1、,31 t0時(shí)，S閉合后uC及各電流的初始值及穩(wěn)態(tài)值。,uC(0 ) 1.5 4V 6V,在S閉合的瞬間，根據(jù)換路定律有： uC(0 )= uC(0+ )= 6V, i1(0+ ) uC(0+ ) /4 =1.5A i2(0+ ) = uC(0+ ) /2 =3A iC(0+ )1.51.533A,S,C,i2,iC,uC,i1,1.5A,2,4,uC( ) 1.5 (4/2)V 2V iC( ) 0 i1( ) 2/4A0.5A , i2( ) 2/2A 1A,32 求開關(guān)閉合后的初始值及穩(wěn)態(tài)值。,iL(0 ) =1243A,在S閉合的瞬間，根據(jù)換路定律有： iL(0+)iL(0) = 3A

2、,4 i1(0+) + uL(0+) =12 uL(0+)= 4.8V,+,S,i2,uL,解：,i1,4,6,iL,12V,i1 (0+) = 3 6/（46） = 1.8A,i2 (0+) = 3 4/（46） = 1.2A,iL( ) 12(4/6) 5A uL( ) 0 i1( ) 12/4 3A i2( ) 12/6 2A,33 電路中,已知R1=5k, R2=10k, C=4F，US= 20V, t =0，打開S，求uC(t) 、iC(t) 、并畫出變化曲線 。,S,uC,R2,R1,US,iC,C,+ ,電路為零輸入響應(yīng),解:,uC(0+)=uC(0) =USR2/（ R1 +R

3、2） =2010/15=13.3V,= R2 C=4102s,uC() =0,iC() =0,uC(t) =13.3e25t V,iC(t) =1.33e25t mA,uC(t),t,iC(t),-1.33mA,13.3V,S,C,uC,i,1mA,10K,34 S打開0.2s后，uC8V，試求C、 i(0+ )、 uC(t)。,解:,uC(0+)=uC(0)=0,= 10103 C,電路為零狀態(tài)響應(yīng),uC() =201010V,uC(t) =10（1et/10000C ）V,t0.2，uC= 8V uC(t) =10（1e1/50000C ）8V e1/50000C0.2， C12.5F,S

4、閉合的瞬間，C相當(dāng)于短路 i(0+ )0,uC(t) =10（1e8t ）,35 電路中,已知R1=3, R2=10, R3=6,C=10F，US= 100V, t=0，閉合S， 求i1(t) 、i2(t) 、i3 (t) 。,S,i3,R2,R3,R1,US,解:,uC(0+)=uC(0)=0,= (R1/R3+R2) C =12105s,i2,i1,C,+ ,電路為零狀態(tài)響應(yīng),i1() =i3() =US /(R1+R3)=11.1A,i2() =0,i3(t) =11.11.85e8333t A,i1(t) =11.1+3.7e8333t A,i2(t) = 5.55e8333t A,C

5、,uC,IS,36 已知R1=6k, R2=2k，C=20F， IS= 4mA，US= 12V, t =0，閉合S ，試求uC(t)。,解:,uC(0+)=uC(0) =ISR2= 8V,= (R1/R2) C = 3102s,9V,uC(t) =9e33.3t V,R2,S,US,+ ,R1,37 電路中,已知R1=4, R2=20, R3=6, C=4F，US= 50V, t =0，閉合S，求uC(t) 、i(t) 、并畫出變化曲線 。,S,R2,R3,R1,US,i,C,+ ,uC,解:,= (R1/R3+R2) C =89.6106s,uC(0+)=uC(0)=50V,uC() =US

6、 R3 /(R1+R3) =30V,t,uC(V).i(A),i(0+)=(US /R1)(R3/R1)uC(0+)/(R2+R3/R1) =0.896A,i() =0,38 電路中,已知R1=R2=4k, R3=2k, C=100F，E1= 10V, E2= 5V，t0，S由a打向b，求uC(t) 、i0(t) 。,解:,uC(0+)=uC(0) = E1R2 /(R1+R2) = 5V,= (R1/R2+R3) C =4101s,uC()= E2R2/(R1+R2) = 2.5V,=1.56mA,i0()= E2/(R1+R2) = 0.625mA,a,b,S,C,uC,i2,2A,3,3

7、9 C= 0.5F，試求i1 (t )、 i2(t)。,解:,uC(0+)=uC(0) =23V 6V,= (6/3) C=1s,電路為零輸入響應(yīng),6,i1,i1 (0+) = 6/6 A= 1A,i2 (0+) = 6/3A = 2A,i2() =0,i1() =0,310 電路中,已知R1=6, R2=4, L=10mH，US= 10V, t =0，閉合S，求uL(t) 、iL(t) 。,US,L,R1,R2,iL,S,uL,解:,= L/R2 = 2.5103s,iL(0+)= iL(0) =US/(R1+R2)=1A,uL(0+) = -R2 iL(0+) = -4V,iL() =0,

8、uL() =0,311 電路中,已知R1=3k, R2=6k, C1=40F， C2=C320F，US= 12V, t0，閉合S，求uC(t) ，畫出曲線。,S,R2,R1,US,解:,uC(0+)=uC(0)=0,= (R1/R2) C =4102s,C1,+ ,電路為零狀態(tài)響應(yīng),C2,C3,uC,C = C2/C3+C1,uC()= USR2/(R1+R2) = 8V,uC(t)= 88e25tV,t,uC(t),8V,IS,312 已知R1=2, R2=R3=3， R4=6, L=10mH, IS= 1mA, US=8V, t0，斷開S ，試求uL(t)、 i(t) 。,解:,R2,S,

9、US,+ ,R1,R3,R4,i,uL,= 0.001/（6+6/2) = 1.33103s,313 已知R1=200, R2=400, C1=0.1F， C20.05F，U= 25V, t=0，斷開S，求uC1(t) 、 uC2(t) 、 i(t) 。,S,R1,R2,U,解:,uC1(0+)=uC1(0) =25400/(200+400) =50/3V uC2(0+)=uC2(0) =25200/(200+400) =25/3V,1= R1 C1=2105s 2= R2 C2=2105s,C1,+ ,C2,uC2,uC2,uC1()= uC2()= 25V,uC1(t)= 258.3e50000tV uC2(t)= 2516.7e50000tV,314 t = 0，閉合S，求iL，畫出曲線。,12V,1H,3,iL,解: iL(0+)= iL(0) =12/(3+3)=2A 依等效變換 i(0+)=(33/62)612/(3+6) =1A = 1/3+(63) =0.2s,33V,S,a,b,iL(t)=3.81.8e5tA， i(t)=0.21.2e5tA,6,3,2,1,315 圖示電路在開關(guān)S閉合前電路已處于穩(wěn)態(tài)，在t =0時(shí)刻開關(guān)閉合。試求開關(guān)閉合后的uC(t)及iL(t)。,+,t=0,10