Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 1 of 36
`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 1 of 36
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`EXHIBIT
`EXHIBIT
`A
`4
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`

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`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 2 of 36
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`e
`
`i=h
`
`o Iu
`
`J ==i
`
`rinciples
`
`of Electric
`
`Circuits
`
`a ae ag
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`

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`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 3 of 36
`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 3 of 36
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`Library of Congress Cataloging-Publication Data
`
`Floyd, Thomas L.
`Principles of electric circuits / Thomas L. Floyd.—Sth ed.
`p.
`cm.
`Includes index.
`ISBN 0-13-232224-2
`1. Electric circuits.
`TK454.F57
`1997
`621.3815—dc20
`
`I. Title.
`
`96-13443
`CIP
`
`Cover photo: © Superstock
`Editor: Linda Ludewig
`Developmental Editor: Carol Hinklin Robison
`Production Editor: Rex Davidson
`
`Text Designer: Anne Flanagan
`Cover Designer: Brian Deep
`Production Buyer: Patricia A. Tonneman
`Marketing Manager: Debbie Yarnell
`Illustrations: Jane Lopez and Steve Botts
`
`This book was set in Times Roman by The Clarinda Company and wasprinted and bound by Von
`Hoffmann Press, Inc. The cover was printed by Von Hoffmann Press,Inc.
`
`© 1997, 1993 by Prentice-Hall, Inc.
`Simon & Schuster/A Viacom Company
`Upper Saddle River, New Jersey 07458
`
`All rights reserved. No part of this book may be reproduced, in any form or by any means, with-
`out permission in writing from the publisher.
`
`Earlier editions © 1989, 1985, 1981 by Merrill Publishing Company
`
`Printed in the United States of America
`
`10098 765 43 2
`
`1
`
`ISBN 0-13-232224-2
`
`Prentice-Hall International (UK) Limited, London
`Prentice-Hall of Australia Pty. Limited, Sydney
`Prentice-Hall Canada Inc., Toronto
`Prentice-Hall Hispanoamericana, S. A., Mexico
`Prentice-Hall of India Private Limited, New Delhi
`Prentice-Hall of Japan, Inc., Tokyo
`Simon & Schuster Asia Pte. Ltd., Singapore
`Editora Prentice-Hall do Brasil, Ltda., Rio de Janeiro
`
`

`

`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 4 of 36
`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 4 of 36
`
`26
`
`@ VOLTAGE, CURRENT, AND RESISTANCE
`
`EXAMPLE 2-1
`
`How many coulombs do 93.75 x 10° electrons represent?
`
`2
`93.75 x 10° electrons
`numberof electrons
`— —Hunmbervekee eeSSeee = ISK 10" C=H0.15C
`
`O= 65 x 10electrons/C~6.25 x 10" electrons/C
`
`Solution
`
`Related Exercise How manyelectrons doesit take to have 3 C of charge?
`
`SECTION 2-2
`REVIEW
`
`1.’ Whatis the symbol for charge?
`2. Whatis the unit of charge, and what is the unit symbol?
`3. What causes positive and negative charge?
`4. How much charge, in coulombs,isthere in 10 x 10’? electrons?
`
`2-3 @ VOLTAGE
`
`As you haveseen, a force ofattraction exists between a positive and a negative charge.
`A certain amount of energy must be exerted in theform of work to overcomethe force
`and move the charges a given distance apart. All opposite charges possess a certain
`potential energy because of the separation between them. The difference in potential
`energy of the chargesis the potentialdifference or voltage. Voltageis the drivingforce
`in electric circuits and is what establishes current.
`
`After completing this section, you should beable to
`
`@ Define voltage and discuss its characteristics
`
`State the formula for voltage
`1 Name anddefine the unit of voltage
`OC) Describe basic sources of voltage
`
`
`
`
`
`Consider a water tank that is supported several feet above the ground. A given amountof
`energy mustbe exerted in the form of work to pump waterupto fill the tank. Once the
`wateris stored in the tank, it has a certain potential energy which,if released, can be used
`to perform work. For example, the water can be allowedto fall down a chute to turn a
`water wheel.
`The difference in potential energy in electrical termsis called voltage (V) and is
`expressed as energy or work (W) per unit charge (Q).
`
`V= Ww
`Q
`
`(2-2)
`
`where W is expressed in joules (J) and Q is in coulombs (C).
`
`Volt: The Unit of Voltage
`The unit of voltage is the volt, symbolized by V.
`One volt is the potential difference (voltage) between two points when
`onejoule of energy is used to move one coulombof charge from one point
`to the other.
`
`

