11Section1.4PowerofTenNotationTABLE1-5CommonPowerofTenMultipliers0.000001=10-61000000=106100 000=1050.00001=10-510 000 = 1040.0001=10-41000=100.001 = 10-3100=1020.01 = 10-210 = 100.1 = 10-11. = 1001 = 100Similarly, the number 0.003 69 may be expressed as 3.69 × 10-3 as illus-trated below.0.00369=0.00369=3.69×10-3123Multiplication and Division Using Powers of TenTo multiplynumbers in powerof ten notation, multiply their base numbers.then add their exponents. Thus,(1.2×103)(1.5× 10°) = (1.2)(1.5)×10(3+4)= 1.8 ×107For division, subtract the exponents in the denominator from those in thenumerator.Thus,4.5 × 10 = 4.5 × 102-(-2) = 1.5 × 103×10-23>EXAMPLE1-33Convert thefollowing numbersto power of ten notation,then perform the operation indicated:a.276x0.009,b.98200/20.Solutiona.276×0.009=(2.76×10)(9×10-3)=24.8×10-1=2.489.82×10498200b.=4.91X103202×10Addition and Subtraction UsingPowers ofTenTo add or subtract, first adjust all numbers tothe samepower of ten.Itdoesnotmatterwhatexponentyou choose,as longas all arethesame
Similarly, the number 0.003 69 may be expressed as 3.69 103 as illustrated below. 0.003 69 0.0 0 3 6 9 3.69 103 1 2 3 Multiplication and Division Using Powers of Ten To multiply numbers in power of ten notation, multiply their base numbers, then add their exponents. Thus, (1.2 103 )(1.5 104 ) (1.2)(1.5) 10(34) 1.8 107 For division, subtract the exponents in the denominator from those in the numerator. Thus, 4 3 .5 10 1 0 2 2 4 3 .5 102(2) 1.5 104 Section 1.4 ■ Power of Ten Notation 11 TABLE 1–5 Common Power of Ten Multipliers 1 000 000 106 0.000001 106 100 000 105 0.00001 105 10 000 104 0.0001 104 1 000 103 0.001 103 100 102 0.01 102 10 101 0.1 101 1 100 1 100 EXAMPLE 1–3 Convert the following numbers to power of ten notation, then perform the operation indicated: a. 276 0.009, b. 98 200/20. Solution a. 276 0.009 (2.76 102 )(9 103 ) 24.8 101 2.48 b. 98 2 2 0 00 9. 2 82 10 1 1 04 4.91 103 Addition and Subtraction Using Powers of Ten To add or subtract, first adjust all numbers to the same power of ten. It does not matter what exponent you choose, as long as all are the same.
12Chapter1Introduction>EXAMPLE1-4Add3.25×10°and5×10a. using 10’ representation,b.using 10 representationSolutiona.5×10=50×102.Thus.3.25×102+50×102=53.25×102b.3.25X102=0.325×103.Thus,0.325X103+5×103=5.325X103which is the same as 53.25×102PowersNOTES..Raisinganumbertoapowerisaform of multiplication(ordivision if theUse common sense when han-exponentisnegative).Forexample,dling numbers. With calculators,(2 × 103)2 = (2 × 103)(2 × 10) = 4 × 106for example, it is often easiertowork directly with numbers inIn general, (N × 10")"=N"×10m.In this notation, (2 ×10°)2= 22×their original form than to con103x2= 4× 10°as before.vert them to power of ten nota-Integer fractional powers represent roots.Thus,42=V4=2 andtion. (As an example, it is more2713 = V27 = 3.sensible to multiply 276×0.009 directly than to convert topower of ten notation as we didinExample1-3(a).)IfthefinalEXAMPLE1-5 Expand thefollowing:result is needed as a power often, you can convert as a lastd. (60)1/3a.(250)3b.(0.0056)2c. (141)-2step.Solutiona.(250)3=(2.5×103)=(2.5)×102x3=15.625×106b.(0.0056)2=(5.6×10-3)2=(5.6)2×10-6=31.36×10-6c. (141)-2 = (1.41 × 103)-2 = (1.41)-2× (10°)-2 = 0.503 × 10-4d. (60)/3=V60=3.915PRACTICEDetermine the following:PROBLEMS2a.(6.9×10°)(0.392×10-2)b. (23.9×10)/(8.15×105)c. 14.6 × 102 + 11.2 × 10' (Express in 10° and 10' notation.)d. (29.6)3e. (0.385)-2Answers:a.2.71×103b.2.93×106c.15.7×102=157×10ld.25.9×10e. 6.75
