125 答案及解析
题12:30; 1.5; 50; 48。
题13: 题14:
du?250t?250t;?1?4m t?0,i;()t?CC??4emA()t?4eVt?0;is;uu(0)?4V(0)?40mACCC?L?dt?200t()?()t?40emAt?0;it?2?5ms;iL12?250t?200t(60?4e?40e)mA?i()t?i()t?t?0。 CL200题15:5; 40; 0.5; 20。 题16:(6?e?t3)V
2t?2t题17:3e?2tA; 51。 (?e?)A ; (10?11e)A题18: 题19:(c) 题20:
?5t()?(6?4e)As ;it;iL(;R;??LR?1;t?0; ??)6A.5?i(0)?2A?i(0)0?2L?L?50?5tut()?10eVt?0;
题21:
(??)6V时;u;u;R0?4?;?0??t4s(09V?)?1?t()?(6?3e4)V,0; ?4s;得ut??t4s1?t?1(??)18V,0;t?4;u;(4)?12?6eV??t4ss时;uu()t?(6?12e4)V?C??6s;
11???(t?4)??(t?4)??166??V?V()?18?(6?6e)e()??18?3.793e得 ut,t?4,t?4s;或 uts;
????????题22:
?1.25t?0.8s ;得 u(t)?12(1?e)V;R0?8?;?,t?0 u(??)12VCC题23:
;i1(0?)?8A;i1(?;iL(??;R0?4?;??)?5A)2AiL(0)?8A??2ti()t?(5?3e)A,t?0。 11?2ts;得i()t?(2?6e)A,t?0;L2题24:
第十三章 拉普拉斯变换答案
【题1】:
;i i(0)?2A(0)?5A1?2?
126 答案及解析
【题2】:c
【题3】:d 【题4】:d 【题5】:c
22K(s?s)?K(s?1)?s112【题6】:A提示:可用比较系数法 K,K1??1 ?2?122s(s?1)s(s?1)【题7】:
3111 ft()?sint?sin(2t)??22222(s?4)(s?1)s?1s?4【题8】:c
【题9】:d 【题10】:
1RRe?t/??(t) ??12C
R1?R2R1R2C【题11】:作s域模型,选用节点法,设节点电压U1(s)(电容电压),和节点电压U2(s)(受控源两端电
U1(s)1s112(s?1)(s?3)(?)U(s)?()U(s)?2I(s) I(s)??U(s)?压),可得:;;解得212222s?12ss(?4s?5)Us)?Us)o(2(112(s?3)?2t (t)?7.2?7.58ecos(t?161.57?)(t)V ?;uos?1ss(?2?j)(s?2?j)???【题12】:u;iL(;复频域模型如图 (0)?40V0)?4AC??
604?030.5890.589140s2?204s?601sU(s)????s?1?)U()s?40?节点方程(得 CCs?5s?5s(s2?6s?6)ss?1.268s?4.732?1.268t?4.732tu(t)?(10?30.589e?0.589e)V , t?0 C【题13】:u u (0)?1V(0)?0VC1?C2?127 答案及解析
81224+1UC1(s)+s321s___ss+8+UC2(s)_
111115(s?1.2)(s?4) (??)(Us)??Us()?2483224818ss8ss(?1)(s?3)2?(12?)s?ss311?4(s?1)(s?3)s3?1?3?t3?3tt3t u?(1?e?e)()tVu?(1?e?e)()tVC1C25s?12.8844UC2(s)?2s(s?1)(s?3)UC1(s)????【题14】:
111?0.4(1??)U()s?? i(0)?0.6ALi(0)?0.41?11?2262s?2ss2??s3333 Us()?(.04s?1)(s?3)
ss(?2)(s?4)U111131??? 6?2s16s40s?280s?411?3?2t4t?(?e?e)(t)A i 2164080(s)? I2?【题15】:
U?i(0)?0.A1 Li(0)?0.2i(0)?0.9ALi(0)?0.61?11?2?22?12.s?3
s(s?2)(s?4)128 答案及解析
31619133?9?2t4t i? I???(?e?e)(t)A8s40s?240s?482040??2t?4t 或i ?0.375?0.e15?0.225e(t)A?+??【题16】:
IL(s)0.50.5sU(s)_2s12.5s0.5_
+
s2?4(s?25)22(s?25)43.960.040412.5s)?2Us()??2 IL( I (s)???Ls2ss?0.5051s?49.495sss(?50s?25)s?50s?252???212.5s?0.505t?49.5t i?(4?3.e96?0.e04)(t)AL?【题17】:
I(s)10?100?s12s13s?40?s?Uc2(s)?
100?4014064246s U I(s)??()s?Is()???C211113ssss?110??s?22s3s12 u()t?(64?24e)()tVC2【题18】:
作s域模型,选用网孔法
t?12?(2?s)I1(s)?sI2(s)??sI1(s)?(s?U(s)?2I1(s)12?2U(s)s1?2)I2(s)?2U(s) 解得: sI2(s)?12(s?4)4s2?13s?66(s?4)U(s)?2I(s)?o2(s?27.)(s?05.6)
?0.56t?2.7t u(t)?(9.e64?3.e64)(t)Vo?【题19】:
i u(0)?1V(0)?2AC?L?复频域模型如图
129 答案及解析
节点方程:(??87s112s?20U(s)??? 得 )Us()?0.1?CC21022.5sss?5s?4s?4s?11?4t?tit()?ut()?(4e?3.e)5A , t?0 C2第十五章 电路方程的矩阵形式答案
题1
(1)?2(2)?4(3)?56(4)?8?(5)1379
(画错一条(包括方向错误)扣2分,错4条以上则无分) 题2:(C) 题3:(D) 题4:(C) 题5:(C) 题6:(A) 题7:
?)(0
题8: