Solution Manual For Electronics Fundamentals: A Systems Approach, 1st Edition

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Online Instructor’s ManualforElectronics FundamentalsA Systems ApproachThomas L. FloydDavid M. Buchla

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1PART ONESolutions to End-of-Chapter Problems

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21.The circuit is first tested with a computer design and simulation program, which can simulatethe performance and look for potential problems. When the simulation is satisfactory, aprototype circuit is constructed, tested, and modified as needed before putting it intoproduction.2.Semiconductor and component manufacturers as well as printed circuit board manufacturers.3.Electronic assemblies have become more complex but also more reliable, so there is less needfor repair. It is generally cheaper for manufacturers to replace a board than troubleshoot it tothe component level. Skills needed by technicians tend to be broader skills than in the past.4.Electrical systems deal primarily with power; electronic systems involve signals and a logicalsequence of processes.5.Advantages are that the digital signal can be processed and stored easily; it is also less subjectto noise.6. A block diagram shows signal flow in a system; a flowchart shows a logical process.7.(a) An electronic oscillator generates a repetitive electronic signal(b) An oscillator does not have a signal input.8.(a) High Voltage Direct CurrentCHAPTER 1SYSTEMS,QUANTITIESAND UNITSSECTION 1-1The Electronics IndustrySECTION 1-2Introduction to Electronic SystemsSECTION 1-3TypesofCircuits

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3(b) HVDC is used for long distance and underwater power transmission.9.A carrier is a high frequency radio wave that can be modulated (changed) by a lowerfrequency signal.10. A stair-step output; each step represents a different digital value.11.(a)3000 =3103(b)75,000 =7.5104(c)2,000,000= 210612.(a)5001= 0.002 =2103(b)20001= 0.0005 =5104(c)000,000,51= 0.0000002 =210713.(a)8400 =8.4103(b)99,000 =9.9104(c)0.2106=210514.(a)0.0002 =2104(b)0.6 =6101(c)7.8102(already in scientific notation)15.(a)2.5106=0.0000025(b)5.0102=500(c)3.9101=0.3916.(a)4.5106=0.0000045(b)8109=0.000000008(c)4.01012=0.000000000004017.(a)9.2106+ 3.4107= 9.2106+ 34106=4.32107(b)5103+ 8.5101= 5103+ 0.00085103=5.00085103(c)5.6108+ 4.6109= 56109+ 4.6109=6.0610818.(a)3.210121.11012= 2.11012(b)2.61081.3107= 261071.3107=24.7107(c)1.5101281013= 15101381013=7101319.(a)(5103)(4105) = 54103 + 5= 20108=2109SECTION 1-4Scientificand EngineeringNotation

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4(b)(1.21012)(3102) = 1.231012 + 2=3.61014(c)(2.2109)(7106) = 2.271096= 15.41015=1.54101420.(a)23105.2100.1= 0.41032= 0.4101=4(b)682.5105.010= 0.5106(8)= 0.5102=50(c)58102102.4= 2.1108(5)=2.1101321.(a)89,000 =89103(b)450,000 =450103(c)12,040,000,000,000 =12.04101222.(a)2.35105=235103(b)7.32107=73.2106(c)1.333109(already in engineering notation)23.(a)0.000345 =345106(b)0.025 =25103(c)0.00000000129 =1.2910924.(a)9.81103=9.81103(b)4.82104=482106(c)4.38107=43810925.(a)2.5103+ 4.6103= (2.5 + 4.6)103=7.1103(b)68106+ 33106= (68 + 33)106=101106(c)1.25106+ 250103= 1.25106+ 0.25106= (1.25 + 0.25)106=1.5010626.(a)(32103)(56103) = 179210(3 + 3)= 1792100=1.792103(b)(1.2106)(1.2106) = 1.4410(66)=1.441012

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5(c)(100)(55103) = 5500103=5.527.(a)3102.250=22.7103(b)631025105= 0.210(3(6))= 0.2109=200106(c)331066010560= 0.84810(33)= 0.848100=84810328.(a)89,000= 89103=89 k(b)450,000= 450103=450 k(c)12,040,000,000,000= 12.041012=12.04 T29.(a)0.000345 A = 345106A =345A(b)0.025 A = 25103A =25 mA(c)0.00000000129 A = 1.29109A =1.29 nA30.(a)31103A =31 mA(b)5.5103V =5.5 kV(c)201012F =20 pF31.(a)3106F =3F(b)3.3106=3.3 M(c)350109A =350 nA32.(a)5A =5106A(b)43 mV =43103V(c)275 k=275103(d)10 MW =10106W33.(a)(5 mA) (1103A/mA) = 5103A =5000A(b)(3200W)(1103W/W) =3.2 mW(c)(5000 kV)(1103) MV/kV =5 MV(d)(10 MW)(1103kW/MW) = 10103kW =10,000 kWSECTION 1-6Metric Unit ConversionsSECTION 1-5Units and Metric Prefixes

