第三章一阶动态电路分析习题分析

1第三章一阶动态电路分析习题分析3-2图3-2所示电路，t0时电路已达稳态，t=0时开关由1扳向2,求iL(0+)，uL(0+)，uR(0+)。解：t0时电路处于稳态（S在1处）iL(0—)=3/(3+6)*3=1AiL(0+)=iL(0—)=1At=0+时刻（S在2处）的等效电路为uR(0+)=-iL(0+)R=(-1)x6=-6V由KVL:6iL(0+)+uL(0+)-uR(0+)=0得:uL(0+)=uR(0+)-6iL(0+)=-12V3-3图3-3所示电路，t=0时开关闭合，已知uc(0—)=4V，求ic(0+)，uR(0+)。解：uc(0+)=uc(0—)=4Vt=0+时刻（S闭合）的等效电路：对节点1由KCL:I1=I2+ic(0+)(1)对回路1由KVL:2I1+4I2-8=0(2)对回路2由KVL:4ic(0+)+uc(0+)-4I2=0(3)将三式联立代入数据得ic(0+)=0.25A,I2=1.25A,uR(0+)=4I2=5V3-5图3-5所示电路t=0时开关S由1扳向2，在t0时电路已达稳态。求初始值i(0+)、ic(0+)和uL(0+)。解：t=0-(S在1)L视为短路，C视为开路il(0+)=iL(0-)=Us/(2+4)=24/6=4AUc(0+)=Uc(0-)=4il(0-)=16VT=0+时刻等效电路12UC(0+)iC(0+)对接点①由kcl:iL(0+)=i(0+)+ic(0+)(1)Ril(0+)Ul(0+)UR(0+)R+-UC(0+)iC(0+)UR(0+)22对回路1由kvl:UL(0+)+4i(0+)+2iL(0+)=0(2)对回路2由kvl:Uc(0+)-4i(0+)=0(3)由（3）得i(0+)=Uc(0+)/4=16/4=4A代入（2）ic(0+)=4-4=0A由（2）式UL(0+)=-4×4-2×4=-24V3-7图3-7所示电路，开关动作前电路已达稳态，t=0时开关S由1扳向2，求t=0+时的iL(t)和uL(t)。解：t=0-时电路处于稳态，电感视为短路iL(0-)=（8/(8+4)）×6=4Ail(0+)=iL(0-)=4A换路后从电感两端看进去等效电阻：R=4+8=12Ωτ=L/R=0.2/12=1/60siL(∞)=0电路响应为iL(t)=4e-60tAUl(t)=LdtiL(t)d=0.2×(-60)×4e-60t=-48e-60tV3-11图示电路，t=0-时电路已达稳态，t=0时开关S打开，求t=0时的电压uc和电流i。解：t=0-电路处于稳态电容视为开路I1=(3/(3+4+2))×6=2AUc(0-)=2I1=2×2=4VUc(0+)=Uc(0-)4V换路后从电容两端看进去等效电阻R=R3+R4=2+1=3Ωτ=RC=3×(1/3)=1s零输入响应Uc（t）=4e-ti=ic=Cdtu(t)d=(1/3)×(-1)×4e-t=-(4/3)4e-t3-10如图所示电路，t=0时开关闭和，求t=0时的iL(t)和uL(t)。解：iL(0+)=iL(0-)=0换路并稳定后，电感视为短路iL(∞)=3从电感两端看进去等效电阻R=6/5Ωτ=L/R=0.3×5/6=0.25s零状态响应为：iL(t)=iL(∞)(1-e-4t)=3(1-e-4t)AuL(t)=LdtiL(t)d=0.3×(-3)×(-4)e-4t=3.6e-4tV33-16求图示电路的阶越响应uc解：uc(0+)=uc(0-)=0加入阶跃函数ε(t)并稳定后，电容视为开路uc(∞)=(1/2)ε(t)V从电容两端看进去等效电阻：R=1+2//2=2Ωτ=RC=2×1=2s阶跃响应uc(t)=uc(∞)(1-e-t)=0.5(1-e-0.5t)ε(t)V3-17图3-17所示电路，开关闭和前电路已达稳态，求开关闭和后的uL。解：用三要素法开关闭合前稳态，电感视为短路①iL(0-)=100/50=2At=0+时刻等效电路图iL(0+)=iL(0-)=2A对接点①由kcl:I1+I2=iL(0+)(1)对回路1由kvl:50I1+Ul(0+)–100=0(2)对回路2由kvl:-50I2+Ul(0+)–50=0(3)上三式联立解得I1=1.5AI2=0.5AUl(0+)=25V换路稳定后，电感视为短路Ul(∞)=0从电感两端看进去的等效电阻R=50//50=25Ωτ=L/R=5/25=0.2sUl=Ul(∞)+[Ul(0+)-Ul(∞)]e-t=25e-5tV3-19图3-19所示电路，已知iL(0-)=6A，试求t=0+时的uL(t)，并定性画出uL(t)的波形。解：用三要素t=0+时等效电路，iL(0+)=iL(0-)=6A10.1Ul(0+)Ul(0+)对节点1由KCLi1(0+)=iL(0+)+0.1uL(0+)=0由上两式联立解得uL(0+)=100V4电路稳定后，电感视为短路uL(∞)=0从电感两端看进去等效电阻R：10.1UpUpIp对回路1由KVL：6Ip+4(Ip+0.1Up)–Up=0R=UP/IP=50/3Ωτ=L/R=0.03SuL(t)=uL(∞)+[uL(0+)-uL(∞)]uL(t)=uL(∞)+[uL(0+)–uL(∞)]e-t=100e-t/0.03V3-20图3-20所示电路，t=0时开关S1闭和、S2打开，t0时电路已达稳态，求t=0+时的电流i(t)。解：三要素法t=0-电路处于稳态电容视为开路，等效电路为：UsRRUc(0-)=（3/（3+1））×8=6Vt=0+时刻等效电路R2R1UC(0+)3A5i(0+)UC(0+)+-Uc(0+)=Uc(0-)=6Vi(0+)=Ul(0+)/3=6/3=2A换路稳定后，电容视为开路R2R13Ai(∞)=(1/2)×3=1.5V从电容两端看进去等效电阻R=3//3=1.5Ωτ=RC=1.5×1=1.5si(t)=i(∞)+[i(0+)-i(∞)]e-t=1.5+0.5e-2t/3A