Chapter1IntroductionFIGURE1-3Laser centering andoptical verification in amanufacturingprocess.(CourtesyVanscoElectronicsLtd.)FIGURE1-4 Sometypicalelec-tronic components.Thesmall compo-nents atthebottomaresurfacemountparts that are installed by the machineshown in Figure 1-3.Surface mountparts
6 Chapter 1 ■ Introduction FIGURE 1–3 Laser centering and optical verification in a manufacturing process. (Courtesy Vansco Electronics Ltd.) FIGURE 1–4 Some typical electronic components. The small components at the bottom are surface mount parts that are installed by the machine shown in Figure 1–3. Surface mount parts
7Section1.2TheSISystemofUnitsTABLE1-1CommonQuantities1.2The SI System of Units1meter=100centimeters=39.37The solution of technical problems requires theuse of units.At present, twoinchesmajor systems-theEnglish(US Customary)and themetric-are in everyday1millimeter=39.37milsuse.For scientific and technical purposes,however,the English systemhas1 inch = 2.54 centimetersbeen largely superseded. In its place the SI system is used. Table 1-1 shows a1foot=0.3048 meterfewfrequentlyencountered quantities with units expressed inbothsystems1 yard = 0.9144 meterThe SI system combines the MKS metric units and the electrical units1 mile = 1.609 kilometersinto one unified system:See Tables 1-2 and 1-3.(Do not worry about the1kilogram=1000grams=2.2poundselectrical units yet. We define them later, starting in Chapter 2.) The units in1 gallon (US) = 3.785 litersTable 1-2 are defined units, while the units in Table 1-3 are derived units,obtainedbycombiningunitsfromTable1-2.Notethat somesymbolsandabbreviations usecapital letters while othersuse lowercaseletters.A few non-SI units are still in use.Forexample,electric motors arecommonly rated inhorsepower,and wires arefrequently specified inAWGsizes(AmericanWireGage,Section3.2).Onoccasion,youwillneed tocon-vert non-SI units to SI units.Table1-4maybe used forthispurpose.Definition of UnitsWhenthemetricsystemcameintobeingin1792,themeterwasdefined asone ten-millionth of the distance from the north pole to the equator and thesecondas1/60X1/60X1/24ofthemean solarday.Later,moreaccuratedef-initionsbased onphysical lawsof naturewereadopted.Themeteris nowTABLE1-2Some SI Base UnitsQuantitySymbolUnitAbbreviationlLengthmmeterMassmkgkilogram1TimeSAsecondLiElectric currentampereTKkelvinTemperatureTABLE1-3Some SI Derived Units*UnitQuantitySymbolAbbreviationFNForcenewtonWJEnergyjouleWPowerP.pwattVVoltageVuE,evoltcCharge0.qcoulomb0RohmResistanceFcfaradCapacitanceLHInductancehenryfeHzhertzFrequencyWbMagnetic fluxweberTBteslaMagnetic flux density*Electrical and magnetic quantities will be explained as you progress through the book. As in Table1-2, the distinction between capitalized and lowercase letters is important
1.2 The SI System of Units The solution of technical problems requires the use of units. At present, two major systems—the English (US Customary) and the metric—are in everyday use. For scientific and technical purposes, however, the English system has been largely superseded. In its place the SI system is used. Table 1–1 shows a few frequently encountered quantities with units expressed in both systems. The SI system combines the MKS metric units and the electrical units into one unified system: See Tables 1–2 and 1–3. (Do not worry about the electrical units yet. We define them later, starting in Chapter 2.) The units in Table 1–2 are defined units, while the units in Table 1–3 are derived units, obtained by combining units from Table 1–2. Note that some symbols and abbreviations use capital letters while others use lowercase letters. A few non-SI units are still in use. For example, electric motors are commonly rated in horsepower, and wires are frequently specified in AWG sizes (American Wire Gage, Section 3.2). On occasion, you will need to convert non-SI units to SI units. Table 1–4 may be used for this purpose. Definition of Units When the metric system came into being in 1792, the meter was defined as one ten-millionth of the distance from the north pole to the equator and the second as 1/60 1/60 1/24 of the mean solar day. Later, more accurate definitions based on physical laws of nature were adopted. The meter is now Section 1.2 ■ The SI System of Units 7 TABLE 1–1 Common Quantities 1 meter 100 centimeters 39.37 inches 1 millimeter 39.37 mils 1 inch 2.54 centimeters 1 foot 0.3048 meter 1 yard 0.9144 meter 1 mile 1.609 kilometers 1 kilogram 1000 grams 2.2 pounds 1 gallon (US) 3.785 liters TABLE 1–2 Some SI Base Units Quantity Symbol Unit Abbreviation Length meter m Mass m kilogram kg Time t second s Electric current I, i ampere A Temperature T kelvin K TABLE 1–3 Some SI Derived Units* Quantity Symbol Unit Abbreviation Force F newton N Energy W joule J Power P, p watt W Voltage V, v, E, e volt V Charge Q, q coulomb C Resistance R ohm Capacitance C farad F Inductance L henry H Frequency f hertz Hz Magnetic flux F weber Wb Magnetic flux density B tesla T *Electrical and magnetic quantities will be explained as you progress through the book. As in Table 1–2, the distinction between capitalized and lowercase letters is important.