`

`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 5 of 36
`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 5 of 36
`
`THE ELECTRIC CIRCUIT #
`
`39
`
`2-6 @ THE ELECTRIC CIRCUIT
`
`
`A basic electric circuit is an arrangement ofphysical components which use voltage,
`current, and resistance to perform some useful function.
`
`After completing this section, you should be able to
`
`™@ Describe a basic electric circuit
`
`C1 Relate a schematic to a physical circuit
`
`
`Define open circuit and closed circuit
`
`Cj Describe various types of protective devices
`
`
`Describe various types of switches
`
`Explain how wiresizes are related to gage numbers
`Define ground
`
`
`
`
`
`
`Direction of Current
`
`For a few years after the discovery of electricity, people assumed all current consisted of
`movingpositive charges. However,in the 1890s, the electron wasidentified as the charge
`carrier for current in conductive materials.
`Today, there are two accepted conventions for the direction of electrical current.
`Electron flow direction, preferred by manyin the fields of electrical and electronics tech-
`nology, assumes for analysis purposes that current is out of the negative terminal of a
`voltage source, through the circuit, and into the positive terminal of the source. Conven-
`tional current direction assumes for analysis purposes that current is out of the positive
`terminal of a voltage source, through the circuit, and into the negative terminal of the
`source. By following the direction of conventional current, there is a rise in voltage across
`a source (negative to positive) and a drop in voltage across a resistor (positive to nega-
`tive).
`
`It actually makes no difference which direction of current is assumed as long as
`it is used consistently. The results of electric circuit analysis are not affected by the
`direction .of current that is assumed for analytical purposes. The direction used for
`analysis is largely a matter of preference, and there are many proponents for each
`approach.
`Conventional current direction is used widely in electronics technology and is used
`almost exclusively at
`the engineering level. Conventional current direction is used
`throughoutthis text. An alternate version of this text that uses electron flow direction is
`also available.
`
`The Basic Circuit
`
`Basically, an electric circuit consists of a voltage source, a load, and a path for current
`between the source and the load. Figure 2-26 showsin pictorial form an example of a
`simple electric circuit: a battery connected to a lamp with two conductors (wires). The
`battery is the voltage source, the lampis the load on the battery because it draws current
`from the battery, and the two wires provide the current path from the positive terminal of
`the battery to the lamp and backto the negative terminal of the battery, as shown in part
`(b). Current goes throughthe filament of the lamp (which hasaresistance), causing it to
`emit visible light. Current through the battery occurs by chemical action. In manypracti-
`cal cases, one terminal of thebattery is connected to a common or ground point. For
`example, in most automobiles, the negative battery terminal is connected to the metal
`chassis of the car. The chassis is the ground for the automobile electrical system and acts
`as a conductor which completesthe circuit.
`
`

`

`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 6 of 36
`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page6 of 36
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`49 = VOLTAGE, CURRENT; AND RESISTANCE
`
`Conductor
`
`FIGURE 2-26
`A simple electric circuit.
`
` {b)
`
`FIGURE 2-27
`Schematic for the circuit in Figure 2-26(a).
`
`The Electric Circuit Schematic
`Anelectric circuit can be represented by a schematic using standard symbols for each
`element, as shown in Figure 2-27 for the simple circuit in Figure 2—26(a). The purpose
`of a schematic is to show in an organized manner how the various components in a given
`circuit are interconnected so that the operation of the circuit can be determined.
`
`Closed and OpenCircuits
`The examplecircuit in Figure 2~26illustrated a closed circuit—thatis, a circuit in which
`the current has a complete path. When the current path is broken,the circuit is called an
`open circuit.
`Switches Switches are commonly used for controlling the opening or closing of cir- ©
`cuits by either mechanical or electronic means. For example, a switch is used to turn a
`lampon oroff as illustrated in Figure 2-28. Each circuit pictorial is shown with its asso-
`ciated schematic. The type of switch indicated is a single-pole-single-throw (SPST) tog-
`gle switch.
`Figure 2-29 shows a somewhat more complicated circuit using a single-pole—
`double-throw (SPDT) type of switch to control the current to two different lamps. When
`one lamp is on, the other is off, and vice versa, as illustrated by the two schematics in
`parts (b) and (c), which represent each of the switch positions.
`
`FIGURE 2-28
`Basic closed and open circuits using an
`
`Closed
`
`me
`switch
`
`SPSTswitch for control.
`switch
`
`Open
`
`(b) There is no current in an open circuit (switch is OFF or in the open position).
`
`

`

`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 7 of 36
`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 7 of 36
`
`GLOSSARY ®
`
`61
`
`= An ohmmeteris connected across a resistor (resistor must be removed from circuit).
`One coulombis the charge of 6.25 x 10" electrons.
`One volt is the potential difference (voltage) between two points when one joule of energy is
`used to move one coulomb from one point to the other.
`= One ampere is the amountof current that exists when one coulomb of charge moves through a
`given cross-sectional area of a material in one second.
`® One ohmis the resistance when there is one ampereof current in a material with one volt applied
`across the material.
`= Figure 2-53 showsthe electrical symbols introduced in this chapter.
`
`IVv
`
`
`
`ti ge Of
`
`Battery
`
`Resistor
`
`Potentiometer
`
`Rheostat
`
`Lamp
`
`Ground
`
`lLoe
`
`NOPB
`switch
`
`oe ee ke
`
`oOo oe o oOo OO om—o
`SPDT
`DPST
`DPDT
`Rotary
`switch
`switch
`switch
`switch
`
`820
`
`o-
`
`NCPB
`switch
`
`SPST
`switch
`
`oe -O -O O-
`
`Circuit breaker
`
`Voltmeter
`
`Ammeter
`
`Ohmmeter
`
`FIGURE 2-53
`
`
`American wire gage (AWG)Astandardization based on wire diameter.
`™ GLOSSARY
`Ammeter An electrical instrument used to measure current.
`Ampere (A)
`Theunit of electrical current.
`Atom The smallest particle of an element possessing the unique characteristics of that element.
`Atomic number The numberof protons in a nucleus.
`Atomic weight The numberofprotons and neutronsin the nucleus of an atom.
`Battery An energy sourcethat uses a chemical reaction to convert chemical energy into electri-
`cal energy.
`Charge An electrical property of matter that exists because of an excess or a deficiency of elec-
`trons. Charge can be either positive or negative.
`Circuit An interconnection ofelectrical components designed to produce a desired result. A basic
`circuit consists of a source, a load, and an interconnecting current path.
`Circuit breaker A resettable protective device used for interrupting excessive current in an elec-
`tric circuit.
`Theunit of the cross-sectional area of a wire.
`Circular mil (CM)
`Closed circuit A circuit with a complete current path.
`Conductance The ability of a circuit to allow current. The unit is the siemens(S).
`Conductor A material in which electric currentis easily established. An example is copper.
`Coulomb (C) The unit of electrical charge.
`Current
`Therate of flow of charge (electrons).
`Electrical Related to the use of electrical voltage and current to achieve desired results.
`Electron The basic particle of electrical charge in matter. The electron possesses negative charge.
`
`

