U N I T
E
364
Electricity and
Magnetism
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Unit Contents
11
The principles of conservation of energy and
charge apply to electrical circuits.
11.1 Electrical Circuits
11.2 Series Circuits
11.3 Parallel and Mixed Circuits
11.4 Power Consumption
12
Properties of magnetic fields apply in nature
and technology.
12.1 Magnetic Forces and Fields
12.2 The Motor Principle
12.3 Using Electromagnetism
13
Electromagnetic induction is used to generate
most of the electrical energy used today.
13.1 Using Magnetism to Induce an Electric Current
13.2 The Generator and Electrical Energy Generation
13.3 The Transmission of Electrical Energy
Unit Task
The first electric motor was invented over 180 years ago.
The design of the motor has vastly improved over the years.
Today, most motors are inexpensive, reliable, and environmentally
friendly. Using concepts learned in this unit, you will research,
design, and build a small electric DC motor that can be powered
using a 9-V battery. You will build your motor using components
found at a hardware store. The function of your motor will be
to lift as much weight as possible without stalling.
DISCOVERING PHYSICS
A city looks beautiful at night. The buildings and signs are lit up and create a wonderful
mosaic of colours. Each light, whether it is in a building, on a sign, or a street light, is part
of the city’s electrical circuit. Electrical engineers help to design and plan the huge electrical
circuit that provides electricity to the buildings and homes in a city. The electrical circuit
must be able to expand as more houses and buildings are added. It must also be designed
to provide electricity to different areas of the city so that a power outage in one area does
not affect other areas in the city. Since the electrical needs of different parts of the city
vary, the amount of electricity provided must be regulated through the different regions.
What principles determine the way in which the electrical circuit of a city is designed?
What devices are used to control the electricity as it flows through the city?
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CHAPTER
11
Learning
Expectations
By the end of this chapter,
you will:
Developing Skills of
Investigation and
Communication
●
●
●
●
use appropriate terminology
related to electricity, including:
direct current, alternating
current, conventional current,
electron flow, electrical
potential difference, electrical
resistance, and power
analyze diagrams of series,
parallel, and mixed circuits
with reference to Ohm’s law
and Kirchhoff’s laws
The principles of conservation
of energy and charge apply to
electrical circuits.
E
lectricity is mysterious and fascinating. You cannot see it move,
you cannot hold it in your hand, it is not even visible — except
when there is a spark — although it is everywhere. Even though
we depend on electricity, most people only think about it when the power
goes out or when they get an electrical shock. We do not stop to think
about all the electrical devices that make our lives easier. Imagine what
your life would be like without TVs, cell phones, computers, MP3 players,
cars, refrigerators, electric stoves and gas furnaces, medical imaging
devices such as magnetic resonance imaging (MRI) and X-ray machines,
and light bulbs (Figure 11.1).
The number of electrical devices that surround us is staggering to
contemplate. And yet, the first practical method of generating electricity
was discovered only 200 years ago when Italian physicist Alessandro
Volta (1745–1827) made the first battery. However, whether it is a cell
phone or a car, all electrical devices contain many of the same components,
and all obey the same physical laws.
design and build real or
computer-simulated mixed
direct current (DC) circuits,
and explain the circuits with
reference to direct current,
potential difference, and
resistance
solve problems involving energy,
power, potential difference,
and current
Understanding Basic Concepts
●
●
●
distinguish between conventional
current and electron flow
distinguish between alternating
current (AC) and direct
current (DC)
explain Ohm’s law and Kirchhoff’s
laws in relation to electricity
Figure 11.1
The Eiffel Tower
looks very impressive
when it is lit up at
night. It has over
20 000 light bulbs,
over 40 km of wires,
and consumes over
120 000 W of power.
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11.1 Electrical Circuits
Section Summary
●
An electrical circuit contains an energy source, conductors, and a load.
●
Ohm’s law describes the relationship between electric current, potential
difference, and resistance in a circuit.
In the late 1700s, Italian scientist Luigi Galvani (1737–1798) performed
a dissection on a dead frog. He discovered that when he touched the leg
muscles of the frog with two different metals, the muscles contracted. He
concluded that these metals somehow released the “animal electricity”
that was stored in the frog. Although Galvani’s conclusion was wrong, his
discovery turned out to be very important.
Alessandro Volta recognized the implications of Galvani’s discovery.
He showed that, instead of the metals releasing electrical energy stored
in the frog, the two metals produced electrical energy, which caused the
legs to contract. He experimented with different metals to see how much
continuous electrical energy he could create. He conducted his experiments
by placing two different metals in contact and then touching them to his
tongue, which could detect a small steady flow of electrical energy.
After much experimentation, Volta placed a pile of copper and zinc
discs on top of one another separated by cardboard discs soaked in a
sulphuric acid solution. He attached wires to each end of the pile and,
for the first time ever, was able to create a large, continuous flow of
electricity. He had created a battery (Figure 11.2). A battery is a device
that converts stored chemical potential energy into electrical energy and
is capable of providing a steady flow of current electricity.
Soon after the invention of the battery, scientists began experimenting
with current electricity and circuits. Current electricity is the flow of
charged particles along a conductor. A charged particle is a particle that has
an electrical charge, such as a proton (positive) or an electron (negative).
⫺
discs
zinc
copper
⫹
Figure 11.2 Volta created the first
battery in 1800.
Circuit Fundamentals
To operate, an electrical device requires electrical energy, which is
provided by a steady flow of charged particles along a closed loop. This
closed loop is called an electrical circuit. The circuit must form a closed
loop so that the charged particles charges moving through the conductor
can return to the battery. A circuit contains:
• a source of electrical energy, such as a battery
• a conductor, such as a wire
• a load that changes electrical energy into light, sound, heat, or motion
As the charged particles travel through a circuit, they carry the
electrical energy from the source (battery) to the load, which converts
the electrical energy into another form of energy. Note that circuits can
be very complex and can contain different types of loads and devices. To
simplify drawing circuits, we use circuit symbols.
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Chapter 11 The principles of conservation of energy and charge apply to electrical circuits.
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Circuit symbols are used to represent the various components in
drawings of electrical circuits (Figure 11.3). Table 11.1 shows the symbols
of some of the most common components.
energy source
electrical load
conducting
wires
switch
Figure 11.3 A simple circuit
containing a battery, a switch,
and a load (the light bulb).
Table 11.1
Circuit Symbols used in Circuit Diagrams
Symbol
Component
Description
wire
conductor; provides a path for current flow
battery
provides electrical energy to the circuit;
longer side is the positive terminal
variable DC
power source
provides a variable amount of energy
to a circuit
ground
electrical connection to ground that prevents
a shock to a person
switch
opens or closes the circuit
light bulb
type of load that converts electrical energy
to light energy
resistor
general load that converts electrical energy
to another form of energy (heat)
ammeter
device that measures the current in a circuit
voltmeter
device that measures the potential difference
in a circuit
To understand what happens as a circuit operates, we need to take a
closer look at the role of charges and the energy they transmit.
Current
Highway 401 in the Toronto area is the busiest highway in North America.
If you were to stand beside the highway and count the cars as they passed,
you would find that about 17 500 cars pass by every hour — that is about
4.8 cars a second or over 420 000 cars a day! City planners find it useful to
know the number of vehicles that use the roads to develop future roadways.
Similarly when looking at electrical circuits, it is useful to know the
quantity of charge that passes by a point in the circuit per second. The
amount of charge transferred per unit time is referred to as current.
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Conventional Current and Electron Flow
PHYSICS INSIGHT
When scientists first began studying circuits, they assumed that positive
charges flowed through the wires in a circuit (Figure 11.4). They called
the flow of positive charges in a circuit conventional current. Since unlike
charges attract, conventional current is considered to flow from the
positive terminal of a battery, around the circuit to the negative terminal.
After scientists learned more about the structure of atoms, they
concluded that current consists of free electrons, which are negatively
charged. During the operation of a circuit, the electrons move from the
negative terminal of the battery to the positive terminal. The movement
of electrons in a circuit is called electron flow. For the purposes of circuit
analysis, it makes no difference whether we talk about conventional
current or electron flow. Throughout the rest of this unit, we will
stay with the convention and use the term conventional current.
Current is represented by the symbol I, and is measured in
amperes (A). Since current is the amount of charge that is
transferred per unit time, the equation for current is:
q
I_
t
Conventional current assumes that
positive charges (protons) move
through the circuit from the positive
terminal of the power supply to
the negative terminal. This model
is incorrect, but has become
entrenched over many years. Electron
flow assumes that electrons move
through the circuit from the negative
terminal to the positive terminal.
point
+
+
+
+
+
+
+
+
wire
+
+
+
+
charges
Figure 11.4 In conventional
current, positive charges move
from the positive terminal to the
negative terminal.
where I is the current in amperes (A), q is the amount of charge in
coulombs (C), and t is the time in seconds (s).
