CHAPTER 18 Instruction and Intervention Support
Circuits and Circuit Elements
1 Core Instruction
Chapter Resources
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18.1
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18.2
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18.3
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626A Chapter 18
Labs and Demonstrations
Go Online for Materials Lists and Solutions Lists.
Textbook: Schematic Diagrams and Circuits
Visual Concepts: Schematic Diagram and Common
Symbols • EMF • Internal Resistance, EMF, and
Terminal Voltage
Teaching Visuals: Schematic Diagram Symbols •
Light Bulb
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Visual Concepts: Current Through Resistors in Series
• Equation for Equivalent Resistance for Resistors in
Series • Equation for Potential Difference Across
Resistors in Series • Equivalent Resistance for
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Resistance for Resistors in Parallel • Equation for
Equivalent Resistance for Resistors in Parallel
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Lab: Resistors in Series and in Parallel
Lab: Resistors in Series and in Parallel (Probeware)
Lab: Series and Parallel Circuits (Probeware)
Textbook: Complex Resistor Combinations
Visual Concepts: Comparing Resistors in Series and
in Parallel • Fuse • Analysis of Complex Circuits
Teaching Visuals: Finding Equivalent Resistance •
Components of a Decorative Light Bulb
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Sample Problem Set II
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Circuits and Circuit Elements 626B
CHAPTER 18
Chapter Overview
Section 1 introduces the concept of an
electric circuit, distinguishes between
open and closed circuits, and describes
the concept of a short circuit.
For strings of decorative lights—
such as these that illuminate the
Riverwalk in San Antonio, Texas—
two types of electric circuits can be
used. In a series circuit, illustrated
on the left, the entire set goes dark
when one bulb is removed from
the circuit. In a parallel circuit,
illustrated on the right, other bulbs
remain lit even when one or more
bulbs are removed.
Section 2 describes the relationships
between equivalent resistance, current,
and potential difference for series
circuits and parallel circuits.
Series circuit
Parallel circuit
Section 3 explores complicated circuits
containing portions in series and
portions in parallel.
About the Image
(c) ©Laurence Parent
The Riverwalk in San Antonio, Texas is a
downtown shopping and entertainment
district built on the banks of the San
Antonio River. The Riverwalk began as a
Works Progress Administration project in
the Great Depression of the 1930s. In the
1970s and 1980s, redevelopment and
expansion of the Riverwalk sparked
an economic revival of downtown
San Antonio.
Lab 626
Preview
The following investigations support the
concepts
presented in this chapter:
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626
Labs
Exploring Circuit Elements
Resistors in Series and in Parallel
Resistors in Series and in Parallel (Probeware)
Series and Parallel Circuits (Probeware)
Design a Circuit (STEM)
626 Chapter 18
QuickLabs
Simple Circuits
Series and Parallel Circuits
Demonstrations
Resistors in Series
Resistors in Parallel
5/26/2011 7:11:50 AM
CHAPTER 18
Circuits
and Circuit
Elements
SECTION 1
Schematic
Diagrams and
Circuits
Focus and Motivate SECTION 2
Activate Prior
Knowledge
Resistors in
Series or in
Parallel
SECTION 3
Complex Resistor
Combinations
Why It Matters
All electric circuits are
wired in series, parallel,
or a combination. The
type of circuit affects the
current and potential
difference of elements
connected to the circuit,
such as decorative light
bulbs on strands or
appliances in your home.
Knowledge to Review
•Potential difference is the change in
electrical potential energy per unit
charge from one point to another.
•Current is the rate at which electric
charges move through a given area.
•ΔV = IR can be used to relate current
and potential difference for specific
electrical devices.
Items to Probe
•Preconceptions about wiring: Ask
students to trace the path of charges
moving in a string of decorative lights.
•Familiarity with schematic
diagrams: Ask students to draw their
own picture representing the wiring of
decorative lights.
ONLINE Physics
HMDScience.com
(br) ©Photodisc/Getty Images
ONLINE LABS
Exploring Circuit Elements
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Resistors in Series and in
Parallel
Series and Parallel Circuits
Design a Circuit
Why It Matters
Connecting to History
Italian physicist Alessandro Volta was the first
person to notice that electric current, like a
water current, needs a medium in which to be
circulated and controlled. He was able to
design a system of conducting elements to
control the flow and path of an electric
current. He invented the first electric battery
in 1800. Volta’s battery included metal discs,
made of metals such as copper and zinc,
PREMIUM
CONTENT
Physics
HMDScience.com
Resistors in Circuits
627
separated by cardboard and connected in a
After briefly explaining this history to
salt solution. Volta’s device was used for
5/26/2011 7:11:58 AM students, ask them to imagine what their lives
electroanalysis and eventually led to the
would be like without electricity. Would
discovery of many new chemical elements.
anything be the same? What would their
science classroom be like?
Since Volta’s time, electric circuits have
become integrated into all aspects of
science and technology. Today they are
everywhere in every electrical or electronic
device we use in our daily lives. The importance of circuits in modern life is undeniable.
Circuits and Circuit Elements 627
SECTION 1
Plan and Prepare
Preview Vocabulary
Latin Word Origins The word
schematic comes from the Latin word
schema, meaning “figure.” This word is
used in technology and science for a
diagram or blueprint, especially of an
electric circuit.
Teach
TEACH FROM VISUALS
SECTION 1
Objectives
Interpret and construct circuit
diagrams.
Identify circuits as open or
closed.
Deduce the potential difference
across the circuit load, given
the potential difference across
the battery’s terminals.
Schematic Diagrams
and Circuits
Key Terms
schematic diagram
Schematic Diagrams
Take a few minutes to examine the battery and light bulb in Figure 1.1(a);
then draw a diagram of each element in the photograph and its connection.
How easily could your diagram be interpreted by someone else? Could the
elements in your diagram be used to depict a string of decorative lights,
such as those draped over the trees of the San Antonio Riverwalk?
schematic diagram a representation
of a circuit that uses lines to represent
wires and different symbols to represent components
FIGURE 1.1 Students should be
encouraged to create alternative
representations of the circuit shown in
the photo. Students should discuss what
their symbols stand for, how convenient
their symbols are for others to use, and
in what way each symbol reflects
relevant information.
TEACH FROM VISUALS
FIGURE 1.1 Students should recognize
that the straight-line symbols connecting the battery symbol with the bulb
symbol in (b) represent not only the
wire but also all parts of the conducting
connection between the bulb and
battery.
Ask Identify the parts of the photo
symbolized by the black straight lines
in the diagrams.
Answer: The black lines symbolize the
conducting path provided by the wires,
clips, and socket.
628 Chapter 18
A diagram that depicts the construction of an electrical apparatus is
called a schematic diagram. The schematic diagram shown in Figure 1.1(b)
uses symbols to represent the bulb, battery, and wire from Figure 1.1(a).
Note that these same symbols can be used to describe these elements in
any electrical apparatus. This way, schematic diagrams can be read by
anyone familiar with the standard set of symbols.
Reading schematic diagrams allows us to determine how the parts in
an electrical device are arranged. In this chapter, you will see how the
arrangement of resistors in an electrical device can affect the current in
and potential difference across the other elements in the device. The
ability to interpret schematic diagrams for complicated electrical equipment is an essential skill for solving problems involving electricity.
Ask Identify information about the
group of elements that is not relevant
to its function and is unnecessary in
a schematic.
Answer: The colors and sizes of the
items shown and whether the wires are
coiled, bent, or straight are irrelevant to
the function of the group of elements.
electric circuit
FIGURE 1.1
As shown in Figure 1.2 on the next page, each element used in a piece
of electrical equipment is represented by a symbol in schematic diagrams
that reflects the element’s construction or function. For example, the
schematic-diagram symbol that represents an
open switch resembles the open knife switch
that is shown in the corresponding photograph. Note that Figure 1.2 also includes
other forms of schematic-diagram symbols; these alternative symbols will not be
used in this book.
(b)
A Battery and Light
Bulb (a) When this battery is
connected to a light bulb, the
potential difference across the
battery generates a current that
illuminates the bulb. (b) The
connections between the light bulb
and battery can be represented in
a schematic diagram.
(a)
628
Chapter 18
Differentiated
Instruction
Below Level
Students may confuse schematic diagrams
with other diagrams, such as geometric or
architectural diagrams. Point out that
schematic diagrams are diagrams in which the
elements and components of a system, such
as an electric circuit or an electric motor, are
illustrated by previously defined symbols and
icons rather than by their real pictures. Tell
students that in an electric schematic diagram,
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for example, in this book, a capacitor is shown
by two horizontal Ts positioned head to head 5/26/2011
and a resistor is shown by a squiggly wire.
Point out or draw the schematic illustrations
of a capacitor and a resistor.
7:13:21 AM
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FIGURE 1.2
SCHEMATIC DIAGRAM SYMBOLS
Component
Symbol used
in this book
Other forms
of this symbol
Wire or
conductor
TEACH FROM VISUALS
Explanation
• Wires that connect elements are
conductors.
• Because wires offer negligible resistance, they are represented by straight
lines
(a)
Resistor or
circuit load
• Resistors are shown having multiple
bends, illustrating resistance to the
movement of charges.
(b)
Bulb or lamp
(d)
(e)
Plug
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(f)
• The plug symbol looks like a container
for two prongs.
• The emf between the two prongs of a
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plug is symbolized by lines of unequal
PH99PE-C20-001-005-A
(g)
length.
Battery
DTSI Graphics
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Multiple cells
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(i)
(h)
Switch
(j)
Capacitor
• Differences in line length indicate a
potential difference between positive
and negative terminals of the battery.
DTSI Graphics
• The longer line represents the positive
HRW • Holt Physics
terminal
of
the battery.
PH99PE-C20-001-005-A
Ask Challenge students to identify
which devices have the following
functions: storing energy, transforming
energy, and conducting current.
Answer: Batteries and capacitors store
energy; Resistors, bulbs, and batteries
transform energy; Wires, resistors, bulbs,
plugs, closed switches, and batteries
conduct current.
The Language of
Physics
Although Figure 1.2 contains several
schematic-diagram symbols, several
stylistic variations exist. For example,
some other symbols for light bulbs are
shown below.
• The small circles indicate the two places
where the switch makes contact with
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• Holt
Physics
the wires.
switches
work by
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• Holt• Physics
only one of the contacts,
PH99PE-C20-001-005-A not both.
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Open
Open
• The multiple bends of the filament
indicate that the light bulb behaves
as a resistor.
• The symbol for the filament of the bulb is
often enclosed in a circle to emphasize
the enclosure of a resistor in a bulb.
FIGURE 1.2 Be sure students recognize
that the different symbols represent
devices with different functions.
Closed
(k)
Closed
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(l)
(m)
• The two parallel plates of a capacitor
are symbolized by two parallel lines of
equal length.
• One curved line indicates that the
capacitor can be used with only direct
current sources with the polarity
as shown.
Because light bulbs behave as resistors
HRW •in
Holt
Physics the
for small changes
voltage,
PH99TE-C20-001-001-A
symbols for resistors are often used
for light bulbs.
DTSI Graphics
Holt Physics
HRW HRW
• Holt• Physics
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Below Level
Create schematic diagrams that students can
mark up. Create your own or make copies of
schematic diagrams shown in this chapter. As
practice, have students place one of the
following labels on each of the symbols shown
in a diagram:
W: wire or connection
R: resistor or circuit load
Bu: bulb or lamp
Circuits and Circuit Elements
P: plug
Ba: battery
S: switch
C: capacitor
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5/26/2011 7:13:22 AM
Circuits and Circuit Elements 629
Electric Circuits
FIGURE 1.3
Teach continued
Answers
Conceptual Challenge
1. Because there is no potential
difference between the bird’s feet,
there is no current in the bird’s body.
2. At first there is no potential difference between the parachutist’s
hands, so there is no current
in the parachutist’s body. If the
parachutist’s feet touch the ground
and the parachutist continues to
hold onto the wire, however, there
will be current in the parachutist’s
body because of the potential
difference between the wire in the
parachutist’s hands and the ground.
A Complete Circuit When all
electrical components are connected,
charges can move freely in a circuit.
The movement of charges in a circuit
can be halted by opening the switch.
Think about how you get the bulb in Figure 1.3 to light up. Will the bulb
stay lit if the switch is opened? Is there any way to light the bulb without
connecting the wires to the battery?
The filament of the light bulb acts as a resistor. When a wire connects
the terminals of the battery to the light bulb, as shown in Figure 1.3,
charges built up on one terminal of the battery have a path to follow to
reach the opposite charges on the other terminal. Because there are
charges moving through the wire, a current exists. This current causes
the filament to heat up and glow.
Together, the bulb, battery, switch, and wire form an electric circuit.
An electric circuit is a path through which charges can flow. A schematic
diagram for a circuit is sometimes called a circuit diagram.
electric circuit a set of electrical
components connected such that they
provide one or more complete paths for
the movement of charges
Any element or group of elements in a circuit that dissipates energy is
called a load. A simple circuit consists of a source of potential difference
and electrical energy, such as a battery, and a load, such as a bulb or
group of bulbs. Because the connecting wire and switch have negligible
resistance, we will not consider these elements as part of the load.
In Figure 1.3, the path from one battery terminal to the other is complete,
a potential difference exists, and electrons move from one terminal to the
other. In other words, there is a closed-loop path for electrons to follow.
This is called a closed circuit. The switch in the circuit in Figure 1.3 must be
closed in order for a steady current to exist.
Without a complete path, there is no charge flow and therefore no
current. This situation is an open circuit. If the switch in Figure 1.3 were
open, as shown in Figure 1.2, the circuit would be open, the current would
be zero, and the bulb would not light up.
Bird on a Wire Why is it possible for a bird to
be perched on a high-voltage wire without being
electrocuted? (Hint: Consider the potential difference between the bird’s two feet.)