`

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`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 8 of 36
`
`62 = VOLTAGE, CURRENT, AND RESISTANCE
`
`Electronic Related to the movement and control of free electrons in semiconductors or vacuum
`devices.
`
`Element Oneof the unique substances that make up the known universe. Each elementis char-
`acterized by a unique atomic structure.
`Free electron A valence electron that has broken away from its parent atom andis free to move
`from atom to atom within the atomic structure of a material.
`
`FuseAprotective device that burns open whenthere is excessive currentin a circuit.
`Generator An energy source that produceselectrical signals.
`Ground The commonor reference point in a circuit.
`Insulator A material that does not allow current under normal conditions.
`
`Joule (J) The unit of energy.
`Load_Thedevice in a circuit upon which work is done.
`Multimeter An instrument that measures voltage, current, and resistance.
`Neutron An atomic particle having no electrical charge.
`NodeAunique pointin a circuit where two or more components are connected.
`Ohm (Q)
`Theunit of resistance.
`Ohmmeter An instrument for measuring resistance.
`Opencircuit A circuit in which there is not a complete current path.
`Photoconductive cell A type of variable resistor that is light-sensitive.
`Potentiometer A three-terminal variable resistor.
`
`Power supply Anelectronic instrument that produces voltage, current, and power from the ac
`powerline orbatteries in a form suitable for use in powering electronic equipment.
`ProtonApositively charged atomic particle.
`Resistance Opposition to current. The unit is the ohm (Q).
`Resistor An electrical component designed specifically to provide resistance.
`Rheostat A two-terminal variableresistor.
`
`Schematic A symbolized diagram of an electrical or electronic circuit.
`Semiconductor A material that has a conductance value between that of a conductor and an insu-
`lator. Silicon and germanium are examples.
`Shell The orbit in which an electron revolves.
`
`SourceAdevice that produceselectrical energy.
`Switch An electrical device for opening and closing a current path.
`Tapered Nonlinear, such as a tapered potentiometer.
`Thermistor A type of variableresistor.
`Tolerance
`Thelimits of variation in the value of a component.
`Valence Related to the outer shell or orbit of an atom.
`Valence electron Anelectron that is present in the outermost shell of an atom.
`Volt The unit of voltage or electromotive force.
`Voltage The amountof energy available to move a certain numberof electrons from one point to
`another in an electric circuit.
`
`Voltmeter An instrument used to measure voltage.
`Wiper
`Thesliding contact in a potentiometer.
`
`w FORMULAS
`
`(2-1)
`
`(2-2)
`
`(2-3)
`
`V=
`
`I=
`
`Q= numberof electrons=? Ch
`6.25 x 10'® electrons/C
`ols
`lo
`
`ange
`
`Voltage equals energy divided by charge.
`
`Current equals charge divided by time.
`
`

`

`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 9 of 36
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`
`72
`
`« OHM’'S LAW
`
`3-1 m OHM’S LAW
`
`
`Ohm’s law describes mathematically how voltage, current, and resistance in a circuit
`are related. Ohm’s law is used in three equivalent forms depending on which quantity
`you-need to determine. In this section, you will learn each of these forms.
`
`After completing this section, you should be able to
`
`@ Explain Ohm’s law
`O Describe how V, £, andRare related
`
`O Express J as a function of V and R
`Ci Express V as a function of J and R
`0 Express R as a function of V and J
`
`
`Ohm determined experimentally that if the voltage across a resistor is increased, the cur-
`rent through theresistor will also increase; and, likewise, if the voltage is decreased, the
`current will decrease. For example, if the voltage is doubled, the current will double. If
`the voltage is halved, the current will also be halved. This relationship is illustrated in
`Figure 3-1, with relative meter indications of voltage and current.
`
`
`
`(a) LessV,less I
`
`(b) More V, more 7
`
`FIGURE 3-1
`Effect of changing the voltage with the resistance at a constant value.
`
`Ohm’slaw alsostates that if the voltage is kept constant, less resistance results in
`more current, and, also, more resistance results in less current. For example, if the resis-
`tance is halved, the current doubles. If the resistance is doubled, the current is halved.
`This conceptis illustrated by the meter indications in Figure 3-2, where the resistance is
`increased and the voltage is held constant.
`
`
`
`(a) Less R, more
`
`(b) MoreR,less I
`
`FIGURE 3-2
`Effect of changing the resistance with the voltage at a constant value.
`
`

