1.电路如图1所示,求图中的电压U。2Ω3Ω3Ω-+4V1A3V+-+-U图1解:根据戴维宁定理,先将3Ω与3V的电压源进行等效代换,将两电流源合并后进行化简,化简后可计算得:2.13//2,8.4OOCRVU故所求电压为VU43.32.已知图2中?,2RVuab求图2解:列KVL方程有:)2/1(321111uRuuuuuababVu3213R解得:3321RVu,.3.以节点4为参考节点,列图3节点电压方程。R2is1R1us3R3R4+_R5R6①②③④is6图3解:613624164111111ssnnniiURURURRR336365325163514254211111011111RUiURRRURURURURURRRssnnnnnn4.列写图4电路的网孔电流方程。图4解:网孔Ⅰ:634221432smmmuiRiRiRRR网孔Ⅱ:033232112mmmiRiRRRiR网孔Ⅲ:235432314smmmuiRRRiRiR5.求图5(a)电路的戴维宁等效电路。+_10V1kΩ1kΩ0.5iiabIILU1A16V+-2R3R4R5RLR842033习题2.7.8图图5(a)图6解:uOC=uab=10V-10+2000isc-500isc=0isc=1/150AR0=uOC/isc=1500Ω戴维宁等效电路如图5(b):图5(b)ⅢⅡⅠR3-us6+R1R2①R4+us2—R5R6①1500Ω_+ba10V6.用戴维宁定理求解图6中LR电阻中的电流LI。IILU1A16V+-2R3R4R5RLR842033习题2.7.8图图6解:将支路断开,得有源二端网络ab,如图(1)所示。先用节点电压法求开路电压:再求ARRUUIcd125.1485.216353于是有VIRRIUUabo4318125.116253将有源二端网络除源得无源二端网络''ba,如图解(2)所示,求等效电阻9382048)204()(25435430RRRRRRRR有戴维宁定理等效电路[图(3)]可求:ARRUILaboL333.0313940II3U1A16V+-2R3R4R5RLR842033习题2.7.8图(1)abcd2R3R4R5R84203习题2.7.8图(2)a'b'7.在图7电路中,已知已知sU=10V,sI=11A,R=2,L=1H,开关S在t=0时合上,闭合前电路处于稳态,求电感电流tLi。_+RL3Ω9ΩV5.220148114816R1RR1IRRUU4353Scd图(1)图(2)4V图(3)sUSRsIL()Lti图7解:(0)(0)11ALLii,()16ALi,0.5LsR,2()(165)AtLtie8.图8电路,电路原处于稳态,t=0时刻,开关由1打向2,求tLi。S448V2H()Lti124图8解:(0)(0)2A3LLii,()2ALi,2142LsR,2()423tLtie9.图9电路中开关k在位置1已久,t=0时开关合向位置2,求换路后响应()ctu,()ti。5VK1225k100k100k10uFcui图9解:500100425100ccuuV,400.04100imA0cuV,0iA,36210010101050102iRC2240.04tctuteVitemA10.图10电路中,已知sU=10V,1R=3k,22Rk,L=10mH,在t=0时开关S闭合,闭合前电路已达稳态。求开关闭合后电感电()Lti和电压()Ltu。sUS1R2RLLuLi图10解:(0)(0)2mALLii,()5mALi,62510LsR,5210()(53)mAtLtie5()210()6VLttLtdiuLedt11.电路如图11知1sU=2sU=10V,1R=10k,2R=5k,C=0.1uF,在0t时开关S处于位置1,电容无初始储能。当t=0时,S与2接通。经过1ms以后S又突然与3接通。用三要素法求0t时()ctu表达式。1sUS1R()ctuC2R2sU123图11解:0t时,(0)0cu01tms时,(0)(0)0ccuu,()10Vcu,11RCms,310()10(1)Vtctue1tms时,1(1)(1)10(1)V=6.32Vccuue,()10Vcu,'20.5RCms3310(1)-210(t-1)0.5(1)()()(1)[]V=(-10+16.32e)Vtctcccuuuue12.两组负载并联,一组KVAS10001,功率因数为0.6,另一组KVAS5002,功率因数为1,求总视在功率和总有功功率。解:根据题意,第一组负载有,rkVSQkWSPa8008.010sin6006.010cos61116111第二组负载有,0sin500105cos2225222SQkWSP所以,总的有功功率kWPPP110050060021总视在功率kVAQPS13608001100222213.图12示正弦交流电路,求负载ZL获得最大功率时ZL,并求获得的最大功率。图12解:利用戴维宁定理对上图进行简化,等效后的电压为:2412242(618)22ocUjVj等效阻抗为:222(11)22RjZjjVj所以当11LZj时候,获得最大功率。最大功率为:222618360904414ocUPWR14.如图13所示电路为日光灯和白炽灯并联的电路,图中1R为灯管电阻,LX为整流器电抗,2R为白炽灯电阻。已知VU220,整流管电阻不计,灯管功率为40W,功率因数为0.5,白炽灯功率为60W,求III,,21及总功率因数。图13解:由111cosUIP01160,5.0cos灯管中的电流为AUPI364.05.022040cos111由,白炽灯的电流为22UIPAUPI7.202206022电路的总功率WPPP100604021日光灯的无功功率为VarUIQ35.69866.0364.0220sin11总视在功率为VAQPS9.61215.3691002222故总电流AUSI5.502209.6121总功率因数822.069.121100cosSP15.如图14,oS123490A,j30Ω,30Ω,45Ω,.IZZZZI已知:求电流图14解:1330(j30)//15j1530j30ZZ,S13132(//)//IZZIZZZZj4(15j15)15j15j3045oo5.657455-36.9o1.1381.9A16.对称三相感性负载接在对称线电压380V上,测得输入线电流为12.1A,输入功率为5.5KW,求功率因数和无功功率?解:由pcos3llIUP可得功率因数为,69.01.12380732.155003cospllIUP电路的无功功率为var5764724.01.12380732.1)69.0sin(arccos3llIUQ17.如图15所示为接成三角形的三相对称负载,已知电源电压UL=220V,电流表读数都是IL=17.3A,三相电路总有功功率P=4500W,试求每相负载的电阻和感抗。图15解:AIILP103221510220345003cosPPIUP1510cosPPRUIUR1610cos12PPXLUIUX18.电路如图16所示,电路激励为i(t)=ε(t),响应为u1(t)、u2(t),用复频域分析法求阶跃响应u1(t)、u2(t)。图16解:电路如图图12-1)65(44221141)()()(21SSSSSSSSGSISU654)(222)(212SSSUSSSUtteetu32138232)(,tteetu32244)(19.电路如图17所示,要求用拉普拉斯变换法求)(tuC。S2.4)UC1s2(S图17图17-1解:运算电路如图17-1所示:ttCCeetuSSSSSSSSU32222)(3222652142//22//2)(,20.动态电路如图12-10所示,开关S在a时电路已达稳态,t=0时将开关S合向b。试用拉氏变换法求电感的电流Li和电容的电压cu。解:换路前:V4)0(u,A22443)0(iCL运算电路如图18-1所示:应用结点电压法5.1S21S63S5S214S82S1S212S2S2S4)S(U2Ct5.1tCe2e6u,5.1S41S6)3S2)(1S(10S42S2)S(U)S(ICLt5.1tLe4e6)t(iAV