E Magnetism Electricity and - SPH3U1sph3u-bosnic.weebly.com/uploads/1/3/5/6/13569308/chapter_11.pdf12.3 Using Electromagnetism ... 13.2 The Generator and Electrical Energy Generation

Electricity and MagnetismE

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11-PHYSICS-11SE-Ch11.indd 36411-PHYSICS-11SE-Ch11.indd 364 7/26/10 10:01:05 AM7/26/10 10:01:05 AM

Unit Contents

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

Properties of magnetic fields apply in nature and technology.

12.1 Magnetic Forces and Fields

12.2 The Motor Principle

12.3 Using Electromagnetism

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

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.

Unit Task

12

13

DISCOVERING PHYSICSA 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?

11


LearningExpectations

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Electricity 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.

The principles of conservation of energy and charge apply to electrical circuits.

C H A P T E R

11

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

● 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.


Chapter 11 The principles of conservation of energy and charge apply to electrical circuits. 367©P

discs

zinc

�

�

copper

Electrical Circuits

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).

Circuit FundamentalsTo 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.

11.1

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.

Figure 11.2 Volta created the first battery in 1800.


368 Unit E Electricity and Magnetism ©P

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.

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.

CurrentHighway 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.

Figure 11.3 A simple circuit containing a battery, a switch, and a load (the light bulb).

energy sourceelectrical load

switch

conductingwires

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

Table 11.1 Circuit Symbols used in Circuit Diagrams



Conventional Current and Electron FlowWhen 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:

I � q _

�t

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.

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.

PHYSICS INSIGHT

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.

Figure 11.4 In conventional current, positive charges move from the positive terminal to the negative terminal.

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.

charges

++

++

++

++

+

++

+

wirepoint

� �

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 10�3.

Given q � 9.00 C �t � 1.00 min � 60.0 s

Requiredcurrent (I)

Practice Problems1. 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?



Figure 11.5 An ammeter in a circuit

Figure 11.7 At the bottom of the water pump, the potential energy of the water is defined as zero.

Measuring CurrentA 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.

ResistanceElectrical 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 DifferenceFor 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 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 take the negative terminal of the battery as the point of zero potential energy.

Suggested Activity● E1 Quick Lab Overview on

page 374

PHYSICS•SOURCE

water pump

water reservoir

Analysis and Solution

I � q _

�t

� 9.00 C __ 60.0 s

� 0.150 C/s

� 0.150 A

Paraphrase

The current delivered by this battery is 0.150 A, or 150 mA.

Figure 11.6 Resistors come in many shapes and sizes to meet different requirements.

Answers1. 3.08 A

2. 1.49 � 103 C

3. 2.4 � 104 s



Electrical PotentialIn 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:

V � E _ 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 DifferenceAs 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

�V � ( Efinal _ q ) � ( Einitial _ q ) �V � �E _ 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.

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.

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.

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.

� � ��

light

VfinalVinitial

lightbulb

Vi � Vf

Figure 11.9 A voltmeter is placed across a component.

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.



Ohm’s LawTwenty-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).

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:

R � �V _ I

Example 11.2

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?

Given�V � 10.0 V q � 20.0 C

Requiredchange in potential energy (�E)

Analysis and Solution

�V � �E _ q

�E � q�V � (20.0 C)(10.0 V) � 200 CV � 200 J

ParaphraseThe amount of electrical energy converted to heat is 200 J.

Practice Problems1. 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?

Answers1. 1.08 � 105 J

2. 20.0 V

Concept Check

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?

Current (A)

Pote

ntia

lDi

ffere

nce

(V)

Potential Difference vs. Current

Figure 11.10 Ohm showed a linear relationship between the potential difference and current.

Suggested Activities● E2 Skill Builder Overview on

page 374

● E3 Inquiry Activity Overview on page 374

PHYSICS•SOURCE

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.

Explore More

What are the effects of Ohm’s law in a simple circuit?

PHYSICS•SOURCE



Take It Further

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.