`

`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 10 of 36
`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 10 of 36
`
`OHM’S LAW a
`
`73
`
`Formula for Current
`
`Ohm’s law can be stated as follows:
`
`omneg
`
`i zi<
`
`(3-1)
`
`This formula describes what was indicated by the circuits of Figures 3-1 and 3-2. For a
`constant value of R, if the value of V is increased, the value of J increases; if V is
`decreased, 7 decreases. Also notice in Equation (3-1) that if V is constant and R is
`increased, J decreases. Similarly, if V is constant and R is decreased, J increases.
`Using Equation (3-1), you can calculate the current if the values of voltage and
`resistance are known.
`
`Formula for Voltage
`Ohm’s law can also be stated another way. By multiplying both sides of Equation (3-1)
`by R and transposing terms, you obtain an equivalent form of Ohm’s law, as follows:
`
`With this equation, you can calculate voltage if the current and resistance are known.
`
`V=IR
`
`(3-2)
`
`Formula for Resistance
`
`Thereis a third equivalent way to state Ohm’s law. By dividing both sides of Equation
`(3-2) by J and transposing terms, you obtain
`
`R=
`
`~I<
`
`(3-3)
`
`This form of Ohm’slaw is used to determineresistance if voltage and current values are
`known.
`Remember,the three formulas—Equations (3-1), (3-2) and (3-3)—areall equiva-
`lent. They are simply three different ways of expressing Ohm’slaw.
`
`SECTION 3-1
`REVIEW
`
`1. Ohm’slaw defines how three basic quantities are related. What are these quantities?
`2. Write the Ohm’s law formula for current.
`
`3. Write the Ohm’s law formula for voltage.
`4, Write the Ohm’s law formulafor resistance.
`
`5. If the voltage across a fixed-value resistor is tripled, does the current increase or
`decrease, and by how much?
`6. If the voltage across a fixed resistor is cut in half, how much will the current
`change?
`7. Thereis a fixed voltage across a resistor, and you measure a currentof 1 A. If you
`replace the resistor with one that has twice the resistance value, how much current
`will you measure?
`8. In a circuit the voltage is doubled and the resistance is cut in half. Would the cur-
`rent increase or decrease, and if so, by how much?
`
`

`

`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 11 of 36
`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 11 of 36
`
`96 = ENERGY AND POWER
`
`4-1 @ ENERGY AND POWER
`
`When there is current through a resistance, energy is released in the form of heat. A
`common example of this is a light bulb that becomes too hot to touch. The current
`through the filament that produces light also produces unwanted heat because the fil-
`ament has resistance. Power is a measure of how fast energy is being used; electrical
`components mustbe able to dissipate a certain amount of energy in a given period of
`time.
`
`After completingthis section, you should be able to
`
`™@ Define energy and power
`C1 Express powerin terms of energy
`C State the unit of power
`O State the common units of energy
`O Perform energy and powercalculations
`
`
`Energy is the fundamental ability to do work.
`
`Poweris the rate at which energy is used.
`
`In other words, power, symbolized by P, is a certain amount of energy used in a certain
`length of time, expressed as follows:
`
`(4-1)
`
`Power = energy
`time
`
`Wt
`
`P=
`
`Energy is measured in joules (J), time is measured in seconds (s), and power is
`measured in watts (W). Note that an italic W is used to represent energy in the form of
`work and a nonitalic W is used for watts, the unit of power.
`Energy in joules divided by time in seconds gives power in watts. For example,if
`50 J of energy are used in 2 s, the power is 50 J/2 s = 25 W.Bydefinition,
`
`Onewatt is the amount of power when onejoule of energy is used in one
`second.
`
`Thus, the number of joules used in one second is always equal to the number of watts.
`For example, if 75 J are used in 1 s, the power is 75 W.
`
`EXAMPLE 4-1
`
`An amountof energy equal to 100 J is used in 5 s. What is the power in watts?
`
`used?
`
`Solution
`
`Related Exercise
`
`If 100 W of poweroccurs for 30 s, how muchenergy, in joules, is
`
`Amounts of power much less than one watt are commonin certain areas of elec-
`tronics. As with small current and voltage values, metric prefixes are used to designate
`small amounts of power. Thus, milliwatts (mW), microwatts (UW), and even picowatts
`(pW) are commonly found in some applications.
`
`

