课件、习题讨论课

1、13. S在 t =0时闭合。求 iR 并定性画出其波形。iR 2k1kS(t=0)+1kuC200pF 6V_9ViR (0+)解: uC (0 ) uC (0 ) 6V2k0+时刻电路:6V0.5kiR (0+)1k+ -1.5V2k-u (0 ) +1k C6V6V i (0 ) 6 (1.5) 3mA9VR2.5diL (2) 求 iC (0+) , uL (0+) 及 dt t 0S(t= 0)1H解: i (0 ) i (0 ) 0LLiLu (0 ) u (0 )CC16V+ 16 (20 / 80) 8 V-16 (20 / 80)i (0 ) i (0 ) i (0 ) i

2、(0 )C123t=0+时刻电路: u (0 ) 16 8 8 8 L0 162080u (0 ) 8 1 4 V+L216V-diL uL (0 ) 4 A / sdt t0L16 i1 (0 )+ -40i2 (0 )20iC (0 )i3 (0 )+40 8V-16+ uL -20iC0.5F+40uC 40-2. 求电压或电流的初始值。(1) 求i1(0+), i2(0+), i3(0+)。4 i1i3解1A+uCu (0 ) u (0 ) 6 1 6 V-CC(0 ) 6 2 At =0+时刻电路:i30.6 4 / 64 i1(0+) i3 (0+)i (0 ) 6i (0 ) 3

3、 1.2 A0.614 6+6Vi (0 ) 4-i (0 ) 3 0.8 A24 6i2 (0+)6i20.60.1F6S(t =0)1. 求时间常数.20.3H3+US0.1H =(0.1+0.3)/5=0.08s-(a)10iu=10i+4i=14i4A42HReq14 =2/Req=2/14=0.143s(b)i-+重点:一阶电路的经典分析方法全响应的的两种分解方法:全响应 = 零输入响应+ 零状态响应全响应 =分量+ 强制分量一阶电路的三要素法。电路原理A(2)习题课1一阶电路25. 图示电路中，已知t 0时， iS =2A，r+=2。求t 0时的 i(t)。iSuC_解： 用三要素

4、法。（1）求 i(0 ) t 0时i 02AS u (0 ) u (0 ) 0CC画出t = 0+时刻的等效电路4i(0 ) ri(0 )i(0 ) 2 4i(0 ) 0.8Ai(0+)44+ ri(0+)_i440.01F+ ri _5Fu (0 ) 16VC+ u -i(0 ) 13ACi() 4AiS 4A 10s2+1i10Vi 4 9e0.1t A-2i (0 ) 3AL43i(0 ) 3 1.75A 846Hi() 5AiL 2 4si 5 6.75e0.25t Ai i i 9 9e0.1t 6.75e0.25t A (t 0)+10V- i+6V-65F+ uC -2i 4AS

5、+S6V2+-6H10V6i -iL求 i :5F求 i :+ uC -2iS 4A2+6Hi10ViL-u (0 ) 16Vi (0 ) 3ACL+10V- i+6V-64. 开关S在 t = 0时刻闭合。 求电流 i 。5F+ uC -2i 4AS+S6V2+-6Hi i10Vi6-iLi i i解：0-时刻电路:+ uC(0-) -uC (0 ) 16V2+i (0 ) 3Ai 4ALS6V2-u (0 ) 16VC6iL(0-)iL (0 ) 3AiR 2k1k S(t=0)+1kuC200pF 6V_9VtiR (t ) 5 (3 5)e mAiR 的定性波形:iR /mA530t/

6、siR 2k1kiR (0 ) 3mAt = 时电路:1ki () 6 9 5mAR3R 1 2 1 5 keq2 13 5 103 200 1012 1 106 s33t t！iR (t ) 5 (3 5)e 5 2e mA （t 0）3习题集10-30SRC1 4F C2 2F R 1k(t=0)u (0 ) 120V u (0 ) 0V+C 1C 2C1uC1C2u_ C2解u (0 ) u (0 ) 120VC 1C 1u (0 ) u (0 ) 0VC 2C 2 RC R C1C2 4 103 seqC C312i 20 F 1 S 2uC1(0-)=0+ +u -+uS1 100

