series-parallelcircuit

• ammeter• voltmeter

Summary

Key Equations

23.1

Reviewing Concepts

23.1 Simple Circuits

• The current is the same everywhere in a simple series circuit.

• The equivalent resistance of a series circuit is the sum of the resistances ofits parts.

• The sum of the voltage drops acrossresistors in series is equal to the potential difference applied across the combination.

• A voltage divider is a series circuit usedto produce a voltage source from ahigher-voltage battery.

• The voltage drops across all branches of a parallel circuit are the same.

• In a parallel circuit, the total current is equal to the sum of the currents inthe branches.

• The reciprocal of the equivalent resis-tance of parallel resistors is equal to the sum of the reciprocals of the indi-vidual resistances.

• If any branch of a parallel circuit isopened, there is no current in that

branch. The current in the other branches is unchanged.


• A fuse or circuit breaker, placed in serieswith appliances, creates an open circuitwhen dangerously high currents flow.

• A complex circuit is often a combina-tion of series and parallel branches.Any parallel branch is first reduced to asingle equivalent resistance. Then anyresistors in series are replaced by a sin-gle resistance.

• An ammeter is used to measure the cur-rent in a branch or part of a circuit. Anammeter always has a low resistanceand is connected in series.

• A voltmeter measures the potential dif-ference (voltage) across any part orcombination of parts of a circuit. Avoltmeter always has a high resistanceand is connected in parallel with thepart of the circuit being measured.

R � RA � RB � . . . I � �Vso

Rurce� �

R

1� � �

R

1

A� � �

R

1

B� � �

R

1

C�

Section 23.1

1. Why is it frustrating when one bulbburns out on a string of holiday treelights connected in series?

2. Why does the equivalent resistancedecrease as more resistors are addedto a parallel circuit?

3. Several resistors with different values are connected in parallel.

How do the values of the individualresistors compare with the equivalentresistance?

4. Why is household wiring done in parallel instead of in series?

5. Why is there a difference in totalresistance between three 60-� resis-tors connected in series and three 60-� resistors connected in parallel?

23 CHAPTER REVIEW

Applying Concepts12. If one of the lamp filaments

burns out, the current willcease and all the lamps will go out.

13. V2 � VR2/(R1 � R2). As R1increases, V2 will decrease.

14. a. There will be no current inthe resistor; b. The current inthe resistor will remain thesame.

15. If one of the filaments burnsout, the resistance and thepotential difference across theother lamps will not change;therefore, their currents willremain the same.

16. No, the 30-� resistors can beused in parallel. Three 30-�resistors in parallel will give a10-� resistance. Two 30-�resistors in parallel will give a15-� resistance.

17. Connect four of the bulbs inseries. The voltage drop acrosseach will be 6 V/4 � 1.5 V.

18. a. The lamp with the lowerresistance: P � IV and I � V/R,so P � V 2/R. Smaller R meanslarger P. b. The lamp with the higherresistance: P � IV and V � IR,so P � I2R. Larger R meanslarger P.

19. a. series; b. series; c. parallel; d. series; e. parallel; f. series; g. series; h. parallel; i. parallel

20. The 30-A fuse allows more cur-rent to flow through the circuit,generating more heat, whichcan be harmful to devicesincluded in the circuit.

ProblemsComplete solutions for all Chapter Review Problemscan be found in the Problems and Solutions Manualaccompanying this text.

Section 23.1

21. a. 25.0 �; b. 2.00 A; c. 4.00 � 101 V; 1.0 � 101 V; d. 8.00 � 101 W; 2.0 � 101 W

22. a. 6.0 � 101 �; b. 6.0 V

23. a. 1.5 A; b. 2.0 � 101 �24. a. 2.3 � 102 �; b. 13 �; c. 3.6 W25. a. 2.2 � 102 �; b. 66 W; c. increased

LEVEL 1

551

Chapter 23 Review 551

CHAPTER 23 REVIEW

6. Compare the amount of current entering ajunction in a parallel circuit with that leavingthe junction. Note: A junction is a point wherethree or more conductors are joined.

Section 23.2

7. Explain the function of a fuse in an electric circuit.

8. What is a short circuit? Why is a short circuitdangerous?

9. Why does an ammeter have a very low resistance?

10. Why does a voltmeter have a very high resistance?

11. How does the way in which an ammeter is con-nected in a circuit differ from the way a volt-meter is connected?

Applying Concepts12. What happens to the current in the other two

lamps if one lamp in a three-lamp series circuitburns out?

13. Suppose that in the voltage divider inFigure 23–4, the resistor RA is made to be avariable resistor. What happens to the voltageoutput, VB, of the voltage divider if the resis-tance of the variable resistor is increased?

14. Circuit A contains three 60-� resistors in series.Circuit B contains three 60-� resistors in paral-lel. How does the current in the second 60-�resistor change if a switch cuts off the currentto the first 60-� resistor ina. circuit A?b. circuit B?