PHYSICS•SOURCE

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

Givenvalues of potential difference (see Table 11.2)values of current (see Table 11.2)

Requiredthe resistance (R)

Analysis and SolutionThe 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.

slope � �y

_ �x

� �V _ �I

� R

Choose two points from the line of best fit.

R � 262 V � 37 V ___ 3.5 A � 0.5 A

� 225 V __ 3.0 A

� 75 �

ParaphraseThe resistor has a resistance of 75 �.

30 1 2 4 5Current (A)

Pote

ntia

l Diff

eren

ce (V

)

500

100150200

300350400

250

Potential Difference vs. Current

Figure 11.12

Practice Problems1. Determine the resistance

of a circuit using the following data.

Current (A)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

Answers1. 55 �

2. Current: 2.25 A and 6.00 A Potential difference: 195.0 V and 520.0 V

variable powersupply

resistor

ammeter

voltmeter

Figure 11.11

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:

�V � IR

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.



PurposeTo create a battery using copper and zinc metals

Activity OverviewIn 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 QuestionsConsider 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?

Creating a Pile Battery

E1 Quick Lab PHYSICS•SOURCE

Figure 11.13 Activity setup

Activity OverviewIn 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.


Using an Ammeter and a Voltmeter

Skill Builder Activity PHYSICS•SOURCEE2

Figure 11.14 Proper circuit setup of an ammeter and a voltmeter

QuestionWhat is the relationship between potential difference, current, and resistance in a simple circuit?

Investigating Ohm’s Law

Inquiry Activity PHYSICS•SOURCEE3

0–10 V powersupply

resistor

ammeter

voltmeter

REQUIRED SKILLS■ Recording and organizing data■ Drawing conclusions

Activity OverviewIn 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.



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?

DI Key Activity

Figure 11.15 Circuit for activity


Chapter 11 The principles of conservation of energy and charge apply to electrical circuits. 375

Check and Reflect11.1

©P

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.

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?

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?

For more questions, go to PHYSICS•SOURCE

Experimental Data



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.

Series Circuits11.2

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.

Current in a Series CircuitIn 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

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 CircuitFigure 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.

Figure 11.16 Decorative lights are sometimes wired in series.

Figure 11.17 Three light bulbs wired in series.

Suggested Activity● E4 Inquiry Activity Overview on

page 379

PHYSICS•SOURCE

Figure 11.18 The total resistance of this circuit can be found by adding the resistors.

R1 �5.00 �

R2 �15.0 �

IT � 2.5 A

�VT �50.0 V



The total resistance in a series circuit is the sum of the resistors. Mathematically this is expressed as:

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:

RT � R1 � R2

� 5.00 � � 15.0 � � 20.0 �

A Short Circuit in a Series CircuitWhat 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

I � �V _ 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.

Potential Difference in a Series CircuitWhat 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)).

Figure 11.19 This is a short circuit because there is no load.

Explore More

How does increasing the number of resistors affect the resistance and current of a series circuit?

PHYSICS•SOURCE

water pump

water reservoir

water wheel(resistor 1)

water wheel(resistor 2)

R1 �5.00 �

R2 �15.0 �

IT � 2.5 A

�VT �50.0 V

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.

(a) (b)



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 �V2 � I2R2

IT � I1 � I2 � 2.5 A

�V1 � (2.5 A)(5.00 �) �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

Figure 11.21 Question 3

�VT �200 V

Concept Check

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.

3. Draw a diagram to represent Figure 11.21 using the water/water pump analogy.

Example 11.4

Determine the current, total resistance, and voltage drops through all the resistors in the circuit shown in Figure 11.22.

Given �VT � 200 V R2 � 50.0 � R1 � 20.0 � R3 � 30.0 �

Requiredcurrent (I) total resistance (RT) voltage drop through all resistors (�V1, �V2, �V3)

Analysis and SolutionFirst, we calculate the total resistance.