Powers Raising a number to a power is a form of multiplication (or division if the exponent is negative). For example, (2 103 ) 2 (2 103 )(2 103 ) 4 106 In general, (N 10n ) m Nm 10nm. In this notation, (2 103 ) 2 22 1032 4 106 as before. Integer fractional powers represent roots. Thus, 41/2 4 2 and 271/3 3 27 3. 12 Chapter 1 ■ Introduction EXAMPLE 1–4 Add 3.25 102 and 5 103 a. using 102 representation, b. using 103 representation. Solution a. 5 103 50 102 . Thus, 3.25 102 50 102 53.25 102 b. 3.25 102 0.325 103 . Thus, 0.325 103 5 103 5.325 103 , which is the same as 53.25 102 Use common sense when handling numbers. With calculators, for example, it is often easier to work directly with numbers in their original form than to convert them to power of ten notation. (As an example, it is more sensible to multiply 276 0.009 directly than to convert to power of ten notation as we did in Example 1–3(a).) If the final result is needed as a power of ten, you can convert as a last step. NOTES. EXAMPLE 1–5 Expand the following: a. (250)3 b. (0.0056)2 c. (141)2 d. (60)1/3 Solution a. (250)3 (2.5 102 ) 3 (2.5)3 1023 15.625 106 b. (0.0056)2 (5.6 103 ) 2 (5.6)2 106 31.36 106 c. (141)2 (1.41 102 ) 2 (1.41)2 (102 ) 2 0.503 104 d. (60)1/3 3 60 3.915 PRACTICE PROBLEMS 2 Determine the following: a. (6.9 105 )(0.392 102 ) b. (23.9 1011)/(8.15 105 ) c. 14.6 102 11.2 101 (Express in 102 and 101 notation.) d. (29.6)3 e. (0.385)2 Answers: a. 2.71 103 b. 2.93 106 c. 15.7 102 157 101 d. 25.9 103 e. 6.75
13Section1.5PrefixesTABLE1-6EngineeringPrefixes1.5PrefixesPrefixSymbolPowerof 101012TScientific and Engineering NotationteraG109gigaIf powerof tennumbersarewrittenwithonedigittotheleftof thedecimal106Mmegaplace,they are said to be in scientific notation.Thus,2.47× 10'is in sci-103kilokentific notation,while24.7×10*and 0.247×10°arenot.However, we10-3millimare more interested in engineering notation. In engineering notation, pre-10-6microufixes areused to represent certain powers of ten;see Table1-6.Thus,a10-9nanon10-12quantity such as 0.045 A (amperes) can be expressed as 45 × 10-3 A, but itpicopis preferable to express it as 45 mA. Here, we have substituted the prefixmilli for the multiplier 10-3. It is usual to select a prefix that results in abase number between 0.1 and 999.Thus, 1.5 × 10-5 s would be expressedas 15 μs.EXAMPLE1-6Expressthefollowinginengineeringnotation:a.10×10*voltsb.0.1×10-3wattsc.250×10-7secondsSolutiona.10×10V=100×103V=100kilovolts=100kVb.0.1×10-3W=0.1milliwatts=0.1mWc.250×10-1 s=25×10-6s=25microseconds=25 μsEXAMPLE1-7Convert0.1MVtokilovolts(kV)Solution0.1MV=0.1X106V=(0.1X10)X103V=100kVRemember that a prefix represents a power of ten and thus the rules forpowerof tencomputationapply.Forexample,whenaddingorsubtracting,adjustto a common base,as illustratedinExample1-8EXAMPLE1-8Computethesumof1ampere(amp)and100milliamperes.SolutionAdjust to a common base, either amps (A) or milliamps (mA)Thus,1A+100mA=1A+100X10-3A=1A+0.1A=1.1AAlternatively.1A+100mA=1000mA+100mA=1100mA
1.5 Prefixes Scientific and Engineering Notation If power of ten numbers are written with one digit to the left of the decimal place, they are said to be in scientific notation. Thus, 2.47 105 is in scientific notation, while 24.7 104 and 0.247 106 are not. However, we are more interested in engineering notation. In engineering notation, prefixes are used to represent certain powers of ten; see Table 1–6. Thus, a quantity such as 0.045 A (amperes) can be expressed as 45 103 A, but it is preferable to express it as 45 mA. Here, we have substituted the prefix milli for the multiplier 103 . It is usual to select a prefix that results in a base number between 0.1 and 999. Thus, 1.5 105 s would be expressed as 15 ms. Section 1.5 ■ Prefixes 13 TABLE 1–6 Engineering Prefixes Power of 10 Prefix Symbol 1012 tera T 109 giga G 106 mega M 103 kilo k 103 milli m 106 micro m 109 nano n 1012 pico p EXAMPLE 1–6 Express the following in engineering notation: a. 10 104 volts b. 0.1 103 watts c. 250 107 seconds Solution a. 10 104 V 100 103 V 100 kilovolts 100 kV b. 0.1 103 W 0.1 milliwatts 0.1 mW c. 250 107 s 25 106 s 25 microseconds 25 ms EXAMPLE 1–7 Convert 0.1 MV to kilovolts (kV). Solution 0.1 MV 0.1 106 V (0.1 103 ) 103 V 100 kV Remember that a prefix represents a power of ten and thus the rules for power of ten computation apply. For example, when adding or subtracting, adjust to a common base, as illustrated in Example 1–8. EXAMPLE 1–8 Compute the sum of 1 ampere (amp) and 100 milliamperes. Solution Adjust to a common base, either amps (A) or milliamps (mA). Thus, 1 A 100 mA 1 A 100 103 A 1 A 0.1 A 1.1 A Alternatively, 1 A 100 mA 1000 mA 100 mA 1100 mA.