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634.(a)A101A101A1mA163= 1103=1000(b)V101V1005.0mV1kV05.033= 0.05106=50,000(c)631011002.0M1k02.0= 0.02103=2105(d)W101W10155kW1mW15533= 155106=1.5510435.(a)50 mA + 680A = 50 mA + 0.68 mA =50.68 mA(b)120 k+ 2.2 M= 0.12 M+ 2.2 M=2.32 M(c)0.02F + 3300 pF = 0.02F + 0.0033F =0.0233F36.(a)k2.12k10k10k2.2k10=0.8197(b)63105010250V50mV250=5000(c)36102101kW2MW1=50037.(a)1.00103has 3 significant digits.(b)0.0057 has 2 significant digits.(c)1502.0 has 5 significant digits.(d)0.000036 has 2 significant digits.(e)0.105 has 3 significant digits.(f)2.6102has 2 significant digits.38.(a)50,50550.5103(b)220.45220(c)46464.65103(d)10.9911.0(e)1.0051.00SECTION 1-7Measured Numbers

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7BASIC PROBLEMS1.Q= (charge per electron)(number of electrons) = (1.61019C/e)(501031e) =801012C2.(6.251018e/C)(80106C) =51014e3.The magnitude of the charge on a proton (p) is equal to the magnitude of the charge on theelectron (e). Therefore, (1.610-19C/p)(29 p) =4.6410-18C4.(1.610-19C/p)(17 p) =2.7210-18C5.(a)C1J10QWV=10 V(b)C2J5QWV=2.5 V(c)C25J100QWV=4 V6.C100J500QWV=5 V7.C40J800QWV=20 V8.W=VQ= (12 V)(2.5 C) =30 J9.2.5 J0.2 CWVQ===12.5 V10.0.2 C10 sQIt===20 mA11.(a)s1C75tQI=75 A(b)s0.5C10tQI=20 A(c)s2C5tQI=2.5 A12.s3C0.6tQI=0.2 ACHAPTER 2VOLTAGE, CURRENT, AND RESISTANCESECTION 2-2 Electrical ChargeSECTION 2-3 VoltageSECTION 2-4 Current

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813.;tQIt=A5C10IQ=2 s14.Q=It= (1.5 A)(0.1 s) =0.15 C15.A:Blue, gray, red, silver:680010%B:Orange, orange, black, silver:3310%C:Yellow, violet, orange, gold:47,0005%16.A:Rmin= 68000.1(6800) = 6800680=6120Rmax= 6800+ 680=7480B:Rmin= 330.1(33) = 333.3=29.7Rmax= 33+ 3.3=36.3C:Rmin= 47,000(0.05)(47,000) = 47,0002350=44,650Rmax= 47,000+ 2350=49,35017.(a)1st band =red, 2nd band =violet, 3rd band =brown, 4th band =gold(b)330;orange, orange, brown, (B)2.2 k:red, red, red (D)39 k:orange, white, orange (A)56 k:green, blue, orange (L)100 k:brown, black, yellow (F)18.(a)36.52%(b)2.74 k0.25%(c)82.5 k1%19.(a)Brown, black, black, gold:10± 5%(b)Green, brown, green, silver: 5,100,000± 10%= 5.1 M± 10%(c)Blue, gray, black, gold:68± 5%20.(a)0.47± 5%:yellow, violet, silver, gold(b)270 k± 5%:red, violet, yellow, gold(c)5.1 M± 5%:green, brown, green, gold21.(a)Red, gray, violet, red, brown: 28,700± 1% =28.7 k± 1%(b)Blue, black, yellow, gold, brown:60.4± 1%(c)White, orange, brown, brown, brown: 9310 ± 1% =9.31 k± 1%22.(a)14.7 k± 1%:brown, yellow, violet, red, brown(b)39.2± 1%:orange, white, red, gold, brown(c)9.76 k± 1%:white, violet, blue, brown, brown23.(a)220 =22(b)472 =4.7 k(c)823 =82 k(d)3K3 =3.3 k(e)560 =56(f)10M =10 M24.500, equal resistance on each side of the contact.SECTION 2-5 Resistance

