一阶电路习题答案1

11-1题11-1图示电路原已达到稳态，当t=0时开关K动作，求t=0+时各元件的电流和电压。(b)12ALKiL3iL3R1R26R3(a)10V+-K3R22R2uC+-C题11-1图iCiR2iR3解：(a)iL(0-)=6A，iL(0+)=iL(0-)=6AAiR46966636)0(2，iR3(0+)=2AiR1(0+)=12iL(0+)=6A(b)VuC41052)0(，uC(0+)=4ViC(0+)=A23410(R2与uC串联)11-2题11-2图示电路原处于稳态，t=0时开关K闭合，求uC1(0+)、uC2(0+)、uL1(0+)、uL2(0+)、i(0+)。i(t=0)5AK题11-2图C2uC2+-3C1uC1+-uL2+-L2L1uL1+-2解:uC1(0-)=uC2(0-)=V632，iL1(0-)=iL2(0-)=A5由换路定则，有uC1(0+)=uC1(0-)=V6，uC2(0+)=uC2(0-)=V6i(0+)3uL2(0+)+-2uL1(0+)+-6V-+6V-+i(0+)5A5A0+网络列网孔电流i(0+)方程:2i(0+)+6+3（-5-5+i(0+)）+6=05i(0+)=30–12=18i(0+)=A6.3518ViuL2.1362.76)0(2)0(1ViuL2.136)0(2)0(211-3求题11-3图示电路的初始值uC(0+)、iL(0+)、iR(0+)、0dtdiL。开关K打开前电路处于稳态。题11-3图1F-uC1+(t=0)K55H5A5iRiL解：AiL5.2)0(，VuC5.125.25)0(由换路定则，有AiL5.2)0(，VuC5.12)0(5A12.5V+-iL(0+)iL(0+)i(0+)i(0+)5A550＋网络0)5)0((5)5.2)0((55.12ii25)0(10i0)0(0)0(5.2)0(0LudtdiiAiLLR11-4题11-4图示电路原处于稳态，求开关开打开后瞬间的iL1(0+)、iL2(0+)。2HiL25(t=0)K10V43HiL1题11-4图解：AiL2)0(1，0)0(2Li换路时满足磁链守恒)0()0()0(3)0(2)0(3)0(2212121LLLLLLiiiiii即4)0(3)0(211LLiiAiAiLL8.0)0(8.054)0(2111-5题11-5图示电路原处于稳态且uC(0-)=5V、uS=10sin(100t+30)V，t=0时开关K闭合，求开关K闭合后的iL(0+)、uL(0+)和iC(0+)。(t=0)K0.5iC题11-5图uS+-uC+-0.01FiL0.01HuL+-0.5解：K闭合前，电路处于正弦稳态，用相量法求电感电流1525.215210130105.05.03010jjILMt0时，)15100sin(25.2)(ttiLAiL83.1)15sin(25.2)0(由换路定则，有iL(0+)=iL(0-)=-1.83A0+等效电路：0.5iC(0+)5V+-iL(0+)uL(0+)+-0.5-5V+AiiViuLCLL83.1)0()0(915.55)183.0(5.05)0(5.0)0(11-6题11-6图示电路，开关K在t=0时打开，开关打开前电路为稳态。求t0时的uC、iC、iR和u1。题11-6图iR6mAu1+-4k3k2FiC2k(t=0)KuC+-解:属于零输入响应VuuCC464242)0()0(sRC26310102105VeeututtCC1004)0()(t0.mAeAeetitttR100100310038.0108.041051)(t0.mAetititRC1008.0)()(t0.VKmAtu2446)(1t0.11-7题11-7图示电路。t0时电路已处于稳态，t＝0时开关K闭合。求使iL(0.003)=0.001A的电源电压US的值。(t=0)K1k题11-7图US+-iL1H1k解：属于零输入响应3102)0(SLUiSLLUii3105.0)0()0(sRL310tSLeUti3103105.0)(001.0105.0003.01033eUS解得：.17.40VUS11-8题11-8图示电路，开关K闭合已很久，t＝0时开关K打开，求t0时的uC(t)和UR(t)。+-0.5uC0.125FK(t=0)25V+-uC+-4uR+-题11-8图解：求uC(0-)5V+-2uC(0-)+-4+-0.5uC(0-)iR(0-))0(5.0)0(4)0(5)0(5.0)0(6CRCCRuiuui5)0(5.0)0(86)0(814)0(5.0)0(CCCCRuuuuiVuuVuCCC4)0()0(425.155.075.05)0(4+-0.5uiu+-uiu5.04iu45.080RsRC1825.1VetutC4)(.0tVetututututCCCR2)(5.0)()(5.0)(.0t11-9求题11-9图示电路的时间常数。(b)(a)10V+-720.5F1Ω--+0.5i112A112Hi1题11-9图解：(a))(5.0),(321FCR.5.15.03sRC(b)--+0.5i111i1i+u-1111)(15.0iuiiiuiuuiuu5.05.0sRLiuR33/22325.11011-10题11-10图示电路。t0时电容上无电荷，求开关闭合后的uC、iR。iR2mA2FuC+-K(t=0)5k题11-10图解：属于零状态响应VuC1025)(sRC26310102105.0,)1(25)()(.0,)1(10)1)(()(100100tmAeKtutitVeeututCRttCC11-11题11-11图示电路原处于稳态，求t0时的iC和uL。iLuL+-K19A(t=0)10Ω5Ω3H4题11-11图解：属于零状态响应.0,36)(.0,)1(4)1)(()(319349544)(33tVedtdiLtutAeeitisRLAitLLttLLL