电工与电子技术 习题答案 第4章习题解.docx

(第4章习题解)选IMR41A4,2C4,3D4.4A4,54.6C4.7C4.8B计算题4.9斛<l)l=10Z60oA,Z2=5Z-45oA相量图见解图4-1而示。<2>/；+2=10cos6(F+j!0si1160o+5cos(-450)+j5sin(-45o)=5+j8.66+3.35-j3.35=8.35+j5.3l<>=9.9Z32o=9.92sin(wz+32。)A/-/.=IoCoS60+j10sin60o-5cosf-45o)-j5sin(-45o)=5+j8.663.35+j3.35=1.65+jl2.0l=12.IZ82o=12.h2sin(/+82。)A<3>/=0.75ms时il=IoTisinWr+60°)=l(2sin(0.7511+)A=-3.66A4.10 解：%=10(h2sin(314/+60o)VM2=3OIsin(314/-30o)V4.11 解；A=-3+j4=5/-53。、A=4+H=5.66/45。4+=-3+j4+4+j4=l+j8A-B=-3+-4-7A×B=5×5.66Z-80=28-j3.9A÷B=六Z-980=-0.12-j.874.12 解：所求正弦量为u=V3O2+40?v2sin(<yr-)1.30=5O2sin(-53.1o)V4.13 解：电流表读数为两电流相汆和l+i-,=6>2sin(ftX+3O3)+4>2sin(ftir+600)6cos3(P+j6sin30o+4cos6(F+j4sin600=5.196+j3+2+j3.464三7.196+j6.464=9.67A电流表读数为9.67A4.14 解：/=50Hzit=HR+j(X1.-XC)=/(300+j(314l5-)314x5XIO6=;(3(X)-j323)=44IZ-47°=44l2sm(314/-47o)V4.15ru22OZ3(TZ=-=i2.2Z-23.1o=100N53.1。=100cos53.F+jl(X)sin53.10=(60+j80)4.16解：<1>U=8/30Z2JZ45=27?4-152(感性)P=UIcosO=822<o*(45n)=16W<2)Z=-=-llOZ165a=IIOZ-I5u=(lO6-j28.5)(容性)R=IO6,X=28.5P1.'lsiR=-220J>s(-15rl=425W4.17解：P=UIcos=1105sif>i>-400W400cos=-,©=43.3。+5。Z=?=5/0°=22Z43.=U+jl5R=15=4XmHX211f4.18 解：未知电表的读数为U=IOZOo+10Z90o+IOZ-90o=IOZOo读数为IOV4.19 解：图a)RC支路：Z1=IO-jl=1(血N-45°KA支路：Z2-10÷jl-l<2Z450Z1+7.20图b1-j三-45°1+j-2Z4507Z-45o×2Z45oZ=I4.20解：VP=ucos.=60.P220C/.1=2AUCOS6022O×O.5z=¥=aa0'60°=110Z60°=(55+j95.2)IJZO0R=55.X95.2.1.=H=0.3H2rf314将功率因数提高到1.需井联电容为：PC=(ta11>1-IanM(M220=三J(tan600-lan0o)3I4x2202=25F