8Chapter1IntroductionTABLE1-4ConversionsMultiply ByToFindWhen YouKnowLengthinches (in)0.0254meters (m)feet (ft)0.3048meters (m)1.609miles (mi)kilometers (km)Forcepounds (lb)4.448newtons (N)Power746horsepower (hp)watts (W)3.6 ×106Energykilowatthour (kWh)joules* ()1.356joules* (J)foot-pound (ft-lb)Note: 1 joule = 1 newton-meter.defined as the distance travelled by light in a vacuum in 1/299792458 of asecond, while the second is defined in terms of the period of a cesium-basedatomic clock. The definition of the kilogram is the mass of a specific plat-inum-iridium cylinder (the international prototype), preserved at the Interna-tional Bureau of Weights and Measures inFrance.Relative Sizeof theUnits*Togain a feel for the SI units and their relative size, refer to Tables 1-I and1-4.Note that1meteris equal to39.37 inches; thus, 1 inch equals1/39.37=0.0254meteror 2.54centimeters.Aforceof onepound isequal to4.448newtons; thus, 1 newton is equal to 1/4.448 =0.225 pound of force, whichis about the force required to lift a /4-pound weight. One joule is the workdone in moving a distance of one meter against a force of one newton. Thisis about equal to the work required to raise a quarter-pound weight onemeter.Raising the weight one meter in one second requires about one wattof power.The watt is also the SI unit for electrical power.A typical electric lamp,for example,dissipatespower at the rate of 60watts, and a toaster at a rateofabout1000watts.The link between electrical and mechanical units can be easily estab-lished.Consider an electrical generator. Mechanical power input produceselectrical power output.If the generator were 100% efficient,then one wattof mechanical power input would yield one watt of electrical power output.This clearlyties the electrical and mechanical systems ofunits together.However, just how big is a watt? While the above examples suggest thatthe watt is quite small, in terms of the rate at which a human can work it isactuallyquitelarge.For example,aperson candomanual labor at arateofabout 60wattswhenaveragedover an8-hourday—justenoughto powerastandard 60-watt electric lamp continuously overthis time!A horse candoconsiderably better. Based on experiment, Isaac Watt determined that a strongdray horse could average 746 watts. From this, he defined the horsepower (hp)as1horsepower=746watts.Thisis thefigurethat we still use today*Paraphrased from Edward C.Jordan and Keith Balmain, Electromagnetic Waves andRadiating Systems,Second Edition. (Englewood Cliffs,New Jersey:Prentice-Hall, Inc,1968)
defined as the distance travelled by light in a vacuum in 1/299 792 458 of a second, while the second is defined in terms of the period of a cesium-based atomic clock. The definition of the kilogram is the mass of a specific platinum-iridium cylinder (the international prototype), preserved at the International Bureau of Weights and Measures in France. Relative Size of the Units* To gain a feel for the SI units and their relative size, refer to Tables 1–1 and 1–4. Note that 1 meter is equal to 39.37 inches; thus, 1 inch equals 1/39.37 0.0254 meter or 2.54 centimeters. A force of one pound is equal to 4.448 newtons; thus, 1 newton is equal to 1/4.448 0.225 pound of force, which is about the force required to lift a 1 ⁄4-pound weight. One joule is the work done in moving a distance of one meter against a force of one newton. This is about equal to the work required to raise a quarter-pound weight one meter. Raising the weight one meter in one second requires about one watt of power. The watt is also the SI unit for electrical power. A typical electric lamp, for example, dissipates power at the rate of 60 watts, and a toaster at a rate of about 1000 watts. The link between electrical and mechanical units can be easily established. Consider an electrical generator. Mechanical power input produces electrical power output. If the generator were 100% efficient, then one watt of mechanical power input would yield one watt of electrical power output. This clearly ties the electrical and mechanical systems of units together. However, just how big is a watt? While the above examples suggest that the watt is quite small, in terms of the rate at which a human can work it is actually quite large. For example, a person can do manual labor at a rate of about 60 watts when averaged over an 8-hour day—just enough to power a standard 60-watt electric lamp continuously over this time! A horse can do considerably better. Based on experiment, Isaac Watt determined that a strong dray horse could average 746 watts. From this, he defined the horsepower (hp) as 1 horsepower 746 watts. This is the figure that we still use today. 