Direct Current (DC)
A battery provides a steady flow of current in one direction, known
as direct current, or DC. Strictly speaking, DC can fluctuate, but cannot
change directions. DC is used in all electrical equipment that requires an
adapter when you plug it into a wall outlet or is powered by a battery.
PHYSICS INSIGHT
A coulomb is equivalent to the charge
on 6.25 1018 electrons or protons.
An ampere is a flow of 1 C of charge
past a point in a conductor in 1 s.
Alternating Current (AC)
A wall socket provides alternating current. Alternating current, or AC,
changes direction periodically. That is, the charges in the wire move back
and forth over the same spot and do not actually move from one terminal to
another. This type of current is used in the wiring of your house. You will
study the reasons why AC is used in chapter 13.
Example 11.1
A battery delivers a charge of 9.00 C in 1.00 min of operation.
What amount of current is generated in mA? The SI prefix
milli- (m) is equal to 103.
Given
q 9.00 C
t 1.00 min 60.0 s
Required
current (I)
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Practice Problems
1. A D-cell battery delivers a charge
of 200.0 C in 65.0 s. Determine the
current produced by this battery.
2. A car battery provides a current of
600.0 A for 2.48 s. Determine the
charge provided by the battery.
3. A battery has a total charge
capacity of 10 800 C. For how
long can this battery deliver a
current of 450 mA?
Chapter 11 The principles of conservation of energy and charge apply to electrical circuits.
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Analysis and Solution
q
I _
t
9.00 C
__
60.0 s
Answers
1. 3.08 A
2. 1.49 103 C
3. 2.4 104 s
0.150 C/s
0.150 A
Paraphrase
The current delivered by this battery is 0.150 A, or 150 mA.
Measuring Current
Figure 11.5 An ammeter in a circuit
A device called an ammeter is used to measure current. Figure 11.5 shows
the proper placement of an ammeter in a circuit. The ammeter is placed
into the circuit so that the current flows through it. Since the current cannot
bypass the meter, the meter measures the entire current.
Resistance
Figure 11.6 Resistors come in
many shapes and sizes to meet
different requirements.
PHYSICS• SOURCE
Suggested Activity
●
E1 Quick Lab Overview on
page 374
Electrical charges move through a circuit with little or no room between
them. If there is something that restricts the flow of current in one spot
in the circuit, the effect is felt throughout the entire circuit.
Resistance is the degree to which the flow of current is opposed in
a circuit. A resistor is a device that resists or restricts the flow of current
(Figure 11.6). As charges move through a resistor, the resistor removes
energy from the charges and converts it to another type of energy. The
energy usually takes the form of heat. For example, a light bulb is a
resistor that converts the energy of the charges into heat and light.
Almost all components offer some resistance in a circuit, even if that is
not their primary role. We will assume that the resistance of a component
is constant.
The Battery and Potential Difference
For a circuit to do anything useful, current must flow through it. But
current does not move by itself — there must be a battery that forces the
electrons through the circuit and provides energy to the components.
To understand fully how circuits work, we need to take a closer look
at the role of the battery.
As the charges pass through a battery, it increases their
potential energy. We can think of a battery as being similar to
water pump
a water pump. The pump increases the potential energy of
water by lifting it to a certain height (Figure 11.7). The potential
energy of the water depends on the mass of the water and the
height to which it is lifted. Therefore it is useful to define a
reference point where the potential energy is zero, which is at
the bottom of the pump. The maximum potential energy is at
the top of the pump. For electrical potential energy we will
water reservoir
take the negative terminal of the battery as the point of zero
potential energy.
Figure 11.7 At the bottom of the water pump, the
potential energy of the water is defined as zero.
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Electrical Potential
In an electrical circuit, it is not practical to refer to the potential energy
of the charges as they move through the circuit. This is because potential
energy depends on the quantity of charge, which changes as the circuit
operates. For example, an electric motor may use more charge per second
under a heavy load than when it has no load. It is more practical to use a
measurement that is independent of the amount of charge flowing in the
circuit. For this reason, we use the term electrical potential.
We define electrical potential as the electrical potential energy per
unit charge. Electrical potential is represented by the symbol V, and its
units are volts (V). It can be written mathematically as:
PHYSICS INSIGHT
The symbol for a battery is shown
below. The longer line is the positive
terminal and the shorter line is the
negative terminal.
E
V_
q
where V is the electrical potential in volts (V), E is the electrical potential
energy in joules (J), and q is the charge in coulombs (C). Note that the
electrical potential is the same whether there are many charges (large
current) or few charges (small current) flowing through the circuit.
Potential Difference
As charges pass through a load in the circuit, they transfer energy to the
load. The charges have a greater electrical potential before they pass
through the load than after they pass through the load. This change
in potential is referred to as the potential difference (V). Potential
difference is measured in volts (V). Potential difference is also referred
to as voltage. The potential difference is always measured between two
points in the circuit (Figure 11.8).
We can determine the change in potential by subtracting the initial
potential from the final potential:
V Vfinal Vinitial
( ) (
Einitial
Efinal
_
V _
q q
light
light
bulb
Vfinal
Vinitial
Vi Vf
Figure 11.8 The potential of the
charges before they pass through a
light bulb is greater than after they
pass through the light bulb. Some
of the energy is converted to light.
)
E
V _
q
where V is the potential difference in volts (V), E is the change in
potential energy of the charges as they pass through a load in joules (J),
and q is the charge in coulombs (C).
A load within a circuit uses energy and decreases the potential.
This creates a negative potential difference, which is also called a voltage
drop, across the component. A voltage drop implies a loss of energy so a
negative sign is not usually used. The combination of all the voltage drops
in a circuit will decrease the potential by exactly the same amount as the
battery increases the potential.
PHYSICS INSIGHT
Do not confuse voltage with volts.
Voltage is potential difference and has
the symbol ΔV. The volt is the unit
for potential difference and has the
symbol V.
Measuring Potential Difference
To measure potential difference, we use a voltmeter. The voltmeter must
be placed across two points in the circuit and it will measure the voltage
drop across that portion of the circuit. Figure 11.9 shows the proper
placement of a voltmeter in a circuit.
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Figure 11.9 A voltmeter is placed
across a component.
Chapter 11 The principles of conservation of energy and charge apply to electrical circuits.
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Example 11.2
Practice Problems
1. A potential difference of
120.0 V is measured across
a light bulb. The light bulb
is left on for 30 min allowing
a charge of 900 C to flow
through it. How much energy
is converted to light and heat?
2. A charge of 50.0 C has a
change in potential energy
of 1.00 103 J as it flows
through a resistor. What is
the potential difference
across the resistor?
1. 1.08 105 J
PHYSICS• SOURCE
Concept Check
Suggested Activities
●
E3 Inquiry Activity Overview on
page 374
Required
change in potential energy (E)
Paraphrase
The amount of electrical energy converted to heat is 200 J.
2. 20.0 V
E2 Skill Builder Overview on
page 374
Given
V 10.0 V
q 20.0 C
Analysis and Solution
E
V _
q
E qV
(20.0 C)(10.0 V)
200 CV
200 J
Answers
●
A potential difference of 10.0 V is measured across a resistor in a
circuit. If a charge of 20.0 C passes through the resistor, how much
electrical energy is dissipated as heat?
1. Explain what current is.
2. Explain the difference between how an ammeter is connected in a circuit with how
a voltmeter is connected.
3. What is the difference between direct current and alternating current?
Ohm’s Law
PHYSICS• SOURCE
Explore More
What are the effects of Ohm’s law in
a simple circuit?
Twenty-seven years after the battery was invented, German scientist Georg
Ohm (1787–1854) determined the relationship between potential difference,
current, and resistance. During his experiments, Ohm applied different
voltages to a resistor and measured the corresponding current and voltage
drop across it. Ohm recorded and made a graph of his data (Figure 11.10).
Potential
Difference (V)
Potential Difference vs. Current
Figure 11.10 Ohm showed a
linear relationship between the
potential difference and current.
Current (A)
PHYSICS INSIGHT
It is common to write Ohm’s
equation as V IR. However, this
equation is not strictly correct because
V represents electrical potential.
372
The slope of this line is the resistance. Because the relationship
between potential difference and current is linear, the resistance does
not change. Ohm wrote this relationship mathematically as:
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V
R_
I
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where R is the resistance in ohms (), V is the potential difference in
volts (V), and I is the current in amperes (A). This mathematical equation
is known as Ohm’s law. Ohm’s law is often rewritten as:
PHYSICS• SOURCE
Take It Further
V IR
Circuits are designed with
different kinds of components.
Research five components. Be
prepared to show the symbol of
each component and provide a brief
description of its purpose.
Note that not all components obey Ohm’s law. For example,
light-emitting diodes (LEDs), diodes, transistors, and fluorescent lights
do not obey Ohm’s law. In this unit, we will assume that all circuit
components obey Ohm’s law.
Example 11.3
A student is asked to determine the value of the resistor in a
circuit (Figure 11.11). Table 11.2 shows the data obtained when
the current was varied.