7、sin 100tVuC1500DS2uS1 u 500 sin(200t 30 )V-u +-S 2C220 F-t = 0时开关从2切换到1. 求uC1或uC2？思考？ i(t ) 0.707 sin(200t 75) 0.0829e200t A (t 0) 5 103 si C1 1 S 2i(0 ) 0.6A+ 20 F+i (t ) 0.707 sin(200t 75)A uS2500uS1st-C20 F -2(5) 根据初值求 A ：i(t ) 0.707 sin(200t 75D ) Ae200t0.6 A 0.707 sin 75A0.0829 i(t ) 0.707 sin(

8、200t 75) 0.0829e200t A (t 0)三要素法:ti(t ) i (t ) i(0 ) i (t )e ststt 0 0.707 sin(200t 75D ) 0.6 0.707 sin(200t 75D )e200tt 0 0.707 sin(200t 75) 0.0829e200t A (t 0)(3) 求 i(0+) i C1 1 S 2i+u + 20 F500+250V500S2uS1-C20 F -50V2+i(0 ) 300 0.6A 500(4) 求 ist (t )i C1u 500 sin(200t 30D )VS2+ 20F1 1 1uS2500CC2

9、00 20 106 250-C220F1250030Dist (t ) 0 ？Ist,m 500 j500 0.707 75 AC26.i 20 F 1 S 2uC1(0-)=0+ +u -+uS1 100 sin 100tVuC1500DS2uS1 u 500 sin(200t 30 )V-u +-S 2C220 Ft = 0时开关从2切换到1. 求 i.-解 : (1) 求uC2(0-)1 1 500C100 20 106 2U j500 100 70.7 45 VC 2m 500 j500u (t ) 70.7 sin(100t 45)Vu (0 ) 50VC 2C 2(2) 求 C1C

10、2 R 5 103 sC1 C2（2）求i()画出t = 时的等效电路+i() 2A2A_uC（3）求时间常数 求从电容两端看过去的等效电阻:加流求压:+uC 8i ri 10i_uC R uC 10eqi Req C 10 0.01 0.1stt i i() 0i ) ( i() e e 2 2 . 10A1 (t ) 0i44+ ri _i44+ ri _4一阶电路的三要素法不是万能的！5etV +S_列写微分方程du5et uu0.5 C C C dt22后续方法：卷积积分，复频域分析EndduC 2u 5etdtC2+uC_20.5F+36V _ 12V若将12V电源改为24V，影响？

11、只影响零输入响应，零状态响应不变。u (0 ) u (0 ) 36 24 (6 4) 24 34VCC2 6 4 0.15su (t ) 34et 0.15Vt 0Cu (t ) u (t ) u (t ) 27(1 et 0.15 ) 34et 0.15CCC 27 7et 0.15V(t 0)2k+uC_6k4k100FS+36V _ 12V再求零输入响应：u (0 ) u (0 ) 36 12 (6 4) 12 32VCC2 6 4 0.15su (t ) 32et 0.15Vt 0C全响应= 零状态响应 + 零输入响应u (t ) u (t ) u (t ) 27(1 et 0.15 ) 32et 0.15CCC 27 5et 0.15V(t 0)2k+uC_6k4k100FS7. 图示电路开关闭合前已处于稳态，t = 0时开关S闭合，求t 0时电容电压的零状态响应、零输入响应和全响应。若将12V电源改为24V，求t 0时电容电压的全响应。+36V _ 12V解：先求零状态响应u () 36 6 27VC2 6 R C (2 / 6) 103 100 106 0.15sequ (t ) 27(1 et 0.15 )V (t 0)C2k+uC_6k4k100FSSR(t=0)u