14Chapter1IntroductionPRACTICE1.Convert1800kVtomegavolts(MV)PROBLEMS32. In Chapter 4, we show that voltage is the product of current times resistance-that is,V=I XR, whereVis in volts, Iis in amperes,and Ris in ohmsGivenI =25mAandR=4k2,convertthesetopoweroften notation,thendetermine V.3.If I, =520μA,I,=0.157mA,and I,=2.75×10-^A,what is li+l2+13in mA?Answers: 1. 1.8 MV.2.100V3.0.952mA1.All conversion factors have a value of what?IN-PROCESSLEARNING2. Convert 14 yards to centimeters.CHECK13.What units does the following reduce to?xmhmin+Xhkmmins4.Express the following in engineering notation:c. 12.3 × 10-4sa.4270msb.0.00153V5.Express the result of each of the following computations as a number times10tothepowerindicated:a. 150 × 120 as a value times 10'; as a value times 10°.b.300×6/0.005 as a valuetimes10°;as a valuetimes10°;as a valuetimes10%c.430+15 as a value times 10°; as a value times 10'd. (3 × 10-2) as a value times 10-6; as a value times 10-56.Express each of the following as indicated.a.752μAinmA.b.0.98mVinμV.c. 270 μs + 0.13 ms in μs and in ms.(Answers are at the end of the chapter.)1.6Significant Digits and Numerical AccuracyThenumber of digits in anumberthat carry actual information aretermedsignificant digits. Thus, if we say a piece of wire is 3.57 meters long, wemeanthat itslength iscloserto3.57mthan it isto3.56mor3.58mand wehave three significant digits.(The number of significant digits includes thefirst estimated digit.)If we say that it is 3.570 m, we mean that it is closer to3.570 m than to 3.569 m or 3.571 m and we have four significant digits.When determining significant digits,zeros usedto locatethe decimal pointare not counted.Thus, 0.00457 has three significant digits; this can be seenifyouexpressitas4.57×10-3
1.6 Significant Digits and Numerical Accuracy The number of digits in a number that carry actual information are termed significant digits. Thus, if we say a piece of wire is 3.57 meters long, we mean that its length is closer to 3.57 m than it is to 3.56 m or 3.58 m and we have three significant digits. (The number of significant digits includes the first estimated digit.) If we say that it is 3.570 m, we mean that it is closer to 3.570 m than to 3.569 m or 3.571 m and we have four significant digits. When determining significant digits, zeros used to locate the decimal point are not counted. Thus, 0.004 57 has three significant digits; this can be seen if you express it as 4.57 103 . 14 Chapter 1 ■ Introduction PRACTICE PROBLEMS 3 1. Convert 1800 kV to megavolts (MV). 2. In Chapter 4, we show that voltage is the product of current times resistance— that is, V I R, where V is in volts, I is in amperes, and R is in ohms. Given I 25 mA and R 4 k, convert these to power of ten notation, then determine V. 3. If I1 520 mA, I2 0.157 mA, and I3 2.75 104 A, what is I1 I2 I3 in mA? Answers: 1. 1.8 MV 2. 100 V 3. 0.952 mA IN-PROCESS LEARNING CHECK 1 1. All conversion factors have a value of what? 2. Convert 14 yards to centimeters. 3. What units does the following reduce to? k h m k m m m h in m s in 4. Express the following in engineering notation: a. 4270 ms b. 0.001 53 V c. 12.3 104 s 5. Express the result of each of the following computations as a number times 10 to the power indicated: a. 150 120 as a value times 104 ; as a value times 103 . b. 300 6/0.005 as a value times 104 ; as a value times 105 ; as a value times 106 . c. 430 15 as a value times 102 ; as a value times 101 . d. (3 102 ) 3 as a value times 106 ; as a value times 105 . 6. Express each of the following as indicated. a. 752 mA in mA. b. 0.98 mV in mV. c. 270 ms 0.13 ms in ms and in ms. (Answers are at the end of the chapter.)
