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（秦曾煌习题A选择题(2)(2)(J第222题2.1.12.2.12.2.22.3.1(1)2.3.2(3)2.4.1(2)2.4.2(3)2.4.3(2)2.4.4(1)2.5.1(2) (4)2.5.2(1)2.7.1(1)2.8.1(3)2.8.2(2)2.8.3(3)2.8.4(3)B基本题2.2.3U=220V, I 1= 10A1 2= 5V2Ai = 10v2sin (wt +90)A i 1 = 10 90A0oi 2= 10sin (wt - 45 )A i 1 = 5v2 - 45 A2.2.4 u = 220e30v u = 220 迓 sin (wt

2、 + 30)Vi = (- 4- j3)A= 5ei43i)j = 5v2sin (wt 143.1 )AU=220/2sin (wt)V U=220 62+( 314L)2=A = ?= 15.8(?)U382.4.8 1=27.7(mA)VR2+( 2 n fL)2 V2oo2+( 314743.3)R 2 000入严WL=314*1.65=518.12.4.9 W=2nf = 314rad/sU22|=.367(A)IZI V(28+2)2+581.1一 2 219Vu J 官=R1?| = 13V (JM=v2+ (518.1) ?.367U= 220vSsin (314t + 20)

3、(V)Uz= - = 3 + j ?4Q Req=3QU= 4.4v?sin (314t - 33) (V)Leq = 127(mH2.5.3(a)A10V2A (b)Vo = 80V(C)2A (d)10V2V (e)A= 10V2A V =100V2V2.5.4(1)A5A(2)Z2为电阻时，A。最大，值为7ARN=cos 1 = z = 0.6p=I Req =580.8(V)2Q= w LeqI = 774.4(var)11159KwC 628070.01 ?106 =U2RRU1R+J (wCR-J (-159?103)301597103tan 60 = ?=r115.5(k Q)U

4、=0.588(v)(3)Z2为电容时，Ao最小，值为12.5.54 U = 220 30%U22= R211 0A 12=u111 60AR+jx1= 1=11+12=11v30。P=Ulcos = 3734.7(W2.5.6 U= 220v2sin (314t)V0 i 仁 22 sin(314?t -45 )A i2= 11v2sin (314?t + 90)A A2 表的读数 12=159uA A1 表的读数 11=11 VU 比 I 超前 45。所以 R=10(Q) L=31.8mH257(吨二。wL令 Ubc=10f 则 I r= 1 0AI c=1 v9OA I =Ir+ Ic=v

5、20AUR=RI=5 V2 135oj Xl=590Xl = 5 Zab=R+ j Xl = 10+j512.6.1 f=2 n vLC1fL=480Khz535KHz2n V(0.3?10-3?360?10-12H=2n V(0.3?10-3?25?10*=1.84Mhz1605KHz124(b) Zab= 1 +j ?10 ?104R4?1=1.5+ 1.5i1-J4r10?100710t62.5.8(a) =aJ+J1(b)_J4 |”30 30 40 -60(A) J7?= v *2.5.9(A)Us=(-j5)*l=5-j*5(b)Us=5*ls+(3+4j)*l=130+j*75解：1 = u2c = 0.5 0Ai_= -5 v 90AI =l ! + I c=0.707 C