8 Chapter 1 ■ Introduction TABLE 1–4 Conversions When You Know Multiply By To Find Length inches (in) 0.0254 meters (m) feet (ft) 0.3048 meters (m) miles (mi) 1.609 kilometers (km) Force pounds (lb) 4.448 newtons (N) Power horsepower (hp) 746 watts (W) Energy kilowatthour (kWh) 3.6 106 joules* (J) foot-pound (ft-lb) 1.356 joules* (J) Note: 1 joule 1 newton-meter. *Paraphrased from Edward C. Jordan and Keith Balmain, Electromagnetic Waves and Radiating Systems, Second Edition. (Englewood Cliffs, New Jersey: Prentice-Hall, Inc, 1968).
9Section1.3 Converting Units1.3Converting UnitsOften quantities expressed in oneunit must be converted to another.Forexample,supposeyouwanttodeterminehowmanykilometersthereareinten miles.Given that 1 mile is equal to 1.609 kilometers, Table 1-1,you canwrite 1 mi = 1.609 km, using the abbreviations in Table 1-4. Now multiplyboth sides by 10. Thus, 10 mi = 16.09 km.Thisprocedure is quite adequate for simple conversions.However,forcomplex conversions, it may be difficult to keeptrack of units.The proce-dure outlined next helps.It involves writing units into the conversionsequence, cancelling where applicable, then gathering up the remaining unitsto ensure that the final result has the correct unitsTo get at the idea, suppose you want to convert 12 centimeters toinches.FromTable1-1,2.54cm=1in.Sincetheseareequivalent,you canwrite2.54cm1in(11)1=1or2.54cm1inNow multiply12 cm by the second ratio and note that unwanted units can-cel.Thus,1in12cmX=4.72 in2.54cmThe quantities in equation l- are called conversion factors.Conver-sion factors have a value of 1 and you can multiply by them without chang-ing the value of an expression. When you have a chain of conversions, selectfactors sothat all unwanted units cancel.This provides an automatic checkon thefinal resultas illustrated inpart (b)of Example1-1.EXAMPLE1-1Givenaspeedof60milesperhour(mph),a.convertittokilometersperhourb..convert it to meters per second.Solutiona. Recall, 1 mi = 1.609 km. Thus,1.609km1 =1 miNowmultiplyboth sides by 60mi/h and cancel units:60mi1.609km60 mi/h ==96.54km/hXh1mib.Given that 1 mi =1.609 km, 1 km =1000 m,1 h=60min, and 1 min=60s,chooseconversionfactorsasfollows:1h1.609km1000m1min1=and 1-1mi1km60min60s
Section 1.3 ■ Converting Units 9 EXAMPLE 1–1 Given a speed of 60 miles per hour (mph), a. convert it to kilometers per hour, b. convert it to meters per second. Solution a. Recall, 1 mi 1.609 km. Thus, 1 1.6 1 0 m 9 k i m Now multiply both sides by 60 mi/h and cancel units: 60 mi/h 60 h mi 1.6 1 0 m 9 k i m 96.54 km/h b. Given that 1 mi 1.609 km, 1 km 1000 m, 1 h 60 min, and 1 min 60 s, choose conversion factors as follows: 1 1.6 1 0 m 9 k i m, 1 10 1 0 k 0 m m, 1 60 1 m h in , and 1 1 6 m 0 i s n 1.3 Converting Units Often quantities expressed in one unit must be converted to another. For example, suppose you want to determine how many kilometers there are in ten miles. Given that 1 mile is equal to 1.609 kilometers, Table 1–1, you can write 1 mi 1.609 km, using the abbreviations in Table 1–4. Now multiply both sides by 10. Thus, 10 mi 16.09 km. This procedure is quite adequate for simple conversions. However, for complex conversions, it may be difficult to keep track of units. The procedure outlined next helps. It involves writing units into the conversion sequence, cancelling where applicable, then gathering up the remaining units to ensure that the final result has the correct units. To get at the idea, suppose you want to convert 12 centimeters to inches. From Table 1–1, 2.54 cm 1 in. Since these are equivalent, you can write 2.5 1 4 in cm 1 or 2.5 1 4 in cm 1 (1–1) Now multiply 12 cm by the second ratio and note that unwanted units cancel. Thus, 12 cm 2.5 1 4 in cm 4.72 in The quantities in equation 1–1 are called conversion factors. Conversion factors have a value of 1 and you can multiply by them without changing the value of an expression. When you have a chain of conversions, select factors so that all unwanted units cancel. This provides an automatic check on the final result as illustrated in part (b) of Example 1–1.
