电路分析基础(英文版)课后答案第一章

《电路分析基础(英文版)课后答案第一章》由会员上传分享，免费在线阅读，更多相关内容在教育资源-天天文库。

1、1CircuitVariablesandCircuitElementsDrillExercisesZ1¡5000tDE1.1q=20edt=4000¹C0dq¡®tdi¡®tdi1DE1.2i==te;=(1¡®t)e;=0whent=;dtdtdt®11Thereforeimax==»=10A®e0:03679eDE1.3[a]Therefore(a)v=¡20V;i=¡4A;(b)v=¡20V,i=4A(c)v=20V,i=¡4A;(d)v=20V,i=4A[b]Usingthereferen

2、cesysteminFig.1.3(a),p=vi=(¡20)(¡4)=80W,sotheboxisabsorbingpower.[c]Theboxisabsorbing80W.12CHAPTER1.CircuitVariablesandCircuitElementsZ14¡10;000t4¡10;000tDE1.4p=vi=20£10eW;w=20£10edt=20J0336DE1.5p=800£10£1:8£10=1440£10=1440MWfromOregontoCaliforniaDE1.

3、6Theinterconnectionisvalid:is=10+15=25Ap100V=100is=2500W(absorbing)p10A=¡100(10)=¡1000W(generating)¡100+vs¡40=0sovs=140Vp15A=¡15(140)=¡2100W(generating)p40V=15(40)=600W(absorbing)Xpdev=p10A+p15A=3100WXpabs=p100V+p40V=3100WXXpdev=pabs=3100WDE1.7[a]vl¡v

4、c+v1¡vs=0;ilRl¡icRc+i1R1¡vs=0isRl+isRc+isR1¡vs=0[b]is=vs=(Rl+Rc+R1)DE1.8[a]24=v2+v5¡v1=3i5+7i5¡(¡2i5)=12i5Thereforei5=24=12=2A[b]v1=¡2i5=¡4V[c]v2=3i5=6V[d]v5=7i5=14VProblems3[e]p24=¡(24)(2)=¡48W;therefore24Vsourceisdelivering48W.DE1.9i2=120=24=5Ai3=12

5、0=8=15Ai1=i2+i3=20A¡200+20R+120=0R=80=20=4−DE1.10[a]PlottingagraphofvtversusitgivesNotethatwhenit=0,vt=25V;thereforethevoltagesourcemustbe25V.Whenvtiszero,it=0:25A,hencetheresistormustbe25=0:25or100−.Acircuitmodelhavingthesamev¡icharacteristicisa25Vso

6、urceinserieswitha100−resistor.4CHAPTER1.CircuitVariablesandCircuitElements[b]252it==0:2A;p=(0:2)(25)=1W:125DE1.11[a]Sinceweareconstructingthemodelfromtwoelements,wehavetwochoicesoninterconnectingthem

7、seriesorparallel.Fromthev¡icharacteristicwerequirev

8、t=25Vwhenit=0.Theonlywaywecansatisfythisrequirementiswithaparallelconnection.Theconstraintthatvt=0whenit=0:25Atellsustheidealcurrentsourcemustproduce0:25A.Thereforetheparallelresistormustbe25=0:25or100−.[b]vtvt¡0:25++=0;5vt=25;vt=5V10025v2tp==1W:25Pro

9、blemsdqP1.1i==24cos4000tdtTherefore,dq=24cos4000tdtProblems5Zq(t)Ztdx=24cos4000ydyq(0)0¯tsin4000y¯q(t)¡q(0)=24¯¯40000Butq(0)=0byhypothesis,i.e.,thecurrentpassesthroughitsmaximumvalueatt=0,soq(t)=6£10¡3sin4000tC=6sin4000tmCP1.2p=(6)(100)£10¡3=0