`

`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 12 of 36
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`
`104 m= ENERGY AND POWER
`
`The rating of the resistor is 1 W, which is insufficient to handle the power. The resis-
`tor has been overheated and may be burned out, making it an open.
`
`Related Exercise A 0.25 W, 1 kQ resistor is connected across a 12 V battery. Willit
`burn out?
`
`
`SECTION 4-3
`REVIEW
`
`1. Name two importantvalues associated with a resistor.
`2. How doesthe physical size of a resistor determine the amount of powerthat it can
`handle?
`
`3. List the standard powerratings of carbon-composition resistors.
`4. A resistor must handle 0.3 W. What minimum power rating of a carbon resistor
`should be used to dissipate the energy properly?
`
`4—4 m ENERGY CONVERSION AND VOLTAGE DROP IN RESISTANCE
`
`As you have seen, when there is current through a resistance, electrical energy is con-
`verted to heat energy. This heat is caused bycollisions of the free electrons within the
`atomic structure of the resistive material. When a collision occurs, heat is given off;
`and the electron loses some of its acquired energy.
`After completing this section, you should be able to
`@ Explain energy conversion and voltage drop
`© Discuss the cause of energy conversion in a circuit
`C1) Define voltage drop
`C Explain the relationship between energy conversion and voltage drop
`
`
`In Figure 4—10, electrons are flowing out of the negative terminal of the battery. They
`have acquired energy from thebattery andare at their highest energy level at the negative
`side of the circuit. As the electrons move through the resistor, they lose energy. The elec-
`trons emerging from the upper endof the resistor are at a lower energy level than those
`entering the lower end because some ofthe energy they had has been converted to heat.
`The drop in energy level of the electrons as they move through the resistor creates a
`potential difference, or voltage drop, across the resistor having the polarity shownin Fig-
`ure 4—10. Notice that the upper endofthe resistor in Figure 4-10 is less negative (more
`positive) than the lower end.
`
`Electrons have lower
`a <——— energy onpositive
`8
`side ofcircuit.
`
`
`y | Energydrop
`in this direction
`Electrons have higher
`f
`
`@ |<—— energy on negative
`4
`side of circuit.
`
`L e—> e-—__>e-—_—_- @__-6
`@ Electron
`
`R
`
`to
`
`\
`
`FIGURE 4-10
`Electron flow in a simple circuit.
`
`

`

`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 13 of 36
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`
`110 = ENERGY AND POWER
`
`SECTION 4-7 2 1. ‘Explain éach of the following .DC controlstatements: a
`(a) DC Vi
`1
`8
`1°
`(b) DC 11
`1M 5M 25M
`REVIEW
`2. Explain each of the following .PRINT statements:
`(a) .PRINT DC V(R2) V(R9)
`
`(b) PRINT DC I(R1)
`
`1(R2) KR3)
`
`
`
`m™@ SUMMARY
`
`@ GLOSSARY
`
`One watt equals one joule per second.
`Watt is the unit of power, joule is a unit of energy, and secondis a unit of time.
`The powerrating of a resistor determines the maximum powerthatit can handlesafely.
`Resistors with a larger physical size can dissipate more powerin the form of heat than smaller
`ones.
`
`A resistor should have a powerrating higher than the maximum powerthatit is expected to han-
`dle in the circuit.
`.
`
`Powerrating is not related to resistance value.
`Aresistor normally opens when it burnsout.
`Energy is equal to power multiplied by time.
`The kilowatt-hour is a unit of energy.
`One kilowatt-hour is one thousand watts used for one hour.
`A powersupply is an energy source used to operate electrical and electronic devices.
`A battery is one type of power supply that converts chemical energy into electrical energy.
`An electronic power supply converts commercial energy (ac from the power company) to regu-
`lated dc or ac at various voltage levels.
`The output powerof a supply is the output voltage times the load current.
`A load is a device that draws current from the power supply.
`The capacity of a battery is measured in ampere-hours (Ah).
`One ampere-hour equals one ampere used for one hour, or any other combination of amperes and
`hours that has a product of one.
`= A power supply with a high efficiency wastes less power than one with a lowerefficiency.
`
`Ampere-hour rating A number given in ampere-hours (Ah) determined by multiplying the cur-
`rent (A) timesthe length of time (h) a battery can deliver that currentto a load.
`Efficiency The ratio of the output power to the input powerof a circuit, expressed as a percent.
`Energy The fundamental ability to do work.
`Joule (J) The unit of energy.
`Kilowatt-hour (kWh) A commonunit of energy used mainly by utility companies.
`Power
`Therate of energy usage.
`Powerrating The maximum amount of powerthat a resistor can dissipate without being dam-
`aged by excessive heat buildup.
`Voltage drop The drop in energy level through a resistor.
`Watt (W) The unit of power. One watt is the power when 1 J of energy is used in 1 s.
`Watt’s law A law that states the relationships of power to current, voltage, and resistance.
`
`@ FORMULAS
`
`(4-1)
`
`(4-2)
`(4-3)
`
`_W
`
`P
`
`W=Pt
`P=IR
`
`Power equals energy divided by time.
`
`Energy equals power multiplied by time.
`Power equals current squared times resistance.
`
`