Table 11.2
Results of Experiment
Current (A)
Potential Difference (V)
0.00
0.0
1.00
70.0
2.00
155.0
4.00
310.0
4.50
327.5
variable power
supply
Current (A)
voltmeter
Figure 11.11
Given
values of potential difference (see Table 11.2)
values of current (see Table 11.2)
Required
the resistance (R)
262 V 37 V
R ___
3.5 A 0.5 A
225 V
__
3.0 A
75 Potential Difference (V)
Analysis and Solution
The current is the independent variable and the potential difference
is the dependent variable. Figure 11.12 shows a graph of the data.
The slope of the line of best fit is the resistance.
Choose two points from
the line of best fit.
Potential Difference vs. Current
400
350
300
250
200
150
100
50
0
0
1
2
3
4
5
Current (A)
Figure 11.12
Paraphrase
The resistor has a resistance of 75 .
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1. Determine the resistance
of a circuit using the
following data.
ammeter
resistor
y
V R
slope _ _
x
I
Practice Problems
Potential
Difference (V)
0.00
0.0
1.00
58.0
2.00
108.0
4.00
220.0
4.50
245.0
2. A student performs an
experiment similar to
Example 11.3. The student
determines the resistor to
have a resistance of 130 .
Copy the following table into
your notebook and fill in the
missing values.
Current (A)
Potential
Difference (V)
0.00
0.0
1.50
292.5
4.00
780.0
Answers
1. 55 2. Current: 2.25 A and 6.00 A
Potential difference: 195.0 V and 520.0 V
Chapter 11 The principles of conservation of energy and charge apply to electrical circuits.
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E1
Quick Lab
PHYSICS• SOURCE
Creating a Pile Battery
Purpose
To create a battery using copper and zinc metals
Activity Overview
In this Quick Lab, you will build a battery using a lemon and two metals.
Your teacher will give you a copy of the full activity.
Prelab Questions
Figure 11.13 Activity setup
Consider the questions below before beginning this activity.
1. Can an electrical potential be created between two dissimilar metals?
2. How does the electrical potential difference change when two batteries are connected together?
E2
Skill Builder Activity
PHYSICS• SOURCE
Using an Ammeter and a Voltmeter
Activity Overview
In this Skill Builder, you will learn how to set up and use an ammeter,
which measures current, and a voltmeter, which measures potential
difference. The ammeter is connected in line with the resistor. The
voltmeter is connected so that the voltmeter’s terminals are on either
side of the resistor.
Your teacher will give you a copy of the full activity.
Figure 11.14 Proper circuit setup of an
ammeter and a voltmeter
REQUIRED SKILLS
DI Key Activity
E3
Inquiry Activity
PHYSICS• SOURCE
■
■
Recording and organizing data
Drawing conclusions
Investigating Ohm’s Law
Question
Activity Overview
What is the relationship between
potential difference, current, and
resistance in a simple circuit?
In this activity, you will build a circuit with a resistor and a power source. You
will then connect an ammeter and a voltmeter, and measure the current as
you increase the voltage.
Your teacher will give you a copy of the full activity.
0–10 V power
supply
ammeter
Prelab Questions
Consider the questions below before beginning this activity.
resistor
voltmeter
Figure 11.15
Circuit for activity
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1. What relationship exists between the voltage, current, and resistance in
a simple circuit?
2. How is the resistance of a circuit affected when the current through the
circuit is changed?
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11.1 Check and Reflect
Key Concept Review
1. What conditions must be met for an electrical
circuit to operate?
2. What advantage did the invention of the
battery bring to the study of electricity?
3. Explain the meaning of the symbols I, q, V,
R, and t.
4. How does conventional current differ from
electron flow?
5. What is DC and how is it different from AC?
6. Explain the term “resistance.”
7. Explain how a light bulb and a resistor are (a)
the same and (b) different.
8. Why is the measurement of electrical potential
more versatile than electrical potential energy?
9. Explain how two batteries can have the same
potential, but different potential energy.
Connect Your Understanding
10. Determine the current generated in a circuit
if 18.0 C of charge flow per minute.
11. What amount of charge is stored in a D-cell
alkaline battery if it provides a current of
450 mA for 45.6 h?
12. The capacity of a 9-V rechargeable battery is
625 mAh. What is the battery’s charge capacity
in coulombs?
13. The potential difference across an electric
motor is 150.0 V. The potential energy of
600.0 J is converted into kinetic energy.
Determine the amount of charge that flowed
through the motor.
14. A rechargeable nickel-metal hydride AA
battery has a charge capacity of 9000 C. How
long can it provide a current of 0.500 A?
15. A light bulb has a potential difference of
120.0 V. If 4.5 103 C of charge pass
through it, how much energy is converted
into light and heat?
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16. A resistor generates 1800.0 J of heat when
200.0 C of charge pass through it. What is
the potential difference across the resistor?
17. A small electric motor requires a potential
difference of 9.00 V and draws a current of
800.0 mA. If it runs for 55.0 s, how much
energy does it consume?
18. A student performs a lab where the voltage
drop across a resistor is measured as a function
of current. Plot a graph from the values given
in the following table, and determine the
resistance of the circuit.
Experimental Data
Current, I (mA)
Potential Difference,
⌬V (V)
0.00
0.00
200
9.00
400
16.0
600
24.0
800
31.0
1000
40.0
19. A potential difference of 2.80 V exists across
a light-emitting diode (LED). If the LED
consumes 42.0 J of energy, what amount of
charge flowed through it?
20. A set of decorative outdoor lights is left on for
12.0 h. The lights draw a current of 2.00 A,
and run on 120.0 V. How much energy will
they consume in this time?
Reflection
21. What concept in this section did you find most
difficult to understand? Why?
For more questions, go to
PHYSICS• SOURCE
Chapter 11 The principles of conservation of energy and charge apply to electrical circuits.
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11.2 Series Circuits
Section Summary
●
Kirchhoff’s voltage law is the law of conservation of energy for circuits.
●
In a series circuit, the current is constant, the total resistance is the sum of the
resistors, and the total voltage of the battery equals the sum of the voltage
drops across the resistors.
During the dark winter months, many trees in parks and in front of
buildings are decorated with strings of decorative lights (Figure 11.16).
Sometimes, an entire string of lights will go dark when one light bulb burns
out. This happens because the light bulbs have been joined together so that
the current flows along one path. Placing all the components in a circuit
along one path is called placing them in series. A series circuit has only
one path for the charges to follow. Figure 11.17 shows a series circuit.
Figure 11.16 Decorative lights
are sometimes wired in series.
Figure 11.17 Three light bulbs wired in series.
PHYSICS• SOURCE
Suggested Activity
●
E4 Inquiry Activity Overview on
page 379
Current in a Series Circuit
In a series circuit, there is only one path for the current to follow. Therefore,
the current in all parts of the circuit will be the same. Mathematically this
is expressed as:
IT I1 In
R1 5.00 VT 50.0 V
R2 15.0 IT 2.5 A
Figure 11.18 The total resistance
of this circuit can be found by
adding the resistors.
376
where IT is the total current in the series circuit and n is the last resistor in
the circuit. Remember that since a circuit forms a closed loop, the number
of charges flowing through the circuit never changes. A break in the path
blocks the current throughout the entire circuit.
Resistance in a Series Circuit
Figure 11.18 shows two resistors in series. Recall that a resistor is a
device that restricts the flow of current. Putting two resistors in series
with one another further restricts the current flow.
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The total resistance in a series circuit is the sum of the resistors.
Mathematically this is expressed as:
PHYSICS• SOURCE
Explore More
RT R1 ... Rn
where RT is the total resistance in the series circuit and n is the last
resistor in the circuit. Using this equation, we can determine the total
resistance for the circuit in Figure 11.18:
How does increasing the number of
resistors affect the resistance and
current of a series circuit?
RT R1 R2
5.00 15.0 20.0 A Short Circuit in a Series Circuit
What would happen to the current if there were no load, or resistance, in
a series circuit? Assume that we remove the resistors in Figure 11.18. We
can use Ohm’s law to calculate the current in the circuit:
V IR
V
I_
R
50.0 V
__
0
A
Ohm’s law suggests that the current would be infinite. Practically
this cannot happen. Either the power supply/battery will burn out or the
wire will heat up and burn out. If this wire were in the wall of a house,
a fire could start. Modern houses have circuit breakers in the electrical
panel that will turn off the electricity to the circuit when the current gets
too large, preventing a fire. A circuit with no load is called a short circuit.
Figure 11.19 shows an example of a short circuit in a series circuit.
Figure 11.19 This is a short circuit
because there is no load.
Potential Difference in a Series Circuit
What is the potential difference across each resistor in series? In a water
pump with two water wheels, the pump lifts the water to a certain height
and then the water falls from that height through the wheels (Figure 11.20(a)).
