2010‐6‐10复习tutorial number 2‐2010•29)SeeFigure7.5.WhatisRTifthe10Ωresistorisshort-circuited?•A)0ΩB)98.6ΩC)100.0ΩD)101.1ΩB虚线左IL侧IL为电I源，sRinRL右侧为负载ab内阻相等，IS*Rin=E时电流源与电压源对负载RL等效。可扩展为交流电中用的复数验证两种情况下IL相等。解释：大小、极性、联接方式Ch4 5 6 7•Ch4 Ohm’s Law: R=V/I（实数）•Z=V/I（复数）•Ch5 KVL:V=V1+V2+V3+…•Ch6 KCL: I=I1+I2+I3+…12)Whatisthetotalresistanceofaparallelcircuitwiththreeresistorswiththevaluesof60,120and180ohmseach?tutorial number 2‐2010A)32.73ohmsB)62.46ohmsC)125.6ohmsD)155ohmsChapter 8•Current and VCurrentandVoltage Source ConversionoltageSourceConversion•Mesh Analysis•Nodal AnalysisTUTORIALNUMBER3,WITHANSWERS1)SeeFigure8.1.Whenconvertedtoavoltagesource,theequivalentvoltageisA)0.3VB)3.33VC)30VD)3000V2)SeeFigure8.1.Whenconvertedtoavoltagesource,theequivalentseriesresistanceis A)0.3ΩB)3.33ΩC)10ΩD)30Ω12010‐6‐1010)seefigure8.5.Whichofthefollowingtermsdescribesthevoltageacross the 7Ω resistor when using the mesh analysis (general) approach?TutorialNumber3,withAnswersA)(7Ω)I1B)(7Ω)I2C)(7Ω)(I1-I2)D)(7Ω)(I1+I2)Fig. 8.29Defining the mesh currents for the network of Fig. 8.28Fig. 8.30 Defining the supermesh current.Chapter 9 Network Theorems•Superposition TheoremSuperpositionTheorem•Thévenin’sTheorem•Norton’s Theorem•Maximum Power Transfer theorem11)SeeFigure8.5.Ifnodalanalysisweretobeusedtosolveforunknown voltages in this circuit, how many nodes would be needed (including the reference node)? A)1b)2c)3d)413))SeeFigureg8.5.Theequationqobtainedfrompperforminggmeshanalysisonmesh#1is:A)5V+3I1+7(I2-I1)=0B)5V+3I1+7(I1-I2)=0C)5V+3I1-7(I1-I2)=0D)5V-3I1+7(I1-I2)=0example9.3a. Using superposition, find the current through the 6-Ωresistor of the network of Fig.9.10.b. Demonstrate that superposition is not applicable to power levels.Fig 9.10 Current source replaced byby openopen-circuitcircuitFig 9.11Fig 9.12 The contribution of E to I2 The contribution of I to I2 22010‐6‐10Example 9.11 Find the Norton equivalent circuit for the network in the shaded area of Fig. 9.60.Solution:Steps 1 and 2 are shown in Fig. 9.61.Fig. 9.61Fig. 9.60Identifying the terminals of particular interest for the network of Fig. 9.60.Step5:See Fig. 9.. This circuit is the same as the first one considered in the development of Thevenin’s theorem. A simple conversion indicates that the Thevenin’s Fig. 9.. circuits are, in fact, the same Substituting the Norton equivalent circuit for the network external to the resistor R(Fig(Fig.9.65).965)Lof Fig.9.60).Fig. 9.65. Converting the Norton equivalent circuit of Fig.9. to a Thevenin equivalent circuit.a+b-a+a+b-RThb-EThEquivalent circuitStep 3 is shown in Fig. 9.62, andRN=R1||R2=3Ω||6Ω=2ΩStep 4 is shown in Fig. 9.63, clearly indicating that the short-circuit connection between terminals a and b Fig. 9.62is in parallel with R2and eliminates its Determining RNeffect.effect. INis therefore the same as isthereforethesameasthrough R1, and the full battery voltage appears across R1since V2=I2R2=(0)6 Ω=0VTherefore,IN=E/R1=9V/3 Ω=3AFig. 9.63Determining RN例如下图所示电路，若负载RL可以任意改变，问负载为何值时其上获得的功率为最大?并求出此时负载上得到的最大功率pLmax。解：(1)求uoc。从从a,b断开断RL，设设uoc如(b)图所示。在图所示在(b)图中，应用电阻并联分图中应用电并联分流公式、欧姆定律及KVL求得u4oc=−4+4+8×4×8+14+33+3+3×18=12V32010‐6‐104)SeeFigure9.3.UsingMillman’stheorem,whatisReq,externaltotheload?A)5ΩB)10ΩC)60ΩD)110ΩTransients in Capacitor or Inductor Networks•ic=dQ/dt=CdvC/dtvL=Ldi/dt•换路定则：换路前后电容的电量及电压不变。换路前换路前后电感的电流不变。