`

`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 14 of 36
`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 14 of 36
`
`SELF-TEST =
`
`111
`
`(4-4)
`
`(4-5)
`
`(4-6)
`(4-7)
`
`P=VI
`2
`
`= ~
`
`
`Pour
`Efficiency = P
`IN
`Pour = Pin — Pruoss
`
`Powerequals voltage times current.
`
`Powerequals voltage squared dividedby resistance.
`
`Power supply efficiency
`
`Output power is input powerless powerloss.
`
`m SELF-TEST
`
`- Power can be defined as
`
`(b) heat
`(a) energy
`(d) the time required to use energy
`(c) the rate at which energy is used
`. Two hundred joules of energy are consumed in 10 s. The poweris
`(a) 2000 W
`(b) 10 W
`(c) 20 W
`(d) 2W
`. If it takes 300 ms to use 10,000 J of energy, the power is
`(a) 33.3 kW
`(b) 33.3 W
`(c) 33.3 mW
`. In 50 kW,there are
`
`(b) 5000 W
`(a) 500 W
`.» In 0.045 W,there are
`
`(c) 0.5 MW
`
`(d) 50,000 W
`
`(c) 4,500 pW (ad) 0.00045 MW
`(b) 45 mW
`(a) 45 kW
`. For 10 V and 50 mA,the poweris
`(d) answers (a), (b), and (c)
`(a) 500 mW
`(b) 0.5 W
`(c) 500,000 pW
`. When the current through a 10 kQ resistor is 10 mA, the poweris
`(a) 1W (b) 10 W
`(c) 100 mW
`(a) 1000 pW
`. A 2.2 kQ resistor dissipates 0.5 W. The currentis
`(a) 15.1 mA
`(b) 0.227 mA
`(c) 1.1mA
`. A330 Q resistor dissipates 2 W. The voltage is
`(a) 2.57 V
`(b) 660 V
`(c) 6.6V
`(d) 25.7 V
`If you used 500 W of powerfor 24 h, you have used
`(a) 0.5 kWh
`(b) 2400 kWh
`(c) 12,000 kWh
`How many watt-hours represent 75 W used for 10 h?
`(a) 75 Wh
`(b) 750 Wh
`(c) 0.75 Wh
`(d) 7500 Wh
`A 100 Q resistor must carry a maximum current of 35 mA.Its rating should be at least
`(a) 35 W
`(b) 35 mW
`(c) 123 mW
`(d) 3500 mW
`The powerrating of a carbon-composition resistor that is to handle up to 1.1 W should be
`(a) 0.25 W
`(b) 1 W
`(c) 2W (d) 5W
`A 22 Q half-watt resistor and a 220 Q half-watt resistor are connected across a 10 V source.
`
`(d) 4.4 mA
`
`(d) 12 kWh
`
`10.
`
`11.
`
`12.
`
`13.
`
`14.
`
`15.
`
`16.
`
`17.
`
`18.
`
`Which one(s) will overheat?
`(d) neither
`(c) both
`(a) 22Q
`(b) 220 2
`When the needle of an analog ohmmeter indicates infinity, the resistor being measured is
`(a) overheated
`(b) shorted
`(c) open
`(d) reversed
`/
`A 12 V battery is connected to a 600 Q load. Under these conditions,it is rated at 50 Ah. How
`long can it supply current to the load?
`;
`(d) 4.16h
`(a) 2500 h
`(b) 50h
`(c) 25h
`A given powersupply is capable of providing 8 A for 2.5 h. Its ampere-hour rating is
`(a) 2.5 Ah
`(b) 20 Ah
`(c) 8 Ah
`A power supply produces a 0.5 W output with an input of 0.6 W. Its percentage of effi-
`ciency is
`(a) 50%
`
`(c) 83.3%
`
`(d) 45%
`
`(b) 60%
`
`