15Section1.6SignificantDigits andNumericalAccuracyMost calculations that you will do in circuit theory will be done using aNOTES...handcalculator.AnerrorthathasbecomequitecommonistoshowmoreWhen working with numbersdigits of "accuracy"in an answer than are warranted, simply because thenumbers appear on the calculator display.The number of digits that youyou will encounter exact num-bers and approximate numbersshould show is related to the number of significant digits in the numbersExact numbers are numbers thatused in the calculation.we know for certain, whileTo illustrate, suppose you have two numbers, A = 3.76 and B = 3.7, toapproximate numbers are num-bemultiplied.Their product is 13.912.If the numbers3.76 and3.7areexactbers that have some uncertaintythis answer is correct.However, if the numbers havebeen obtained bymea-For example, when we say thatsurementwherevalues cannotbe determined exactly,they will have somethere are 60minutes in one hour,uncertainty and theproduct must reflect this uncertainty.Forexample, sup-the 60here is exact.However.ifpose Aand Bhave an uncertainty of 1 in their first estimated digitthat is.we measure the length of a wireA =3.76±0.0l and B =3.7±0.1.This means thatA can be as small asand state it as 60 m, the 60 in3.75oras largeas3.77,whileB canbeas small as3.6oraslargeas3.8this case carries some uncer-Thus.theirproduct canbeas small as3.75×3.6=13.50 or aslargeastainty (depending on how good3.77 × 3.8=14.326.The best that we can say about the product is that it isour measurement is), and is thusL4.i.e..thatvouknowitonlytothenearestwholenumber.Youcannotevenan approximatenumber.Whensaythat it is 14.0 sincethis implies that youknowtheanswer tothenearestan exact number is included in atenth,which,as you can seefrom theabove,you do not.calculation, there is no limit toWe can nowgive a“rule of thumb"for determining significant digits.howmany decimal placesyouThe number of significant digits in a result due to multiplication or divisioncan associate with it-the accuis the sameas the number of significant digits inthenumberwiththeleasracy of the result is affected onlynumber of significant digits.In theprevious calculation,forexample,3.7hasby the approximate numberstwosignificantdigits sothattheanswercanhaveonlytwo significantdigitsinvolved in the calculationaswell.Thisagrees withourearlier observationthatthe answeris14,notMany numbers encountered in14.0 (which has three).technical work are approximateWhen adding or subtracting, you must also use common sense. Foras they have been obtained byexample,supposetwocurrentsaremeasuredas24.7A(oneplaceknownmeasurement.after the decimal point)and 123 mA (i.e., 0.123A).Their sum is 24.823AHowever, the right-hand digits 23 in the answer are not significant. Theycannot be, since,if you don'tknow what the second digit after the decimalNOTES...Epoint is for thefirst current, it is senseless to claim that youknow their sumIn this book, given numbers areto thethird decimal place!Thebestthat you can sayabout the sum is that itassumedto beexact unless oth-also has one significant digit after the decimal place, that is,erwise noted. Thus, when avalue is given as 3 volts, take it24.7A(Oneplaceafterdecimal)tomeanexactly3volts.notsim-ply that it has one significant+0.123Afigure. Since our numbers are24.823A→24.8A (One place after decimal)assumed to be exact, all digitsare significant, and we use asmany digits as are convenient inTherefore,whenadding numbers,add thegiven data,then round theresulttoexamples and problems: Finalthe last column where all given numbers have significant digits.Theprocessanswers are usuallyrounded to 3issimilarforsubtraction.digits