RT � R1 � R2 � R3

� 20.0 � � 50.0 � � 30.0 � � 100 �

Figure 11.22

R2 �50.0 �

R3 � 30.0 �

R1 � 20.0 �

�VT �200 V

Practice Problems1. 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 �.



Series Circuit SummaryThis section introduced three equations that can be used to determine the current, resistance, and potential difference in a series circuit (Table 11.3).

We can calculate the current using Ohm’s law.

�VT � IRT

I � �VT _ RT

� 200 V __ 100 �

� 2.00 A

We can now calculate the voltage drops using Ohm’s law. �V1 � IR1

� (2.00 A)(20.0 �) � 40.0 V

�V2 � IR2

� (2.00 A)(50.0 �) � 100 V

�V3 � IR3

� (2.00 A)(30.0 �) � 60.0 V

ParaphraseThe 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.

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.

Table 11.3 Series Circuit Equations

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.

PHYSICS•SOURCE

QuestionWhat are the current and voltage drops across the resistors in a series circuit?

Activity OverviewIn 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.


Measuring Current and Potential Difference in a Series Circuit

Inquiry Activity PHYSICS•SOURCE

REQUIRED SKILLS■ Using appropriate equipment and tools■ Reporting results

E4

R1

R2

R3

Answers1. RT � 20.0 � IT � 0.500 A �V1 � 2.0 V �V2 � 5.0 V �V3 � 3.0 V

2. RT � 120.0 � IT � 0.100 A �V1 � 0.500 V �V2 � 1.50 V �V3 � 10.0 V

Figure 11.23 Series circuit for activity


1. What is the current at different positions in a series circuit?

2. Use Kirchhoff’s voltage law to predict what happens to the voltage drops across the components in a closed circuit.


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.


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.

Question 8

9. Determine the value of the third resistor shown in the following circuit diagram.

Question 9

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.

Question 11

12. A series circuit is sometimes compared witha 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?

�VT � ?

R1 � 50.0 �

I � 5.00 A R3 � 200 �

R2 � 100 �


11.2

R1 � 80.0 �

R3 � ?

R2 �90.0 �

I � 200 mA

�VT �40.0 V

�VT � 10 V

�VT � 10 V



water pump

water reservoir

water wheel

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.

Current in a Parallel CircuitIn 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 must be equal to the current leaving the junction. The same is true at junction B. Kirchhoff’s currentlaw is written mathematically as:

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.

Parallel and Mixed Circuits11.3

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.

Figure 11.24 Three light bulbs wired in parallel. The dots in the circuit diagram represent the four junctions.

Figure 11.25 In this parallel circuit, the current has two paths to follow.

IT

IT

R2R1

A

�

�

B

Figure 11.26 A water pump with two parallel water wheels.

junctions



Resistance in a Parallel CircuitFigure 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 _ RT

� 1 _ R1

� ... � 1 _ 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 CircuitFigure 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.

Potential Difference in a Parallel CircuitRecall 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:

�VT � �V1 � �Vn

where �VT is the potential difference provided by the battery.


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PHYSICS•SOURCE

Example 11.5

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

Requiredtotal resistance (RT) current through branch 1 (I1)total current (IT) current through branch 2 (I2)

Analysis and SolutionWe need to determine the total resistance to find the total current.

1 _ RT

� 1 _ R1

� 1 _ R2

1 _ RT

� 1 __ 400 �

� 1 __ 100 �

1 _ RT

� 0.0125 ��1

RT � 80.0 �

�VT � ITRT

IT � �VT _ RT

� 240 V __ 80.0 �

� 3.00 A

R2 �100 �

R1 �400 �

�VT �240 V

Figure 11.28

Practice Problems1. Determine the total resistance, total

current, and the currents through each branch of the following circuit.

2. Determine the total resistance, total current, and the currents through each branch of the following circuit.

Answers1. RT � 30.0 �, IT � 3.00 A

I1 � 2.00 A I2 � 1.00 A

2. RT � 2.73 Ω IT � 44.0 A I1 � 24.0 A I2 � 12.0 A I3 � 8.00 A

R1 �45.0 �

R2 �90.0 �

�VT �90.0 V

R1 �5.00 �

R2 �10.0 �

R3 �15.0 �

�VT �120 V

Figure 11.27 A short circuit in a parallel circuit

Figure 11.29

Figure 11.30



Parallel Circuit SummaryThis 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.