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925.There is current throughLamp 2.26.See Figure 2-1.Figure 2-127.See Figure 2-2(a).Figure 2-2Figure 2-328.See Figure 2-2(b).29.Position 1: V1 =0 V, V2 =VSPosition 2: V1 =VS, V2 =0 V30.See Figure 2-3.31.On the 600 V DC scale:250 V32.R= (10)(10) =10033.(a)2(100) =200(b)15(10 M) =150 M(c)45(100) =450034.See Figure 2-4.SECTION 2-6 The Electric CircuitSECTION 2-7 Basic Circuit MeasurementsFigure 2-4

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10ADVANCED PROBLEMS35.I=tQQ=It= (2 A)(15 s) = 30 CV=C30J1000QW=33.3 V36.I=tQQ= (number of electrons) / (number of electrons/coulomb)Q=e/C1025.6e105741815= 9.184102CI=s10250C10184.932tQ=0.367 A37.Total wire length = 100 ftResistance per 1000 ft = (1000 ft)(6/100 ft) = 60Smallest wire size isAWG 27which has 51.47/1000 ft38.(a)4R7J =4.75%(b)560KF =560 k1%(c)1M5G =1.5 M2%39.The circuit in (b) can have both lamps on at the same time.40.There is always current throughR5.41.See Figure 2-5.Figure 2-542.See Figure 2-5.

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1143.See Figure 2-6.Figure 2-644.See Figure 2-7.

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12BASIC PROBLEMS1.Iis directly proportional toVand will change the same percentage asV.(a)I= 3(1 A) =3 A(b)I= 1 A(0.8)(1 A) = 1 A0.8 A =0.2 A(c)I= 1 A + (0.5)(1 A) = 1 A + 0.5 A =1.5 A2.(a)When the resistance doubles, the current is halved from 100 mA to50 mA.(b)When the resistance is reduced by 30%, the current increases from 100 mA toI=V/0.7R= 1.429(V/R) = (1.429)(100 mA)143 mA(c)When the resistance is quadrupled, the current decreases from 100 mA to25 mA.3.Tripling the voltage triples the current from 10 mA to 30 mA, but doubling the resistancehalves the current to15 mA.4.(a)I=1V5RV=5 A(b)I=10V15RV=1.5 A(c)I=50 V100VR=0.5 A(d)I=k15V30RV= 2mA(e)I=M7.4V250RV=53.2A5.(a)I=k7.2V9RV=3.33 mA(b)I=k10V5.5RV=550A(c)I=k68V40RV=588A(d)I=k2kV1RV=500 mA(e)I=M10kV66RV=6.60 mA6.I=10V12RV=1.2 ACHAPTER 3OHM’S LAW, ENERGY, AND POWERSECTION 3-1 Ohm’s LawSECTION 3-2 Application of Ohm’s Law

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137.(a)I=k10V25RV=2.50 mA(b)I=M2.2V5RV=2.27A(c)I=k8.1V15RV=8.33 mA8.Orange, violet, yellow, gold, brown37.4± 1%I=S12 V= 37.4ΩVR=0.321 A9.I=24 V37.4Ω= 0.642 A0.642 A is greater than 0.5 A, sothe fuse will blow.10.(a)V=IR= (2 A)(18) =36 V(b)V=IR= (5 A)(47) =235 V(c)V=IR= (2.5 A)(620) =1550 V(d)V=IR= (0.6 A)(47) =28.2 V(e)V=IR= (0.1 A)(470) =47 V11.(a)V=IR= (1 mA)(10) =10 mV(b)V=IR= (50 mA)(33) =1.65 V(c)V=IR= (3 A)(4.7 k) =14.1 kV(d)V=IR= (1.6 mA)(2.2 k) =3.52 V(e)V=IR= (250A)(1 k) =250 mV(f)V=IR= (500 mA)(1.5 M) =750 kV(g)V=IR= (850A)(10 M) =8.5 kV(h)V=IR= (75A)(47) =3.53 mV12.V=IR= (3 A)(20 m) =60 mV13.(a)V=IR= (3 mA)(27 k) =81 V(b)V=IR= (5A)(100 M) =500 V(c)V=IR= (2.5 A)(47) =117.5 V14.(a)R=A2V10IV=5(b)R=A45V90IV=2(c)R=A5V50IV=10(d)R=A10V5.5IV=0.55(e)R=A0.5V150IV=30015.(a)R=A5kV10IV=2 k(b)R=mA2V7IV=3.5 k(c)R=mA250V500IV=2 k(d)R=A500V50IV=100 k(e)R=mA1kV1IV= 1M