`

`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 15 of 36
`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 15 of 36
`
`262
`
`@
`
`SERIES-PARALLEL CIRCUITS
`
`= Theload resistor should be large comparedto the resistance across which it is connected, in order
`that the loading effect may be minimized. A /0-times value is sometimes usedas a rule of thumb,
`but the value depends on the accuracy required for the output voltage.
`= Tofindtotal resistance of a ladder network,start at the point farthest from the source and reduce
`the resistance in steps.
`= A Wheatstone bridge can be used to measure an unknownresistance.
`® A bridge is balanced when the output voltage is zero. The balanced condition produces zero cur-
`rent through a load connected across the output terminals of the bridge.
`® Open circuits and short circuits are typical circuit faults.
`= Resistors normally open when they burn out.
`
`
`
`m GLOSSARY
`
`Bleeder current The currentleft after the total load current is subtracted from the total current
`into the circuit.
`
`Load An element(resistor or other component) connected across the output terminals of a circuit
`that draws current from thecircuit.
`
`Sensitivity factor The ohms-per-volt rating of a voltmeter.
`
`(7-1)
`
`Runx=Ry(z)
`
`Unknownresistance in aWheatstonebridge
`
`4 m
`
`™ FORMULA
`
`@ SELF-TEST
`
`1. Which of the following statements are true concerning Figure 7-63?
`(a) R, and R, are in series with R3, Ry, and Rs
`(b) R, and R, are in series
`(c) R3, Ra, and Rs; are in parallel
`(a) The series combination of R, and R, is in parallel with the series combination of R3, Ra,
`and R,
`(e) answers (b) and (d)
`
`FIGURE 7-63
`
`2. Thetotal resistance of Figure 7-63 can be found with which of the following formulas?
`(a) Ry + Ro + Rs] Rall Rs
`(b) Ry || Ro + Rs ll Rall Rs
`(c) (R, + R2) || (R3 + Ry + Rs)
`(d) none of these answers
`3. If all of the resistors in Figure 7-63 have the same value, when voltage is applied across ter-
`minals A and B, the current is
`
`(b) greatest in R3, R,, and Rs
`(a) greatest in Rs
`(d) the samein all the resistors
`(c) greatest in R, and R,
`4. Two 1 kQ resistors are in series and this series combination is in parallel with a 2.2 kQ
`resistor. The voltage across one of the 1 kQ resistors is 6 V. The voltage across the 2.2 kQ
`resistor is
`
`(d) 13.2 V
`(c) 12 V
`(b) 3 V
`(a) 6V
`5, The parallel combination of a 330 Q resistor and a 470 Q resistor is in series with the parallel
`combination of four 1 kQ resistors. A 100 V source is connected across the circuit. The resis-
`tor with the most current has a value of
`
`(a) 1kQ
`
`(b) 330 Q
`
`(c) 470 Q
`
`

`

`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 16 of 36
`Case 6:20-cv-00945-ADA Document 37-4 Filed 09/20/21 Page 16 of 36
`278 = CIRCUIT THEOREMS AND CONVERSIONS
`uses
`
`8-1 # THE VOLTAGE SOURCE
`
`The voltage source is the principal type of energy source in electronic applications, so
`it is important to understand its characteristics. The voltage source ideally provides
`constantvoltage to a load even when the load resistance varies.
`
`After completing this section, you should be able to
`
`® Describe the characteristics of a voltage source
`
`C1 Compare a practical voltage source to an ideal source
`
`
`Discuss the effect of loading on a practical voltage source
`
`
`Figure 8-1(a) is the familiar symbolfor an ideal de voltage source. The voltage across
`its terminals A and B remainsfixed regardless of the value of load resistance that may be
`connected across its output. Figure 8—1(b) shows a loadresistor, R;, connected. All of the
`source voltage, Vs, is dropped across R;. Ideally, R, can be changed to any value except
`zero, and the voltage will remain fixed. The ideal voltage source has an internal resistance
`of zero.
`
`FIGURE 8-1
`Ideal de voltage source.
`
`+
`
`‘5 ==
`
`A
`
`B
`
`
`
`(a) Untoaded
`
`(b) Loaded
`
`In reality, no voltage source is ideal. That is, all voltage sources have some inher-
`ent internal resistance as a result of their physical and/or chemical makeup, which can
`be represented by a resistor in series with an ideal source, as shown in Figure 8—2(a).
`Rg is the internal source resistance and Vs is the source voltage. With no load, the out-
`put voltage (voltage from A to B) is Vs. This voltage is sometimes called the