Parachutist on a Wire Suppose a parachutist
lands on a high-voltage wire and grabs the wire
in preparation to be rescued. Will the parachutist
be electrocuted? If the wire breaks, why should
the parachutist let go of the wire as it falls to
the ground? (Hint: First consider the potential
difference between the parachutist’s two hands
holding the wire. Then consider the potential
difference between the wire and the ground.)
(b) ©blickwinkel/Alamy
Conceptual Challenge
630
Chapter 18
Differentiated
Instruction
Inclusion
Students with kinesthetic learning styles may
benefit from using a fluid model for electric
current in a circuit. In this model, charges
moving due to potential difference are
analogous to water moving to a level of lower
gravitational potential energy. Wires are
analogous to horizontal pipes, and resistors are
analogous to water wheels, which transform
the energy to another form. Batteries and
generators act like pumps in that they lift
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630 Chapter 18
water up, increasing its potential energy. If
possible, have kinesthetic learners build a fluid 5/26/2011
model of a basic circuit.
7:13:25 AM
Why It Matters
CFLs and LEDs
CFLs and LEDs
Many electrical products, such as
decorative lights, extension cords, and
appliances, have a prominent tag labeled
“UL.” This mark, from Underwriters
Laboratories, indicates that the product
has been tested by UL engineers for
electrical, fire, and other hazards.
T
he most familiar of light bulbs, incandescent
bulbs, may soon be a relic of the past. Thomas
Edison first invented these bulbs in 1879 and they
have been in use ever since. They work by heating a
small metal filament that glows and produces light.
Although incandescent bulbs give off very warm and
pleasant light, they are extremely inefficient. Nearly 90%
of the energy they use is converted into heat and only
10% is converted into light. New federal law requires that
by 2014 all bulbs be at least 30% more efficient. Two new
types of bulbs look to replace incandescent bulbs.
(br) ©Gustoimages/Photo Researchers, Inc.; (tr) ©GIPhotoStock/Photo Researchers, Inc.
The first type of light bulb is called compact fluorescent
light (or CFL for short). CFLs work by running an electrical
current through a tube that contains a mixture of gases.
The atoms of gas absorb energy from the electricity and
emit ultraviolet light. Humans, however, cannot see
ultraviolet light. What happens next is that the ultraviolet
light hits the surface of the tube that has been coated
with a chemical that absorbs the ultraviolet light and
emits visible light.
Untitled-699 631
electrons here release no energy, so LEDs are more
energy efficient than both incandescent and CFLs. In
addition, because LEDs are made of solid material, they
can be very small and are very durable so they last a
long time.
Although both CFLs and LEDS cost considerably more
than incandescent bulbs, they use much less energy to
produce the same amount of light. In addition, they have
a much longer life span. When both of these factors are
taken into account, replacing your incandescent bulbs
with CFLs or LEDs may cost more up front, but they end
up saving money over the life of the bulb.
The second type of light bulb is called light-emitting
diode (or LED for short). LEDs work by moving electrons
and protons in a solid piece of material called a
semiconductor. As the electrons move through this
material they lose energy and release light. The
Short circuits can be hazardous.
Without a load, such as a bulb or other resistor, the circuit contains little
resistance to the movement of charges. This situation is called a short
circuit. For example, a short circuit occurs when a wire is connected from
one terminal of a battery to the other by a wire with little resistance. This
commonly occurs when uninsulated wires connected to different terminals come into contact with each other.
When short circuits occur in the wiring of your home, the increase in
current can become unsafe. Most wires cannot withstand the increased
current, and they begin to overheat. The wire’s insulation may even melt
or cause a fire.
Circuits and Circuit Elements
Below Level
Because batteries are said to run down, many
students believe that current is consumed by a
circuit. To check for this misconception, ask
students to draw arrows representing the
current in a simple circuit.
Some may believe that current is used up in
the resistor. Their diagrams will show charges
moving only from the battery to the bulb.
Others may think that the current comes back
to the battery but has decreased in magnitude.
631
Arrows representing current in their diagrams
may get smaller after the resistor. 5/26/2011 7:13:29 AM
Point out that the number of charges
entering a part of the circuit in some time
interval equals the number of charges leaving it
in the same time interval.
Explain that the chemicals in the battery
react to produce a potential difference.
Eventually, most of these reacting chemicals are
converted to other substances, and the battery
no longer produces a potential difference.
Circuits and Circuit Elements 631
Teach continued
The Language of
Physics
The term emf originally stood for
electromotive force. This term may be
misleading because emf is not a force.
Rather, it refers to a potential difference
measured in volts. The voltage value on
a battery label denotes its emf.
In this text, internal resistance will be
disregarded unless specifically noted.
The value of the terminal voltage, ΔV,
can be found from the emf (ε ), the total
current (I ), and the internal resistance (r )
with the following equation:
ΔV = ε − Ir
QuickLab
The source of potential difference and electrical energy is the
circuit’s emf.
Will a bulb in a circuit light up if you remove the battery? Without a
potential difference, there is no charge flow and no current. The battery
is necessary because the battery is the source of potential difference and
electrical energy for the circuit. So, the bulb must be connected to the
battery to be lit.
MATERIALS
• 1 miniature light bulb
• 1 D-cell battery
• wires
• rubber band or tape
Any device that increases the potential energy of charges circulating
in a circuit is a source of emf, or electromotive force. The emf is the energy
per unit charge supplied by a source of electric current. Think of such a
source as a “charge pump” that forces electrons to move in a certain
direction. Batteries and generators are examples of emf sources.
SAFETY
Do not perform this lab
with any batteries or
electrical devices other
than those listed here.
Never work with electricity near
water. Be sure the floor and all work
surfaces are dry.
For conventional current, the terminal voltage is less than the emf.
Look at the battery attached to the light bulb in the circuit shown in
Figure 1.4. As shown in the inset, instead of behaving only like a source
of emf, the battery behaves as if it contains both an emf source and a
resistor. The battery’s internal resistance to current is the result of moving
charges colliding with atoms inside the battery while the charges are
traveling from one terminal to the other. Thus, when charges move
conventionally in a battery, the potential difference across the battery’s
terminals, the terminal voltage, is actually slightly less than the emf.
SIMPLE CIRCUITS
Connect the bulb to the battery
using two wires, using a rubber
band or tape to hold the wire
to the battery. Once you have
gotten the bulb to light, try
different arrangements to see
whether there is more than one
way to get the bulb to light.
Can you make the bulb light
using just one wire? Diagram
each arrangement that you try,
and note whether it produces
light. Explain exactly which
parts of the bulb, battery, and
wire must be connected for the
light bulb to produce light.
Unless otherwise stated, any reference in this book to the potential
difference across a battery should be thought of as the potential difference measured across the battery’s terminals rather than as the emf of the
battery. In other words, all examples and end-of-chapter problems will
disregard the internal resistance of the battery.
FIGURE 1.4
Teacher’s Notes
To light the bulb, students should
connect the bottom of the bulb to one
terminal of the battery and the side of
the bulb’s base to the other terminal.
The bulb can be lit with one wire by
holding the base of the bulb to one of
the battery’s terminals and using the
wire to connect the side of the bulb’s
base to the other terminal
A Battery’s Internal Resistance (a) A battery in a circuit behaves
as if it contains both (b) an emf source and (c) an internal resistance. For
simplicity’s sake, in problem solving it will be assumed that this internal
resistance is insignificant.
(a)
(c)
(b)
632
Chapter 18
Differentiated
Instruction
Small internal
resistance
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PH99PE-C20-001-008a-A
Below Level
Comparing electromotive force with voltage
drop from the electrons' perspective could
provide students with a basis for understanding
fluctuations in potential energy. Point out that
raising the potential energy of electrons in a
source yields electromotive force, while
decreasing the potential energy of electrons in
a load results in a voltage drop.
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5/26/2011 7:13:30 AM
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Potential difference across a load equals the terminal voltage.
Key Models and
Analogies
When charges move within a battery from one terminal to the other,
the chemical energy of the battery is converted to the electrical potential
energy of the charges. As charges move through the circuit, their
electrical potential energy is converted to other forms of energy. For
instance, when the load is a resistor, the electrical potential energy of the
charges is converted to the internal energy of the resistor and dissipated
as thermal energy and light energy.
From an energy-transformation
perspective, think of batteries as
electrical-energy-supply devices and of
resistors and light bulbs as electricalenergy-consuming devices. The electric
current conveys this energy from the
battery to the resistor.
Because energy is conserved, the energy gained and the energy lost
must be equal for one complete trip around the circuit (starting and
ending at the same place). Thus, the electrical potential energy gained in
the battery must equal the energy dissipated by the load. Because the
potential difference is the measurement of potential energy per amount
of charge, the potential increase across the battery must equal the
potential decrease across the load.
Assess and Reteach SECTION 1 FORMATIVE ASSESSMENT
Reviewing Main Ideas
Assess Use the Formative Assessment
on this page to evaluate student
mastery of the section.
FIGURE 1.5
1. Identify the types of elements in the schematic diagram
illustrated in Figure 1.5 and the number of each type.
Reteach For students who need
additional instruction, download the
Section Study Guide.
2. Using the symbols listed in Figure 1.2, draw a schematic
diagram of a working circuit that contains two resistors,
an emf source, and a closed switch.
3. In which of the circuits pictured below will there be
no current?
FIGURE 1.6
FIGURE 1.7
FIGURE 1.8
FIGURE 1.9
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Response to Intervention To reassess
students’ mastery, use the Section Quiz,
available to print or to take directly
online at HMDScience.com.
HRW • Holt Physics
PH99PE-C20-001-009-A
HRW • Holt Physics
PH99PE-C20-001-011-A
4. If the potential difference across the bulb in a certain flashlight is 3.0 V,
HRW • Holt Physics
what is the potential difference across the combination of batteries
used
PH99PE-C20-001-013-A
HRW • Holt Physics
to power it? PH99PE-C20-001-012-A
Critical Thinking
5. In what forms is the electrical energy that is supplied to a string of decorative lights dissipated?
Answers to Section Assessment
1. one battery, one closed switch, two
resistors, and three bulbs
2. Students' diagrams should include the
circuit elements as they appear in Figure
1.2.
3. Figure 1.7 and Figure 1.9 will have no
current in them.
4. 3.0 V
5. It is converted to thermal energy and
light energy.
Circuits and Circuit Elements
633
5/26/2011 7:13:30 AM
Circuits and Circuit Elements 633
Transistors and
Integrated Circuits
The branch of physics that studies the
properties of semiconductors and
related technologies is called
solid-state physics.
The transistor was invented at Bell
Labs in 1947. The integrated circuit was
invented by Jack Kilby of Texas
Instruments in 1958. In 1959, Robert
Noyce received a patent for the
silicon-based integrated circuit. Noyce
later founded Intel, the company
responsible for the creation of the
microprocessor. The invention of the
integrated circuit brought about an
enormous boom in technology because
large, complex circuits could be contained in a small area. The microprocessor chip in a typical personal computer
contains tens of millions of transistors.
WHY IT MATTERS
Transistors and
Integrated Circuits
Y
ou may have heard about objects called
semiconductors. Semiconductors are materials that
have properties between those of insulators and
conductors. They play an important role in today’s world, as
they are the foundation of circuits found in virtually every
electronic device.
Most commercial semiconductors are made primarily of
either silicon or germanium. The conductive properties of
semiconductors can be enhanced by adding impurities to the
base material in a process called doping. Depending on how
a semiconductor is doped, it can be either an n-type
semiconductor or a p-type semiconductor. N-type
semiconductors carry negative charges (in the form of
electrons), and p-type semiconductors carry positive charges.
The positive charges in a p-type semiconductor are not
actually positively charged particles. They are “holes” created
by the absence of electrons.
The most interesting and useful properties of semiconductors
emerge when more than one type of semiconductor is used in
a device. One such device is a diode, which is made by placing
a p-type semiconductor next to an n-type semiconductor. The
junction where the two types meet is called a p-n junction.
A diode has almost infinite resistance in one direction and
Motherboards, such as the one pictured
above, include multiple transistors.
nearly zero resistance in the other direction. One useful
application of diodes is the conversion of alternating current to
direct current.
A transistor is a device that contains three layers of
semiconductors. Transistors can be either pnp transistors or
npn transistors, depending on the order of the layers.
A transistor is like two diodes placed back-to-back. You
might think this would mean that no current exists in a
transistor, as there is infinite resistance at one or the other of
the p-n junctions. However, if a small voltage is applied to
the middle layer of the transistor, the p-n junctions are
altered in such a way that a large amount of current can be
in the transistor. As a result, transistors can be used as
switches, allowing a small current to turn a larger current on
or off. Transistor-based switches are the building blocks of
computers. A single switch turned on or off can represent a
binary digit, or bit, which is always either a one or a zero.
An integrated circuit is a collection of transistors, diodes,
capacitors, and resistors embedded in a single piece of
silicon, known as a chip. Much of the rapid progress in the
computer and electronics industries in the past few decades
has been a result of improvements in semiconductor
technologies. These improvements allow smaller and smaller
transistors and other circuit elements to be placed on chips.
A typical computer motherboard, such as the one shown
here, contains several integrated circuits, each one
containing several million transistors.
(tr) ©Ariel Skelley/Getty Images; (bl) ©Mitch Hrdlicka/Photodisc/Getty Images
W h y i t M at t e r s
634
Differentiated
Instruction
Below Level
Students may not be able to figure out the
position of semiconductors in comparison with
insulators and conductors. Explain that semiconductors are devices between insulators and
conductors. They are used in integrated circuits
and in devices called transistors, by which we
can switch or amplify electronic signals.
