电工技术参考答案

电工技术作业答案作业1答案1-1E-RI-U=0o£=0.6x5+230=2337P=£/=233x5=U65W(发出)EP=hR=25x0.6=15W(吸收)宠0°P=230x5=1650W(吸收)u尸=P+P(功率平衡)该支路吸收电能ERqUp7001-2(1)P=UI/=200=4AR=°==12.5。NNNN50N1216(2)U=50+0.5x4=52V52⑶I=_=104ASho"0.51-3(a)U=5x40=4VI=-5=-0AA10+4050(b)U=lx(10+40)=507I=IA1-4U=7—(—16)—14=9VAB/18u…1-5S断开时U=6-x5=-4VTOC\o"1-5"\h\za9S闭合时U=6X5=\Va5+1作业2答案2-1I+1+1=0123—£+(R+R)/—(R+R)1+E=01101120222—E+(R+R)I—(R+R)1+E=Q11()1130333

2211代入数据解方程可得，1=4/=/=—A132332-2—£+(/+/)R+/R=0S3233U-IR+1R=0

5133代入数据解方程可得/=-0.5AU=3.5V32-3略2-4/=0A2-5该电路中只有两个结点，选左边结点为参考结点，结点电压方程如下:EEE1+-2+-3U二101010111++EEE1+-2+-3U二101010111++10101010一440310-V再根据KVL可得三个电流/=12-6将10。支路断开，左边回路电流为1015-13=一11Ai23311A3=1A2"'右边回路电流为U=13+1-1x2=1270C1x11x1+1+11+1=1Q12=%io+7Ti2-7(a)R=1Q

eqUoc=1+10=117(b)32eq4+8123=4+2=54sc4作业3答案3-1X=3£=314x1.59=500Q差？=0.44A500z=0.44V2sin(314r-90°)X106=100。cWC314x31.81063-2x==/X.ccoC31.8x314220=1°°Q/=亚一=2.2Ai=22j2sin(314z+90°)c3-3Z=1.5+/X=1.5+/2KQL吃匕35212.5AQcos<p=f0.62.5P=U/cos(p=380x0.152x0.6=34.6561VQ=U/sin(p=380x0.152x0.8=46.2VARS=ui=380x0.152=57.76E4Z=R-jXZ=R-jXZ=R-jX3-4X=」_=_」?6__=31.80ccoC314x100Z=R-jX220二百_220-31.8x^2=4.9A)ZXZ

3-5Z+U231Z+Z

23-5000=；10+j33.3Q=J43.3J150_220Z0o一丽5=5.08/—90。AZ=Z=口3/z+z23z=□2z+z23I=3.39/—90°A71501J50•=口/,=1.69/—90。A/15013-63-7A=50mAc=—(氏4-W)=

1/2C012——竺V3一次(arccos0.96^=3.792202x314作业4答案4-1略4-2略4-3相电压:U=380/邪=220电流和线电流:〃=/=220/^62+82=22AP14-4⑴各相电流分别为：/=220/10=224I=220/20=114UVI=220/30=7.334W中线电流为三个相电流的和：I=16.8AN(2)(2)(2)U=0UN'I=0U(2)U=0UN'I=0U=220V=UUN'VN'WN'I=220/20=114V(3)U相负载的电压和电流均为0,V相负载的电压为U=380/50x207=152VVNyV相负载和W相负载的电流均为：19AW相负载的电压为U=380/50x3()V=228VU相负载电压为0,V和W相负载电压均为380VV相负载电流为380/20=19A,W相负载电流为：380/30=12.67A,U负载电流为二者的相量和19/3Ao4-5(1)15A(2)15A(23)22.54-6p=y/3UIcos(p=1.73x380x17.3xco钳=4501U可得cos(p=0.395ii|Z=380/17.3/J?=38Q=JR2+X2X/K=物)=2.33可解得415。，

*LLX=35。L4-7星形连接时：p=y[3UIcos^=1.73x380x22=14462.8Wii三角形连接时：p=j3UIcos(p=1.73x380x38x小=43388.4Wii作业5答案5-1答案略5-2答案略5-3答案略5-4(a)5-4(a)无过渡过程5-4(a)无过渡过程5-4(a)无过渡过程因为S动作后C的能量不发生改变。(b)无过渡过程因为S动作后L的能量不发生改变。(c)有过渡过程因为S动作后L的能量发生改变。(e)有过渡过程因为S动作后C的能量发生改变。5-5U(0)=U(0)=2VUQ)=0工=lx2=2STOC\o"1-5"\h\zc+c-c-tdu(t)_tu(t)=2e2vi(£)=c_c_=-eiAcdt6391295-6i(0)=--Ai(0)=i(°)=，(oo)=_.Ai(co)=/+7l+l--j77l74T-L-312u1———QR7/47eq、1218u.qioz(t)=_「一巴2AI(0=2—77l775-7可以看作两个过渡过程的叠加：对左边回路：i(0=0.5+e-4()tAs

对右边回路：i(0=1.5e-iootA

si(t)=O.5+e-4ot+1.5e-iootAs作业6答案6-16-16-1(|)=BS=lx20x10-4=2x10-3Wbmm2206-1220220二495220二495匝4.44力4.44

m4.44力4.44

4.44力4.44

mx50x2xl0-3I=100/40=2.2542I=100/40=2.254I=100/40=2.2542I=2.25/5.5=0.41A6-26-2N=10000/60=166IINN'=m0Q=3.33AI6-2N=10000/60=166IINN'=m0Q=3.33AI30002n10000x0.45-c二1124010000=45.5A220(1)152x450=307I=包蚂=227.34]=2蹩220in220zn150(1)15(1)152x450=307I=包蚂=227.34]=2蹩220in220zn150333.3A6-46-4Q)叫=0.32同名端：1-3-6及2-4-5(1)副边可输出110V,6-4Q)叫=0.32同名端：1-3-6及2-4-5(1)副边可输出110V,及220V两种电压。(2)可以，接线方法为：把3和6接到一起，从4、5两端输出。作业7答案7-1略7-2略7-3略7-4略7-51=35000IJ3x380x0.85x0.88=71.1AP=35/.85=41.2KWS=20/3000=0.00677-6⑴

Q)

⑶

(4)T=9550X55/1480=354.9N・mNS=20/l500=0.003T=2.2x354.9=780.8N・mmaxT=2x354.9=709.8Ne