Principles and Applications of Electrical Engineering 5th Edition Solution Manual

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5thEdition Problem solutions, Chapter 1Chapter 1:Introduction to Electrical Engineering –Instructor NotesChapter 1 is introductory in nature, establishing some rationale for studying electrical engineering methods,even though the students' primary interest may lie in other areas. The material in this chapter should be included inevery syllabus, and can typically be thoroughly covered in a single-day introductory lecture. Oftentimes, reading ofthis material is left up to the discretion of the student.1.1

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5thEdition Problem solutions, Chapter 1Problem 1.1Solution:A few examples are:Bathroomventilation fanelectric toothbrushhair dryerelectric shaverelectric heater fanKitchenmicrowave fanmicrowave turntablemixerfood processorblendercoffee grindergarbage disposalceiling fanelectric clockexhaust fanrefrigerator compressordish washerUtility Roomclothes washerdryerair conditionerfurnace blowerpumpFamily RoomVCR drivecassette tape drivereel-to-reel tape driverecord turntable drivecomputer fantreadmillMiscellaneouslawn toolspower tools1.2

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5thEdition Problem solutions, Chapter 1Problem 1.2Solution:Several examples are listed below for each system:a)A shipCircuit Analysisdesign of the ship's electrical systemElectromagneticsradarSolid-State ElectronicsradiosonarElectric MachinespumpelevatorElectric Power SystemslightinggeneratorsDigital Logic Circuitselevator controlComputer SystemsnavigationCommunication SystemsradiotelephoneElectro-OpticsMorse lightbridge displaysInstrumentationcompassspeed indicatorControl SystemsrudderHVACb) A Commercial Passenger AircraftCircuit AnalysisDesign of the plane's electrical systemElectromagneticsradarmicrowave ovenSolid-State ElectronicsradioElectric MachinesturbinesfansElectric Power SystemslightingHVACDigital Logic Circuitsseat beltsComputer Systemsnavigation1.3

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5thEdition Problem solutions, Chapter 1Communication SystemsradiotelephoneElectro-Opticscockpit displaysInstrumentationcompassair speed indicatorinclinometeraltimeterControl Systemsrudderflapsc)HouseholdCircuit Analysisdesign of the home's electrical systemElectromagneticsmicrowave ovenstereo speakersSolid-State ElectronicstelevisionstereoVCRElectric Machinesappliancespower toolsfansElectric Power SystemslightingHVACreceptaclesDigital Logic CircuitsclockstimersComputer Systemsmicrowave ovenprogrammableVCRCommunication SystemstelephoneCB radiotelevisionradioElectro-Opticsdigital clocksInstrumentationelectric meterControl Systemsthermostat1.4

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5thEdition Problem solutions, Chapter 1Problem 1.3Solution:Some examples are:a)HVAClightingoffice equipmenttypewritercomputercopy machineclockstaplershredderelevatorb)conveyorpunch presslightingventilationdrill presshoistlathec)power sawdrilllightingelevatorpumpcompressor1.5