15. What happens to the current in the other twolamps if one lamp in a three-lamp parallel cir-cuit burns out?

16. An engineer needs a 10-� resistor and a 15-�resistor. But there are only 30-� resistors instock. Must new resistors be purchased?Explain.

17. If you have a 6-V battery and many 1.5-V bulbs,how could you connect them so that they light but do not have more than 1.5 V acrosseach bulb?

18. Two lamps have different resistances, one largerthan the other.

a. If they are connected in parallel, which isbrighter (dissipates more power)?

b. When connected in series, which is brighter?19. For each of the following, write the form of

circuit that applies: series or parallel.a. The current is the same throughout.b. The total resistance is equal to the sum of

the individual resistances.c. The voltage drop is the same across each

resistor.d. The voltage drop is proportional to the

resistance.e. Adding a resistor decreases the total

resistance.f. Adding a resistor increases the total

resistance.g. If the current through one resistor goes to

zero, there is no current in the entire circuit.h. If the current through one resistor goes to

zero, the current through all other resistorsremains the same.

i. This form is suitable for house wiring.20. Why is it dangerous to replace a 15-A fuse in a

circuit with a fuse of 30 A?

ProblemsSection 23.1

21. A 20.0-� lamp and a 5.0-� lamp are connectedin series and placed across a potential differ-ence of 50.0 V. What isa. the equivalent resistance of the circuit?b. the current in the circuit?c. the voltage drop across each lamp?d. the power dissipated in each lamp?

22. The load across a battery consists of two resis-tors, with values of 15 � and 45 �, connectedin series.a. What is the total resistance of the load?b. What is the voltage of the battery if the cur-

rent in the circuit is 0.10 A?23. A lamp having a resistance of 10.0 � is con-

nected across a 15-V battery.a. What is the current through the lamp?b. What resistance must be connected in series

with the lamp to reduce the current to0.50 A?

Texas TEKS

Pages 548–549: 3(D), 6(E) Pages 550–551: 3(B), 6(E)

26. a. 0.63 A; b. 190 �; c. 210 �27. a. 57 V; 63 V; b. 17 W; c. 19 W28. a. 8.89 �; b. 4.50 A; c. 2.50 A

29. 66 �30. 2.00 � 102 �31. a. 52 �; b. 110 V; c. 9.8 �;

d. 96 V

Section 23.2

32. a. 2.0 A; b. 3.0 A; c. 15 A33. Yes34. 22.5 A; 0.750 A35. 50.6 W; 16.9 W

36a. Refer to Problems andSolutions Manual.

b. 1.0 � 103 s

LEVEL 2

LEVEL 1

LEVEL 2

552

23 CHAPTER REVIEW


CHAPTER 23 REVIEW

24. A string of 18 identical holiday tree lights isconnected in series to a 120-V source. Thestring dissipates 64.0 W.a. What is the equivalent resistance of the

light string?b. What is the resistance of a single light?c. What power is dissipated by each lamp?

25. One of the bulbs in problem 24 burns out. The lamp has a wire that shorts out the lampfilament when it burns out. This drops theresistance of the lamp to zero.a. What is the resistance of the light string now?b. Find the power dissipated by the string.c. Did the power go up or down when a bulb

burned out?26. A 75.0-W bulb is connected to a 120-V source.

a. What is the current through the bulb?b. What is the resistance of the bulb?c. A lamp dimmer puts a resistance in series

with the bulb. What resistance would beneeded to reduce the current to 0.300 A?

27. In problem 26, you found the resistance of alamp and a dimmer resistor.a. Assuming that the resistances are constant,

find the voltage drops across the lamp andthe resistor.

b. Find the power dissipated by the lamp.c. Find the power dissipated by the dimmer

resistor.28. A 16.0-� and a 20.0-� resistor are connected in

parallel. A difference in potential of 40.0 V isapplied to the combination.a. Compute the equivalent resistance of the

parallel circuit.b. What is the current in the circuit?c. How large is the current through the

16.0-� resistor?29. Amy needs 5.0 V for some integrated circuit

experiments. She uses a 6.0-V battery and tworesistors to make a voltage divider. One resistoris 330 �. She decides to make the other resistorsmaller. What value should it have?

30. Pete is designing a voltage divider using a 12.0-V battery and a 100.0-� resistor as RB.What resistor should be used as RA if the output voltage across RB is to be 4.00 V?

31. A typical television dissipates 275 W when it isplugged into a 120-V outlet.

a. Find the resistance of the television.b. The television and 2.5-� wires connecting

the outlet to the fuse form a series circuitthat works like a voltage divider. Find thevoltage drop across the television.

c. A 12-� hair dryer is plugged into the sameoutlet. Find the equivalent resistance of thetwo appliances.

d. Find the voltage drop across the televisionand hair dryer. The lower voltage explainswhy the television picture sometimes shrinkswhen another appliance is turned on.

Section 23.2

32. A circuit contains six 240-� lamps (60-W bulbs)and a 10.0-� heater connected in parallel. Thevoltage across the circuit is 120 V. What is thecurrent in the circuita. when four lamps are turned on?b. when all lamps are on?c. when six lamps and the heater are operating?