Similarly, in a series circuit, the battery increases the potential of the
charges by a certain amount, and the components in the circuit must
reduce the potential by the same amount (Figure 11.20(b)).
(a)
water pump
(b)
water wheel
(resistor 1)
water wheel
(resistor 2)
water reservoir
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11-PHYSICS-11SE-Ch11.indd 377
R1 5.00 VT 50.0 V
R2 15.0 IT 2.5 A
Figure 11.20 (a) A water pump increases the potential of the
water. (b) A battery increases the potential of the charges while
the resistors in the circuit decrease the potential of the charges.
Chapter 11 The principles of conservation of energy and charge apply to electrical circuits.
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The reason for this has to do with the law of conservation of energy.
In 1845, German physicist Gustav Kirchhoff (1824–1887) recognized that,
in any closed circuit loop, the sum of the potential differences through all
the components must be zero. This is referred to as Kirchhoff’s voltage
law. Kirchhoff’s voltage law is written mathematically as:
0 V1 ... Vn
For example, in Figure 11.20(b), the sum of the voltage drops across
both resistors should equal the sum of the voltage increases. We will refer
to the potential difference caused by the battery as VT. Note that VT is
the total voltage increase in the circuit. We can modify Kirchhoff’s voltage
law slightly and write it mathematically as:
VT V1 ... Vn
where VT is the potential difference provided by the battery. We can use
Ohm’s law to calculate V1 and V2.
V1 I1R1
IT I1 I2 2.5 A
V1 (2.5 A)(5.00 )
V2 I2R2
V2 (2.5 A)(15.0 )
12.5 V
37.5 V
The voltage drops across R1 and R2 are 12.5 V and 37.5 V, respectively.
We can check that V1 and V2 are correct using Kirchhoff’s voltage law.
VT V1 V2
12.5 V 37.5 V
50.0 V
Concept Check
VT 200 V
1. Draw a circuit diagram showing a battery and four light bulbs in series.
2. Explain what happens to the current and potential difference in a series circuit.
Figure 11.21 Question 3
3. Draw a diagram to represent Figure 11.21 using the water/water pump analogy.
Example 11.4
Practice Problems
1. Determine the current, total
resistance, and voltage drops
in a series circuit in which
VT 10.0 V, R1 4.0 ,
R2 10.0 , and R3 6.0 .
2. Determine the current, total
resistance, and voltage drops
in a series circuit in which
VT 12.0 V, R1 5.0 ,
R2 15.0 , and R3 100 .
Determine the current, total resistance, and voltage drops through
all the resistors in the circuit shown in Figure 11.22.
Given
VT 200 V
R1 20.0 R2 50.0 R3 30.0 Required
current (I)
total resistance (RT)
voltage drop through all resistors
(V1, V2, V3)
R1 20.0 VT 200 V
R2 50.0 R3 30.0 Figure 11.22
Analysis and Solution
First, we calculate the total resistance.
RT R1 R2 R3
20.0 50.0 30.0 100 378
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We can calculate the current using Ohm’s law.
Answers
VT IRT
1.
RT 20.0 V1 2.0 V
V3 3.0 V
IT 0.500 A
V2 5.0 V
2.
RT 120.0 V1 0.500 V
V3 10.0 V
IT 0.100 A
V2 1.50 V
V
I _T
RT
200 V
__
100 2.00 A
We can now calculate the voltage drops using Ohm’s law.
V2 IR2
V3 IR3
V1 IR1
(2.00 A)(20.0 )
(2.00 A)(50.0 )
(2.00 A)(30.0 )
40.0 V
100 V
60.0 V
Paraphrase
The total resistance is 100 , the current is 2.00 A, and the voltage
drops across R1, R2, and R3 are 40.0 V, 100 V, and 60.0 V, respectively.
Series Circuit Summary
This section introduced three equations that can be used to determine
the current, resistance, and potential difference in a series circuit
(Table 11.3).
Series Circuit Equations
Table 11.3
Current
IT I1 In
Current remains the same
throughout the entire circuit.
Resistance
RT R1 ... Rn
The total resistance is the sum of
all the resistances in the circuit.
Potential Difference
VT V1 ... Vn
The sum of the voltage drops
through the circuit is equal to
the voltage increase provided
by the battery.
PHYSICS• SOURCE
Take It Further
To see how the number of light bulbs
in a circuit affects the brightness
of the bulbs, use circuit simulation
software to create two circuits with
a different number of identical light
bulbs. Be prepared to present your
findings by showing the circuit, the
current, and the voltage drop across
each light bulb.
REQUIRED SKILLS
E4
Inquiry Activity
PHYSICS• SOURCE
■
■
Using appropriate equipment and tools
Reporting results
Measuring Current and Potential Difference in a Series Circuit
Question
Prelab Questions
What are the current and voltage drops across the resistors in
a series circuit?
Consider the questions below before beginning
this activity.
R1
Activity Overview
In this activity, you will measure the
current and the potential difference
in a series circuit. You will need to
correctly connect an ammeter and
a voltmeter.
Your teacher will give you a copy
of the full activity.
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11-PHYSICS-11SE-Ch11.indd 379
1. What is the current at different positions in a
series circuit?
R2
R3
2. Use Kirchhoff’s voltage law to predict what
happens to the voltage drops across the
components in a closed circuit.
Figure 11.23 Series
circuit for activity
Chapter 11 The principles of conservation of energy and charge apply to electrical circuits.
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11.2 Check and Reflect
Key Concept Review
1. What is a series circuit?
2. Draw a circuit diagram that shows four
resistors in series.
3. Explain how Kirchhoff’s voltage law is the
same as the law of conservation of energy.
4. What happens to the current when there is a
short circuit?
5. Explain what causes a short circuit.
10. Two resistors are in series. Determine
the resistance (in k) provided by R1 if
VT 20.0 V, I 5.00 mA, and R2 1.00 k.
The SI prefix kilo- (k) is equal to 103.
11. All of the light bulbs shown in the circuits
below are identical and have the same
resistance.
(a) Explain which circuit would have a larger
current.
(b) In which circuit would the light bulbs
glow brightest? Explain your answer.
Connect Your Understanding
6. Explain what happens to a series circuit if one
component in the circuit breaks.
7. A series circuit contains three resistors:
R1 is 12.0 , R2 is 18.0 , and R3 is 45.0 .
A battery provides a potential difference
of 100.0 V.
(a) What is the total resistance of the circuit?
(b) What current flows through the circuit?
(c) What is the voltage drop across each resistor?
8. Determine the potential difference across the
battery and the three light bulbs shown in the
following circuit diagram.
R1 50.0 VT ?
R2 100 I 5.00 A
R3 200 Question 8
9. Determine the value of the third resistor shown
in the following circuit diagram.
VT 10 V
Question 11
12. A series circuit is sometimes compared with
a closed water pipe. The components of a
circuit are similar to components placed along
the water pipe, and electrical potential is
similar to the pressure in the water pipe caused
by a water pump. Explain how this analogy
might make sense in terms of potential
difference, current, and resistance.
Reflection
13. What did you learn about series circuits that
you did not know?
For more questions, go to
R1 80.0 VT 40.0 V
VT 10 V
PHYSICS• SOURCE
R2 90.0 I 200 mA
R3 ?
Question 9
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11.3 Parallel and Mixed Circuits
Section Summary
●
Kirchhoff’s current law states that the current entering a junction must be
equal to the current leaving the junction.
●
In a parallel circuit, the total current equals the sum of the currents through
each branch, the total resistance decreases as the number of branches
increases, and the voltage is constant across each branch.
Studying a series circuit helps to develop an understanding of the
fundamentals of electrical circuits. However, in practice, a series
circuit has a drawback — if one component malfunctions or a
wire breaks, the entire circuit stops working. This is the electrical
equivalent of a city that has only one road that passes by every
house and business. If a traffic jam occurs anywhere along the
road, all traffic will stop.
The solution is to have many roads that are connected to
each other, which allows traffic to branch out in many directions.
Electrical circuits are designed in a similar fashion. A parallel
circuit is a closed circuit in which the current has more than one
path, or branch, to follow (Figure 11.24). The point at which the
path splits is called a junction.
junctions
Figure 11.24 Three light bulbs wired in
parallel. The dots in the circuit diagram
represent the four junctions.
Current in a Parallel Circuit
In a parallel circuit, there are two or more branches for the current
to follow. Look at the parallel circuit shown in Figure 11.25. If we
assume the flow of conventional current, charges exit the positive
terminal of the battery and move through the circuit to junction A.
At junction A, the current splits into two paths and recombines at
junction B as it flows back to the battery.
Gustav Kirchhoff recognized that electrical charge is conserved
in any closed electrical circuit. This is because of the law of
conservation of charge, which states that the total electrical
charge of a closed system remains constant. Kirchhoff was
able to describe the law of conservation of charge in terms
of current. This is known as Kirchhoff’s current law,
which states that the current entering junction A
water pump
must be equal to the current leaving the junction.