Untitled-703 634
634 Chapter 18
5/26/2011 7:16:18 AM
ntitled-700 635
SECTION 2
SECTION 2
Resistors in Series
or in Parallel
Key Terms
series
parallel
Resistors in Series
In a circuit that consists of a single bulb and a battery, the potential
difference across the bulb equals the terminal voltage. The total current
in the circuit can be found using the equation ∆V = IR.
What happens when a second bulb is added to such a circuit, as
shown in Figure 2.1? When moving through this circuit, charges that pass
through one bulb must also move through the second bulb. Because all
charges in the circuit must follow the same conducting path, these bulbs
are said to be connected in series.
Objectives
Plan and Prepare Calculate the equivalent
resistance for a circuit of
resistors in series, and find the
current in and potential
difference across each resistor
in the circuit.
Preview Vocabulary
Calculate the equivalent
resistance for a circuit of
resistors in parallel, and find the
current in and potential
difference across each resistor
in the circuit.
series describes two or more components of a circuit that provide a single
path for current
Scientific Meanings The word
equivalent is used in daily language to
express equal things or amounts. But in
science, particularly in physics, this word
is used in a different way. Equivalent is a
term for expressing two quantities that
are the same with respect to all their
attributes. For example, in the case of
two forces, we use the term equivalent
instead of equal, since force is described
by two different attributes: magnitude
and direction.
Resistors in series carry the same current.
Light-bulb filaments are resistors; thus, Figure 2.1(b) represents the two
bulbs in Figure 2.1(a) as resistors. Because charge is conserved, charges
cannot build up or disappear at a point. For this reason, the amount of
charge that enters one bulb in a given time interval equals the amount of
charge that exits that bulb in the same amount of time. Because there is
only one path for a charge to follow, the amount of charge entering and
exiting the first bulb must equal the amount of charge that enters and
exits the second bulb in the same time interval.
Teach Demonstration
Resistors in Series
Purpose Demonstrate that series
circuits require all elements to conduct.
Because the current is the amount of charge moving past a point per unit
of time, the current in the first bulb must equal the current in the second
bulb. This is true for any number of resistors arranged in series. When many
resistors are connected in series, the current in each resistor is the same.
Materials two flashlight bulbs, bulb
holders, battery, battery holder, three
short pieces of wire
FIGURE 2.1
Two Bulbs in Series These two light bulbs are connected in series. Because
light-bulb filaments are resistors, (a) the two bulbs in this series circuit can be
represented by (b) two resistors in the schematic diagram shown on the right.
(a)
(b)
R1
Differentiated Instruction
Pre-AP
Procedure Wire the bulbs in series with
the battery, and point out the lit bulbs.
Trace the path for the movement of
charges. Ask students to predict what
will happen if you unscrew the second
bulb. Unscrew it. Point out that the
charges no longer have a complete path.
R2
Circuits and Circuit Elements
635
HRW • Holt Physics
PH99PE-C20-002-001-A
Explain the use of ammeters, which measure
the magnitude of current in a circuit. Point out
that using an ammeter is simple. An ammeter
can be connected (in series) to any point on a
circuit to read the magnitude of the electric
current.
5/26/2011 7:14:34 AM
TEACH FROM VISUALS
FIGURE 2.1 Point out that even though
the resistors are different and must be
labeled R1 and R2, there is only one value
for current.
Ask Is the current within the battery
less than, equal to, or greater than the
circuit current?
Answer: The current within the battery
is the same as the circuit current.
Circuits and Circuit Elements 635
The total current in a series circuit depends on how many resistors are
present and on how much resistance each offers. Thus, to find the total
current, first use the individual resistance values to find the total resistance of the circuit, called the equivalent resistance. Then the equivalent
resistance can be used to find the current.
Teach continued
TEACH FROM VISUALS
The equivalent resistance in a series circuit is the sum of the
circuit’s resistances.
FIGURE 2.2 Be certain students
understand what is meant by the idea
that the resistor labeled Req can replace
the other two resistors. The current in
and potential difference across the
equivalent resistor is the same as if the
two resistors are taken together.
As described in Section 1, the potential difference across the battery,
∆V, must equal the potential difference across the load, ∆V1 + ∆V2,
where ∆V1 is the potential difference across R1 and ∆V2 is the potential
difference across R2.
∆V = ∆V1 + ∆V2
According to ∆V = IR, the potential difference across each resistor is
equal to the current in that resistor multiplied by the resistance.
Ask Explain why it is not necessary to
label the current in Figure 2.2(b) as Ieq.
Answer: The current is the same in this
equivalent resistor as in the original
circuit.
∆V = I1R1 + I2 R2
Because the resistors are in series, the current in each is the same. For this
reason, I1 and I2 can be replaced with a single variable for the current, I.
∆V = I(R1 + R2)
FIGURE 2.2
Equivalent Resistance for
a Series Circuit (a) The two
resistors in the actual circuit have
the same effect on the current in the
circuit as (b) the equivalent resistor.
(a)
R1
∆V = I(Req)
Now set the last two equations for ∆V equal to each other, and divide
by the current.
R2
I
Finding a value for the equivalent resistance of the circuit is now
possible. If you imagine the equivalent resistance replacing the original
two resistors, as shown in Figure 2.2, you can treat the circuit as if it
contains only one resistor and use ∆V = IR to relate the total potential
difference, current, and equivalent resistance.
∆V = I(Req) = I(R1 + R2)
I
Req = R1 + R2
Req
(b)
I
Thus, the equivalent resistance of the series combination is the sum of
the individual resistances. An extension of this analysis shows that the
equivalent resistance of two or more resistors connected in series can be
calculated using the following equation.
Resistors in Series
Req = R1 + R2 + R3 . . .
HRW • Holt Physics
PH99PE-C20-002-002-A
Equivalent resistance equals the total of individual
resistances in series.
Because R eq represents the sum of the individual resistances that have
been connected in series, the equivalent resistance of a series combination
of resistors is always greater than any individual resistance.
636
Chapter 18
Differentiated
Instruction
Below Level
Students may confuse the terms current and
resistance when describing circuits. Remind
them that in a single-loop circuit, the current is
the same at every point. For resistors in series,
the same current passes through each resistor.
Untitled-700 636
636 Chapter 18
5/26/2011 7:14:35 AM
ntitled-700 637
To find the total current in a series circuit, first simplify the circuit to a
single equivalent resistance using the boxed equation above; then use
∆V = IR to calculate the current.
∆V
I=_
R eq
Classroom Practice
Resistors in Series
Calculate the equivalent resistance, the
current in each resistor, and the potential
difference across each resistor if a 24.0 V
battery is connected in series to the
following:
a.five 2.0 Ω resistors
Because the current in each bulb is equal to the total current, you can
also use ∆V = IR to calculate the potential difference across each resistor.
∆V1 = IR1
and ∆V2 = IR2
The method described above can be used to find the potential difference
across resistors in a series circuit containing any number of resistors.
Resistors in Series
b.50 2.0 Ω resistors
Sample Problem A A 9.0 V battery is
connected to four light bulbs, as shown at
right. Find the equivalent resistance for
the circuit and the current in the circuit.
4.0 Ω
Given:
Answers:
a. 1.0 × 101 Ω, 2.4 A, 4.8 V
b. 1.0 × 102 Ω, 0.24 A, 0.48 V
c. 1.0 × 103 Ω, 0.024 A, 0.048 V
7.0 Ω
2.0 Ω
ANALYZE
c.500 2.0 Ω resistors
5.0 Ω
∆V = 9.0 V
R1 = 2.0 Ω
R2 = 4.0 Ω
R3 = 5.0 Ω
R4 = 7.0 Ω
Unknown:
Req = ?
Diagram:
4.0 Ω
I=?
5.0 Ω
2.0 Ω
PLAN
7.0Ω
9.0 V
• Holt Physics
Choose an equationHRW
or situation:
PH99PE-C20-002-008-A
Because the resistors are connected end to end, they are in series.
Thus, the equivalent resistance can be calculated with the equation for
resistors in series.
Req = R1 + R2 + R3 . . .
The following equation can be used to calculate the current.
∆V = IReq
Rearrange the equation to isolate the unknown:
No rearrangement is necessary to calculate Req, but ∆V = IReq must be
rearranged to calculate current.
∆V
I=_
Req
Continued
Problem Solving
Alternative Approaches
Because the resistors are in series, the current
is the same in each resistor. If I is the value of
current, the problem can also be solved by
first applying ΔV = IR to each resistor and
then using the sum of potential differences to
calculate I:
ΔV1 = R1I = 2.0I
Circuits and Circuit Elements
637
ΔV4 = R4I = 7.0I
ΔV1 + ΔV2 + ΔV3 + ΔV4 = ΔV 5/26/2011
ΔV = 18.0I
Now substitute the given value for ΔV:
18.0I = 9.0 V
I = 0.5 A
7:14:36 AM
ΔV2 = R2I = 4.0I
ΔV3 = R3I = 5.0I
Circuits and Circuit Elements 637
Resistors in Series
Teach continued
SOLVE
(continued)
Substitute the values into the equation and solve:
Req = 2.0 Ω + 4.0 Ω + 5.0 Ω + 7.0 Ω
Req = 18.0 Ω
PROBLEM guide A
Substitute the equivalent resistance value into the equation for
current.
Use this guide to assign problems.
SE = Student Edition Textbook
PW = Sample Problem Set I (online)
PB = S ample Problem Set II (online)
Solving for:
Req
I
R
9.0 V
∆V = _
I=_
Req 18.0 Ω
I = 0.50A
SE Sample, 1–2, 4, 6;
Ch. Rvw. 16–17
PW Sample, 1–2, 5
PB 4–6
CHECK YOUR
WORK
18.0 Ω > 7.0 Ω
SE Sample, 1–2, 4;
Ch. Rvw. 17
PW Sample, 4
PB 7–10
1. A 12.0 V storage battery is connected to three resistors, 6.75 Ω, 15.3 Ω, and 21.6 Ω,
respectively. The resistors are joined in series.
a. Calculate the equivalent resistance.
b. What is the current in the circuit?
SE 5
PW 3, 6
PB Sample, 1–3
∆V
SE 3, 4
PW 4
P
PW 7
2. A 4.0 Ω resistor, an 8.0 Ω resistor, and a 12.0 Ω resistor are connected in series with
a 24.0 V battery.
a. Calculate the equivalent resistance.
b. Calculate the current in the circuit.
c. What is the current in each resistor?
3. Because the current in the equivalent resistor of Sample Problem A is 0.50 A, it
must also be the current in each resistor of the original circuit. Find the potential
difference across each resistor.
*Challenging Problem
4. A series combination of two resistors, 7.25 Ω and 4.03 Ω, is connected to a 9.00 V
battery.
Answers
Practice A
1. a. 43.6 Ω
b. 0.275 A
2. a. 24.0 Ω
b. 1.00 A
c. 1.00 A
3. 1.0 V, 2.0 V, 2.5 V, 3.5 V
4. a. 11.28 Ω, 0.798 A
b. 5.79 V, 3.22 V
5. 0.5 Ω
6. a. 67.6 Ω
b. 45 bulbs
638 Chapter 18
For resistors connected in series, the equivalent resistance should be
greater than the largest resistance in the circuit.
a. Calculate the equivalent resistance of the circuit and the current.
b. What is the potential difference across each resistor?
5. A 7.0 Ω resistor is connected in series with another resistor and a 4.5 V battery.
The current in the circuit is 0.60 A. Calculate the value of the unknown resistance.
6. Several light bulbs are connected in series across a 115 V source of emf.
a. What is the equivalent resistance if the current in the circuit is 1.70 A?
b. If each light bulb has a resistance of 1.50 Ω, how many light bulbs are in the
circuit?
638
Chapter 18
Differentiated
Instruction
Below Level
Students may rely on the diagrams of circuits
to determine whether a circuit is in series or in
parallel. Tell students that relying on the
diagrams is not always useful. They should use
the rule that a circuit is in series when the
same current runs through every resistor.
Untitled-700 638
5/26/2011 7:14:36 AM
ntitled-700 639
Series circuits require all elements to conduct.
What happens to a series circuit when a single bulb burns out? Consider
what a circuit diagram for a string of lights with one broken filament
would look like. As the schematic diagram in Figure 2.3 shows, the broken
filament means that there is a gap in the conducting pathway used to
make up the circuit. Because the circuit is no longer closed, there is no
current in it and all of the bulbs go dark.
TEACH FROM VISUALS
FIGURE 2.3 Have students examine a
burned-out bulb to see the broken
filament. Tell students that a bulb is said
to burn out when its filament breaks.
Point out that when the filament is
broken, charges no longer have a
complete pathway from the base of
the bulb to the threads.
Why, then, would anyone arrange resistors in series? Resistors can be
placed in series with a device in order to regulate the current in that
device. In the case of decorative lights, adding an additional bulb will
decrease the current in each bulb. Thus, the filament of each bulb need
not withstand such a high current. Another advantage to placing resistors
in series is that several lesser resistances can be used to add up to a single
greater resistance that is unavailable. Finally, in some cases, it is important to have a circuit that will have no current if any one of its component
parts fails. This technique is used in a variety of contexts, including some
burglar alarm systems.
Ask Would the other bulbs light if a
wire were attached from the base of the
burned-out bulb to its threads?
FIGURE 2.3
Answer: Yes, charges would have a
complete path to follow.
Burned-Out Filament in a Series Circuit A burned-out
filament in a bulb has the same effect as an open switch. Because this
series circuit is no longer complete, there is no current in the circuit.
TEACH FROM VISUALS
FIGURE 2.4 Have students trace each of
the alternative pathways with their
fingers. Point out that as long as any of
these pathways remain intact, there will
be current in the circuit.
Resistors in Parallel
As discussed above, when a single bulb in a series light set burns out, the
entire string of lights goes dark because the circuit is no longer closed.