33. If the circuit in problem 32 has a fuse rated at12 A, will the fuse melt if everything is on?

34. Determine the reading of each ammeter andeach voltmeter in Figure 23–15.

35. Determine the power used by each resistanceshown in Figure 23–15.

36. During a laboratory exercise, you are suppliedwith a battery of potential difference V, twoheating elements of low resistance that can beplaced in water, an ammeter of very small resis-tance, a voltmeter of extremely high resistance,wires of negligible resistance, a beaker that iswell insulated and has negligible heat capacity,and 100.0 g of water at 25˚C.a. By means of a diagram and standard symbols,

show how these components should be con-nected to heat the water as rapidly as possible.

45.0 V

10 �

30 �

30 �

30 �l1

l2

A

V1

A

V2 V3

FIGURE 23–15

Texas TEKS

Pages 552–553: 3(B), 6(E)

23 CHAPTER REVIEW

37. a. 240 �; b. 0.50 A; c. 6.0 � 101 W

38. a. 5.9 �; b. 21 A; c. 11 V; d. 110 V; Yes

Critical Thinking ProblemsComplete solutions for all ChapterReview Critical Thinking Problems canbe found in the Problems and SolutionsManual accompanying this text.39. Refer to Problems and

Solutions Manual.40. a. 0.134 A; b. 0.395 W;

c. 0.409 W

Going Furthera. 6000 �b. 6000 �c. 18 000 �d. 2000 �

553

Program Resources

Chapter Assessment, pp. 107–110 TestCheck Software, Chapter 23 MindJogger Videoquizzes, Chapter 23

Alternate Assessment in the ScienceClassroom

Performance Assessment in the ScienceClassroom

Supplemental Problems, Chapter 23 L2

L1

ELLL1

L1

L1

PHYSICSBe sure to check the GlencoeScience Web site for links to chapter material: science.glencoe.com

Chapter 23 Review 553

b. If the voltmeter reading holds steady at50.0 V and the ammeter reading holdssteady at 5.0 A, estimate the time in secondsrequired to completely vaporize the water inthe beaker. Use 4200 J/kg˚C as the specificheat of water and 2.3 � 106 J/kg as the heatof vaporization of water.

37. A typical home circuit is shown in Figure 23–16.The lead lines to the kitchen lamp each has avery low resistance of 0.25 �. The lamp has aresistance of 240.0 �. Although it is a parallelcircuit, the lead lines are in series with each ofthe components of the circuit.

a. Compute the equivalent resistance of the cir-cuit consisting of just the light and the leadlines to and from the light.

b. Find the current to the bulb.c. Find the power dissipated in the bulb.

38. A power saw is operated by an electric motor.When electric motors are first turned on, theyhave a very low resistance. Suppose that akitchen light in problem 37 is on and a powersaw is turned on. The saw and lead lines havean initial total resistance of 6.0 �.a. Compute the equivalent resistance of the

light-saw parallel circuit.b. What is the total current flowing in the circuit?c. What is the total voltage drop across the

two leads to the light?d. What voltage remains to operate the light?

Will this cause the light to dim temporarily?

Critical Thinking Problems39. A 50-200-250-W three-way bulb has three ter-

minals on its base. Sketch how these terminalscould be connected inside the bulb to providethe three brightnesses. Explain how to connect120 V across two terminals at a time to obtaina low, medium, and high level of brightness.

40. Batteries consist of an ideal source of potentialdifference in series with a small resistance. Theelectrical energy of the battery is produced bychemical reactions that occur in the battery.However, these reactions also result in a smallresistance that, unfortunately, cannot be com-pletely eliminated. A flashlight contains twobatteries in series. Each has a potential differ-ence of 1.50 V and an internal resistance of0.200 �. The bulb has a resistance of 22.0 �. a. What is the current through the bulb? b. How much power does the bulb dissipate? c. How much greater would the power be if thebatteries had no internal resistance?

Going FurtherUsing What You’ve Learned An ohmmeter ismade by connecting a 6.0-V battery in series withan adjustable resistor and an ideal ammeter. Theammeter deflects full-scale with a current of1.0 mA. The two leads are touched together andthe resistance is adjusted so that 1.0 mA flows.a. What is the resistance of the adjustable resistor?b. The leads are now connected to an unknown

resistance. What resistance would produce a cur-rent of half-scale, 0.50 mA?

c. What resistance would produce a reading ofquarter-scale, 0.25 mA?

d. What resistance would produce a reading ofthree-quarters-scale, 0.75 mA?

CHAPTER 23 REVIEW

Switch box 120 V Wall outlets

Power saw

0.25 �

0.25 �

Kitchen light240 �

FIGURE 23–16

Extra Practice For more practice solving problems, go to Extra Practice Problems, Appendix B.

PHYSICSTo review content, do the interactive quizzes on theGlencoe Science Web site atscience.glencoe.com



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