The same is true at junction B. Kirchhoff’s current
law is written mathematically as:
A
IT
R1
IT
R2
B
Figure 11.25 In this parallel circuit, the
current has two paths to follow.
water wheel
IT I1 ... In
where 1 through n are the branches of the circuit.
A parallel circuit is similar to a water pump
with two water wheels (Figure 11.26). The water
splits into two paths and recombines in the reservoir
at the bottom.
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water reservoir
Figure 11.26 A water pump with two parallel water wheels.
Chapter 11 The principles of conservation of energy and charge apply to electrical circuits.
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Resistance in a Parallel Circuit
Figure 11.25 on the previous page shows a parallel circuit with two
resistors. As the number of resistors in parallel increases, the total circuit
resistance decreases. In fact, the total circuit resistance is always less than
the lowest resistance found in the branches. The resistance of a parallel
circuit is determined by the following equation:
1 ... _
1
1 _
_
RT
R1
Rn
where RT is the total resistance in the parallel circuit and n is the last
resistor in the circuit.
A Short Circuit in a Parallel Circuit
Figure 11.27 shows a short circuit in a parallel circuit. This is an example
of a short circuit because the current does not split equally. There is a
path for the current to follow that bypasses the resistor.
Figure 11.27 A short circuit in a
parallel circuit
Potential Difference in a Parallel Circuit
Recall that Kirchhoff’s voltage law states that in any closed circuit loop,
the sum of the potential difference (voltage drop) through the resistors
must equal the voltage gain across the battery. However, in a parallel
circuit, the potential difference through each load is the same and is equal
to the voltage increase of the battery. Mathematically this is expressed as:
PHYSICS• SOURCE
Suggested Activity
●
E5 Inquiry Activity Overview on
page 388
VT V1 Vn
where VT is the potential difference provided by the battery.
Example 11.5
Practice Problems
1. Determine the total resistance, total
current, and the currents through
each branch of the following circuit.
VT 90.0 V
R1 45.0 R2 90.0 Figure 11.29
2. Determine the total resistance, total
current, and the currents through
each branch of the following circuit.
VT 120 V
R1 5.00 R2 10.0 R3 15.0 Figure 11.30
Answers
1. RT 30.0 ,
I1 2.00 A
IT 3.00 A
I2 1.00 A
2. RT 2.73 Ω
I2 12.0 A
IT 44.0 A
I3 8.00 A
382
I1 24.0 A
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Analyze the parallel circuit shown in Figure 11.28 to
determine the total resistance, total current, and current
through each branch.
Given
R1 400 R2 100 VT 240 V
Required
total resistance (RT)
total current (IT)
VT 240 V
R1 400 R2 100 Figure 11.28
current through branch 1 (I1)
current through branch 2 (I2)
Analysis and Solution
We need to determine the total resistance to find the
total current.
1 _
1
1 _
_
VT ITRT
RT
R1
R2
V
1 __
1
1 __
_
IT _T
RT
RT
400 100 240
V
__
1 0.0125 1
_
80.0
RT
3.00
A
R 80.0 T
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We can use Ohm’s law to find the current through each branch.
Remember, the potential difference through each branch is the same
and equal to the voltage provided by the battery.
V1 VT
V2 VT
V1 I1R1
V2 I2R2
V
I1 _1
R1
240
V
__
400 0.600 A
V
I2 _2
R2
240
V
__
100 2.40 A
We can check that I1 and I2 are correct using Kirchhoff’s current law.
I T I1 I 2
0.600 A 2.40 A
3.00 A
Paraphrase
The total resistance of the circuit is 80 , the total current is 3.00 A, the
current through branch 1 is 0.600 A, and the current through branch 2
is 2.40 A.
Parallel Circuit Summary
This section introduced three equations that can be used to determine the
current, resistance, and potential difference in a parallel circuit. Table 11.4
summarizes these equations.
Table 11.4
Parallel Circuit Equations
Current
IT I1 ... In
The sum of the currents through each
branch must equal the total current
of the circuit.
Resistance
1
1
1
_
_
_
... RT
R1
Rn
The total resistance of the circuit decreases
as the number of branches increases.
Potential Difference
VT V1 Vn
The voltage drops across the loads
are the same and are equal to the
voltage increase of the battery.
Concept Check
1. Draw a circuit diagram showing a battery and four light bulbs in parallel.
2. Explain what happens to the current and potential difference in a parallel circuit.
3. Describe two situations in which a parallel circuit would be preferable to a
series circuit.
Mixed Circuits
Almost all electric devices contain a combination of series and parallel
circuits. These circuits are called mixed circuits. For example, the circuits
in a car, computer, house, and cell phone are mixed circuits (Figure 11.31).
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Figure 11.31 A cell phone circuit
board contains many components
that are wired in series and parallel.
Chapter 11 The principles of conservation of energy and charge apply to electrical circuits.
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Simplifying a Mixed Circuit
R1 24.0 VT 72.0 V
R2 20.0 R3 30.0 We can use Ohm’s law to analyze a mixed circuit. However, because a
mixed circuit contains a combination of series and parallel circuits, it
can be challenging to determine the resistance, current, and potential
difference. The best approach is to look at the parallel part of the circuit
separately from the series part of the circuit. We can also reduce the
complexity of the circuit by, wherever possible, replacing several resistors
with an equivalent resistor.
For example, we can analyze the mixed circuit shown in Figure 11.32.
We begin by simplifying the circuit. We can calculate the combined resistance
of R2 and R3, and substitute an equivalent resistor (R) in their place.
1 _
1 _
1
_
R
R2
R3
1 __
1
1
_
__
R
20.0 30.0 1 0.083 1
_
R
Figure 11.32 In this mixed circuit,
R1 is in series with R2 and R3, which
are in parallel with each other.
R 12.0 R1 24.0 VT 72.0 V
R 12.0 We can redraw the circuit to show that the two parallel resistors have
an effective resistance of 12.0 . We can use the equivalent resistor R to
represent the two parallel resistors in our circuit diagram (Figure 11.33).
Since the circuit is now a series circuit, we can calculate the total
resistance of the circuit using the sum of the resistors:
Figure 11.33 Replacing the
two parallel resistors with the
equivalent resistor R makes
this a series circuit.
RT R1 R
24.0 12.0 36.0 We can use Ohm’s law to calculate the total current:
V
IT _T
RT
72.0 V
__
36.0 2.00 A
We can now use Ohm’s law to determine the voltage drop across R1 and R:
V1 ITR1
(2.00 A)(24.0 )
48.0 V
R1 24.0 V1 48.0 V
R2 20.0 V2 24.0 V
VT 72.0 V
R3 30.0 V3 24.0 V
IT 2.00 A
Figure 11.34 The voltage drop
across each resistor is known.
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V ITR
(2.00 A)(12.0 )
24.0 V
Since R is really the equivalent resistor for R2 and R3, which are
in parallel, the voltage drop across each of the parallel branches
must be 24.0 V. The original circuit can be redrawn with the
voltage drops written beside all the resistors (Figure 11.34).
To determine the current through each branch of the
mixed circuit, we use the voltage drop through each branch.
V
I2 _2
R2
24.0 V
__
20.0 1.20 A
V
I3 _3
R3
24.0 V
__
30.0 0.80 A
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Table 11.5 outlines the steps you can use to analyze mixed circuits
when the resistances are known.
Table 11.5
PHYSICS• SOURCE
Explore More
Simplifying a Mixed Circuit
How does increasing the number of
resistors affect the resistance and
current of a mixed circuit?
Step
Procedure
Step 1
Reduce the circuit to a simple series circuit by using equivalent resistors.
Step 2
Determine the total resistance and total current of the series circuit using the
equation for RT for series circuits and Ohm’s law.
Step 3
Determine the voltage drop across each resistor in the circuit using Ohm’s law.
Step 4
Redraw the original circuit with the voltage drops beside each resistor.
Remember that the voltage drop across the parallel resistors will be the same.
Step 5
Determine the current through the parallel resistors using Ohm’s law.
Example 11.6
Analyze the circuit diagram shown in Figure 11.35 to solve for the
current and potential difference through each resistor.
Given
R1 400 R2 500 R3 1800 R4 1200 R5 600 VT 200 V
Required
current and voltage drop through each resistor
Analysis and Solution
This mixed circuit must be simplified to a series circuit so that
we can solve for total resistance and total current. The first step is
to combine R1 and R2 into an equivalent resistor called R12. Since
these two resistors are in series, the total resistance is the sum of the
resistors. We can redraw the circuit diagram to look like Figure 11.36.
R12 R1 R2
400 500 900 R1 400 VT 200 V
R2 500 R3 1800 R4 1200 R3 1800 R4 1200 R5 600 Figure 11.35
VT 200 V
R12 900 R5 600 Figure 11.36
The next step is to combine the three resistors in parallel into one
equivalent resistor called R:
1 __
1
1
1 __
_
__
R
900 1800 1200 1 0.00250 1
_
R
R 400 We can redraw the circuit diagram (Figure 11.37) and determine RT:
RT R R5
400 600 1.00 103 RⲐⲐ 400 VT 200 V
R5 600 We can determine the total current of the circuit using Ohm’s law:
Figure 11.37
VT ITRT
V
IT _T
RT
200 V
___
1.00 103 0.200 A
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Chapter 11 The principles of conservation of energy and charge apply to electrical circuits.