What would happen if thereHRW
were• alternative
Holt Physicspathways for the movement
PH99PE-C20-002-004-A
2.4?
of charge, as shown in Figure
A wiring arrangement that provides alternative pathways for the
movement of a charge is a parallel arrangement. The bulbs of the decorative light set shown in the schematic diagram in Figure 2.4 are arranged in
parallel with each other.
parallel describes two or more
components of a circuit that provide
separate conducting paths for current
because the components are connected
across common points or junctions
FIGURE 2.4
Ask What parts of the circuit would
have current if all of the bulbs except
the last one on the right had broken
filaments?
Answer: There would be current in the
intact bulb and in the wires that connect
the bulb across the potential difference.
A Parallel Circuit These decorative lights are wired in parallel. Notice that in a
parallel arrangement there is more than one path for current.
HRW • Holt Physics
PH99PE-C20-002-012-A
Circuits and Circuit Elements
639
5/26/2011 7:14:37 AM
Circuits and Circuit Elements 639
Resistors in parallel have the same potential
differences across them.
FIGURE 2.5
Teach continued
A Simple Parallel Circuit
(a) This simple parallel circuit with two bulbs connected to a battery can
be represented by (b) the schematic diagram shown on the right.
TEACH FROM VISUALS
(a)
(b)
R1
FIGURE 2.5 Working through a diagram
like Figure 2.5(b) with numerical
examples may help students understand
the relationships for current in a parallel
circuit.
R2
Ask Assume that I from the battery is 5 A
and I1 = 2 A. What must I2 be?
HRW • Holt Physics
PH99PE-C20-002-005b-A
Answer: 3 A
Resistors in Parallel
Purpose Demonstrate that parallel
circuits do not require all elements to
conduct.
Because charge is conserved, the sum of the currents in each bulb equals
the current I delivered by the battery. This is true for all resistors in parallel.
I = I1 + I2 + I3 . . .
The parallel circuit shown in Figure 2.5 can be simplified to an equivalent
resistance with a method similar to the one used for series circuits. To do
this, first show the relationship among the currents.
Materials two flashlight bulbs, bulb
holders, battery, battery holder, four
short pieces of wire
QuickLab
Teacher’s Notes
For this lab to be effective, it is very
important that the straws be taped
together. Crimping one end of a straw
and stuffing it into another straw will
not work well.
Homework Options This QuickLab can
easily be performed outside of the
physics lab room.
I = I1 + I2
SERIES AND PARALLEL CIRCUITS
Cut the regular drinking straws
and thin stirring straws into equal
lengths. Tape them end to end in
long tubes to form series combinations. Form parallel combinations by taping the straws together
side by side.
openings to compare the airflow
(or current) that you achieve with
each combination.
Rank the combinations according
to how much resistance they offer.
Classify them according to the
amount of current created in each.
Try several combinations of like
and unlike straws. Blow through
each combination of tubes,
holding your fingers in front of the
Straws in series
MATERIALS
• 4 regular drinking straws
• 4 stirring straws or coffee stirrers
• tape
Straws in series
Straws in parallel
Straws in parallel
640
Chapter 18
Problem
Solving
Deconstructing Problems
Show students how the last formula on this
page is obtained. First, write the formula for
each potential difference across each resistor:
ΔV1 = I1R1 and ΔV2 = I2R2
Untitled-700 640
Divide both sides of the first formula by R1
and the second formula by R2:
ΔV1
(1)​ _
​  = I1
R1
640 Chapter 18
The sum of currents in parallel resistors equals the total current.
In Figure 2.5, when a certain amount of charge leaves the positive terminal
and reaches the branch on the left side of the circuit, some of the charge
moves through the top bulb and some moves through the bottom bulb. If
one of the bulbs has less resistance, more charge moves through that bulb
because the bulb offers less opposition to the flow
of charges.
Demonstration
Procedure Connect the bulbs in parallel
with the battery as shown in Figure 2.5.
Trace the path for the movement of the
charges. Ask students to predict what
will happen if you unscrew the second
bulb. Unscrew it. Point out that the
charges still have a complete path in the
other bulb.
To explore the consequences of arranging
resistors in parallel, consider the two bulbs
connected to a battery in Figure 2.5(a). In this
arrangement, the left side of each bulb is connected to the positive terminal of the battery, and
the right side of each bulb is connected to the
negative terminal. Because the sides of each bulb
are connected to common points, the potential
difference across each bulb is the same. If the
common points are the battery’s terminals, as
they are in the figure, the potential difference
across each resistor is also equal to the terminal
voltage of the battery. The current in each bulb,
however, is not always the same.
020-QKL
ΔV ​
​ _
2
= I020-QKL-001-A
2
R2
On the other hand, we have the following
formula for the total current in the circuit:
(2)
ΔV
(3)​ _ ​  = I
Req
Since I = I1 + I2, replace the equivalent of I
from (1) and (2) in (3):
ΔV1 _
ΔV2
_
​ ΔV ​  = _
​   ​
+ ​   ​
R1
R2
Req
5/26/2011 7:14:38 AM
Then substitute the equivalents for current according to ∆V = IR.
Conceptual Challenge
∆V1 _
∆V2
∆V = _
_
+
Req
R1
R2
Car Headlights How can you
tell that the headlights on a car
are wired in parallel rather than in
series? How would the brightness of the bulbs differ if they
were wired in series across the
same 12 V battery instead of in
parallel?
Because the potential difference across each bulb in a parallel arrangement equals the terminal voltage (∆V = ∆V1 = ∆V2), you can
divide each side of the equation by ∆V to get the following equation.
1 +_
1
1 =_
_
Req R1 R2
An extension of this analysis shows that the equivalent resistance of
two or more resistors connected in parallel can be calculated using the
following equation.
Simple Circuits Sketch as
many different circuits as
you can using three light bulbs—
each of which has the same
resistance—and a battery.
Resistors in Parallel
1 =_
1 +_
1 +_
1 ...
_
Req R1 R2 R3
The equivalent resistance of resistors in parallel
can be calculated using a reciprocal relationship.
Answers
Conceptual Challenge
1. Car headlights must be wired in
parallel so that if one burns out, the
other will stay lit. If they were wired
in series, they would be less bright.
2. There are four possible circuits: all
resistors in series, all resistors in
parallel, one resistor in series with
two others in parallel, and one
resistor in parallel with two others
in series.
Notice that this equation does not give the value of the equivalent
resistance directly. You must take the reciprocal of your answer to obtain
the value of the equivalent resistance.
Because of the reciprocal relationship, the equivalent resistance for a
parallel arrangement of resistors must always be less than the smallest
resistance in the group of resistors.
The conclusions made about both series and parallel circuits are
summarized in Figure 2.6.
FIGURE 2.6
RESISTORS IN SERIES OR IN PARALLEL
Series
Parallel
schematic diagram
(tr) ©KRT/NewsCom
current
ntitled-700 641
HRW • Holt Physics
= I2 = I3 . . .
I = I1PH99PE-C20-002-010-A
= same for each resistor
I = I1 + I2 + I3 . . .
HRW • Holt Physics
= sum of currents
potential difference
∆V = ∆V1 + ∆V2 + ∆V3 . . .
= sum of potential differences
∆V = ∆V1 + ∆V2 + ∆V3 . . .
= same for each resistor
equivalent resistance
Req = R1 + R2 + R3 . . .
= sum of individual resistances
1 =_
1 +_
1 +_
1
_
Req R1 R2 R3
= reciprocal sum of resistances
Differentiated Instruction
PH99PE-C20-002-011-A
Circuits and Circuit Elements
641
Below Level
Use a simple numerical example to demonstrate that mathematically adding the inverses
is not the same as taking the inverse of the
sum. The example below uses resistances that
have values of 2 and 3 in parallel.
Correct: ​ _21  ​+ ​ _31 ​ = __
​ 65  ​, Req = ​ __65 ​
5/26/2011 7:14:39 AM
Incorrect: 2 + 3 = 5, Req ≠ ​ _51 ​
Circuits and Circuit Elements 641
Resistors in Parallel
Teach continued
Sample Problem B A 9.0 V battery is
connected to four resistors, as shown at right.
Find the equivalent resistance for the circuit and
the total current in the circuit.
Classroom Practice
Resistors in Parallel
Find the equivalent resistance, the
current in each resistor, and the current
drawn by the circuit load for a 9.0 V
battery connected in parallel to three
30.0 Ω resistors.
Answer: 10.0 Ω, 0.30 A, 0.90 A
ANALYZE
Given:
2.0 Ω
4.0 Ω
5.0 Ω
7.0 Ω
∆V = 9.0 V
R1 = 2.0 Ω
R2 = 4.0 Ω
R3 = 5.0 Ω
R4 = 7.0 Ω
PROBLEM guide B
Unknown:
Use this guide to assign problems.
SE = Student Edition Textbook
PW = Sample Problem Set I (online)
PB = S ample Problem Set II (online)
Solving for:
Diagram:
Req
I
9.0 V
PLAN
∆V
SE 4b
I=?
4.0 Ω
7.0 Ω
Choose an equationHRW
or situation:
• Holt Physics
Because both sidesPH99PE-C20-002-009-A
of each resistor are connected to common points,
they are in parallel. Thus, the equivalent resistance can be calculated
with the equation for resistors in parallel.
1 +_
1 +_
1 =_
1 . . . for parallel
_
Req R1 R2 R3
The following equation can be used to calculate the current.
∆V = IReq
SE Sample, 1, 3–4;
Ch. Rvw. 18–19
PW Sample, 6–7
PB 7–10
PW 3
PB Sample, 1–3
2.0 Ω
5.0 Ω
SE Sample, 2–4;
Ch. Rvw. 18–19
PW Sample, 1–2, 4–6
PB 4–6
R
Req = ?
Rearrange the equation to isolate the unknown:
No rearrangement is necessary to calculate Req ;
rearrange ∆V = IReq to calculate the total current
delivered by the battery.
∆V
I=_
Req
SOLVE
Tips and Tricks
The equation for resistors
in parallel gives you the
reciprocal of the equivalent
resistance. Be sure to take
the reciprocal of this value
in the final step to find the
equivalent resistance.
Substitute the values into the equation and solve:
*Challenging Problem
1 +_
1 +_
1 =_
1 +_
1
_
Req 2.0 Ω 4.0 Ω 5.0 Ω 7.0 Ω
0.20 + _
0.14 = _
0.5 + _
0.25 +_
1.09
1 =_
_
Req 1 Ω
1Ω
1Ω
1Ω
1Ω
1Ω
Req = _
1.09
Req = 0.917 Ω
642
Chapter 18
Problem
Solving
Take It Further
Modify the sample problem to provide an
example with 5 resistors. The total of their
resistance magnitude is still 18 Ω, as in the
sample problem. Provide the following data:
First resistor: 6 Ω
Second resistor: 1 Ω
Third resistor: 5 Ω
Fourth resistor: 4 Ω
Fifth resistor: 2 Ω
Untitled-700 642
642 Chapter 18
Continued
Have students calculate the total current in
this circuit and compare it with the result in
the sample problem.
Answer: I = 4.25 A;
the current decreased by 5.55 A.
5/26/2011 7:14:40 AM
Untitled-700 643
Resistors in Parallel
(continued)
Substitute that equivalent resistance value
in the equation for current.
Practice B
1. 4.5 A, 2.2 A, 1.8 A, 1.3 A
2. 50.0 Ω
3. a. 2.2 Ω
b. 6.0 A, 3.0 A, 2.00 A
4. a. 2.99 Ω
b. 36.0 V
c. 2.00 A, 6.00 A
The calculator answer is
9.814612868, but because the
potential difference, 9.0 V, has
only two significant digits, the
answer is reported as 9.8 A.
∆Vtot _
I=_
= 9.0 V
Req
0.917 Ω
I = 9.8 A
CHECK YOUR
WORK
Answers
Calculator Solution
For resistors connected in parallel, the equivalent resistance should be
less than the smallest resistance.
0.917 Ω < 2.0 Ω
1. The potential difference across the equivalent resistance in Sample Problem B
equals the potential difference across each of the individual parallel resistors.
Calculate the value for the current in each resistor.
2. A length of wire is cut into five equal pieces. The five pieces are then connected in
parallel, with the resulting resistance being 2.00 Ω. What was the resistance of the
original length of wire before it was cut up?
3. A 4.0 Ω resistor, an 8.0 Ω resistor, and a 12.0 Ω resistor are connected in parallel
across a 24.0 V battery.
a. What is the equivalent resistance of the circuit?
b. What is the current in each resistor?
4. An 18.0 Ω, 9.00 Ω, and 6.00 Ω resistor are connected in parallel to an emf source.
A current of 4.00 A is in the 9.00 Ω resistor.
a. Calculate the equivalent resistance of the circuit.
b. What is the potential difference across the source?
c. Calculate the current in the other resistors.
Parallel circuits do not require all elements to conduct.
What happens when a bulb burns out in a string of decorative lights that
is wired in parallel? There is no current in that branch of the circuit, but
each of the parallel branches provides a separate alternative pathway for
current. Thus, the potential difference supplied to the other branches and
the current in these branches remain the same, and the bulbs in these
branches remain lit.
When resistors are wired in parallel with an emf source, the potential
difference across each resistor always equals the potential difference
across the source. Because household circuits are arranged in parallel,
appliance manufacturers are able to standardize their design, producing
Circuits and Circuit Elements
Alternative Approaches
The problem can also be solved by applying
ΔV = IR to each resistor to find its current,
then adding these to get the total
current. Finally, use Req = ___
​ ΔV
​ to find Req.