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1416.R=mA2V6IV=3 k17.(a)R=A2V8IV=4(b)R=mA4V12IV=3 k(c)R=A150V30IV= 0.2 M=200 k18.I=3.2 V3.9VR==0.82 A19.P=26 J10 sWt==2.6 W20.Since 1 watt = 1 joule,P= 350 J/s =350 W21.P=h5J7500tWs18,000J7500s3600h1h5J7500= 0.417 J/s =417 mW22.(a)1000 W = 1103W =1 kW(b)3750 W = 3.750103W =3.75 kW(c)160 W = 0.160103W =0.160 kW(d)50,000 W = 50103W =50 kW23.(a)1,000,000 W = 1106W =1 MW(b)3106W =3 MW(c)15107W = 150106W =150 MW(d)8700 kW = 8.7106W =8.7 MW24.(a)1 W = 1000103W =1000 mW(b)0.4 W = 400103W =400 mW(c)0.002 W = 2103W =2 mW(d)0.0125 W = 12.5103W =12.5 mW25.(a)2 W =2,000,000W(b)0.0005 W =500W(c)0.25 mW =250W(d)0.00667 mW =6.67W26.(a)1.5 kW = 1.5103W =1500 W(b)0.5 MW = 0.5106W =500,000 W(c)350 mW = 350103W =0.350 W(d)9000W = 9000106W =0.009 WSECTION 3-3 Energy and Power

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1527.WPtin wattsWVQQItWPVItSo, (1 V)(1 A) = 1 W28.1 J1 sWPt= 1 W1 kW = 1000 W =1000 J1 s1 kW-second = 1000 J1 kWh = 36001000 J1 kWh = 3.6106J29.P=VI= (5.5 V)(3 mA) =16.5 mW30.P=VI= (115 V)(3 A) =345 W31.P=I2R= (500 mA)2(4.7 k) =1.18 kW32.P=I2R= (5.0 A)2(2010-3) =500 mW33.P=620)V60(22RV=5.81 W34.P=56)V5.1(22RV= 0.0402 W =40.2 mW35.P=I2RR=22A)(2W100IP=25SECTION 3-4 Power in an Electric Circuit

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1636.5106watts for 1 minute = 5103kWminhrmin/160kWmin1053=83.3 kWh37.6700 W/s(1000 W/kW)(3600 s/h)=0.00186 kWh38.(50 W)(12 h) =600 Wh50 W = 0.05 kW(0.05 kW)(12 h) =0.6 kWh39.1.25 V10LVIR= 0.125 AP = VI= (1.25 V)(0.125 A) = 0.156 W =156 mW40.P=Wt156 mW =156 mJ1 stot(156 mJ/s)(90 h)(3600 s/h)W=50,544 J41.P=I2R= (10 mA)2(6.8 k) = 0.68 WUse the next highest standard power rating of1 W.42.If the 8 W resistor is used, it will be operating in a marginal condition.To allow for asafety margin of 20%, use a12 Wresistor.43.(a)+at top,at bottom of resistor(b)+at bottom,at top of resistor(c)+on right,on left of resistor44.VOUT=)50)(W1(LLRP=7.07 V45.Ampere-hour rating = (1.5 A)(24 h) =36 AhSECTION 3-5 The Power Rating of ResistorsSECTION 3-6Energy Conversion and Voltage Drop in a ResistanceSECTION 3-7 Power Suppliesand Batteries

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1746.I=h10Ah80=8 A47.I =h48mAh650=13.5 mA48.PLOST=PINPOUT= 500 mW400 mW =100 mW% efficiency =%100mW500mW400%100INOUTPP=80%49.POUT= (efficiency)PIN= (0.85)(5 W) =4.25 W50.The 4th bulb from the left is open.51.If should takefive(maximum) resistance measurements.ADVANCED PROBLEMS52.Assume that the total consumption of the power supply is the input power plus the power lost.POUT= 2 W% efficiency =%100INOUTPPPIN=%100%60W2%100efficiency%OUTP= 3.33 WThe power supply itself usesPINPOUT= 3.33 W2 W = 1.33 WEnergy= W = Pt= (1.33 W)(24 h) = 31.9 Wh0.032 kWh53.Rf=A0.8V120IV=150SECTION 3-8 Introduction to TroubleshootingTroubleshootinhg