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Practice Problems
1. Determine the current and potential
difference through each resistor in
Figure 11.38.
R1 10.0 VT 10.0 V
R3 100 R2 30.0 R4 200 V ITR
(0.200 A)(400 )
80.0 V
Figure 11.38
2. Determine the current and potential
difference through each resistor in
Figure 11.39.
R3 4.50 R1 30.0 R2 10.0 R4 8.00 Figure 11.39
Answers
1. I1 0.0625 A, I2 0.0625 A,
I3 0.0250 A, I4 0.0125 A, I5 0.100 A,
V1 0.625 V, V2 1.88 V, V3 2.50 V,
V4 2.50 V, and V5 7.50 V
2. I1 1.50 A, I2 4.50 A, I3 6.00 A, I4 6.00 A,
V1 45.0 V, V2 45.0 V, V3 27.0 V,
and V4 48.0 V
PHYSICS• SOURCE
V5 ITR5
(0.200 A)(600 )
120 V
We can use the potential difference to determine the current
through each branch:
V1 I12R12
R5 75.0 VT 120 V
Now we can determine the voltage drop across R and R5.
Remember, the voltage drop across R is the voltage drop
across the three parallel branches in the original circuit.
V3 I3R3
V4 I4R4
V
I3 _3
R3
80.0
V
__
1800 0.04444 A
V12
I12 _
R12
80.0
V
__
900 0.08889 A
V
I4 _4
R4
80.0
V
__
1200 0.06667 A
Note that, since R1 is in series with R2, I12 I1 I2.
The final step is to determine the voltage drop across the
resistors in branch one. The current through branch one is
0.0889 A, and the voltage drop across the entire branch is
80.0 V. Resistors 1 and 2 are in series with each other in this
branch, so the sum of their voltage drops must equal 80.0 V.
We can determine the voltage drop across the resistors using
Ohm’s law:
V1 I1R1
(0.08889 A)(400 )
35.6 V
V2 I2R2
(0.08889 A)(500 )
44.4 V
Paraphrase
We can summarize our results in the following table:
R1
R2
R3
R4
R5
ΔV
35.6 V
44.4 V
80.0 V
80.0 V
120 V
I
0.0889 A
0.0889 A
0.0444 A
0.0667 A
0.200 A
Suggested Activity
●
E6 Inquiry Activity Overview on
page 388
Application of a Mixed Circuit Design
VT 30 V
R1
R2
R3
R4
Figure 11.40 R1 is wired in series
with the battery and the light
bulbs, which are in parallel
with each other.
386
The steps that you take to analyze a mixed circuit depend on the
information that you are given and what you need to determine. For
example, imagine we have three identical light bulbs. Each one operates
with 10 V and has a resistance of 60 . The light bulbs are to be wired
together so that if one bulb burns out, the other two light bulbs remain
working. We also have a 30-V battery and some resistors. If we place
the light bulbs in parallel, there will be a potential difference of
30 V across each light bulb, which will cause the bulbs to burn out.
One solution is to use a resistor in series with the battery to drop
the voltage to 10 V (Figure 11.40). To determine the value of this resistor,
we first calculate the resistance of the parallel portion of the circuit only.
Since we know the resistance of each bulb, we will use the equation
to calculate the resistance of a parallel circuit using the symbol R to
represent the parallel portion of the circuit.
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1 _
1 _
1 _
1
_
R
R2
R3
R4
1 _
1 _
1 _
1
_
R
60 60 60 1 0.050 1
_
R
R 20 We can redraw the circuit to show that the three light bulbs
wired in parallel have an effective resistance of 20 . We can use the
equivalent resistor R// to represent the three light bulbs in our circuit
diagram (Figure 11.41). The circuit has been simplified to a series circuit.
Kirchhoff’s voltage law states that the sum of the voltage drops must
equal the increase in potential of the battery. If the voltage drop across
the equivalent resistor (V) is 10 V — because the light bulbs require
10 V — we can calculate the voltage drop across R1:
VT 30 V
R1 ?
RⲐⲐ 20 Figure 11.41 The equivalent
resistor, R represents the
combined resistance of the
three light bulbs.
VT V1 V
V1 VT V
30 V 10 V
20 V
Since we are dealing with a series circuit, the current is the same
throughout the circuit. In other words, the current through R is the
same as the current through R1. We can calculate the current through R
using Ohm’s law:
V IR
V
I _
R
10 V
_
20 0.50 A
We can now determine the value of R1 using Ohm’s law:
V1 I1R1
PHYSICS• SOURCE
Take It Further
There are many types of circuit
designs that are fundamental
to electronics. For example, a
Wheatstone bridge is a common
circuit design that has a simple
purpose: to find the exact resistance
of a resistor. Use a circuit simulation
tool to experiment with a
Wheatstone bridge. Identify how
the circuit is designed and the
components are used.
V
R1 _1
I1
20
V
__
0.50 A
40 Therefore, R1 must have a resistance of 40 if the voltage across the light
bulbs is to be 10 V. The effect of R1 in our mixed circuit is to lower the
electrical potential across the light bulbs so the potential difference across
the bulbs will be correct.
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REQUIRED SKILLS
E5
Inquiry Activity
PHYSICS• SOURCE
■
■
Measuring
Drawing conclusions
Measuring Current and Potential Difference
in a Parallel Circuit
Question
What are the current and voltage drops across the resistors in a parallel circuit?
Activity Overview
In this activity, you will measure the current and the potential difference in a
parallel circuit. You will need to correctly connect an ammeter and a voltmeter.
You will analyze your results and compare them with the values obtained using
the parallel circuit equations.
Your teacher will give you a copy of the full activity.
Prelab Questions
Consider the questions below before
beginning this activity.
1. What does Kirchhoff’s voltage law
predict about the voltage drop through
each branch of a parallel circuit?
2. What does Kirchhoff’s current law
predict about the current through each
branch of a parallel circuit?
Figure 11.42 Connect the parallel circuit as shown.
REQUIRED SKILLS
E6
Inquiry Activity
PHYSICS• SOURCE
■
■
Recording and organizing data
Reporting results
Measuring Current and Potential Difference
in a Mixed Circuit
Question
R1
What are the current and voltage drops across the resistors in a mixed circuit?
Activity Overview
In this activity, you will measure the current and the potential difference in a
mixed circuit. You will need to correctly connect an ammeter and a voltmeter.
You will analyze your results and compare them with your calculated values.
Your teacher will give you a copy of the full activity.
R2
R3
Figure 11.43 A mixed circuit
Prelab Questions
Consider the questions below before beginning this activity.
1. Which components of this mixed circuit are in series?
2. Which components of this mixed circuit are in parallel?
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11.3 Check and Reflect
Key Concept Review
1. Copy the following table into your notebook, and
fill in the cells with the appropriate equations.
Circuit Summary
Type of Circuit
Potential
Difference
10. Determine the current through the ammeter
in the circuit with (a) two light bulbs in
parallel and (b) after a third light bulb has
been added in parallel. All light bulbs have
a resistance of 4.00 .
(a)
Resistance
Current
VT 12.0 V
Series circuit
Parallel circuit
(b)
2. What is a parallel circuit?
3. Draw a circuit diagram that shows three
resistors in parallel.
4. What effect does increasing the number of
paths in a parallel circuit have on (a) the total
resistance and (b) the total current?
VT 12.0 V
Question 10
11. Determine the total current and the current
through all the branches of the following
parallel circuit.
5. If a parallel circuit develops a short circuit
in one of the paths, what will happen to the
current flow through the other paths?
VT 180 V
6. If a parallel circuit contains three paths, each
containing resistors of exactly the same value,
explain what will happen if a resistor in one
of the paths burns out and does not allow
current to flow.
Question 11
7. If two pathways in a parallel circuit have
different resistances, will the current in each
pathway be the same? Explain your answer.
R1 60 k
R2 20 k
R3 30 k
R4 45 k
12. Determine the voltage drops and current through
all the resistors in the following circuit diagram.
R1 10.0 VT 90.0 V
R2 30.0 R3 30.0 R4 20.0 Connect Your Understanding
Question 12
8. Determine the value of R2 in the following
circuit diagram.
R1 700 VT 112 V
R2 ?
R3 600 I 0.720 A
Reflection
13. What new insights or strategies did you
develop for solving mixed circuits?
For more questions, go to
Question 8
PHYSICS• SOURCE
9. Determine the total resistance and the current
through the branches of the following circuit.