I
tot
9.0 V
I1 = _
​ ΔV ​  = ​ _  ​ = 4.5 A
R1
2.0 Ω
9.0 V
​ ΔV ​  = ​ _  ​ = 2.2 A
I2 = _
R2
4.0 Ω
​ ΔV
​  ​ 9.0 V  ​
I3 = _ = _ = 1.8 A
R3
5.0 Ω
643
5/26/2011 7:14:41 AM
9.0 V
ΔV ​ =
​ _  ​ = 1.3 A
I4 =​ _
R4
7.0 Ω
Itot = I1 + I2 + I3 + I4
Itot = 9.8 A
9.0 V
Req = ​ _ ​  = 0.92 Ω
9.8 A
The slight difference in the answer obtained this
way is due to rounding.
Circuits and Circuit Elements 643
devices that all operate at the same potential difference. As a result,
manufacturers can choose the resistance to ensure that the current will
be neither too high nor too low for the internal wiring and other components that make up the device.
Did YOU Know?
Assess and Reteach
Assess Use the Formative Assessment
on this page to evaluate student
mastery of the section.
Reteach For students who need
additional instruction, download the
Section Study Guide.
Response to Intervention To reassess
students’ mastery, use the Section Quiz,
available to print or to take directly
online at HMDScience.com.
Because the potential difference
provided by a wall outlet in a home in
North America is not the same as the
potential difference that is standard on other continents, appliances
made in North America are not always
compatible with wall outlets in homes
on other continents.
Additionally, the equivalent resistance of several parallel resistors is
less than the resistance of any of the individual resistors. Thus, a low
equivalent resistance can be created with a group of resistors of
higher resistances.
SECTION 2 FORMATIVE ASSESSMENT
Reviewing Main Ideas
1. Two resistors are wired in series. In another circuit, the same two resistors
are wired in parallel. In which circuit is the equivalent resistance greater?
2. A 5 Ω, a 10 Ω, and a 15 Ω resistor are connected in series.
a. Which resistor has the most current in it?
b. Which resistor has the largest potential difference across it?
3. A 5 Ω, a 10 Ω, and a 15 Ω resistor are connected in parallel.
a. Which resistor has the most current in it?
b. Which resistor has the largest potential difference across it?
4. Find the current in and potential difference across each of the resistors in
the following circuits:
a. a 2.0 Ω and a 4.0 Ω resistor wired in series with a 12 V source
b. a 2.0 Ω and a 4.0 Ω resistor wired in parallel with a 12 V source
Interpreting Graphics
5. The brightness of a bulb depends only on the bulb’s resistance and on
the potential difference across it. A bulb with a greater potential difference dissipates more power and thus is brighter. The five bulbs shown
in Figure 2.7 are identical, and so are the three batteries. Rank the bulbs
in order of brightness from greatest to least, indicating if any are equal.
Explain your reasoning. (Disregard the resistance of the wires.)
FIGURE 2.7
(a)
(b)
(c)
(d)
(e)
644
Chapter 18
Answers
to Section Assessment
1. in the series circuit
Untitled-700
2. a.644All have equal I.
b. 15 Ω
3. a. 5 Ω
b. All have equal ΔV.
4. a. 2.0 Ω: 2.0 A, 4.0 V
4.0 Ω: 2.0 A, 8.0 V
b. 2.0 Ω: 6.0 A, 12 V
4.0 Ω: 3.0 A, 12 V
644 Chapter 18
5. Because the resistance of each bulb is the
same, the brightness depends only on the 5/26/2011
potential difference. Bulbs (a), (d), and (e)
have equal potential difference across
them and thus equal brightnesses. Because
bulbs (b) and (c) have the same resistance
and are in series, they have equal but lesser
potential differences and are equally bright
but less bright than bulbs (a), (d), and (e).
7:14:41 AM
ntitled-783 645
SECTION 3
SECTION 3
Complex Resistor
Combinations
Objectives
Plan and Prepare Calculate the equivalent
resistance for a complex circuit
involving both series and
parallel portions.
Preview Vocabulary
Scientific Meanings The word fuse is
used in everyday language to describe
the melting and blending of different
things. In physics, a fuse is a protective
component in electric devices.
Calculate the current in and
potential difference across
individual elements within a
complex circuit.
Resistors Combined Both in Parallel and in Series
Series and parallel circuits are not often encountered independent of one
another. Most circuits today employ both series and parallel wiring to
utilize the advantages of each type.
A common example of a complex circuit is the electrical wiring typical
in a home. In a home, a fuse or circuit breaker is connected in series to
numerous outlets, which are wired to one another in parallel. An example
of a typical household circuit is shown in Figure 3.1.
Teach As a result of the outlets being wired in parallel, all the appliances
operate independently; if one is switched off, any others remain on.
Wiring the outlets in parallel ensures that an identical potential difference
exists across any appliance. This way, appliance manufacturers can
produce appliances that all use the same standard potential difference.
TEACH FROM VISUALS
FIGURE 3.1 Explain that the circuit
breaker is shown as a switch because it
contains a switch that opens when the
current in the circuit becomes too large.
To prevent excessive current, a fuse or circuit breaker must be placed
in series with all of the outlets. Fuses and circuit breakers open the circuit
when the current becomes too high. A fuse is a small
FIGURE 3.1
metallic strip that melts if the current exceeds a certain
value. After a fuse has melted, it must be replaced. A
A Household Circuit (a) When all of these
circuit breaker, a more modern device, triggers a switch
devices are plugged into the same household circuit,
when current reaches a certain value. The switch must be
(b) the result is a parallel combination of resistors in
reset, rather than replaced, after the circuit overload has
series with a circuit breaker.
been removed. Both fuses and circuit breakers must be in
series with the entire load to prevent excessive current
from reaching any appliance. In fact, if all the devices in
Figure 3.1 were used at once, the circuit would be overloaded. The circuit breaker would interrupt the current.
Fuses and circuit breakers are carefully selected to
meet the demands of a circuit. If the circuit is to carry
currents as large as 30 A, an appropriate fuse or circuit
breaker must be used. Because the fuse or circuit breaker
is placed in series with the rest of the circuit, the current in
the fuse or circuit breaker is the same as the total current
in the circuit. To find this current, one must determine the
equivalent resistance.
When determining the equivalent resistance for a
complex circuit, you must simplify the circuit into groups
of series and parallel resistors and then find the equivalent
resistance for each group by using the rules for finding the
equivalent resistance of series and parallel resistors.
Differentiated Instruction
Pre-AP
Point out that each device added in parallel
to a circuit draws more current from the
emf source.
Mathematically, because of the following
equation, the more resistors that are added in
parallel, the more current there will be in the
main wires of the circuit.
ΔV ​  . . .
Itot = _
​ ΔV ​  + ​ _
R1
R2
Ask How much of the total current is
in the circuit breaker when it is in series
with the parallel combination of
devices shown?
Answer: All of the total current is in the
circuit breaker when it is in series.
(a)
Microwave: 8.0 Ω
Blender: 41.1 Ω
Toaster: 16.9 Ω
∆V = 120 V
(b)
Circuit
breaker: 0.01 Ω
HRW • Holt Physics
PH99PE-C20-003-001-A
Circuits and Circuit Elements
645
If the current is too great, the main wires,
plugs, and outlet connections will heat
up.4:03:40 PM
6/3/2011
Excessive current can damage equipment and
even cause fires.
Circuits and Circuit Elements 645
PREMIUM CONTENT
Interactive Demo
Equivalent Resistance
Teach continued
Classroom Practice
ANALYZE
Equivalent Resistance
Use the following values with the circuit
in Figure 3.2. What is the equivalent
resistance for each circuit?
a.Ra = 5.0 Ω, Rb = 3.0 Ω, Rc = 6.0 Ω
b.Ra = 6.0 Ω, Rb = 8.0 Ω, Rc = 2.0 Ω
I
R
P
6.0 Ω
6.0 Ω
2.0 Ω
4.0 Ω
3.0 Ω
1.0 Ω
9.0 V
Tips and Tricks
Redraw the circuit as a group of
HRW • Holt Physics
PH99PE-C20-003-003-A
For now, disregard
the emf source,
resistors along one side of
and work only with the resistances.
the circuit.
Because bends in a wire do not
affect the circuit, they do not need
to be represented in a schematic
6.0 Ω 2.0 Ω
1.0 Ω
3.0 Ω 6.0 Ω
diagram. Redraw the circuit
without the corners, keeping the
arrangement of the circuit elements
4.0 Ω
the same, as shown at right.
SOLVE
Identify components in series, and
calculate their equivalent resistance.
Resistors in groups (a) and (b) are in series.
For group (a): Req = 3.0 Ω + 6.0 Ω = 9.0 Ω
For group (b): Req = 6.0 Ω + 2.0 Ω = 8.0 Ω
PROBLEM guide C
Use this guide to assign problems.
SE = Student Edition Textbook
PW = Sample Problem Set I (online)
PB = S ample Problem Set II (online)
Solving for:
The best approach is to divide the
circuit into groups of series and
parallel resistors. This way, the
methods presented in Sample
Problems A and B can be used to
calculate the equivalent resistance
for each group.
PLAN
Answers:
a. 7.0 Ω
b. 7.6 Ω
Req
HMDScience.com
Sample Problem C Determine the equivalent resistance of
the complex circuit shown below.
SE Sample, 1–2;
Ch. Rvw. 23–24
PW Sample, 1, 4–5
PB 4–6
3.0 Ω
6.0 Ω
6.0 Ω
(b)
(a)
2.0 Ω
PW 2–3
PB 7–10
4.0 Ω
PW 3
PB Sample, 1–3
PB 4,6
*Challenging Problem
1.0 Ω
4.0 Ω
8.0 Ω
9.0 Ω
HRW • Holt Physics
PH99PE-C20-003-004-A
9.0 Ω
2.7 Ω
(d)
12.7 Ω
1.0 Ω
(c)
1.0 Ω
9.0 V
Continued
646
Chapter 18
Problem
Solving
Take It Further
In the given sample problem, change the
magnitudes of the resistors to 6 Ω and ask
students to find the equivalent resistance of
the circuit. Then ask students to draw two
different circuits, each with the same number
of resistors—one in series and the other in
parallel. Have them calculate the equivalent
resistance of each circuit, given that the
magnitude of each resistor is 6 Ω.
Untitled-783 646
646 Chapter 18
Ask students to compare the results and
state their finding as a general fact.
Answer: The resistance of the circuit in parallel
has the least magnitude.
6/3/2011 4:03:41 PM
ntitled-783 647
Equivalent Resistance
(continued)
Identify components in parallel, and
calculate their equivalent resistance.
Resistors in group (c) are in parallel.
For group (c):
Answers
Tips and Tricks
It doesn’t matter in what order
the operations of simplifying
the circuit are done, as long as
the simpler equivalent circuits
still have the same current
in and potential difference
across the load.
0.37
0.12 Ω + _
0.25 = _
1 =_
1 =_
1 +_
_
1
Req 8.0 Ω 4.0 Ω
1Ω
1Ω
Req = 2.7 Ω
Repeat steps 2 and 3 until the resistors in the circuit are
reduced to a single equivalent resistance.
The remainder of the resistors, group (d), are in series.
For group (d): Req = 9.0 Ω + 2.7 Ω + 1.0 Ω
Practice C
1. a. 27.8 Ω
b. 26.6 Ω
c. 23.4 Ω
2. a. 50.9 Ω
b. 57.6 Ω
Req = 12.7 Ω
1. For each of the following sets of values, determine the equivalent
resistance for the circuit shown in Figure 3.2.
a. Ra = 25.0 Ω
Rb = 3.0 Ω
Rc = 40.0 Ω
b. Ra = 12.0 Ω
Rb = 35.0 Ω
Rc = 25.0 Ω
c. Ra = 15.0 Ω
Rb = 28.0 Ω
Rc = 12.0 Ω
Ra
40.0 V
Rb = 3.0 Ω
Re = 18.0 Ω
Rc = 40.0 Ω
b. Ra = 12.0 Ω
Rd = 50.0 Ω
Rb = 35.0 Ω
Re = 45.0 Ω
Rc = 25.0 Ω
Rc
Figure 3.2
HRW • Holt Physics
Ra
PH99PE-C20-003-007-A
2. For each of the following sets of values, determine the equivalent
resistance for the circuit shown in Figure 3.3.
a. Ra = 25.0 Ω
Rd = 15.0 Ω
Rb
25.0 V
Rb
Rd
Rc
Re
Figure 3.3
HRW • Holt Physics
PH99PE-C20-003-008-A
Work backward to find the current in and potential difference across
a part of a circuit.
Now that the equivalent resistance for a complex circuit has been determined, you can work backward to find the current in and potential
difference across any resistor in that circuit. In the household example,
substitute potential difference and equivalent resistance in ∆V = IR to
find the total current in the circuit. Because the fuse or circuit breaker is
in series with the load, the current in it is equal to the total current. Once
this total current is determined, ∆V = IR can again be used to find the
potential difference across the fuse or circuit breaker.
There is no single formula for finding the current in and potential
difference across a resistor buried inside a complex circuit. Instead,
∆V = IR and the rules reviewed in Figure 3.4 must be applied to smaller
pieces of the circuit until the desired values are found.
Circuits and Circuit Elements
647
Alternative Approaches
Students should be encouraged to examine
alternative ways to group the resistors. For
example, they could first find the equivalent
resistance of the 6.0 Ω and 2.0 Ω resistors
shown as group (b) in Sample Problem C, then
find the equivalent resistance of the 8.0 Ω and
4.0 Ω resistors shown as group (c). The result
would be four resistors in series: 6.0 Ω, 3.0 Ω,
2.7 Ω, and 1.0 Ω. The equivalent resistance of
these four resistors is 12.7 Ω.
6/3/2011 4:03:42 PM
Circuits and Circuit Elements 647
FIGURE 3.4
SERIES AND PARALLEL RESISTORS
Teach continued
Classroom Practice
Current in and Potential
Difference Across a Resistor
Use the following values with the circuit
in Figure 3.5 after Sample Problem D.