电感的电流变VC=Vf+(Vi–Vf)e-t/τ10.23式（三要素法）可由P13导出iC=if+(ii–if)e-t/ττ=RThCτ=L/RThW=CV2/2W=LI2/21)SeeFigure9.1.WhatistheTheveninresistanceexternaltotheresistorR?A)3.4ΩB)12ΩC)14ΩD)15.4Ω2)SeeFigure9.1.WhatistheTheveninvoltageexternaltotheresistorR?A)3.3ΩB)4.3ΩC)6.7ΩD)10ΩTutorialNumber4,withAnswers5)SeeFigure9.4.WhatistheNortonequivalentresistanceRNexternaltotheresistorR?A)3.33ΩB)5ΩC)15ΩD)infinity6)SeeFigure9.4.WhatistheNortonequivalentcurrentIN?A)0.33AB)0.67AC)1.0AD)2.0ATutorialNumber5,withAnswers7)SeeFigure10.2.Thevoltageacrossa10μFcapacitoris changing asshowninthediagram.Whatisthecurrentthroughthecapacitoratt=1 ms?A)0AB)10μAC)10mAD)50mA8)SeeFigure10.2.Thevoltageacrossa10μFcapacitorischangingasshowninthediagram.Whatisthecurrentthroughthecapacitoratt=6 ms?A)0AB)10μAC)10mAD)50mA42010‐6‐10tutorial number 6‐2010‐answers17)SeeFigure10.8.Assumingthecapacitorisinitiallydischarged,whatistheinstantaneouscurrentthroughthe1kΩresistor0.2safterthrowingtheswitchtoposition1?18)SeeFigure10.8.AssumethatthecapacitorhaschargedtoVC=6V.HowlongwillittakeforthecapacitortodischargetoVC=4Vaftertheswitchisthrowntoposition2?17)1.35mA18)0.41stutorial number 6‐2010‐answers14)SeeFigure11.7.Whatisthetimeconstantforthiscircuit?15)SeeFigure11.7.Aftertheclosingoftheswitch,whatisthefinal steady-state voltage value across the 10kΩresistor?14)1.8μs15)90.9VQUESTION 4A一（15 marks) In the circuit given above，the switch has been kept open for a long time before t=0 seconds．The switch is closed at t=0 seconds and then opened at t＝25 μseconds again.着重说明初末状态的求解iLvL12)SeeFigure11.3.Whatisthe steady-state current through the coil aftertheswitchcloses?13)SeeFigure11.3.Assumethatthe10kΩresistorischangedtoa 10MΩresistorandthatsteady-state conditions are present before the change.Whatwillthemaximumcoilvoltagereach after the switch opens?12)10mA13)100kV16)Howmuchenergyisstoredina1mHinductorif3Aofcurrentflowthroughit?17)Whatisthetimeconstantofa500millihenrycoilanda3,300ohmresistorinseries?16)4.5mJ17)0.015secondsa.Determine the magnitude of the current iLand the voltage vLin steady state before the switch is closed at t＝0 seconds．（2 marks）b.Give the magnitude of it=0 seconds．（2 marksLimmediately after the closing the switch at ）c.Calculate the magnitude of v1 marks）Limmediately after closing the switch at t=0 seconds．（d.Calculate the time constant of the circuit for the period 25 μs＞t≥ 0 s，and give an expression for iL．（4 marks）e.Calculate the magnitude of iswitch at t＝25 μs.（1 marksLimmediately before opening the ）f.Give the magnitude of vLimmediately after opening the switch at t＝25 μseconds．（1 marks）g.Calculate the time constant of the circuit for the period t＞25 μs（ie．after opening the switch）and give an expression for iL．In your expression the instant at which the switch is opened can be considered as the time origin. （4 marks）52010‐6‐10Solutions:a.iL＝12/(100＋200)A=0.04A vL＝0（2 marks）b.iL=0.04A at t=0 seconds．（2 marks）c.vL=12－0.04×200=4V immediately after closing the switch at t=0 seconds．（1 marks）d.τ1＝5×10－3/200 s=25μs 25 μs ＞t≥ 0 s iL=If+(Ii－If)e－t/τ1=12/200+(0.04－12/200) e－t/τ1．（4 marks）e.iL=0.06－0.02e－1=0.053A .（1 marks）f.vL＝12－0.053(100+200)=－3.79V．（1 marks）g.τ2＝5×10－3/(100＋200) s=50/3 μs iL ＝12/(100＋200) +(0.053－12/300) e－t/τ2．（4 marks）eg: P624图15.35 （似某次考题）FIG. 15.37FIG. 15.38Impedance diagram for the series R‐L‐C circuit Phasor diagram for the series R‐L‐C of Fig. 15.35.circuit of Fig. 15.35.Chapter 13 •Sinusoidal Alternating Waveforms•Phase relationships•Effective or Root Mean Square (RMS) ValueChapter 14• Phasornotation Phasorsand PhasorDiagrams• Sinusoidal response of resistor, inductor and capacitor• Low and high frequency response of inductors and capacitors• Average power and Power FactorChapter 15 16Impedance and Admittance diagramsSemiconductors DiodesZTZT＝Z1+Z2+Z3 ＝R∠0°+XL∠90°+XC∠－90°＝3Ω+j7Ω－j3Ω＝3Ω+j4Ωand ZT＝5Ω∠53.13°ImpedanceImpedance diagram: As shown in Fig. 15.37.diagram:AsshowninFig1537II＝E/ZT＝(50 V∠0°)/(5Ω∠53.13°) ＝10 A∠－53.13 °VR,VLandVCVR=IZR=(I∠θ)(R∠0°) =(10A∠-53.13°)(3Ω∠0°) =30V∠-53.13°VL=IZL=(I∠θ)(XL∠90°) =(10A∠-53.13°)(7Ω∠90°) =70V∠36.87°VC=IZC=(I∠θ)(XC∠-90°) =(10A∠-53.13°)(3Ω∠-90°) =30V∠-143.13°Kirchhoff’s voltage law:∑V＝E－VR－VL－VC ＝0E=VR+VL+VC说明分压原理i=2(10)sin(ωt−53.13°)=14.14sin(ωt−53.13°)vR=2(30)sin(ωt−53.13°)=42.42sin(ωt−53.13°)vL=2(70)sin(ωt+36.87°)=98.98sin(ωt+36.87°)vC=2(30)sin(ωt−143.13°)=42.42sin(ωt−143.13°)62010‐6‐10P659 Eg. 15.1872010‐6‐108