Mixed CircuitsAlmost 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).

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

�V1 � I1R1

I1 � �V1 _ R1

� 240 V __ 400 �

� 0.600 A

�V2 � �VT

�V2 � I2R2

I2 � �V2 _ R2

� 240 V __ 100 �

� 2.40 A

We can check that I1 and I2 are correct using Kirchhoff’s current law.

IT � I1 � I2

� 0.600 A � 2.40 A � 3.00 A

ParaphraseThe 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.

Current IT � I1 � ... � In The sum of the currents through each branch must equal the total current of the circuit.

Resistance 1 _ RT

� 1 _ R1

� ... � 1 _ 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.

Table 11.4 Parallel Circuit Equations

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.

Figure 11.31 A cell phone circuit board contains many components that are wired in series and parallel.



Simplifying a Mixed Circuit 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 _ R��

� 1 _ R2

� 1 _ R3

1 _ R��

� 1 __ 20.0 �

� 1 __ 30.0 �

1 _ R��

� 0.083 ��1

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:

RT � R1 � R�� 24.0 � � 12.0 � � 36.0 �

We can use Ohm’s law to calculate the total current:

IT � �VT _ 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 �V�� ITR��

� (2.00 A)(24.0 �) � (2.00 A)(12.0 �) � 48.0 V � 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.

I2 � �V2 _ R2

I3 � �V3 _ R3

� 24.0 V __ 20.0 �

� 24.0 V __ 30.0 �

� 1.20 A � 0.80 A

R1 � 24.0 �

R2 �20.0 �

R3 �30.0 �

�VT �72.0 V

Figure 11.32 In this mixed circuit, R1 is in series with R2 and R3 , which are in parallel with each other.

Figure 11.33 Replacing the two parallel resistors with the equivalent resistor R�� makes this a series circuit.

R1 � 24.0 �

R�� 12.0 �

�VT �72.0 V

R1 � 24.0 ��V1 � 48.0 V

R2 � 20.0 � R3 � 30.0 ��VT �72.0 V �V2 � 24.0 V �V3 � 24.0 V

IT � 2.00 A

Figure 11.34 The voltage drop across each resistor is known.



Table 11.5 outlines the steps you can use to analyze mixed circuits when the resistances are known.

Example 11.6

Analyze the circuit diagram shown in Figure 11.35 to solve for the current and potential difference through each resistor.

GivenR1 � 400 � R3 � 1800 � R5 � 600 �R2 � 500 � R4 � 1200 � �VT � 200 V

Requiredcurrent and voltage drop through each resistor

Analysis and SolutionThis 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 �

The next step is to combine the three resistors in parallel into one equivalent resistor called R��:

1 _ R��

� 1 __ 900 �

� 1 __ 1800 �

� 1 __ 1200 �

1 _ R��

� 0.00250 ��1

R�� 400 �

We can redraw the circuit diagram (Figure 11.37) and determine RT:

RT � R�� R5

� 400 � � 600 � � 1.00 � 103 �

We can determine the total current of the circuit using Ohm’s law:

�VT � ITRT

IT � �VT _ RT

� 200 V ___ 1.00 � 103 �

� 0.200 A

Figure 11.37

R5 � 600 �

R�� 400 �

�VT �200 V

Figure 11.35

Figure 11.36

R1 �400 �R2 �500 �

R5 � 600 �

R3 �1800 �

R4 �1200 �

�VT �200 V

R12 �900 �

R5 � 600 �

R3 �1800 �

R4 �1200 �

�VT �200 V

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.

Table 11.5 Simplifying a Mixed Circuit Explore More

How does increasing the number of resistors affect the resistance and current of a mixed circuit?