What is the current in and potential
difference across each of the resistors?
Ra = 8.0 Ω, Rb = 4.0 Ω, Rc = 6.0 Ω,
Rd = 3.0 Ω, Re = 9.0 Ω, Rf = 7.0 Ω
Answers:
Ia = 0.35 A, ΔVa = 2.8 V
Series
Parallel
current
same as total
add to find total
potential difference
add to find total
same as total
PREMIUM CONTENT
Current in and Potential Difference Across a Resistor
HMDScience.com
Sample Problem D Determine the current in and potential
difference across the 2.0 Ω resistor highlighted in the figure below.
ANALYZE
Ib = 0.35 A, ΔVb = 1.4 V
Ic = 0.70 A, ΔVc = 4.2 V
Id = 0.80 A, ΔVd = 2.4 V
PLAN
Ie = 0.27 A, ΔVe = 2.4 V
If = 1.05 A, ΔVf = 7.4 V
Tips and Tricks
It is not necessary to solve
for R eq first and then work
backward to find current
in or potential difference
across a particular resistor,
as shown in this Sample
Problem, but working
through these steps keeps
the mathematical operations
at each step simpler.
First determine the total circuit
current by reducing the resistors to a
single equivalent resistance. Then
rebuild the circuit in steps, calculating
the current and potential difference
for the equivalent resistance of each
group until the current in and
potential difference across the 2.0 Ω
resistor are known.
Determine the equivalent
resistance of the circuit.
The equivalent resistance of the
circuit is 12.7 Ω; this value is
calculated in Sample Problem C.
Calculate the total current in the
circuit.
Substitute the potential difference
and equivalent resistance in ∆V = IR,
and rearrange the equation to find the
current delivered by the battery.
9.0 V = 0.71 A
∆V = _
I=_
Req 12.7 Ω
Alternative Approaches
Remind students that they can check each
step by using ΔV = IR for each resistor in a
set, as discussed in the Tip on this student
page. They can also check the sum of ΔV for
series circuits and the sum of I for parallel
circuits.
For A, the potential difference across the
2.7 Ω resistor is 1.9 V. For the other two
resistors in series in group (d):
Untitled-783 648
6.0 Ω
6.0 Ω
2.0 Ω
1.0 Ω
4.0 Ω
3.0 Ω
9.0 V
HRW • Holt Physics
PH99PE-C20-003-003-A
3.0 Ω
6.0 Ω
6.0 Ω
(b)
(a)
2.0 Ω
4.0 Ω
8.0 Ω
9.0 Ω
4.0 Ω
9.0 Ω
1.0 Ω
2.7 Ω
1.0 Ω
(c)
1.0 Ω
(d)
12.7 Ω
Determine a path from the equivalent
resistance found in step 1 to the 2.0 Ω
resistor.
9.0 V
Review the path taken to find the equivalent
resistance in the figure at right, and work backward
through this path. The equivalent resistance for the entire circuit
is the same as the equivalent resistance for group (d). The center
resistor in group (d) in turn is the equivalent resistance for group
(c). The top resistor in group (c) is the equivalent resistance for
group (b), and the right resistor in group (b) is the 2.0 Ω resistor.
648
Chapter 18
Problem
Solving
648 Chapter 18
Interactive Demo
Continued
ΔV = (0.71 A)(9.0 Ω) = 6.4 V
ΔV = (0.71 A)(1.0 Ω) = 0.71 V
The total ΔV across group (d) matches the
terminal voltage.
1.9 V + 6.4 V + 0.71 V = 9.0 V
For B, the current across the 8.0 Ω resistor
is 0.24 A. For the other resistor in group (c):
1.9 V
I = ​ _  ​ = 0.48 A
4.0 Ω
6/3/2011 4:03:43 PM
ntitled-783 649
Current in and Potential Difference Across a Resistor
SOLVE
(continued)
Teaching Tip
Follow the path determined in step 3, and calculate the current in
and potential difference across each equivalent resistance. Repeat
this process until the desired values are found.
Tell students that it is good problemsolving technique to check each step
of the solution before proceeding with
the next step. This prevents students
from having to rework the entire
solution in case they make an error in
one of the steps.
A. Regroup, evaluate, and calculate.
Replace the circuit’s equivalent resistance with group (d). The resistors
in group (d) are in series; therefore, the current in each resistor is the
same as the current in the equivalent resistance, which equals 0.71 A.
The potential difference across the 2.7 Ω resistor in group (d) can be
calculated using ∆V = IR.
Given:
I = 0.71 A
Unknown:
∆V = ?
R = 2.7 Ω
∆V = I = (0.71 A)(2.7 Ω) = 1.9 V
B. Regroup, evaluate, and calculate.
Replace the center resistor with group (c).
The resistors in group (c) are in parallel; therefore, the potential
difference across each resistor is the same as the potential difference
across the 2.7 Ω equivalent resistance, which equals 1.9 V. The current
in the 8.0 Ω resistor in group (c) can be calculated using ∆V = IR.
Given:
∆V = 1.9 V R = 8.0 Ω
Unknown:
I=?
1.9 V = 0.24 A
∆V = _
I=_
R
80 Ω
C. Regroup, evaluate, and calculate.
Replace the 8.0 Ω resistor with group (b).
Tips and Tricks
You can check each step
in problems like Sample
Problem D by using ∆V = IR
for each resistor in a set. You
can also check the sum of
∆V for series circuits and the
sum of I for parallel circuits.
The resistors in group (b) are in series; therefore, the current in each
resistor is the same as the current in the 8.0 Ω equivalent resistance,
which equals 0.24 A.
I = 0.24 A
The potential difference across the 2.0 Ω resistor can be calculated
using ∆V = IR.
Given:
I = 0.24 A
Unknown:
∆V = ?
R = 2.0 Ω
∆V = IR = (0.24 A) (2.0 Ω) = 0.48 V
∆V = 0.48 V
Continued
The total of these currents is 0.72 A, which
differs from 0.71 A because of rounding.
Circuits and Circuit Elements
649
6/3/2011 4:03:43 PM
For C, the potential difference across the
2.0 Ω resistor is 0.48 V. For the other resistor:
ΔV = (0.24 A)(6.0 Ω) = 1.4 V
The total of these potential differences is 1.9 V,
which was given in the previous step.
Circuits and Circuit Elements 649
Current in and Potential Difference Across a Resistor
Teach continued
1. Calculate the current in and potential difference across each of the resistors
shown in the schematic diagram in Figure 3.5.
PROBLEM guide D
Ra = 5.0 Ω
Use this guide to assign problems.
SE = Student Edition Textbook
PW = Sample Problem Set I (online)
PB = S ample Problem Set II (online)
Solving for:
I
SE Sample, Practice;
Ch. Rvw. 25–26
PW Sample, 3
PB 1–10
∆V
SE Sample, Practice;
Ch. Rvw. 25–26
PW Sample, 1–3
PB 1–10
14.0 V
R c = 4.0 Ω
Rd = 4.0 Ω
R e = 4.0 Ω
Figure 3.5
HRW • Holt Physics
PH99PE-C20-003-014-A
Decorative Lights
and Bulbs
Filament
L
ight sets arranged in series cannot remain lit if a
bulb burns out. Wiring in parallel can eliminate
this problem, but each bulb must then be able to
withstand 120 V. To eliminate the drawbacks of either
approach, modern light sets typically contain two or
three sections connected to each other in parallel, each
of which contains bulbs in series.
Answers
Practice D
Ra: 0.50 A, 2.5 V
When one bulb is removed from a modern light set,
half or one-third of the lights in the set go dark because
the bulbs in that section are wired in series. When a bulb
burns out, however, all of the other bulbs in the set
remain lit. How is this possible?
Rb: 0.50 A, 3.5 V
Rc: 1.5 A, 6.0 V
Rd: 1.0 A, 4.0 V
Re: 1.0 A, 4.0 V
Modern decorative bulbs have a short loop of insulated
wire, called the jumper, that is wrapped around the wires
connected to the filament, as shown at right. There is no
current in the insulated wire when the bulb is functioning
properly. When the filament breaks, however, the current
in the section is zero and the potential difference across
the two wires connected to the broken filament is then
120 V. This large potential difference creates a spark
Rf: 2.0 A, 4.0 V
Why It Matters
650 Chapter 18
Rb = 7.0 Ω
R f = 2.0 Ω
*Challenging Problem
Decorative Lights and Bulbs
Although decorative lights are an
excellent topic during classroom
discussion of series and parallel circuits,
many decorative light sets use the
jumpers described in this feature to
avoid the pitfalls of each type of circuit.
In effect, the jumper functions like a
switch that remains open while the
filament conducts and closes to connect
the wires when the filament burns out.
(continued)
Jumper
Glass insulator
across the two wires that burns the insulation off the
small loop of wire. Once that occurs, the small loop
closes the circuit, and the other bulbs in the section
remain lit.
Because the small loop in the burned out bulb
has very little resistance, the equivalent resistance
of that portion of the light set decreases; its current
increases. This increased current results in a slight
increase in each bulb’s brightness. As more bulbs
burn out, the temperature in each bulb
increases and can become a fire hazard;
thus, bulbs should be replaced soon
after burning out.
650
Chapter 18
Differentiated
Instruction
Below Level
Students may still not be aware of the
importance of the order of the resistors in
calculations. Have them review the diagram in
the practice problem on this page and ask
them to combine Ra with Rc in series. Then
have them combine the result with Rb once in
series and once in parallel. Have them compare
their result with the result they obtain for
resistors Ra, Rb, and Rc in the problem.
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SECTION 3 FORMATIVE ASSESSMENT
Assess and Reteach Reviewing Main Ideas
1. Find the equivalent resistance of the complex circuit
shown in Figure 3.6.
FIGURE 3.6
2. What is the current in the 1.5 Ω resistor in the complex
circuit shown in Figure 3.6?
5.0 Ω
3. What is the potential difference across the 1.5 Ω resistor
in the circuit shown in Figure 3.6?
18.0 V
4. A certain strand of miniature lights contains 35 bulbs
wired in series, with each bulb having a resistance of
15.0 Ω. What is the equivalent resistance when three
such strands are connected in parallel across a potential
difference of 120.0 V?
1.5 Ω
Assess Use the Formative Assessment
on this page to evaluate student
mastery of the section.
5.0 Ω
Reteach For students who need
additional instruction, download the
Section Study Guide.
5.0 Ω
5.0 Ω
Response to Intervention To reassess
students’ mastery, use the Section Quiz,
available to print or to take directly
online at HMDScience.com.
HRW • Holt Physics
PH99PE-C20-003-015-A
5. What is the current in and potential difference across
each of the bulbs in the strands of lights described in item 4?
6. If one of the bulbs in one of the three strands of lights in item 4 goes out
while the other bulbs in that strand remain lit, what is the current in and
potential difference across each of the lit bulbs in that strand?
Interpreting Graphics
7. Figure 3.7 depicts a household circuit containing several appliances and a
circuit breaker attached to a 120 V source of potential difference.
a. Is the current in the toaster equal to the current in the microwave?
b. Is the potential difference across the microwave equal to the potential
difference across the popcorn popper?
c. Is the current in the circuit breaker equal to the total current in all of
the appliances combined?
d. Determine the equivalent resistance for the circuit.
e. Determine how much current is in the toaster.
FIGURE 3.7
Toaster: 16.9 Ω
Microwave: 8.0 Ω
Popcorn popper: 10.0 Ω
Circuit breaker: 0.01 Ω
120 V
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PH99PE-C20-003-016-A
Answers to Section Assessment
1. 9.8 Ω
2. 1.8 A
3. 2.7 V
4. 175 Ω
5. 0.229 A, 3.44 V
6. 0.235 A, 3.52 V
7. a. No, the current in the toaster is less
than the current in the microwave.
b. Yes, the potential differences are equal
because they are in parallel.
Circuits and Circuit Elements
651
c. yes, because it is in series with the rest
of the circuit
6/3/2011 4:03:45 PM
d. 3.6 Ω
e. 7.1 A
Circuits and Circuit Elements 651
Careers in physics
Semiconductor
Technician
Brad Baker credits his father’s example
with helping guide him into a career in
technology. “My father was a systems
analyst and into computers,” says Baker.
“That piqued my interest growing up.”
When Baker uses a cooking analogy to
describe his work, it is more than mere
wordplay. After receiving guidance from
an engineer, Baker and his team begin a
tinkering process that tweaks the
different variables to achieve the desired
result. The variables include power,
wattage, chemistry gas flow, temperature,
and the actual time that the chip is in the
process.
Baker works with another process
called photo engineering, which helps
devices do more work while getting
smaller. In Baker’s words, engineers “print
the design on a wafer, and then we etch
away the parts that aren’t needed.” After
the etch process, another team adds
material to connect different layers, and
the process is repeated.
CAREERS IN PHYSICS
Semiconductor
Technician
E
lectronic chips are used in a wide variety of devices,
from toys to phones to computers. To learn more
about chip making as a career, read the interview
with etch process engineering technician Brad Baker, who
works for Motorola.
What training did you receive in order to
become a semiconductor technician?
My experience is fairly unique. My degree is in psychology.
You have to have an associate’s degree in some sort of
electrical or engineering field or an undergraduate degree in
any field.
What about semiconductor manufacturing
made it more interesting than other fields?
While attending college, I worked at an airline. There was not
a lot of opportunity to advance, which helped point me in
other directions. Circuitry has a lot of parallels to the
biological aspects of the brain, which is what I studied in
school. We use the scientific method a lot.
What is the nature of your work?
I work on the etch process team. Device engineers design the
actual semiconductor. Our job is to figure out how to make
what they have requested. It’s sort of like being a chef. Once
you have experience, you know which ingredient to add.
What is your favorite thing about your job?
I feel like a scientist. My company gives us the freedom to
try new things and develop new processes.
