電工學簡明教程第二版答案第二章

1、電工學簡明教程第二版習題A選擇題2.1.1(2)2.2.1(2)2.2.2(1)2.3.1(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)(秦B基本題秦曾煌第222題2.2.3U=220V,I1 = 10A, I2 = 5v2AU=220V2sin(wt)VU = 220 O0Vi1 = 10v2sin(wt+ 900)A i1 = 10 90Ai2 = 10sin(wt- 450)A i1 = 5v2 62+( 314L)2= ?= 1

2、5.8(?)380V2002+( 314?43.3)啟c。1 R =200013742.5W=2 n = 314rad sU 燈管=R1?I = 103VU= 220V sin(314t+ 200)(V)U= 4.42 sin(314t- 33)(V)=27.7(mA)wL=314*1.65=518.1,U220I=|Z|/21 128+ 20) +581.10.367(A)U鎮(zhèn)=V202 + (518.1)2? 0.367= 190Vu燈管+ u鎮(zhèn)=203220uZ= ; = 30 + j ? 40 QReq = 30 QR2= cos(1= - = 0.61wCP=l2Req= 580.8

3、(W)Q= wLeqI2 = 774.4(var)U2 _ R _R1 =3U1 R+j(- wC)R- j(- 159?103)2.5.3tan6O03159?103=r?= 1155( k Q)U2 =R/ 2 1 2Vr2+( w_)wCU1=0.588(v)(a)A0 = 10v2A(b) V0= 80V(C)A0 = 2A (d) V0 = 10v2V(e) A0 = 10V2AV0 = 1002V2.5.4(1) A。= 5A (2)z2為電阻時，A 0最大，值為7A(3)Z2為電容時，a 0最小，值為12.5.5令u = 220 30v,I2=S=1100Al2= =11600A

4、I=I1+I2=11 v3 - 300P=Ulcos= 37347(w)2.5.6U = 220v2si n(3141)Vi1=22sin(314? t - 45D)A, i2=11v2s in (314?t+92)AA2表的讀數(shù) l2=159uA A1表的讀數(shù) I仁11 v2U 比 I 超前 45 所以 R=10(Q) L=31.8mHI=I1+I2=11 A 讀數(shù)為 112.5.7R(a) Zab= 1 = - j10( Q)jwc+ jwLR(b) Zab=1+ j?104?10_4? 二 =1.5 + 1.5jj104?100?10- 625.8(a)匸詈2?R + l1?R1= P

5、且 I = 8 所以 R=10VR2 + (X)2 ?8 = 113X= + 10QZ=10：j10Q+j1(b)匸冊？30 3(f = 40 - 6O0(A)2.5.9(A)Us=(-j5)*l=5-j*5(b) Us=5*ls+(3+4j)*l=130+j*75解：IU=0500A 1。=存0.59弘I =I1 + I =0.707450ACU=(2+j2)*I+Uc=j2+1令 Ubc=100則 lR = 1 (fAIC=1 900AI =Ir+ Ic=遷00aUR=RI=5 v2 1353 jXL=590XL = 5 Zab=R+ jXL = 10+ j52.6.1fL =12 鈕0.3?10-3?360?10-祀=