PHYSICS•SOURCE



Application of a Mixed Circuit DesignThe 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.

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.

�V�� ITR�� V5 � ITR5

� (0.200 A)(400 �) � (0.200 A)(600 �) � 80.0 V � 120 V

We can use the potential difference to determine the current through each branch:

�V1 � I12R12

I12 � �V12 _ R12

� 80.0 V __ 900 �

� 0.08889 A

�V3 � I3R3

I3 � �V3 _ R3

� 80.0 V __ 1800 �

� 0.04444 A

�V4 � I4R4

I4 � �V4 _ 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 �V2 � I2R2

� (0.08889 A)(400 �) � (0.08889 A)(500 �) � 35.6 V � 44.4 V

ParaphraseWe 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

Practice Problems1. Determine the current and potential

difference through each resistor in Figure 11.38.

2. Determine the current and potential difference through each resistor in Figure 11.39.

Answers1. 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

R1 �10.0 �R2 �30.0 �

R5 � 75.0 �

R3 �100 �

R4 �200 �

�VT �10.0 V

R3 � 4.50 �

R4 � 8.00 �

R1 �30.0 �

R2 �10.0 �

�VT �120 V


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PHYSICS•SOURCE

Figure 11.40 R1 is wired in series with the battery and the light bulbs, which are in parallel with each other.

R1R2 R3 R4

�VT �30 V

Figure 11.38

Figure 11.39



1 _ R��

� 1 _ R2

� 1 _ R3

� 1 _ R4

1 _ R��

� 1 _ 60 �

� 1 _ 60 �

� 1 _ 60 �

1 _ R��

� 0.050 ��1

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 � �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�� I��R��

I�� V�� _ R��

� 10 V _ 20 �

� 0.50 A

We can now determine the value of R1 using Ohm’s law:

�V1 � I1R1

R1 � �V1 _ 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.

Figure 11.41 The equivalent resistor, R�� represents the combined resistance of the three light bulbs.

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.

PHYSICS•SOURCE

�VT �30 V

R1 � ?R�� 20 �



QuestionWhat are the current and voltage drops across the resistors in a parallel circuit?

Activity OverviewIn 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.



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?

Measuring Current and Potential Difference in a Parallel Circuit


REQUIRED SKILLS■ Measuring■ Drawing conclusions

E5

Figure 11.42 Connect the parallel circuit as shown.

QuestionWhat are the current and voltage drops across the resistors in a mixed circuit?

Activity OverviewIn 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.



1. Which components of this mixed circuit are in series?

2. Which components of this mixed circuit are in parallel?

Measuring Current and Potential Difference in a Mixed Circuit


REQUIRED SKILLS■ Recording and organizing data■ Reporting results

E6

R1

R2 R3

Figure 11.43 A mixed circuit



Check and Reflect

©P

11.3

Key Concept Review

1. Copy the following table into your notebook, and fill in the cells with the appropriate equations.

Type of CircuitPotential Difference Resistance Current

Series circuit

Parallel circuit

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?

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?

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.

7. If two pathways in a parallel circuit have different resistances, will the current in each pathway be the same? Explain your answer.


8. Determine the value of R2 in the following circuit diagram.

Question 8

9. Determine the total resistance and the current through the branches of the following circuit.

Question 9

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)

(b)

Question 10

11. Determine the total current and the current through all the branches of the following parallel circuit.

Question 11

12. Determine the voltage drops and current through all the resistors in the following circuit diagram.

Question 12

Reflection

13. What new insights or strategies did you develop for solving mixed circuits?

R1 �1.250 k�

R2 �1.500 k�

R3 �2.200 k�

�VT �75.0 V

�VT �12.0 V

�VT �12.0 V

R1 �60 k�

R2 �20 k�

R3 �30 k�

R4 �45 k�

�VT �180 V

R1 � 10.0 �

R4 � 20.0 �

R2 �30.0 �

R3 �30.0 �

�VT �90.0 V


R1 �700 � R2 �?