Brad Baker is creating a recipe on the plasma
etch tool to test a new process.
What advice do you have for students who
are interested in semiconductor
engineering?
The field is very science
oriented, so choose chemical
engineering, electrical
engineering, or material
science as majors. Other
strengths are the ability to
understand and meet
challenges, knowledge of
trouble-shooting
techniques, patience,
and analytical skills.
Also, everything is
computer automated,
so you have to know
how to use
computers.
Has your job changed since you started it?
Each generation of device is smaller, so we have to do more
in less space. As the devices get smaller, it becomes more
challenging to get a design process that is powerful enough
but doesn’t etch too much or too little.
Brad Baker
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CHAPTER 18
SECTION 1
C h a p t e r s u m m a ry
Summary
Teaching Tip
Schematic Diagrams and Circuits
KEY TERMS
• Schematic diagrams use standardized symbols to summarize the contents
of electric circuits.
Ask students to prepare a concept map
for the chapter. The concept map
should include most of the vocabulary
terms, along with other integral terms
or concepts.
schematic diagram
electric circuit
• A circuit is a set of electrical components connected so that they provide
one or more complete paths for the movement of charges.
• Any device that transforms nonelectrical energy into electrical energy, such
as a battery or a generator, is a source of emf.
• If the internal resistance of a battery is neglected, the emf can be considered equal to the terminal voltage, the potential difference across the
source’s two terminals.
SECTION 2
Resistors in Series or in Parallel
KEY TERMS
• Resistors in series have the same current.
series
parallel
• The equivalent resistance of a set of resistors connected in series is the
sum of the individual resistances.
• The sum of currents in parallel resistors equals the total current.
• The equivalent resistance of a set of resistors connected in parallel is
calculated using an inverse relationship.
SECTION 3
Complex Resistor Combinations
• Many complex circuits can be understood by isolating segments that are in
series or in parallel and simplifying them to their equivalent resistances.
VARIABLE SYMBOLS
Quantities
I
current
Units
A
DIAGRAM SYMBOLS
Conversions
amperes
= C/s
= coulombs of charge
per second
R
resistance
Ω
ohms
= V/A
= volts per ampere of
current
∆V
potential
difference
V
volts
= J/C
= joules of energy per
coulomb of charge
Wire or conductor
Resistor or circuit load
(a)
(b)
Bulb or lamp
(d)
Plug
Battery / direct-current
emf source
Switch
(f)
HRW • Holt Physics
(h)
PH99PE-C20-001-005-A
Capacitor
( j)
(l)
Problem Solving
HRW • Holt Physics
See Appendix D: Equations for a summary
PH99PE-C20-001-005-A
of the equations introduced in this chapter. If
HRW problem-solving
• Holt Physics practice,
you need more
HRW • Holt Physics
HRW • Holt Physics
PH99PE-C20-001-005-A
see Appendix
I: Additional Problems.PH99PE-C20-001-005-A
DTSI Graphics
PH99PE-C20-001-005-A
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HRW • Holt Physics
PH99PE-C20-001-005-A
PH99PE-C20-001-005-A
Chapter Summary
653
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Circuits and Circuit Elements 653
C HAPTER RE V I E W
Answers
1. Schematic diagrams are useful
because they summarize the
contents of an electric circuit.
2. Accept any schematic diagram that
contains three resistors, a battery,
and a switch.
3. B, C
4. 12.0 V
5. b
6. Charges move through both the emf
source and the load.
7. When the circuit is open, there is no
complete path for charge flow and
hence no current.
8. Some of the electrical energy is
dissipated by heat, and the remainder
is converted into light energy.
9. A potential difference across the
body from contact with a faulty wire
can generate a current in the body.
10.all; D
11. b
12. a
13. Because the resistance is very low,
the current in a short circuit is high
​). Such high currents can
(I = ___
​ ΔV
R
cause wires to overheat, causing
a fire.
14.A fuse will not work in parallel
because there is an alternative path
for the current.
CHAPTER 18
Schematic Diagrams
and Circuits
10. Which of the switches in the circuit below will
complete a circuit when closed? Which will cause
a short circuit?
B
REVIEWING MAIN IDEAS
1. Why are schematic diagrams useful?
2. Draw a circuit diagram for a circuit containing three
5.0 Ω resistors, a 6.0 V battery, and a switch.
C
A
D
3. The switch in the circuit shown below can be set to
connect to points A, B, or C. Which of these connections will provide a complete circuit?
HRW
• Holt Physics or
Resistors in
Series
PH99PE-C20-CHR-002-A
in Parallel
A
B
C
4. If the batteries in a cassette recorder provide a
terminal voltageHRW
of 12.0
V, what
is the potential
• Holt
Physics
PH99PE-C20-CHR-001-A
difference across
the entire recorder?
5. In a case in which the internal resistance of a battery
is significant, which is greater?
a. the terminal voltage
b. the emf of the battery
CONCEPTUAL QUESTIONS
6. Do charges move from a source of potential difference into a load or through both the source and the
load?
7. Assuming that you want to create a circuit that has
current in it, why should there be no openings in the
circuit?
8. Suppose a 9 V battery is connected across a light bulb.
In what form is the electrical energy supplied by the
battery dissipated by the light bulb?
REVIEWING MAIN IDEAS
11. If four resistors in a circuit are connected in series,
which of the following is the same for the resistors in
the circuit?
a. potential difference across the resistors
b. current in the resistors
12. If four resistors in a circuit are in parallel, which of the
following is the same for the resistors in the circuit?
a. potential difference across the resistors
b. current in the resistors
CONCEPTUAL QUESTIONS
13. A short circuit is a circuit containing a path of very
low resistance in parallel with some other part of the
circuit. Discuss the effect of a short circuit on the
current within the portion of the circuit that has very
low resistance.
14. Fuses protect electrical devices by opening a circuit if
the current in the circuit is too high. Would a fuse
work successfully if it were connected in parallel with
the device that it is supposed to protect?
9. Why is it dangerous to use an electrical appliance
when you are in the bathtub?
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C HAPTER RE V I E W
15. What might be an advantage of using two identical
resistors in parallel that are connected in series with
another identical parallel pair, as shown below,
instead of using a single resistor?
21. The technician in item 20 finds another resistor,
so now there are three resistors with the same
resistance.
a. How many different resistances can the technician
achieve?
b. Express the effective resistance of each possibility
in terms of R.
PRACTICE PROBLEMS
22. Three identical light bulbs are connected in circuit to
a battery, as shown below. Compare the level of
brightness of each bulb when all the bulbs are
illuminated. What happens to the brightness of each
bulb if the following changes are made to the circuit?
a. Bulb A is removed from its socket.
b. Bulb C is removed from its socket.
c. A wire is connected directly between points D
and E.
d. A wire is connected directly between points D
and F.
CONCEPTUAL QUESTIONS
18 Ω
15. Because the resistors in each group
are in parallel, a broken resistor does
not open the circuit.
16.0.75 Ω
17. a. 24 Ω
b. 1.0 A
18.a. 2.2 Ω
b. 11 A
19. a. 2.99 Ω
b. 4.0 A
20.a. three combinations
b. R, 2R, __
​ R2 ​
21. a. seven combinations
R __
3R
R ___
2R ___
b. R, 2R, 3R, ​ __
2 ​,  ​ 3  ​,  ​  3   ​, ​  2   ​
22.Bulb A is brighter than bulbs B and
C. Bulbs B and C have the same
brightness.
a. Bulbs B and C stay the same.
b. Bulb A stays the same and bulb
B goes dark because the circuit
is open at C.
c. There is no change in any of the
bulbs.
d. No bulbs light because there is a
short circuit across the battery.
20. A technician has two resistors, each of which has the
same resistance, R.
a. How many different resistances can the technician
achieve?
b. Express the effective resistance of each possibility
in terms of R.
9.0 Ω
23.15 Ω
HRW • Holt Physics
PH99PE-C20-CHR-013-A
For problems 16–17,
see Sample Problem A.
16. A length of wire is cut into five equal pieces. If each
piece has a resistance of 0.15 Ω, what was the resistance of the original length of wire?
17. A 4.0 Ω resistor, an 8.0 Ω resistor, and a 12 Ω resistor
are connected in series with a 24 V battery.
Determine the following:
a. the equivalent resistance for the circuit
b. the current in the circuit
18. The resistors in item 17 are connected in parallel
across a 24 V battery. Determine the following:
a. the equivalent resistance for the circuit
b. the current delivered by the battery
19. An 18.0 Ω resistor, 9.00 Ω resistor, and 6.00 Ω resistor
are connected in parallel across a 12 V battery.
Determine the following:
a. the equivalent resistance for the circuit
b. the current delivered by the battery
A
D
For problems 18–19, see Sample Problem B.
B
C
E
F
9.0 volts
PRACTICE PROBLEMS
For problems 23–24, see Sample Problem C.
Complex Resistor
Combinations
23. Find the equivalent resistance of the circuit shown in
the figure below.
30.0 V
12 Ω
6.0 Ω
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Circuits and Circuit Elements 655
C HAPTER RE V I E W
2 4.13.3 Ω
25.3.0 Ω: 1.8 A, 5.4 V
6.0 Ω: 1.1 A, 6.5 V
9.0 Ω: 0.72 A, 6.5 V
26.a. 1.7 A
b. 3.4 V
c. 5.1 V
d. 0.42 A
27.28 V
28.2.2 V
29.3.8 V
30.3.0 × 101 V
31. a. 33.0 Ω
b. 132 V
c. 4.00 A, 4.00 A
32.a. Place one 20 Ω resistor in series
with two parallel 50 Ω resistors.
b. Place two parallel 50 Ω resistors in
series with two parallel 20 Ω
resistors, or place two circuits,
each composed of a 20 Ω resistor
in series with a 50 Ω resistor,
in parallel.
33.10.0 Ω
34.1875 Ω
CHAPTER REVIEW
24. Find the equivalent resistance of the circuit shown in
the figure below.
7.0 Ω
12.0 V
7.0 Ω
7.0 Ω
1.5 Ω
7.0 Ω
For problems 25–26, see Sample Problem D.
HRW • Holt Physics
25. For the circuitPH99PE-C20-CHR-005-A
shown below, determine the current in
each resistor and the potential difference across each
resistor.
6.0 Ω
9.0 Ω
3.0 Ω
12 V
26. For the circuit shown in the figure below, determine
the following: HRW • Holt Physics
PH99PE-C20-CHR-007-A
6.0 Ω
3.0 Ω
3.0 Ω
6.0 Ω
4.0 Ω
2.0 Ω
12.0 Ω
18.0 V
29. A 9.0 Ω resistor and a 6.0 Ω resistor are connected in
series to a battery, and the current through the 9.0 Ω
resistor is 0.25 A. What is the potential difference
across the battery?
30. A 9.0 Ω resistor and a 6.0 Ω resistor are connected in
series with an emf source. The potential difference
across the 6.0 Ω resistor is measured with a voltmeter
to be 12 V. Find the potential difference across the
emf source.
31. An 18.0 Ω, 9.00 Ω, and 6.00 Ω resistor are connected
in series with an emf source. The current in the
9.00 Ω resistor is measured to be 4.00 A.
a. Calculate the equivalent resistance of the three
resistors in the circuit.
b. Find the potential difference across the emf
source.
c. Find the current in the other resistors.
32. The stockroom has only 20 Ω and 50 Ω resistors.
a. You need a resistance of 45 Ω. How can this
resistance be achieved using three resistors?
b. Describe two ways to achieve a resistance of 35 Ω
using four resistors.
33. The equivalent resistance of the circuit shown
below is 60.0 Ω. Use the diagram to determine the
value of R.
R
a.
b.
c.
d.
the current in the 2.0 Ω resistor
• Holt Physics
the potentialHRW
difference
across the 2.0 Ω resistor
PH99PE-C20-CHR-006-A
the potential difference across the 12.0 Ω resistor
the current in the 12.0 Ω resistor
Mixed Review
REVIEWING MAIN IDEAS
27. An 8.0 Ω resistor and a 6.0 Ω resistor are connected in
series with a battery. The potential difference across
the 6.0 Ω resistor is measured as 12 V. Find the
potential difference across the battery.
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656 Chapter 18
28. A 9.0 Ω resistor and a 6.0 Ω resistor are connected in
parallel to a battery, and the current in the 9.0 Ω
resistor is found to be 0.25 A. Find the potential
difference across the battery.
90.0 Ω
10.0 Ω
10.0 Ω
90.0 Ω
34. Two identical parallel-wired strings of 25 bulbs are
connected to each other in series. If the equivalent
HRW • Holt Physics
resistance of the combination
is 150.0 Ω and it is
PH99PE-C20-CHR-013-A
connected across a potential difference of 120.0 V,
what is the resistance of each individual bulb?
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CHAPTER REVIEW
35. The figures (a)–(e) below depict five resistance
diagrams. Each individual resistance is 6.0 Ω.
(a)
(d)
(b)
39. A resistor with an unknown resistance is connected
in parallel to a 12 Ω resistor. When both resistors are
connected to an emf source of 12 V, the current in the
unknown resistor is measured with an ammeter to be
3.0 A. What is the resistance of the unknown resistor?
40. The resistors described in item 37 are reconnected in
parallel to the same 18.0 V battery. Find the current in
each resistor and the potential difference across each
resistor.
(e)
(c)
a. Which resistance combination has the largest
equivalent resistance?
HRW combination
• Holt Physics
b. Which resistance
has the smallest
PH99PE-C20-CHR-011-A
equivalent
resistance?
c. Which resistance combination has an equivalent
resistance of 4.0 Ω?
d. Which resistance combination has an equivalent
resistance of 9.0 Ω?
41. The equivalent resistance for the circuit shown below
drops to one-half its original value when the switch,
S, is closed. Determine the value of R.
