电工学简明教程第二版答案(第二章)

1、电工学简明教程第二版(秦曾煌 主编)习题A诜择颍2.1.1(2)2.2.12.3.1(1)2.3.22.4.2(3)2.4.32.5.1(2) (4)2.5.22.8.1(3)2.8.22.8.4(3)B基本题2.2.3U=220vJl=10A -U=22O,，2sin(2)2.2.2(1)(3)2.4.1(2)(2)2.4.4(1)(1)2.7.1(1)(2)2.8.3(3)=5 丿2A0=2200v第2.2.2题ij = 10、任sin (wt + 90)A il = 10 /2 - 45A2.2.4 u = 220v u = 220pQsin (wt + 30)vi = ( - 4 -

2、j3)A = 5e“ 143l) i = 5p吃sin (wt - 143.1)Ai =(4 - j3)A = 5e*( 一 问)j _ sp/in (wt - 36.9)A2.4.5(l)220V交流电压s闭合 Ur = 220V l= = 22A UL = OV Uc = OVS打开 I = 0A UR = UL = OV Uc = 220VS闭合= OV Z = R + JwL = 10 +J*314 7 io non 220-llA UB- Rl- HOvV U -wU-HOV KL也开U=22O湮sin (314t)vi1U 220Z=R*|(wL- - 10 + K10- ) =

3、10 fl = =22%314 31401Uo-wLl - 10 22 220V U一J-220VRC wc请浏览后下载，资料供参考，期待您的好评与关注!1 I22.4.61 6K j2nfL |R2 + (wL)2=TU=380VJ=30mA 380J1602 + (314L)2=i30* 103l=40(H)R = r=?=6 6? + (314L)2 = -y- = L = 15.8(ni/)2.4.82.4.92.4.102.4.11|=Jr2 + (2nfL)2=j2002 + (314 43.3)=27.7(mA)R 2000X 二 COS ./2sin (314t 4 2O)(V

4、)1U(U = 4.4.sin (314t-33)(V)JZ=J = 30 + j * 40Q Req = 30flJeq = 40Leq=127(mH)R22X /OS Z? = 115.5(kl)U2 = i 2 押 + &)*1=0588(v)(a)A0 = 10V2 A(b)V()= 80V(C) A0 = 2A (d) v“oQvAo = 10*2 Av0 = iooQv2.5.4(1)Ao 二 5A 花为电阻时，A。最大,值为7A(3)花为电容时,Ao小，值为12.5.5令u = 220 30v2 =驚=11OCA12= R 4：)x1=ii60(a= 6280.0.01.1O-6

5、=159Ku广 R + j( 士)-R_j(mo3)P=UlcosP = 3734.7(w)2.5.6U = 220vsin (314t)V i*22Sin (314 * t - 45G)A, i2=llV2sin (314 * t + 90U)AA2表的读数I2=159uA Al表的读数I1=11V2U 比 I 超前45 所以 R二 10(Q) L=31.8mH1=11+12=11 A 读数为 11R=jlO(d)2.5.7(a) Zab=wc + U1 +j* 104* 10 4 ；= 1.5 + 1.5j(b) Zab=holoo-io65亠 O A / o fa r za 2.5.8(

6、a) 1=5-)5=l+ji(b)l=(3 + j4) - j4 * 30 30 - 40- 60 (A)2.S.9(A)Us=(-j5)*I=5-j*5(b) Us=5*Is+(3+4j)*I=130+j*752.S.10Uco解.h =乏=。5 VOAUAlc 一 - j2 一 J V 90 Ai=【i +【c=0.707v45AU=(2+j2)*I+Uc=j2+l2.5.11 令 Ubc=lO0 则 Ir=1V0A Ic=i 90A i=】R +【瓦。%UR=RI=5 2 1351 jXL=590CXL= 5 Zab=R+ jXL = 10 + J52.6.1f=2n1 L= 2nJ(0.3tl0 3 * 360 10 =480KhzV535KHz1fH=2n7(0.3 10 3 *25 10=1.84Mhz1605KHz频率可训范期为480至1840KH乙 能满足中波段535至1605KHZ的要求2n1000*L2.6.2(2n*5OO* L2tt500 C)2n - 1000 C2.6.3设 Z=R+jXI2*R + I1*R1 = PHI =