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5thEditionProblem solutions,Chapter 2Chapter 2:Fundamentals of Electrical Circuits – Instructor NotesChapter 2 develops the foundations for the first part of the book.Coverage of the entire chapter would betypical in an introductory course.The first four sections provide the basic definitions (Section 2.1) and coverKirchoff’s Laws and the passive sign convention (Sections 2.2, 2.3 And 2.4); A special feature,Focus onMethodology: The Passive Sign Convention(p. 39) and two examples illustrate this very important topic.A secondfeature, that will recur throughout the first six chapters, is presented in the form of sidebars.Make The Connection:Mechanical Analog of Voltage Sources(p. 24) andMake The Connection: Hydraulic Analog of Current Sources(p.26) present the concept of analogies between electrical and other physical domains.Sections 2.5and 2.6 introduce thei-vcharacteristic and the resistance element.Tables 2.1 and 2.2 on p. 45summarize the resistivity of common materials and standard resistor values; Table 2.3 on p. 48 provides theresistance of copper wire for various gauges.The sidebarMake The Connection: Electric Circuit Analog ofHydraulic Systems – Fluid Resistance(p. 44) continues the electric-hydraulic system analogy.Finally, Sections 2.7 and 2.8 introduce some basic but important concepts related to ideal and non-idealcurrent sources, and measuring instruments.In the context of measuring instruments, Chapter 2 introduces a thirdfeature of this book, theFocus on Measurementsboxes, referenced in the next paragraph.The Instructor will find that although the material in Chapter 2 is quite basic, it is possible to give anapplied flavor to the subject matter by emphasizing a few selected topics in the examples presented in class.Inparticular, a lecture could be devoted toresistance devices, including the resistive displacement transducer ofFocuson Measurements: Resistive throttle position sensor(pp. 56-59), the resistance straingauges ofFocus onMeasurements: Resistance strain gauges(pp. 58-59),andFocus on Measurements: The Wheatstone bridge andforce measurements(pp. 59-60). The instructor wishing to gain a more in-depth understanding of resistance straingauges will find a detailed analysis in the references1.Early motivation for the application of circuit analysis to problems of practical interest to the non-electricalengineer can be found in theFocus on Measurements: The Wheatstone bridge and force measurements.TheWheatstone bridge material can also serve as an introduction to a laboratory experiment on strain gauges and themeasurement of force (see, for example2).Finally, the material on practical measuring instruments in Section2.8bcan also motivate a number of useful examples.The homework problems include a variety of practical examples, with emphasis on instrumentation.Problem 2.51 illustrates analysis related to fuses; problems 2.65 relates to wire gauges; problem 2.70 discusses thethermistor; problems 2.71 discusses moving coil meters; problems 2.79 and 2.80 illustrate calculations related tostrain gauge bridges; a variety of problems related to practical measuring devices are also included in the lastsection.The 5thEdition of this book includes 26 new problems; some of the 4thEdition problems were removed,increasing the end-of-chapter problem count from 66 to 80.It has been the author's experience that providing the students with an early introduction to practicalapplications of electrical engineering to their own disciplines can increase the interest level in the coursesignificantly.Learning Objectives for Chapter 26.Identify the principalelements of electrical circuits: nodes, loops, meshes, branches, and voltageand current sources.7.ApplyKirchhoff’s Lawsto simple electrical circuits and derive the basic circuit equations.8.Apply thepassive sign conventionandcompute powerdissipated by circuit elements.9.Apply thevoltage and current divider lawsto calculate unknown variables in simple series,parallel and series-parallel circuits.10. Understand the rules for connectingelectrical measuring instrumentsto electrical circuits for themeasurement of voltage, current, and power.2.11E. O. Doebelin, Measurement Systems – Application and Design, 4thEdition, McGraw-Hill, New York, 1990.2G. Rizzoni, A Practical Introduction to Electronic Instrumentation, 3rdEdition, Kendall-Hunt, 1998.

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5thEditionProblem solutions,Chapter 2Section 2.1: DefinitionsProblem 2.1Solution:Known quantities:Initial Coulombic potential energy,; initial velocity,UVi=17kJ/Ci=93M ms; final Coulombic potential energy,.Vf=6kJ/CFind:The change in velocity of the electron.Assumptions:∆PEg<< ∆PEcAnalysis:Using the first law of thermodynamics, we obtain the final velocity of the electron:Qheat−W= ∆KE+ ∆PEc+ ∆PEg+...Heat is not applicable to a single particle. W=0 since no external forces are applied.∆KE= −∆PEc12me(Uf2−Ui2)= −Qe(Vf−Vi)Uf2=Ui2−2Qeme(Vf−Vi)=93M ms⎛⎝⎜⎞⎠⎟2−2−1.6×10−19C()9.11×10−37g6kV−17kV()=8.649×1015m2s2−3.864×1015m2s2Uf=6.917×107msUf−Ui=93M ms−69.17M ms=23.83M ms.2.2

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5thEditionProblem solutions,Chapter 2Problem 2.2Solution:Known quantities:MKSQ units.Find:Equivalent units of volt, ampere and ohm.Analysis:Voltage=Volt=JouleCoulombV=JCCurrent=Ampere=Coulombseconda=CsResistance=Ohm=VoltAmpere=Joule×secondCoulomb2Ω =J⋅sC2Conductance=Siemens or Mho=AmpereVolt=C2J⋅sProblem 2.3Solution:Known quantities:Battery nominal rate of 100 A-h.Find:a)Charge potentially derived from the batteryb)Electrons contained in that charge.Assumptions:Battery fully charged.Analysis:a)100A×1hr=100Cs⎛⎝⎜⎞⎠⎟1hr()3600shr⎛⎝⎜⎞⎠⎟ =360000Cb)charge on electron:−1.602×10−19Cno. of electrons =360×103C1.602×10−19C=224.7×10222.3