10Chapter1IntroductionThus,60mi60mi1.609Km1000m1h1min=26.8m/sXhhI mi1km60min60sYoucanalso solvethisproblembytreatingthenumeratoranddenomi-nator separately.For example,you can convert milesto meters and hours toseconds, then divide (see Example 1-2). In the final analysis, both methodsare equivalent.EXAMPLE1-2DoExample 1-1(b)byexpanding thetopand bottom sepa-rately.Solution1.609 km1000m=96540m60mi=60mi×1km1mi60s60min=3600s1h=lhx41h1minThus,velocity=96540m/3600s=26.8m/sasabovePRACTICE1. Area = rr. Given r = 8 inches, determine area in square meters (m).PROBLEMS12.Acartravels60feet in 2 seconds.Determinea. its speed in meters per second,b. its speed in kilometers per hour.For part (b), use the method of Example 1-1, then check using the method ofExample 1-2.Answers:1.0.130m22.a.9.14m/sb.32.9km/h1.4Powerof Ten NotationElectricalvaluesvarytremendouslyinsize.Inelectronicsystems,forexamplevoltages may range from a few millionths of a volt to several thousand volts,whileinpower systems,voltages ofupto several hundred thousand are com-mon.Tohandlethislargerange,thepoweroftennotation(Tablel-5)isused.Toexpress a number inpoweroftennotation,movethedecimalpointtowhereyou wantit,thenmultiplytheresultbythepower of tenneededtorestore the number to its original value.Thus,247000 =2.47X 105.(Thenumber 10 is called thebase, and its power is called the exponent.)An easyway to determine the exponent is to count the number of places (right or left)that you moved the decimal point.Thus,247000=247000=2.47×10554321
You can also solve this problem by treating the numerator and denominator separately. For example, you can convert miles to meters and hours to seconds, then divide (see Example 1–2). In the final analysis, both methods are equivalent. 10 Chapter 1 ■ Introduction Thus, 60 h mi 60 h mi 1.6 1 0 m 9 k i m 10 1 0 k 0 m m 60 1 m h in 1 6 m 0 i s n 26.8 m/s EXAMPLE 1–2 Do Example 1–1(b) by expanding the top and bottom separately. Solution 60 mi 60 mi 1.6 1 0 m 9 k i m 10 1 0 k 0 m m 96 540 m 1 h 1 h 60 1 m h in 1 6 m 0 i s n 3600 s Thus, velocity 96 540 m/3600 s 26.8 m/s as above. PRACTICE PROBLEMS 1 1. Area pr2 . Given r 8 inches, determine area in square meters (m2 ). 2. A car travels 60 feet in 2 seconds. Determine a. its speed in meters per second, b. its speed in kilometers per hour. For part (b), use the method of Example 1–1, then check using the method of Example 1–2. Answers: 1. 0.130 m2 2. a. 9.14 m/s b. 32.9 km/h 1.4 Power of Ten Notation Electrical values vary tremendously in size. In electronic systems, for example, voltages may range from a few millionths of a volt to several thousand volts, while in power systems, voltages of up to several hundred thousand are common. To handle this large range, the power of ten notation (Table 1–5) is used. To express a number in power of ten notation, move the decimal point to where you want it, then multiply the result by the power of ten needed to restore the number to its original value. Thus, 247 000 2.47 105 . (The number 10 is called the base, and its power is called the exponent.) An easy way to determine the exponent is to count the number of places (right or left) that you moved the decimal point. Thus, 247 000 2 4 7 0 0 0 2.47 105 5 4 3 2 1