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1854.Measure the current with an ammeter connected as shown in Figure 3-1. Then calculate theunknown resistance with the formula,R= 12 V/I.Figure 3-155.CalculateIfor each value ofV:I1=100V0=0 AI2=100V10=100 mAI3=100V20=200 mAI4=100V30=300 mAI5=100V40=400 mAI6=100V50=500 mAI7=100V60=600 mAI8=100V70=700 mAI9=100V80=800 mAI10=100V90=900 mAI11=100V100=1 AFigure 3-2The graph is a straight line as shown inFigure 3-2. This indicates alinearrelationship betweenIandV.

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1956.R=mA5V1SIV=200(a)I=200V1.5SRV=7.5 mA(b)I=200V2SRV=10 mA(c)I=200V3SRV=15 mA(d)I=200V4SRV=20 mA(e)I=200V10SRV=50 mA57.R1=A2V1IV=0.5R2=A1V1IV=1R3=A0.5V1IV=258.mA50V10mA302VV2=mA50mA)V)(30(10= 6 Vnew valueThe voltage decreased by 4 V, from 10 V to 6 V.59.The current increase is 50%, so the voltage increase must be the same; that is, the voltage mustbe increased by (0.5)(20 V) =10 V.The new value of voltage isV2= 20 V + (0.5)(20 V) = 20 V + 10 V =30 V60.Wire resistance:RW=CM1624.3ft)/ft)(24CM(10.4= 0.154(a)154.100V6WRRVI=59.9 mA(b)VR= (59.9 mA)(100) =5.99 V(c)WRV= 6 V5.99 V = 0.01 VFor one length of wire,V=2V01.0=0.005 V61.300 W = 0.3 kW30 days = (30 days)(24 h/day) = 720 hEnergy = (0.3 kW)(720 h) =216 kWh62.days31kWh1500= 48.39 kWh/dayP=h/day24kWh/day48.39=2.02 kW

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2063.The minimum power rating you should use is12 Wso that the power dissipation does notexceed the rating.64.(a)P=10V)(1222RV=14.4 W(b)W=Pt= (14.4 W)(2 min)(1/60 h/min) =0.48 Wh(c)Neither, the power is the same because it is not time dependent.65.VR(max)= 120 V100 V = 20 VImax=8V20min(max)RVR= 2.5 AA fuse with a rating of less than 2.5 A must be used.A 2 A fuse is recommended.66.0.5 W0.030PIR4.08 A67.Power will increase by four times.66.The materials required for the Load Test Box are as follows:ItemComponentQty1Resistor: 5.0, 10 W12Resistor: 16, 5 W13Resistor: 100, 2.0 W14Resistor: 150, 3.0 W151 pole, 4 position rotary switch16Knob17Enclosure (4” x 4”2”Al)18Banana plug terminals29Fuse (1.5 A) and fuse holder110PC board (etched with pattern)111Screws, washers, nuts412Standoffs4

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2169.See Figure 3-3.Figure 3-370.Ris open.71.No fault72.R1is shorted.73.Lamp 4 is shorted.74.Lamp 6 is open.Multisim Troubleshooting ProblemsTroubleshootinhg

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22BASIC PROBLEMS1.See Figure 4-1.Figure 4-12.The groups of series resistors areR1,R2,R3,R9R4;R13,R7,R14,R16;R6,R8,R12;R10,R11,R15,R5See Figure 4-2.Figure 4-23.1 81371416RRRRR= 68 k+ 33 k+ 47 k+ 22 k=170 k4.2 3128610Ω+ 18 Ω + 22 ΩRRRR=50CHAPTER 4SERIES CIRCUITSSECTION 4-1 Resistors in Series