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5thEditionProblem solutions,Chapter 2Problem 2.4Solution:Known quantities:Two-rate change charge cycle shown in Figure P2.4.Find:a)The charge transferred to the batteryb)The energy transferred to the battery.Analysis:a) To find the charge delivered to the battery during the chargecycle, we examine the charge-current relationship:i=dqdtordq=i⋅dtthus:Q=i(t)dtt0t1∫Q=50mAdt05hrs∫+20mAdt5hrs10hrs∫=0.05dt018000s∫+0.02dt1800036000s∫=900+360=1260Cb) To find the energy transferred to the battery, we examine the energy relationshipp=dwdtordw=p(t)dtw=p(t)dtt0t1∫=v(t)i(t)dtt0t1∫observing that the energy delivered to the battery is the integral of the power over the charge cycle. Thus,w=0.05(1 + 0.75t18000) dt018000∫+0.02(1 +0.25t18000) dt1800036000∫=(0.05t+0.7536000t2)018000+(0.02t+0.2536000t2)1800036000J1732.5=w2.4

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5thEditionProblem solutions,Chapter 2Problem 2.5Solution:Known quantities:Rated voltage of the battery; rated capacity of the battery.Find:a)The rated chemical energy stored in the batteryb)The total charge that can be supplied at the rated voltage.Analysis:a)∆V≡ ∆PEc∆QI=∆Q∆tChemical energy= ∆PEc= ∆V⋅ ∆Q=∆V⋅I⋅∆t()=12V350 A−hr 3600 shr=15.12 MJ.As the battery discharges, the voltage will decrease below the rated voltage. The remaining chemical energy storedin the battery is less useful or not useful.b)is the total charge passing through the battery and gaining 12 J/C of electrical energy.∆Q∆Q=I⋅ ∆t=350a hr=350 Cshr⋅3600 shr=1.26 MC.Problem 2.6Solution:Known quantities:Resistance of external circuit.Find:a)Current supplied by an ideal voltage sourceb)Voltage supplied by an ideal current source.Assumptions:Ideal voltage and current sources.Analysis:a) An ideal voltage source produces a constant voltage at or below its rated current. Current is determined by thepower required by the external circuit (modeled as R).IVPRVIss⋅==b) An ideal current source produces a constant current at or below its rated voltage. Voltage is determined by thepower demanded by the external circuit (modeled as R).V=Is⋅RP=V⋅IsA real source will overheat and, perhaps, burn up if its rated power is exceeded.2.5

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5thEditionProblem solutions,Chapter 2Problem 2.7Solution:Known quantities:Rated discharge current of the battery; rated voltage of the battery; rated discharge time of the battery.Find:a)Energy stored in the battery when fully rechargingb)Energy stored in the battery after dischargingAnalysis:a)Energy=Power×time=1A()12V()120hr()60 minhr⎛⎝⎜⎞⎠⎟60 secmin⎛⎝⎜⎞⎠⎟J105.184=w6×b)Assume that 150 W is the combined power rating of both lights; then,J104.32sec3600)8)(150(w6used×=⎟⎠⎞⎜⎝⎛=hrhrsWJ108643×=−=usedstoredwwwProblem 2.8Solution:Known quantities:Recharging current and recharging voltageFind:a)Total transferred chargeb)Total transferred energyAnalysis:a)C48,600)(4)(60)(60214(90)(60))(2)(90)(60216(30)(60))(4)(30)(6021curvetime-currentunder theareaQ=++++=∫==IdtC48,600=Qb)pdtdw=so∫∫==vidtpdtws10800t0027A,t36004-12=is0027t1800A,t54002-6=is1800t0A,t18004-10=is10800t0V,t108003+9=v321≤≤≤≤≤≤≤≤2.6

