电路与电子技术基础第四章习题答案

14-14-24-124Vt=0SiL(t)=AS12iL+24V4H6-4-14-20S600SiL(t)=24V/12=2AiL(0+)=2AiL()=2A=L/R=4/(12//6)=1sτtLLLLeiiiti−⋅∞−++∞=)]()0([)()(=2A4-34-2t=0t0i(t)10i1-++i14A4uOC-10i1-+i1i4A42HS4-24-3tKVL0)0()0(==−+ii10i1+4i1=0i1=0Ai()=4A5614410111==+=iiiuOCVi()14456)(0==∞==iuiuROCSCOC2710==RLτs)1(4)40(4)(77tteeti−−−=−+=A4-44-3i(t)=10mAR=10kL=1mHat=0abt0u(t)iR(t)iL(t)abiR(t)iL(t)+i(t)Ru(t)L-(a)4-34-4uLi10mA0t0t-100V-10mA(b)(c)ab(a)t000≥=+tRidtdiLLLiL(0)=i(t)=10(mA)0)(1010)(710≥==−−tmAeetittLτSRL73310101010−−=×==τ0)(10010101010))(0()1)(0()(77101033≥−=×××−=−=−==−−−−−tVeeeLRLieLidtdiLtutttLtLLLτττ0)(10)()(710≥−=−=−tmAetititLR(b)(c)4-54-4at=0b31abi(t)3A21F34-44-5u)(623)0(Vc=×=12b)(236)0()0(ARuic===SRCi3130)(=×===∞τ30)(2)(31≥=−tAetit4-64-5t0t=0u(t)u(t)=uc(t)uc(t)RCSCR23)1//2(0=×==τ)(32)2//1(1)()(221)0(VuVucc=×=∞=×=0)1(322)322(32))()0(()()(5.05.05.0≥−+=−+=∞−+∞=−−−−teeeeuuututtttcccc43421321τ02)(215.05.00'≥=×==−−−teVeeUutttcτ0))(1(32)1(11212)1(5.05.0≥−=−×+×=−=−−−tVeeeRIutttscτ3++0.5+us(t)23Fuc(t)uo(t)--2uc(t)-4-64-7+1Au(t)23F1-4-54-64-74-6u≥=0100)(tttsuc(0)=5Vuo(t)uo(t)uo(t)=-0.52uc(t)uc(t)uc(t)t0SCR5183)2//3(0=×==τ4uc(t)uc(0+)=uc(0-)=5V0)(55)(185'≥==−−tVeetttcτuuo(t)055.0)(2)(185''≥−=×−=−tetututcouc(t)0))(1(52)1(1322)(185185≥−=−××+=−−tVeetuttcuo(t)0))(1(525.0)(2185≥−−=×−=−tVetuutco4-84-7C=0.2FuVC=0.05Fu)1(20)(5.0tCet−−=C(0-)=5Vt0uC(t)RO+CuC+-UOC-+uCC-4-74-8C)1()(τtOCCeUtu−−=VUOC=20V=2s=R0CR0=2/0.2=10C=0.05F=100.05=0.5suC(0+)=5VuC()=20VttCCCCeeuuutu21520)]()0([)()(−−+−=∞−+∞=τV4-94-8t=0Si(t)24.0)150100/(60)0()0(=+==−+LLiiA2424.0100)0(100)0(=×=×=−+LCiuV50.1H20F+-iLiuCiC100100150+60V-4-80.1H20F+-iLiuCiC100100150S+60V-4-84-9RLtttLLeeeiti10001.010024.024.0)0()(−−−+===τARCtRCtCCeeutu50024)0()(−−+==VtCCedtduCti50024.0)(−−==AKCL)(24.0)]()([)(1000500ttCLeetititi−−−=+−=A4-104-9t=0uc(t)iL(t))(03.010)41(150_)0(3AiL=×+=)(120150414_)0(Vuc=×+=t=0)(120)0()0()(03.0)0()0(VuuAiiccLL====−+−+1k4k+4Fuc(t)100H-150ViL(t)4-94-10t00=−+cLLudtdiLRidtduCicL−=02=++cccudtduRCdtudLCLCr2+RCr+1=002500400122−=−LCLR620100210423=××==LRα501041001160=××==−LCω2110205022220=−=−=αωωd52arcsin5020arcsinarcsin0===ωαθ)2110sin(21750)52arcsin2110cos(21600)2110sin(104211003.0)52arcsin2110cos(211050120sin)0()cos()0()(2020206200tetetetetCiteututtttddLdtdcc−−−−−−−−=××−+−×=+−=ωωθωωωαdtduCticL−=)(iLiS(t)+GuCCL-4-104-114-114-10≥=0100)(tttisG=5SL=0.25HC=1FiL(t)0)0(=+CuViA0)0(=+Lt0122=++LLLidtdiGLdtidLC0.25s2+1.25s+1=0s1=-1s2=-41)(421++=−−ttLekekti0)0(=+Li0)0()0(===++dtdiLuLLCu=+=++04012121kkkk341−=k312=kttLeeti431341)(−−+−=A