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235.RT= 82+ 56=1386.(a)RT= 560+ 1.0 k=1560(b)RT= 47+ 33=80(c)RT= 1.5 k+ 2.2 k+ 10 k=13.7 k(d)RT= 1.0 k+ 1.8 k+ 100 k+ 1.0 M=1,102,800(round to 1.10 M)7.(a)RT= 1.0 k+ 4.7 k+ 2.2 k=7.9 k(b)RT= 10+ 10+ 12+ 1.0=33(c)RT= 1.0 M+ 560 k+ 1.0 M+ 680 k+ 10 M=13.24 MSee Figure 4-3.Figure 4-38.RT= 12(5.6 k) =67.2 k9.RT= 6(47) + 8(100) + 2(22) = 282+ 800+ 44=112610.RT=R1+R2+R3+R4+R5R5=RT(R1+R2+R3+R4)= 20 k(4.7 k+ 1.0 k+ 2.2 k+ 3.9 k)= 20 k11.8 k=8.2 k11.(a)R18=R13+R7+R14+R16= 68 k+ 33 k+ 47 k+ 22 k=170 k(b)R23=R12+R8+R6= 10+ 18+ 22=50(c)R47=R10+R11+R15+R5= 2.2 k+ 8.2 k+ 1.0 k+ 1.0 k=12.4 k(d)R56=R1+R2+R3+R9+R4= 220+ 330+ 390+ 470+ 560=1.97 k12.RT=R18+R23+R47+R56= 170 k+ 50+ 12.4 k+ 1.97 k=184.42 kSECTION 4-2Total Series Resistance

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2413.I=120V12TSRV=0.1 A14.I=5 mAat all points in the circuit.15.(a)RT= 2.2 k+ 5.6 k+ 1.0 k= 8.8 kI=k8.8V5.5TRV=625A(b)RT= 1.0 M+ 2.2 M+ 560 k= 3.76 MI=M76.3V16=4.26AThe ammeters are connected in series. See Figure 4-4.Figure 4-416.(a)V1=V5.5k8.8k.22ST1VRR=1.375 VV2=V5.5k8.8k6.5ST2VRR=3.5 VV3=V5.5k8.8k.01ST3VRR=625 mV(b)V1=V16M76.3M0.1ST1VRR=4.26 VSECTION 4-4Application of Ohm’s LawSECTION 4-3 Current in a Series Circuit

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25V2=V16M76.3M2.2ST2VRR=9.36 VV3=V16M3.76k056ST3VRR=2.38 V17.(a)RT= 3(470) = 1410I=48 V1410=34.0 mA(b)(34.0 mA)(470Ω)RVIR=16 V(c)22min(34.0 mA) 470ΩPI R=0.543 W18.RT=mA1V5TSIV= 5 kReach=4k5=1.25 k19.See Figure 4-5.Figure 4-520.The total voltage is 6 V + 6 V + 6 V6 V =12 V21.VS= 5.5 V + 8.2 V + 12.3 V =26 V22.VS=V1+V2+V3+V4+V5V5=VS(V1+V2+V3+V4) = 20 V(1.5 V + 5.5 V + 3 V + 6 V) = 20 V16 V =4 V23.(a)By Kirchhoff’s voltage law:15 V = 2 V +V2+ 3.2 V + 1 V + 1.5 V + 0.5 VSECTION 4-5 Voltage Sources in SeriesSECTION 4-6 Kirchhoff’s Voltage Law

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26V2= 15 V(2 V + 3.2 V + 1 V + 1.5 V + 0.5 V) = 15 V8.2 V =6.8 VSee Figure 4-6(a).(b)VR=8 V;V2R=16 V;V3R=24 V;V4R=32 VSee Figure 4-6(b).24.10050022=4.4%25.(a)VAB=V1214747=3.84 V(b)VAB=V8k6.5k5.5V8k3.3k2.2k0.1k3.3k2.2=6.77 V26.SAVV=15 V23S12313.3 kΩ15 V18.9 kBRRVVRRR=10.6 V3S1233.3 kΩ15 V18.9 kCRVVRRR=2.62 V27.3minS123680Ω12 V2150RVVRRR=3.80 V23maxS1231680Ω12 V2150RRVVRRR=9.38 VSECTION 4-7 Voltage DividersFigure 4-6