R
36. Three small lamps are connected to a 9.0 V battery, as
shown below.
R 1 = 4.5 Ω
R 2 = 3.0 Ω
R 3 = 2.0 Ω
9.0 V
a.
b.
c.
d.
What is the equivalent resistance of this circuit?
What is the current in the battery?
HRW • Holt Physics
What is the
current in each bulb?
PH99PE-C20-CHR-012-A
What is the potential difference across each bulb?
10.0 Ω
10.0 Ω
90.0 Ω
3 5.a. a
b. c
c. d
d. e
36.a. 5.7 Ω
b. 1.6 A
c. 1.6 A (R1), 0.63 A (R2), 0.95 A (R3)
d. 7.2 V (R1), 1.9 V (R2), 1.9 V (R3)
37. 18.0 Ω: 0.750 A, 13.5 V
6.0 Ω: 0.750 A, 4.5 V
38.a.
30.0 Ω
15.0 Ω
42. You can obtain only four 20.0 Ω resistors from the
HRW • Holt Physics
stockroom.
PH99PE-C20-CHR-008-A
a. How can you achieve a resistance of 50.0 Ω under
these circumstances?
b. What can you do if you need a 5.0 Ω resistor?
43. Four resistors are connected to a battery with a
terminal voltage of 12.0 V, as shown below.
Determine the following:
30.0 Ω
50.0 Ω
90.0 Ω
37. An 18.0 Ω resistor and a 6.0 Ω resistor are connected
in series to an 18.0 V battery. Find the current in and
the potential difference across each resistor.
38. A 30.0 Ω resistor is connected in parallel to a 15.0 Ω
resistor. These are joined in series to a 5.00 Ω resistor
and a source with a potential difference of 30.0 V.
a. Draw a schematic diagram for this circuit.
b. Calculate the equivalent resistance.
c. Calculate the current in each resistor.
d. Calculate the potential difference across each
resistor.
90.0 Ω
S
C HAPTER RE V I E W
20.0 Ω
12.0 V
a.
b.
c.
d.
e.
the equivalent resistance for the circuit
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• Holt Physics
the current in the
battery
the current inPH99PE-C20-CHR-010-A
the 30.0 Ω resistor
the power dissipated by the 50.0 Ω resistor
the power dissipated by the 20.0 Ω resistor
(∆V)2
(Hint: Remember that P = _ = I∆V.)
R
Chapter Review
657
5.00 Ω
30.0 V
b. 15.00 Ω
• Holt
Physics
HRW
c. 5.00
Ω: 2.00
A;
PH99TE-C20-CHR-001-A
15.0 Ω: 1.33 A;
30.0 Ω: 0.667 A
d. 5.00 Ω: 10.0 V;
15.0 Ω: 20.0 V;
30.0 Ω: 20.0 V
39.4.0 Ω
40.18.0 Ω: 1.00 A, 18.0 V
6.0 Ω: 3.0 A, 18.0 V
41.13.96 Ω
42.a. two resistors in series with two
parallel resistors
b. four parallel resistors
43.a. 62.4 Ω
b. 0.192 A
c. 0.102 A
d. 0.520 W
e. 0.737 W
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Circuits and Circuit Elements 657
C HAPTER RE V I E W
4.6.0 Ω (A), 3.0 Ω (B)
4
45.The circuit must contain three
groups of resistors—each containing
three resistors in parallel—that are
connected to one another in series.
46.a. 14.0 Ω
b. 2.0 A
47.a. 5.1 Ω
b. 4.5 V
48.no; Assuming devices are wired in
parallel, total current is 20 A. The
circuit breaker will open when the
devices are both on.
49.a. 11 A (heater), 9.2 A (toaster),
12 A (grill)
b. The total current is 32.2 A, so the
30.0 A circuit breaker will open
the circuit if these appliances are
all on.
CHAPTER REVIEW
44. Two resistors, A and B, are connected in series to a
6.0 V battery. A voltmeter connected across resistor
A measures a potential difference of 4.0 V. When the
two resistors are connected in parallel across the
6.0 V battery, the current in B is found to be 2.0 A.
Find the resistances of A and B.
45. Draw a schematic diagram of nine 100 Ω resistors
arranged in a series-parallel network so that the total
resistance of the network is also 100 Ω. All nine
resistors must be used.
46. For the circuit below, find the following:
28 V
5.0 Ω
3.0 Ω
3.0 Ω
10.0 Ω
10.0 Ω
4.0 Ω
4.0 Ω
2.0 Ω
3.0 Ω
a. the equivalent
resistance of the circuit
HRW • Holt Physics
b. the current
in the 5.0 Ω resistor
PH99PE-C20-CHR-014-A
47. The power supplied to the circuit shown below is
4.00 W. Determine the following:
10.0 Ω
4.0 Ω
3.0 Ω
5.0 Ω
3.0 Ω
a. the equivalent resistance of the circuit
b. the potential difference across the battery
48. Your toaster oven and coffee maker each dissipate
1200 W of power. Can you operate both of these
HRW • Holt Physics
appliances at the
same time if the 120 V line you use
PH99PE-C20-CHR-009-A
in your kitchen has a circuit breaker rated at 15 A?
Explain.
49. An electric heater is rated at 1300 W, a toaster is rated
at 1100 W, and an electric grill is rated at 1500 W. The
three appliances are connected in parallel across a
120 V emf source.
a. Find the current in each appliance.
b. Is a 30.0 A circuit breaker sufficient in this situation? Explain.
Parallel Resistors
Electric circuits are often composed of combinations of series
and parallel circuits. The overall resistance of a circuit is
determined by dividing the circuit into groups of series and
parallel resistors and determining the equivalent resistance of
each group. As you learned earlier in this chapter, the equivalent resistance of parallel resistors is given by the following
equation:
In this graphing calculator activity, you will determine the
equivalent resistance for various resistors in parallel. You will
confirm that the equivalent resistance is always less than the
smallest resistor, and you will relate the number of resistors
and changes in resistance to the equivalent resistance.
Go online to HMDScience.com to find this graphing
calculator activity.
1 +_
1 +_
1 +···
1 =_
_
Req R1 R2 R3
One interesting consequence of this equation is that the
equivalent resistance for resistors in parallel will always be
less than the smallest resistor in the group.
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Chapter 18
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CHAPTER REVIEW
ALTERNATIVE ASSESSMENT
1. How many ways can two or more batteries be connected in a circuit with a light bulb? How will the
current change depending on the arrangement? First
draw diagrams of the circuits you want to test. Then
identify the measurements you need to make to
answer the question. If your teacher approves your
plan, obtain the necessary equipment and perform
the experiment.
2. Research the career of an electrical engineer or
technician. Prepare materials for people interested in
this career field. Include information on where
people in this career field work, which tools and
equipment they use, and the challenges of their field.
Indicate what training is typically necessary to enter
the field.
4. You and your friend want to start a business exporting
small electrical appliances. You have found people
willing to be your partners to distribute these appliances in Germany. Write a letter to these potential
partners that describes your product line and that
asks for the information you will need about the
electric power, sources, consumption, and distribution in Germany.
5. Contact an electrician, builder, or contractor, and ask
to see a house electrical plan. Study the diagram to
identify the circuit breakers, their connections to
different appliances in the home, and the limitations
they impose on the circuit’s design. Find out how
much current, on average, is in each appliance in the
house. Draw a diagram of the house, showing which
circuit breakers control which appliances. Your
diagram should also keep the current in each of these
appliances under the performance and safety limits.
3. The manager of an automotive repair shop has been
contacted by two competing firms that are selling
ammeters to be used in testing automobile electrical
systems. One firm has published claims that its
ammeter is better because it has high internal
resistance. The other firm has published claims that
its ammeter is better because it has low resistance.
Write a report with your recommendation to the
manager of the automotive repair shop. Include
diagrams and calculations that explain how you
reached your conclusion.
Chapter Review
C HAPTER RE V I E W
Alternative
Assessment Answers
1. Students’ plans should be safe and
should test series and parallel
combinations of batteries.
2. Students should recognize that the
principles of circuits are applied by
electrical engineers and technicians.
3. An ammeter is connected in series
in a circuit, so it must have low
internal resistance in order to
measure all parts of the circuit
without interfering with it.
4. Students’ letters should formulate
clear direct questions and request
meaningful information.
5. Students should draw several parallel
sets of appliances, each of which
shows the appliances wired in series
to a circuit breaker.
659
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Circuits and Circuit Elements 659
S TA N D A R D S - B A S E D
ASSESSMENT
Answers
1. C
2. J
3. B
4. F
5. B
6. J
7. D
8. G
Standards-Based Assessment
MULTIPLE CHOICE
1. Which of the following is the correct term for a
circuit that does not have a closed-loop path for
electron flow?
A. closed circuit
B. dead circuit
C. open circuit
D. short circuit
2. Which of the following is the correct term for a
circuit in which the load has been unintentionally
bypassed?
F. closed circuit
G. dead circuit
H. open circuit
J. short circuit
5. Which of the following is the correct equation for
the current in the resistor?
A. I = IA + IB + IC
∆V
B. IB = _
Req
C. IB = Itotal + IA
∆V
D. IB = _
RB
Use the diagram below to answer questions 6–7.
A
B
C
Use the diagram below to answer questions 3–5.
A
B
6. Which of the following is the correct equation for
the equivalent resistance of the circuit?
F. Req = RA + RB + RC
1 =_
1 +_
1 +_
1
G. _
Req RA RB RC
C
3. Which of the circuit elements contribute to the load
of the circuit?
A. Only A
B. A and B, but not C
C. Only C
D. A, B, and C
4. Which of the following is the correct equation for
the equivalent resistance of the circuit?
F. Req = RA + RB
1 =_
1 +_
1
G. _
Req RA RB
H. Req = I∆V
1 =_
1 +_
1 +_
1
J. _
Req RA RB RC
660
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660 Chapter 18
H. Req = I∆V
(
)
1 +_
1
J. Req = RA + _
RB RC
-1
7. Which of the following is the correct equation for
the current in resistor B?
A. I = IA + IB + IC
∆V
B. IB = _
Req
C. IB = Itotal + IA
∆VB
D. IB = _
RB
8. Three 2.0 Ω resistors are connected in series to a
12 V battery. What is the potential difference across
each resistor?
F. 2.0 V
G. 4.0 V
H. 12 V
J. 36 V
Chapter 18
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TEST PREP
Use the following passage to answer questions 9–11.
EXTENDED RESPONSE
Six light bulbs are connected in parallel to a 9.0 V
battery. Each bulb has a resistance of 3.0 Ω.
15. Using standard symbols for circuit elements, draw a
diagram of a circuit that contains a battery, an open
switch, and a light bulb in parallel with a resistor.
Add an arrow to indicate the direction of current if
the switch were closed.
9. What is the potential difference across each bulb?
A. 1.5 V
B. 3.0 V
C. 9.0 V
D. 27 V
Use the diagram below to answer questions 16–17.
10. What is the current in each bulb?
F. 0.5 A
G. 3.0 A
H. 4.5 A
J. 18 A
11. What is the total current in the circuit?
A. 0.5 A
B. 3.0 A
C. 4.5 A
D. 18 A
SHORT RESPONSE
12. Which is greater, a battery’s terminal voltage or
the same battery’s emf? Explain why these two
quantities are not equal.
13. Describe how a short circuit could lead to a fire.
14. Explain the advantage of wiring the bulbs in a string
of decorative lights in parallel rather than in series.
1.5 Ω
6.0 Ω
12 V
R = 3.0 Ω
16. For the circuit shown, calculate the following:
a. the equivalent resistance of the circuit
b. the current in the light bulb.
Show all your work for both calculations.
9. C
10.G
11. D
12. A battery’s emf is slightly greater
than its terminal voltage. The
difference is due to the battery’s
internal resistance.
13. In a short circuit, the equivalent
resistance of the circuit drops very
low, causing the current to be very
high. The higher current can cause
wires still in the circuit to overheat,
which may in turn cause a fire in
materials contacting the wires.
14.If one bulb is removed, the other
bulbs will still carry current.
15.
17. After a period of time, the 6.0 Ω resistor fails and
breaks. Describe what happens to the brightness of
the bulb. Support your answer.
18. Find the current in and potential difference across
each of the resistors in the following circuits:
a. a 4.0 Ω and a 12.0 Ω resistor wired in series with
a 4.0 V source.
b. a 4.0 Ω and a 12.0 Ω resistor wired in parallel
with a 4.0 V source.
Show all your work for each calculation.
19. Find the current in and potential difference across
each of the resistors in the following circuits:
a. a 150 Ω and a 180 Ω resistor wired in series with a
12 V source.
b. a 150 Ω and a 180 Ω resistor wired in parallel
with a 12 V source.
Show all your work for each calculation.
10
9
8
11 12 1
7 6 5
Test Tip
2
3
4
Prepare yourself for taking an important
test by getting plenty of sleep the
night before and by eating a healthy
breakfast on the day of the test.
Standards-Based Assessment
661
I
16.a. 4.2 Ω b. 2.9 A (Go online to see
the full solution.)
17. The bulb will grow dim. The loss of
the 6.0 Ω resistor causes the
equivalent resistance of the circuit to
increase to 4.5 Ω. As a result, the current in the bulb drops to 2.7 A, and
the brightness of the bulb decreases.
18.a. 4.0 Ω: 0.25 A, 1.0 V
12.0 Ω: 0.25 A, 3.0 V
b. 4.0 Ω: 1.0 A, 4.0 V
12.0 Ω: 0.33 A, 4.0 V
19. a. 150 Ω: 0.036 A, 5.4 V
180 Ω: 0.036 A, 6.5 V
b. 150 Ω: 0.080 A, 12 V
180 Ω: 0.067 A, 12 V
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Circuits and Circuit Elements 661