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5thEditionProblem solutions,Chapter 2where321iiii++=Therefore,333331080072006322720018006322180006322108007200372001800218000110456610648104105108380109132107292006001081016296001080601086410072090××+××+×=⎟⎟⎠⎞⎜⎜⎝⎛×−−++⎟⎟⎠⎞⎜⎜⎝⎛×−−++⎟⎟⎠⎞⎜⎜⎝⎛×−−+=∫+∫+∫=.-.-...tttt.tttt.ttttdtvidtvidtviwkJ.Energy =9489Problem 2.9Solution:Known quantities:Current-time curveFind:a)Amount of charge during 1stsecondb)Amount of charge for 2 to 10 secondsc)Sketch charge-time curveAnalysis:a)11043ti−×=Coulombs102secamp1022104104033102310311−−−−×=×=×=∫×=∫=ttdtidtQb) The charge transferred fromtotis the same as fromt=1=2t=0tot=1.Q2=4×10−3CoulombsThe charge transferred fromtotis the same in magnitude at=2=3nd opposite in direction to that fromt=1tot=2.Q3=2×10−3Coulombst=4Q4=2×10−3−4×10−3dt34∫=2×10−3−4×10−3= −2×10−3Coulombst=5,6, 7Q5= −2×10−3+2×10−345∫dt=0Q6=0+2×10−356∫dt=2×10−3CoulombsQ7=2×10−3+2×10−367∫dt=4×10−3Coulombst=8,9,10sQ=4×10−3Coulombs2.7

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5thEditionProblem solutions,Chapter 22.8Problem 2.10Known quantities:urve and voltage-time curve of battery rechargingTotal transferred energyAnaa)Solution:Current-time cFind:a)Total transferred chargeb)lysis:AmA1.0100=- time curve=Idt∫=0.1()2( )3600()+3600()0.1e- t-2()/0.82dt26∫=1,011 CQ=area under the currentQ= 1, 011 Cdwdt=Pso()b)()()()()()()()8,114J0.1e936001.01.89e6.8136003600360042/0.822-t-20t/0.824220=∫+∫+=∫+∫=∫∫=vidtPdtw=dtdtdtvidtviEnergy=8,114Jroblem 2.11Known quantities:urve and voltage-time curve of battery rechargingTotal transferred energyAnaa)PSolution:Current-time cFind:c)Total transferred charged)lysis:AmA04.0=40Q=area under the current - time curve=Idt∫=0.04()6()3600()=864CC648=Qb)Pdtdw=so./t--.-t/1671040305136000404502136003600004240220404220=∫−+∫−=∫+∫=(dtPdtw36=∫∫=)()() ()()()()()()J,dt.e..dt.e..dtvidtviviJ,Energy1671=

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5thEditionProblem solutions,Chapter 22.9Problem 2.12Solution:Known quantities:Current-time curve and voltage-time curve of battery rechargingFind:edeedya05t/12=∫=∫=Idte)Total transferrchargf)Total transferrenergAnalysis:a)()12 -areQ=C8,564dte3600curvetime-currentunder theC8,564=Qdwdt=Pb)so()()()8,986Jee3123600205t/12-5t/12-=∫−=∫∫==dtvidtPdtwEnergy=8,986J

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5thEditionProblem solutions,Chapter 2Section 2.2, 2.3 KCL, KVLProblem 2.13Solution:Known quantities:Circuit shown in Figure P2.13 with currentsI0=−2A,A,A, and voltage sourceI1= −4IS=8VS=12V.Find:The unknown currents.Analysis:Applying KCL to node (a) and node (b):I0+I1+I2=0I0+IS+I1-I3=0⎧⎨⎩⇒I2= −(I0+I1)=6 AI3=I0+IS+I1=2 A⎧⎨⎪⎩⎪Problem 2.14Solution:Known quantities:Circuit shown in Figure P2.14.Find:The unknown currents.Analysis:Applying KCL at the node:0562=++--ithus i = 3 A which means that a 3-A current is leaving the node.Problem 2.15Solution:Known quantities:Circuit shown in Figure P2.15.Find:The unknown currents.Analysis:Applying KCL at the node:0256=++-ithus i = -3 A which means that a 3-A current is leaving the node.2.10

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