一阶电路习题

1、第五章电路的暂态分析三.习题详解5-1 在图 51 所示电路中，己知E = 2V, R = 10G, t/c(0_) = 0,匚(0_) = 0,开关=0时闭合，试求换路后电流i、J、及电压“c的初始值和稳态值。解r = 0+时，电感相当于断路，电容相当于短路Q(OJ = ZL(O_) = OA wc(O+) = mc(O_) = OV根据欧姆定律E 2心于訂2Ac(0+) = j(0+) = 0.2A 稳态时i(oo) = z(0+) = 0.2AiL (co) = i(oo) = 0.2Auc(oo) = OVic (oo) = OA图51题5 1图5-2图5-2所示电路己处于稳定状态，/

2、 = 0时开关闭合，求电流论、i和电压叫的初始值。E/(0-)=,i(0-)=;Uc(0-)=rJr；e心0+时Eur (0 ) = wr (0 )=E八“ C &+/?3E uc (0 ) R,ic (0 ) =ER, R2ERJD 1 D图5-2题5-2图r2R皿mz曲&5-3电路如图5-3所示。求在开关Q闭合瞬间(t = 0+)各元件中的电流及两端电 压；当电路到达稳态时又各等于多少设在=0一时，电路中的储能元件均未储能。解 t = Q_时，电路中储能元件均未储能4(0) = 0匚 2 (0一)= 0ci(O-) = %2(。一)= t = Q+时，电感支路相当于开路，电 容支路相当于短

3、路/Ll(+) = r，Ll(0-)= 0 匚2(。+)=匚2 3) = 0 wci (+ ) = uci (_ ) = 0 %2(+) = %2(0_) = 0ZC1(+) = G(J= R ：r = 1AUL1 (+ )=如1 (- ) = ZC1 (+ )尺2 =1x8= 8V41 =，R2 =，C1 (。+ ) =，C2 (。+ ) = 1AuRi = iRiR =1x2 = 2VuR =诂 R, =1x8 = 8V稳态时，电感支路相当于短路，电容支路相当于开路 () = lL2 () = R : R = 1Awl1(o) = wL2(co) = 0Vici(8)= ic2(8)= 0

4、VUcl (oo) = uc2 (oo) = 8V55 在图 55 中，E = 20V,= 12kl,= 6kQ,= 10/F,C2 = 20/F :电 容元件原先均未储能。当开关闭合后，试求电容元件两端电压“c。解 因电容原先未储能，所以f = 0+时uc(9+) = u c (9-) = 0V4-OE旳UC稳态后uc(oo) = E = 20 Vt = R.一一图5 5 题5_5图2 cx + c270= 6xl03x xl0-6=4xl0-2s3所以% = uc ()+% (+) 一 uc ()k= 20 + (0 = 20(1 25jv5-6在图5-6中电路原己稳定，已知E = 100

5、V,/? = 10kQ,C = 4/zF,求开关换 接后lOOnis时的电容电圧和放电电流。解 开关换接前(一)=丘=1 V开关换接后陀(0+ ) = % (0_ ) = 100 V时间常数r=/?C = 10xl03x4xl0-6 =4x10_2s电容电压的零输入响应为图5-6 题5-6图换接后lOOnis时uc = 100 e-2 5 Vi = 0.01严 A5-7电路如图5-7所示，在开关Q闭合前电路己处于稳态，求开关闭合后的电压Il Q Q解 mc(0_) = 6x103 x9x10-3 = 54V”c(+) = %(-)= 54 V稳态时,%为3kG电阻电压W(oO)=图5-7题5-

6、7图时间常数 r = (6xl03/3xl03)2xl0_6 = 4xl0 s 所以uc (0 = uc (oo) + uc (0+) - uc (oo)_7 = 18 + (54-lS)e25t = 18 + 36 訂血 v5-8图5-8所示电路原己稳定，求开关闭合后电容两端电圧和流经开关的电流 i(t) o解开关闭合前%(。一)= E 开关闭合后wc(+)= c(-)= E流经开关的电流i为图5-8题5-8图uc(t) = Ee7 =100e-25tV其中i2为RC电路零输入响应电流T = R2CE - i2 (0 =e r 2 R2开关闭合后电容电压为uc(t) = Ee T=1 + (

7、0.75 -1)八血=1-0.25八叫nA5-10 图 5-10 中电路己稳定，己知E = 30V, R, = 10Q, 7?2 = 2R, C=2“F , C2 =CJ2,求开关断开后流经电源支路的电流7(。o解Q断开前电容G和C?相当于开路E30z(0 )*(0 ) = -= 1A2R1 + R2 10 + 20ucl (0_) = i2 (0_)/?2 = lx 20 = 20VuC2 (0_) = i2 (0_ )7?1 = 1 x 10 = 10V图5-10 题5-10图Q断开后Mci(+)= wci(-)= 20VWC2(0+) = WC2(0-)=10VE-.cl(0+) = 3

8、0-20 =1Alk & 10q =&C =10x2x10 =2xioR220r2 = R2C2 =20x1x10 =2x10s 稳态时，电容C和C2相当于开路L (oo) = 0 i2(8)= 0，2(0+)= 匕如2 =卫二厶1A根据三要素法可得认)= 0 + (0 + l) 2x10-5 =e 2x10-5A i2 (0 = 0 + (0 + l)e 2x10-3 = e 2x10-5 Ai(t) = h (t) + i2 (0 = 2e 2xioA5-12 图5 - 12所示/?厶电路中，己知图5-12题5-12图电流初始值为f = o时开关闭合，经过多少时间电流为匚(+) = J(_