R3 �600 �

I � 0.720 A

�VT �112 V

Circuit Summary



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:

P � �E _ �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 EquationsRecall 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:

P � �Vq

_ �t

Since q � I�t, we can further simplify the equation as:

P � �V(I�t)

__ �t

� �V(I�t)

__ �t

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.

Power Consumption11.4

Section Summary

● Power is the rate at which energy is transferred.

● The power consumed by an electrical appliance can be determined.

PHYSICS INSIGHT

Remember that work is equal to change in energy. The equation for power can also be written as:

P � W _ �t

PHYSICS INSIGHT

It is common practice to omit the � from an equation. This is not correct and the power equation should be written as:

P � �VI



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:

�V � IR

I � �V _ R

P � �V ( �V _ R )

P � �V2

_ R

P � �V2

_ R �V and R are known

I is not known

Table 11.6 Derivation of Power Equations

Explore More

How can you arrange four identical resistors in a circuit to consume the most power? The least power?

PHYSICS•SOURCE

Example 11.7

Determine the power consumed by the light bulb in Figure 11.44.

Given�VT � 120 VRL � 240 �

Requiredpower consumed by the light bulb (PL)

Analysis and SolutionSince the light bulb is the only component in the circuit,�VT � �VL � 120 V

PL � �VL

2

_ RL

� (120 V)2

__ 240 �

� 60.0 W

ParaphraseThe light bulb consumes 60.0 W of power.

�VT �120 V

RL �240 �

Practice Problems1. Determine the power

dissipated by the resistor in Figure 11.45.

2. Determine the resistance and power consumption of the resistor in Figure 11.46.

Answers1. 0.022 W

2. R � 4.50 � P � 35.3 W

R �4.50 k�

�VT �10.0 V

R � ?

I � 2.80 A

�VT �12.6 V

Figure 11.44

Power Consumption in CircuitsAs 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.


page 394

PHYSICS•SOURCE

Figure 11.45

Figure 11.46



Given�VT � 120 V R1 � R2 � R3 � 240 �

Requiredpower consumed by the light bulbs in a series circuit and in a parallel circuit (P1, P2, P3)

Analysis and SolutionWe will analyze the power consumed by each light bulb in the series circuit.

RT � R1 � R2 � R3

� 240 � � 240 � � 240 �

� 720 �

P1 � I12 R1 P2 � I 22 R2 P3 � I 32R3

� (0.167 A)2(240 �) � (0.167 A)2(240 �) � (0.167 A)2(240 �) � 6.67 W � 6.67 W � 6.67 W

We can now analyze the power consumed by each light bulb in the parallel circuit:

P1 � �V1

2

_ R1

P2 � �V2

2

_ R2

P3 � �V3

2

_ R3

� (120 V)2

__ 240 �

� (120 V)2

__ 240 �

� (120 V)2

__ 240 �

� 60.0 W � 60.0 W � 60.0 W

ParaphraseEach light bulb consumes 6.67 W of power when placed in series, but 60.0 W of power when placed in parallel.

Practice Problems1. Determine the power consumed by

the resistors in Figure 11.49.

2. Determine the power through each resistor in Figure 11.50.

Answers1. P1 � 5.56 W P2 � 5.00 W P3 � 2.78 W

2. P1 � 56.3 kW P2 � 37.5 kW P3 � 150 kW

�VT �80.0 V

R2 �180 �

R3 � 100 �

R1 � 200 �

R1 �4.00 k�

R2 �6.00 k�

R3 �1.50 k�

�VT �15.0 kV

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.

�VT �120 V

R2 �240 �

R3 � 240 �

R1 � 240 �

Figure 11.47

�VT �120 V

R1 �240 �

R2 �240 �

R3 �240 �

Figure 11.48

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).

�VT � ITRT

IT � �VT _ RT

� 120 V __ 720 �

� 0.167 A

Figure 11.49

Figure 11.50



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.