Section 1.6 ■ Significant Digits and Numerical Accuracy 15 When working with numbers, you will encounter exact numbers and approximate numbers. Exact numbers are numbers that we know for certain, while approximate numbers are numbers that have some uncertainty. For example, when we say that there are 60 minutes in one hour, the 60 here is exact. However, if we measure the length of a wire and state it as 60 m, the 60 in this case carries some uncertainty (depending on how good our measurement is), and is thus an approximate number. When an exact number is included in a calculation, there is no limit to how many decimal places you can associate with it—the accuracy of the result is affected only by the approximate numbers involved in the calculation. Many numbers encountered in technical work are approximate, as they have been obtained by measurement. NOTES. In this book, given numbers are assumed to be exact unless otherwise noted. Thus, when a value is given as 3 volts, take it to mean exactly 3 volts, not simply that it has one significant figure. Since our numbers are assumed to be exact, all digits are significant, and we use as many digits as are convenient in examples and problems. Final answers are usually rounded to 3 digits. NOTES. Most calculations that you will do in circuit theory will be done using a hand calculator. An error that has become quite common is to show more digits of “accuracy” in an answer than are warranted, simply because the numbers appear on the calculator display. The number of digits that you should show is related to the number of significant digits in the numbers used in the calculation. To illustrate, suppose you have two numbers, A 3.76 and B 3.7, to be multiplied. Their product is 13.912. If the numbers 3.76 and 3.7 are exact this answer is correct. However, if the numbers have been obtained by measurement where values cannot be determined exactly, they will have some uncertainty and the product must reflect this uncertainty. For example, suppose A and B have an uncertainty of 1 in their first estimated digit—that is, A 3.76 0.01 and B 3.7 0.1. This means that A can be as small as 3.75 or as large as 3.77, while B can be as small as 3.6 or as large as 3.8. Thus, their product can be as small as 3.75 3.6 13.50 or as large as 3.77 3.8 14.326. The best that we can say about the product is that it is 14, i.e., that you know it only to the nearest whole number. You cannot even say that it is 14.0 since this implies that you know the answer to the nearest tenth, which, as you can see from the above, you do not. We can now give a “rule of thumb” for determining significant digits. The number of significant digits in a result due to multiplication or division is the same as the number of significant digits in the number with the least number of significant digits. In the previous calculation, for example, 3.7 has two significant digits so that the answer can have only two significant digits as well. This agrees with our earlier observation that the answer is 14, not 14.0 (which has three). When adding or subtracting, you must also use common sense. For example, suppose two currents are measured as 24.7 A (one place known after the decimal point) and 123 mA (i.e., 0.123 A). Their sum is 24.823 A. However, the right-hand digits 23 in the answer are not significant. They cannot be, since, if you don’t know what the second digit after the decimal point is for the first current, it is senseless to claim that you know their sum to the third decimal place! The best that you can say about the sum is that it also has one significant digit after the decimal place, that is, 24.7 A (One place after decimal) 0.123 A 24.823 A → 24.8 A (One place after decimal) Therefore, when adding numbers, add the given data, then round the result to the last column where all given numbers have significant digits. The process is similar for subtraction.