9、)=人/? = 200G 上= 0.25H,其初始值的一半解开关闭合后时间常数所以设经过时间f时电流为初始值的一半，则-1(0 = 0e r = 2解得45-14在图5-14中，开关Q先合在位置1,电路处于稳定状态。=0时，将开关 从位置1合到位置2,试求！ = ：时心之值。在r = z时，又将开关合到位置1 ,试求 r = 2xio_2s时c之值。此时再将开关合到2,做出“C的变化曲线。充电电路和放电电 路的时间常数是否相等解 开关合在位置1时uc (0_) = 10V开关合到位置2,电容放电r2 = (10x 103 + 20x 103) x | x 10-6 = 0.015%(/) =

10、%(。+)幺=10, =2- 2e_100/A 闭合Q讥 0+)*(+) = 2Aii() = # =号=3Ai2(8)= 0A则和2的变化规律为如= = 5x103$1 & 2如=空=2乂10-2$ 尺2 1W)#(8)+百(0+) i】(8)k “ =3 + (2-3)尸叫=3_-200么i2 (0 = z2(qo) + i2 (0+) - i2 (oo) =0 + (2-0)严=2e5QtA5-21电路如图5-22所示，试用三要素法求f 20时的人、必及。换路前电路处 于稳态。解 换路前电路处于稳态，所以有12h （+ ） = （- ） = = 2 Aor-l=l换路后+Oi2v1H+?

11、vO(0+) + f2(0+) = iL(0+) = 2AH0+）x6“2（0+）x3 + 9-12 氈 P 题 5-2】图联立以上两式可以得到讯0+) = 3，2(+) = ia处于稳态时，i(8)= = 2A9Z2() = 3 = 3A(co) = i(oo) + G(oo)= 5Ar = = 0.5s6/3根据三要素法可求得iL (0 = iL S) + UL (0+) - iL (oo)/7 = 5 + (2-5)严=5 - 3严 an (0 = n (s) + L (0+) - L ()e7 =2 + (1-2)e2t =2-e2tAi2 (r) = i2 (oo) + i2 (0+

12、 ) -i2 (s)厂 =3 + (1-3)e2f =3-2e2,A四、自测题及参考答案5-1图5-23所示电路己在稳定状态下断开开关S,则该电路（）A）因为有储能元件厶要产生过渡过程B）因为电路有储能元件且发生换路，要产生过渡过程0因为换路时元件厶的电流储能不发生变化，不产生过渡过程5-2图5-24所示电路在达到稳态状态后移动&上的滑动的触点，该电路将产生 过渡过程。这时因为（）oA）电路发生换路为（）。A) -6VB) 6VC) OV5-8 AT串联电路与电压为8V的恒压源于t = O瞬间接通，如图5-29a所示，且知 比（0）=0V,当电容器的电容值分别为10“F、50“F、30“F、2

13、0“F时得到的四根你曲 线如图5-29b所示。其中20/F电容所对应的你（f）曲线是（）5-9血串联电路与电压为8V的恒压源于接通，如图5-31a所示。在t = 0瞬间将开关S闭合。当电感分别为1H、3H、6H、4H时得到的四根你（f）曲线如图531b所示。 其中1H电感所对应的柿（0曲线是（）。图5-29题5-7图a)b)图5-30 题5-8图5-10在换路瞬间，下列说法正确的是（A）电感电流不能跃变B）电感电压必然跃变C）电容电流必然跃变lOOpF4-5kQA20kQ-6V图5-55 题5-34图参考答案51C52B53A54C55C56C57A58B59A510 A511wL(0+) =

14、 5V, A(0+)=l. 5A, wc(0+) = 15V , zc(0+) = 0.5A ,伽(0+) = 10V,他(O+) = 15V,他(0+) = 5V,(。+) = 1A ,(0+) = 1.5A ,入(0+) = 0.5A。512 ic(0+ ) = 0, wL(0+ ) = 2V ；ic(co) = 0 , z(oo)=0, 2l(qo)=1/12A,比(8)二5V。5_13 ”c(+) = 3V , c(+) = 75A ,匚(0+) = 1A , uL(0+ ) = 9ViR (0+) = 0, Ur (0+) = 0, uc(8)= 9V , ic (oo) = 0fL

15、(oo) = 0.75 A t wL(oo) = 0, z(oo) = 0.5A , uR (oo) = 3V514 uc(0+) = 6V , i()+) = 1.5A, i2(0+ ) = 3A, ;c(0+) = 3A ；L (co) = 0.5A , i2 (oo) = 1A, ic (oo) = 0。5-15 uc(t) = 20_10rV, i=10/inA。5-16 wc(r) = (2O + 3Oe_/o l2)V,波形如图 5-56 所示；wc(3r) = 21.5V 5-17 Mc(r) = 12(l-30e-100/)V , i严(2 + 1.8100r)niA , i2 = (2 -1.2100/)niA 5-18 uc(t) = 10宀V , t = 8x103s。519= 1/2S , r2 = 2.5/zs5-20 c的波形如图5-57所示，其变化周期T=o5-21 uc(0 = 10(l-e-105,)V, %(/-15盟)=7.77严()o5-22 uRi = 15.4e_10(/-01)V o5-23 uc=2-2y曲V,冷=4 + 2訂汕冋V, i = 0.2 + 0.1小火叭nA5-24 uc = 12e_5/V , uR = -2eStN5_25 uc = 6(1 e25t )V。5-26 uc = (-6 + 24e_1000