House Wiring and Power ConsumptionA 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 PC � �VIC

IH � PH _ �V

IC � PC _ �V

� 1200 W __ 120 V

� 180 W __ 120 V

� 10.00 A � 1.50 A

The total current drawn by the hair dryer and computer is:

IT � IH � IC

� 10.00 A � 1.50 A � 11.50 A

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).

IV � PV _ �V

IT � IH � IC � IV

� 600 W __ 120 V

� 10.0 A � 1.50 A � 5.00 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.

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.

PHYSICS•SOURCE

PH � 1200 W

PC � 180 W

�VT �120 V

PHYSICS INSIGHT

The symbol represents a fuse in a circuit diagram.



Example 11.9

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?

GivenPK � 900 W PV � 600 W PTV � 100 W PM � 1500 WPCM � 800 W PC � 150 W IT � 15.0 A VT � 120 V

Requiredcombination of four appliances that will not exceed the maximum power possible in the circuit

Analysis and SolutionThe 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

ParaphraseThere are two combinations of four appliances running simultaneously that will not overload the circuit.

Practice Problems1. 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.

Answers1. air conditioner and the washing machine

2. I1 � 13.33 A I2 � 7.50 A I3 � 5.83 A

space heaterstove

PH � 900 W PS �

700 W

�VT �120 V

A1

A2 A3

QuestionHow does the power consumption of a series circuit compare with the power consumption in a parallel circuit?

Activity OverviewIn 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.



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?

Comparing the Power Consumption of Series and Parallel Circuits


REQUIRED SKILLS■ Predicting■ Analyzing patterns

E7

Figure 11.53 Series circuit for the activity

Figure 11.52



Check and Reflect

©P

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.

Appliance Power Consumption

toaster oven 1100 W

slow cooker 600 W

blender 300 W

popcorn maker 250 W

Question 12

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?

dishwasher garburator microwave

PD � 1500 W

PG � 450 W

PM � 1200 W

�VT �120 V


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?


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.

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.

Question 10

heating coil

I � 2.40 A R � 6.80 k�

R1 � 33.33 �

R2 �100 �

R3 �200 �

�VT �500 V

11.4

Power Consumption of Appliances


396 Unit E Electricity and Magnetism

11 CHAPTER REVIEWC H A P T E R

©P

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

Question 12

13. Write a paragraph explaining how power and energy are related. c

14. Determine the power dissipated by the light bulb in the following diagram. a

Question 14


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

Question 20

R1 � 30 �

R2 � 15 �

I �0.200 A

�VT � ?

�V � 50.0 V

light bulb

I � 5.00 A

R1 � 8.00 �

R3 � ?

R2 �20.0 �

I � 5.00 A

�VT �200 V


Chapter 11 Review 397©P

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

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

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

Question 23

24. Determine the voltage drop and current through each resistor in the following circuit. t

Question 24

25. Determine the value of resistor 2 in the following circuit. t

Question 25

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

Question 27

28. Determine the power consumed by each resistor in the following circuit. t

Question 28

Reflection

29. What is the most interesting thing that you learned in this chapter? c

R1 � 15.0 � R2 � 38.0 �

R4 �22.0 �

R5 �8.00 �

R3 �5.00 �

I �568.2 mA

�VT � ?

R1 �4.50 k�

R2 �6.50 k�

�VT �90.0 V

�VT �120 V

R1 �80 �

R3 � ?

R2 �80 �

R1 �5.00 �R2 �15.0 �

R3 �2.00 �

R4 �28.0 �

�VT �50.0 V

R1 � 120.0 �

I � 1.25 A

R2 � ? R3 �600 �

�VT �400 V

circuit breaker (fuse)Imax � 15.0 A

table saw

drill presshedge

trimmer

PTS � 1400 W

PHT � 450 W

PDP � 890 W

�VT �120 V

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.

Unit Task Link


ACHIEVEMENT CHART CATEGORIESk Knowledge and understanding t Thinking and investigation

c Communication a Application

R1 � 200 �

R2 � 1000 �

R3 � 1500 �

I � 6.0 A


Documents

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