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2728.RT=R+ 2R+ 3R+ 4R+ 5R= 15RVR=V915RR=0.6 VVR=V9152RR=1.2 VVR=V9153RR=1.8 VVR=V9154RR=2.4 VVR=V9155RR=3.0 V29.V5.6k=10 V(by measurement);I=k6.5V10= 1.79 mA;V1k= (1.79 mA)(1 k) =1.79 V;V560(1.79 mA)(560) =1 V;V10k= (1.79 mA)(10 k) =17.9 V30.PT= 5(50 mW) =250 mW31.RT= 5.6 k+ 1 k+ 560+ 10 k= 17.16 kP=I2RT= (1.79 mA)2(17.16 k) = 0.055 W =55 mW32.Voltage from pointAto ground (G):VAG=10 VResistance betweenAandG:RAG= 5.6 k+ 5.6 k+ 1.0 k+ 1.0 k= 13.2 kResistance betweenBandG:RBG= 5.6 k+ 1.0 k+ 1.0 k= 7.6 kResistance betweenCandG:RCG= 1.0 k+ 1.0 k= 2 kVBG=V10k2.13k6.7V10AGBGRR=5.76 VVCG=V10k2.13k2V10AGCGRR=1.52 VVDG=V10k2.13k0.1V10AGDGRR=0.758 V33.Measure the voltage at pointAwith respect to ground and the voltage at pointBwith respect toground. The difference of these two voltages isVR2.VR2= VAVBSECTION 4-8 Power in Series CircuitsSECTION 4-9Voltage Measurements

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2834.RT=R1+R2+R3+R4+R5= 560 k+ 560 k+ 100 k+ 1.0 M+ 100 k= 2.32 MVT= 15 VVA=V15M32.2M76.1TTVRRAG=11.4 VVB=V15M32.2M2.1TTVRRBG=7.76 VVC=V15M32.2M1.1TTVRRCG=7.11 VVD=V15M32.2k100TTVRRDG=647 mV35.11.38 V7.11 VACACVVV=-=-=4.27 V36.7.11 V11.38 VCACAVVV=-=-=4.27 V37.(a)Zero current indicates an open.R4is opensince all the voltage is dropped across it.(b)300V10321SRRRV= 33.3 mAR4andR5have no effect on the current. There is ashort fromAtoB.38.RT= 10 k+ 8.2 k+ 12 k+ 2.2 k+ 5.6 k= 38 kThe meter reads about 28 k. It should read 38 k.The 10 kresistor is shorted.ADVANCED PROBLEMS39.V1=IR1= (10 mA)(680) = 6.8 VV2=IR2= (10 mA)(1.0 k) = 10 VV4=IR4= (10 mA)(270) = 2.7 VV5=IR5= (10 mA)(270) = 2.7 VV3=VS(V1+V2+V4+V5)V3= 30 V(6.8 V + 10 V + 2.7 V + 2.7 V) = 30 V22.2 V = 7.8 VR3=mA10V7.83IV= 0.78 k=78040.RT= 3(5.6 k) + 1.0 k+ 2(100) = 18 kThree 5.6 kresistors, one 1 kresistor, and two 100resistors41.VA=10 V,RT= 22 k+ 10 k+ 47 k+ 12 k+ 5.6 k= 96.6 kSECTION 4-10 Troubleshooting

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29VB=VAV22k= 10 VV10k6.96k22= 10 V2.28 V =7.72 VVC=VBV10k= 7.72 VV10k6.96k10= 7.72 V1.04 V =6.68 VVD=VCV47k= 6.68 VV10k6.96k47= 6.68 V4.87 V =1.81 VVE=VDV12k= 1.81 VV10k6.96k12= 1.81 V1.24 V =0.57 VVF=0 V42.V2=IR2= (20 mA)(100) =2 VR5=mA20V6.6SIV=330R6=226mA)(20mW112IP=280V6=IR6= (20 mA)(280) =5.6 VV1=VS(20 V +V6) = 30 V(20 V + 5.6 V) =4.4 VR1=mA20V4.41IV=220V3+V4= 20 VV2V5= 20 V2 V6.6 V = 11.4 VV3=V4=2V11.4=5.7 VR3=R4=mA20V5.73IV=28543.VS=IRT= (250 mA)(1.5 k) = 375 VInew= 250 mA0.25(250 mA) = 250 mA62.5 mA = 188 mARnew=mA188V375newSIV2000500must be added to the existing 1500to reduceIby 25%.44.P=I2RImax=120W5.0RP= 0.0645 A = 64.5 mASince all resistors in series have the same current, use the largestRto determine the maximumcurrent allowable because the largestRhas the greatest power.Thus, the120resistor burns out first.45.(a)PT=W21W41W81= 0.125 W + 0.25 W + 0.5 W = 0.875 WI =2400W875.0TTRP=19.1 mA(b)VS=ITRT= (19.1 mA)(2400) =45.8 V(c)R=2IP

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