VT 75.0 V
R1 1.250 k
R2 1.500 k
R3 2.200 k
Question 9
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11.4 Power Consumption
Section Summary
PHYSICS INSIGHT
Remember that work is equal to
change in energy. The equation for
power can also be written as:
W
P_
t
●
Power is the rate at which energy is transferred.
●
The power consumed by an electrical appliance can be determined.
People often refer to the awesome power of nature. They might talk about
a powerful hurricane or the powerful explosion of a volcano. However, in
physics, power is defined as the rate at which energy is transferred or the
change in energy per unit time. In other words, it is the amount of energy
generated or used each second. The unit for power is the watt (W). The
equation for power is:
E
P_
t
where P is the power in watts (W), E is the change in potential energy
in joules (J), and t is the time in seconds (s). Note that 1 W 1 J/s.
An electrical circuit has components, such as resistors or light bulbs,
that consume power. The power consumed by a component is the amount
of energy it uses every second. For example, a 60-W incandescent light
bulb consumes 60 J of energy each second.
A battery does not consume energy — it produces electrical energy
from the chemical compounds contained inside. It is capable of generating
power. This means that it can increase the energy of the charges by a certain
amount each second. The total power consumed by all the components in
the circuit must equal the power generated by the battery or power supply.
Power Equations
Recall from section 11.1 that the change in energy can be determined by
the equation
E Vq
where E is the change in potential energy in joules (J) of the charges
as they pass through a load and q is the charge in coulombs (C). We can
substitute this equation into the equation for power:
Vq
P_
t
PHYSICS INSIGHT
Since q It, we can further simplify the equation as:
V(It)
P __
t
V(It)
__
t
It is common practice to omit the from an equation. This is not correct
and the power equation should be
written as:
P VI
P VI
where P is the power in watts (W), V is the potential difference in
volts (V), and I is the current in amperes (A). Note that 1 W 1 VA.
We can also combine Ohm’s law with the power equation. Table 11.6
shows the derivation of other forms of the power equation.
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Table 11.6
Derivation of Power Equations
PHYSICS• SOURCE
Derivation
Equation
When to Use
Substitute Ohm’s law equation for V
into the power equation:
V IR
P VI
P (IR)I
P I2R
P I2R
I and R are known
V is not known
Substitute Ohm’s law equation for I
into the power equation:
P
V IR
V
_
I
R
P V
P
V2
_
R
Explore More
How can you arrange four identical
resistors in a circuit to consume the
most power? The least power?
V and R are known
I is not known
V
(_
R )
V2
_
R
Example 11.7
Determine the power consumed by the light bulb in Figure 11.44.
Given
VT 120 V
RL 240 VT 120 V
Required
power consumed by the light bulb (PL)
RL 240 Figure 11.44
Analysis and Solution
Since the light bulb is the only component in the circuit,
VT VL 120 V
VL2
PL _
RL
Practice Problems
1. Determine the power
dissipated by the resistor
in Figure 11.45.
VT 10.0 V
R
4.50 k
Figure 11.45
2. Determine the resistance and
power consumption of the
resistor in Figure 11.46.
I 2.80 A
(120 V)2
__
240 60.0 W
VT 12.6 V
Paraphrase
The light bulb consumes 60.0 W of power.
R?
Figure 11.46
Answers
1. 0.022 W
Power Consumption in Circuits
As you have seen, there are differences between series and parallel
circuits. The power consumption depends on whether the circuit is
a series or parallel circuit. In a series circuit, the voltage drop across a
component depends on the number of components in series with it. As
the number of components in a series circuit increases, the resistance also
increases, and the power consumed by each component is less. As the
number of components in a parallel circuit increase, the total resistance
decreases and, as a result, the power consumed by the circuit increases.
©P
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2. R 4.50 P 35.3 W
PHYSICS• SOURCE
Suggested Activity
●
E7 Inquiry Activity Overview on
page 394
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Example 11.8
Compare the power consumed by the light bulbs in a series
circuit (Figure 11.47) with the power consumed by the light
bulbs in a parallel circuit (Figure 11.48).
R1 240 VT 120 V
R2 240 VT 120 V
R1 240 R2 240 R3 240 R3 240 Figure 11.48
Figure 11.47
Given
VT 120 V
Practice Problems
1. Determine the power consumed by
the resistors in Figure 11.49.
R1 200 VT 80.0 V
R1 R2 R3 240 Required
power consumed by the light bulbs in a series circuit and in a
parallel circuit (P1, P2, P3)
Analysis and Solution
We will analyze the power consumed by each light bulb in the
series circuit.
R2 180 RT R1 R2 R3
240 240 240 720 R3 100 Figure 11.49
2. Determine the power through each
resistor in Figure 11.50.
VT 15.0 kV
R1 4.00 k
R2 6.00 k
Figure 11.50
Answers
1. P1 5.56 W
P2 5.00 W
P3 2.78 W
2. P1 56.3 kW
P2 37.5 kW
P3 150 kW
R3 1.50 k
VT ITRT
V
IT _T
RT
120
V
__
720 0.167 A
P1 I R
P2 I R 2
P3 I 32R3
2
2
(0.167 A) (240 )
(0.167 A) (240 )
(0.167 A)2(240 )
6.67 W
6.67 W
6.67 W
2
1 1
2
2
We can now analyze the power consumed by each light bulb in
the parallel circuit:
V12
P1 _
R1
(120 V)2
__
240 60.0 W
V22
P2 _
R2
(120 V)2
__
240 60.0 W
V32
P3 _
R3
(120 V)2
__
240 60.0 W
Paraphrase
Each light bulb consumes 6.67 W of power when placed in series, but
60.0 W of power when placed in parallel.
Concept Check
1. What component in a circuit generates power?
2. Which type of circuit would consume more power: a circuit containing two resistors
in series or a parallel circuit with the same resistors?
3. Is it possible for a circuit component to consume more energy than is generated by
the battery? Explain your answer.
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House Wiring and Power Consumption
A house is wired so that the outlets are in parallel with each other.
Appliances plugged into the wall will be in parallel with each other and
will draw the power they need. One appliance will not interfere with
the operation of another appliance. In other words, each appliance will
always receive the same potential difference: 120 V.
Imagine that you have three power outlets in your bedroom. They are
wired in parallel so that up to three different appliances can be plugged into
the wall. A hair dryer and a computer are plugged into two separate outlets
in the room. What is the current through the ammeter in Figure 11.51?
The voltage drop across each branch of this circuit is 120 V because
it is a parallel circuit. Each branch draws current, and the sum of the
currents in these branches equals the total current through the ammeter.
To determine the current through the ammeter, calculate the current drawn
by the hair dryer (IH) and computer (IC).
PH VIH
PH
IH _
V
1200 W
__
120 V
10.00 A
PC VIC
PC
IC _
V
180 W
__
120 V
1.50 A
VT 120 V
The total current drawn by the hair dryer and computer is:
IT IH IC
10.00 A 1.50 A
11.50 A
PH 1200 W
PC 180 W
Figure 11.51 Two appliances are
plugged into two outlets in this
room and turned on: a 1200-W
hair dryer and a 180-W computer.
The other outlet is not used and
no current flows through it.
Since house wiring can carry up to 15 A safely, running these two
appliances at the same time poses no hazard. But what happens if a
600-W vacuum cleaner is plugged into the third outlet and turned on?
Calculate the current of the vacuum cleaner (IV) and then the total
current (IT).
PV
IV _
V
600 W
__
120 V
5.00 A
IT IH IC IV
10.0 A 1.50 A 5.00 A
16.5 A
The total current will be 16.5 A, which exceeds the maximum current
allowed through the wires.
To prevent this much current, a circuit breaker is placed into the
circuit. A circuit breaker is a switch that opens when a current higher than
a certain amount flows through it. In this case the breaker would “trip,”
or open, when a current greater than 15 A flows through the circuit. The
circuit breaker protects the wires from carrying too much current and
possibly overheating and starting a fire. For most rooms in a house, the
maximum power rating of the circuit is 1800 W (120 V 15 A). Some
rooms — such as the kitchen and laundry room — use 240 V and 20 A
because the appliances, such as the stove and clothes dryer, consume
more power. The maximum power consumption in that case is 4800 W
(240 V 20 A).
Some houses use fuses instead of circuit breakers. A circuit breaker
can be reset and reused. A fuse burns out and can only be used once.
©P
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PHYSICS INSIGHT
The symbol
represents
a fuse in a circuit diagram.
PHYSICS• SOURCE
Take It Further
Every house contains different
appliances. A typical house has a
refrigerator, a stove, and a washer
and dryer. Each appliance consumes
a different amount of power. Make
a list of ten household appliances in
your home and research the typical
power consumption of each.
Chapter 11 The principles of conservation of energy and charge apply to electrical circuits.
393
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Example 11.9
Practice Problems
1. There are five appliances: an
air conditioner (1000 W), a
refrigerator (450 W), a washing
machine (500 W), a popcorn
popper (250 W), and a stereo
(20 W). What are the two
highest-power appliances
that can be plugged into two
outlets and be running at the
same time? The maximum
current capacity of the circuit
is 15.0 A and the outlets
provide 120 V.
