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=1A iL(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: 6 iL (0+)+uL(0+)-uR(0+)=0 得: uL(0+)=uR (0+)-6 iL (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
2、)=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=4A il(0+) = iL(0-) =4A换路后从电感两端看进去等效电阻:R=4+8=12 =L/R=0.2/
3、12=1/60siL()=0电路响应为 iL(t) =4e-60t A Ul(t) =L=0.2(-60) 4e-60t = - 48e-60t V3-11 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=2A Uc(0-) =2 I1=22=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=3=RC=3(1/3)=1s零输入响应 Uc(t)=4e-ti = ic =C=(1/3) (-1) 4e-t =-(4/3)4e-t 3-10 如
4、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=6/5=L/R=0.35/6=0.25s零状态响应为: iL(t) = iL() (1- e-4t )=3(1-e-4t )A uL(t) =L=0.3(-3)(-4)e-4t =3.6e-4t V3-16 求图示电路的阶越响应uc解:uc(0+) =uc(0-) =0加入阶跃函数(t) 并稳定后,电容视为开路uc() =(1/2)(t) V从电容两端看进去等效电阻:R=1+2/2=2=RC=21=2s阶跃响应 uc (
5、t)= uc() (1- e-)=0.5(1- e-0.5t ) (t) V 3-17 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 iL(0-) =100/50=2At=0+ 时刻等效电路图 iL(0+) = iL(0-) =2A对接点由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 =1.5A I2 =0.5A Ul(0+) =25V换路稳定后,电感视为短路 Ul() =0从电
6、感两端看进去的等效电阻 R=50/50=25 =L/R=5/25=0.2s Ul = Ul() + Ul(0+) - Ul() e-=25 e-5t V3-19 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-)=6A 对节点1由KCL i1(0+)= iL(0+)+0.1 uL(0+)=0由上两式联立解得 uL(0+)=100V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:对回路1由KVL:6Ip+4(Ip+0.1Up) Up=0 R= UP/IP=50/3 =L
7、/R=0.03S uL(t)= uL()+ uL(0+)- uL() uL(t)= uL() + uL(0+) uL() e-=100 e-t/0.03 V3-20 图3-20所示电路,t=0时开关S1闭和、S2打开,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:Uc(0-) =(3/(3+1)8=6Vt=0+ 时刻等效电路Uc(0+) = Uc(0-) =6Vi(0+) = Ul(0+) /3=6/3=2A换路稳定后,电容视为开路i()=(1/2) 3=1.5V从电容两端看进去等效电阻R=3/3=1.5=RC=1.51=1.5s i(t) = i() + i(0+) - i() e-=1.5+0.5 e-2t/3 A5