2. Determine the current
through each of the ammeters
in Figure 11.52.
space heater
A2
VT 120 V
stove
A3
PH 900 W
A1
PS 700 W
Figure 11.52
Answers
1. air conditioner and the washing machine
2. I1 13.33 A
I2 7.50 A
I3 5.83 A
A student rents a small apartment that has four 120-V outlets wired
in parallel. She moves in with six appliances: a kettle (900 W), a
vacuum cleaner (600 W), a TV (100 W), a microwave oven (1500 W),
a coffee maker (800 W), and a computer (150 W). What combination
of four appliances can she operate simultaneously if the maximum
current allowed in the circuit is 15.0 A?
Given
PK 900 W
PCM 800 W
PV 600 W
PC 150 W
PTV 100 W
IT 15.0 A
PM 1500 W
VT 120 V
Required
combination of four appliances that will not exceed the maximum
power possible in the circuit
Analysis and Solution
The total current cannot exceed 15.0 A. Since we know that the
wiring of the apartment is in parallel and the potential difference of
the circuit is 120 V, we can determine the total power of the circuit.
PT VTIT
(120 V)(15.0 A)
1800 W
The power of four components running simultaneously cannot
exceed 1800 W. Possible combinations are:
PTV PC PV PCM 100 W 150 W 600 W 800 W 1650 W
PTV PC PV PK 100 W 150 W 600 W 900 W 1750 W
Paraphrase
There are two combinations of four appliances running
simultaneously that will not overload the circuit.
REQUIRED SKILLS
E7
Inquiry Activity
PHYSICS• SOURCE
■
■
Predicting
Analyzing patterns
Comparing the Power Consumption
of Series and Parallel Circuits
Question
How does the power consumption of a series
circuit compare with the power consumption in
a parallel circuit?
Activity Overview
In this activity, you will investigate the power
consumption of two resistors when they are wired in
series and then in parallel. You will measure the current
and voltage drop across each resistor, and determine
the power consumption.
Your teacher will give you a copy of the full activity.
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Figure 11.53 Series
circuit for the activity
Prelab Questions
Consider the questions below before beginning
this activity.
1. How does the voltage drop compare among branches
in a parallel circuit?
2. Does the current in a series circuit change through
each component?
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11.4 Check and Reflect
Key Concept Review
1. Define power.
2. How does the power consumed in a series circuit
differ from that consumed in a parallel circuit?
3. Explain, in terms of electrical energy, how a
60-W light bulb differs from a 100-W light bulb.
4. What effect does a lower resistance have on the
power consumed by a component in a circuit?
5. (a) What is the purpose of a circuit breaker
or fuse?
(b) How are they different?
Connect Your Understanding
6. A block heater of a car is designed to keep
the oil in the engine block warm during
cold weather. If a heater has a voltage of
120 V and draws 3.30 A, how much power
does it consume?
7. Determine the power consumed (in kW) by
the heating coil in the following diagram.
I 2.40 A
11. There are two circuits with two identical
resistors. One is a series circuit and one is a
parallel circuit. Compare the amount of power
used in the series circuit and the parallel
circuit. Justify your answer.
12. A kitchen of a house is wired to accept a
maximum of 30.0 A. Determine which of the
following appliances can be plugged into the
fourth outlet in the circuit below and used
simultaneously with the other appliances.
Power Consumption of Appliances
Appliance
Power Consumption
toaster oven
1100 W
slow cooker
600 W
blender
300 W
popcorn maker
250 W
dishwasher
heating coil
VT 120 V
R 6.80 k
Question 12
PD 1500 W
garburator microwave
PG 450 W
PM 1200 W
Question 7
8. A power generation facility produces 455 MW
of power. If the transmission line has a voltage
of 500 000 V, what current does it carry?
The SI prefix mega- (M) is equal to 106.
9. A typical lightning strike during a
thunderstorm can carry 400 kA and have
a potential difference of 1.1 GV. How much
power does it generate? The SI prefix giga- (G)
is equal to 109.
10. Determine the power consumed by each
resistor in the following circuit.
R1 33.33 VT 500 V
R2 100 R3 200 13. A 60-W incandescent light bulb and a 15-W
compact fluorescent light bulb (CFL) produce
the same amount of light. Both use 120 V.
Determine the current used by the
incandescent bulb and the CFL.
14. A 60-W light bulb and a 100-W light bulb
are connected in parallel. Which one will
be brighter?
Reflection
15. What topic discussed in this section did you
find the most interesting? Why?
For more questions, go to
PHYSICS• SOURCE
Question 10
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11
CHAPTER
CHAPTER REVIEW
Key Concept Review
1. Draw a circuit diagram of a circuit that
contains an energy source, an ammeter,
a light bulb, and a voltmeter. k
2. Distinguish between conventional current
and electron flow. k
3. What is the difference between alternating
current and direct current? k
4. What device is used to measure current and
how is it connected to a circuit? k
5. State two effects a resistor has on a circuit. k
6. Explain how potential difference is different
from potential energy. Use an analogy to
explain the difference. k
7. What device is used to measure potential
difference, and how is it connected to a
circuit? k
8. Explain the significance of Ohm’s law and
Kirchhoff’s laws. k
9. Use the proper symbols to draw the following
circuits.
(a) A battery is connected to two light bulbs
in series. k
(b) A variable power source is connected to
three resistors in parallel. k
10. Distinguish between current in a series circuit
and a parallel circuit. k
11. Distinguish between potential difference in a
series circuit and a parallel circuit. k
12. Determine the potential difference across the
battery in the following circuit. t
R1 30 I
0.200 A
VT ?
R2 15 14. Determine the power dissipated by the
light bulb in the following diagram. a
light bulb
I 5.00 A
V 50.0 V
Question 14
Connect Your Understanding
15. Does a circuit component that has no voltage
drop across it consume power? Explain
your answer. t
16. A rechargeable “AA” battery has a charge
capacity of 2500 mAh. Determine the total
charge contained in the battery. t
17. A 4.50-V flashlight draws a current of 400.0 mA
when the light is on. If the flashlight is on for
50.0 s, what quantity of electrical potential
energy is converted into light and heat? t
18. A computer laptop battery contains a total
potential energy of 262 800 J and delivers
14.4 V. Determine how long the laptop can
operate if it draws a current of 1.69 A. t
19. The potential difference across a compact
fluorescent light bulb is 120.0 V. If 350.0 J
of electrical potential energy are converted
to light and heat, how much charge flowed
through the bulb? t
20. Determine the value of resistor 3 in the
following circuit diagram. t
R1 8.00 R2 20.0 VT 200 V
I 5.00 A
R3 ?
Question 20
Question 12
13. Write a paragraph explaining how power and
energy are related. c
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ACHIEVEMENT CHART CATEGORIES
k Knowledge and understanding
t Thinking and investigation
c Communication
a Application
21. (a) What is the potential difference of the
battery in the circuit diagram below? t
(b) Determine the voltage drop across all the
resistors in the circuit diagram below. t
R1 15.0 R2 38.0 R3 5.00 VT ?
R5 8.00 R4 22.0 26. Draw a mixed circuit of your own design.
Choose values for your resistors and determine
the voltage drop and current through each
resistor. t
27. Determine which appliances can be turned on
simultaneously in the following diagram of a
garage circuit without making the circuit
breaker trip. t
I
568.2 mA
drill press
hedge
trimmer
table saw
Question 21
22. (a) Determine the total resistance of the circuit
shown below. t
(b) Determine the total current of the circuit. t
(c) Determine the current through each
resistor. t
VT 90.0 V
R1 4.50 k
R2 6.50 k
PTS 1400 W
PHT 450 W
PDP 890 W
circuit breaker (fuse)
Imax 15.0 A
Question 27
28. Determine the power consumed by each
resistor in the following circuit. t
R1 200 Question 22
23. Determine the value that R3 must have so that
the light bulbs in the circuit have a potential
difference of 50 V across them. t
R3 ?
R2 1000 I 6.0 A
R3 1500 VT 120 V
R1 80 R2 80 Question 28
Reflection
Question 23
24. Determine the voltage drop and current
through each resistor in the following circuit. t
VT 50.0 V
VT 120 V
R1 5.00 R3 2.00 R2 15.0 R4 28.0 Question 24
25. Determine the value of resistor 2 in the
following circuit. t
29. What is the most interesting thing that you
learned in this chapter? c
Unit Task Link
An electric motor is made up of a small number of
components. Determine the types of components that are
used in a motor. Research the best type of wire to use for
your motor. You should think about the thickness (gauge)
of the wire. Determine whether the wire should be insulated
or not, and if so, the type of insulation that is best.
R1 120.0 VT 400 V
R2 ?
R3 600 For more questions, go to
PHYSICS• SOURCE
I 1.25 A
Question 25
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Chapter 11 Review
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