电工电子技术机械工业出版社习题答案王鼎王桂琴市公开课一等奖省赛课微课金奖PPT课件.pptx

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,习 题 精 选,第 一 章,第 二 章,第 三 章,第 四 章,第 五 章,第 七 章,返回,第1页,第 一 章,11,12,13,14,15,16,17,18,19,110,111,112,第2页,第 二 章,29,210,211,212,213,214,215,216,21,22,23,24,25,26,27,28,217,218219,220,第3页,第 三 章,31,32,33,34,35,36,37,38,39,310,311,312,313,314,315,316,第4页,第 四 章,49,410,411,412,413,414,415,416,41,42,43,44,45,46,47,48,417,418,419,420,421,422,423,第5页,第 五 章,51,52,53,54,55,56,57,58,59,510,511,512,513,514,515,516,第6页,第 七 章,71,72,73,74,75,76,78,79,710,711,第7页,11 确定元件电压电流实际方向，计算功率、指出元件时发出功率还是吸收功率。,N,1,10V,2.5A,N,2,60V,1.5A,实际方向：如,依照参考方向,P,N1,U I,10V2.5A,25W 发出,P,N1,U I,60V1.5A,90W 吸收,第8页,12 计算电流、电压。,+,+,60V,U,10,20V,+,60,u,200,U,40V,I,40V/104A,+,1A,U,2,I,10,4A,I,I,4A1A3A,U,3A（210）,36V,第9页,13,试求图中线性电阻两端电压；,各电源及线性电阻功率，并判断其性质。,（a）,解：,+,+,3V,u,2,1A,u,=3V,I,R,=,u,/,R,=3/2=1.5A,P,R,1.53=4.5W,P,Is,=13=3W,P,Us,=(1.51)3,=1.5W,（b）,解：,+,+,3V,u,2,1A,u,=12=2V,P,R,=(1)(2)=2W,P,Us,=13=3W,P,Is,=(3+,2)=5W,第10页,+,+,3V,u,1,2,1A,+,10,（c）,解：,u,1,=3V,I,R,=,u,/,R,=3/2=1.5A,u,2,=110=10V,u,2,P,R1,=1.53=4.5W,P,R2,=(1)(10)=10W,P,Us,=(1.51)3=1.5W,P,Is,=1(3+10)=13W,第11页,14 电压表、电流表都正偏，输出功率是,；电压表正偏，电流表反偏，输出功率是,。,A,1,A,2,V,A,电压表正偏：所测电压与电压表极性一致；,电流表正偏：所测电流由电流表正极流入负 极流出。,A,1,A,2,第12页,15 计算各支路电流。,2,2,1,1,1V,3V,2V,I,1,I,2,I,3,I,4,I,5,I,6,I,2,221A,I,3,313A,21,I,4,30,I,4,1A,1,I,4,2,I,1,10,I,1,1A,对a：,I,5,I,1,I,2,I,4,1A,对b：,I,6,I,3,I,4,I,1,5A,a,b,第13页,16,已知,U,S1,=6V，,U,S2,=12V，,I,S,=2A；,R,1,=2,，,R,2,=1,试计算,各电压源及电流,源功率，并判断其性质。,+,U,S1,I,S,R,1,+,U,S2,R,2,解：,U,R1,=,I,S,R,1,=4V,U,IS,=,U,R1,+,U,S1,=10V,U,R2,=,U,S2,U,S1,=6V,I,2,=,U,R2,/,R,2,=6A,I,1,=,I,S,+,I,2,=8A,I,2,I,1,P,US1,=68W=48W,P,US2,=6 12W=72W,P,Is,=102W=20W,U,IS,第14页,17 计算,R,0，,R,=时电流,I,与,I,E,。,I,E,7A,R,6,2A,4,3,I,12V,R,0,电路被分为两部分,6、3电阻两端电压为12V,I,12V62A,I,1,12V34A,I,E,（247）A,1A,I,1,R,=,I,1,12V34A,I,E,4A2A2A,I,7A2A4A5A,第15页,15A,12A,18 计算,I,、,U,S,、,R,。,U,S,1,6A,12,3,R,I,6A,9A,18A,3A,123V9,R,151V0,R,73,U,S,123V 183V0,U,S,90V,第16页,19 求图中A点电位及恒流源功率。,对A：,I,211mA,U,A,=,I,11V,A,1k,6V,1k,1k,1mA,2mA,I,U,1,U,2,U,2,121,I,0,U,2,3V,P,2,2(3)6mW,6,U,111,U,A,0,U,1,6V,P,1,166mW,第17页,3V,110 求图中a、b、c各点电位。,电路中存在两个独立电流，如图,10,a,1A,U,C,110 10V,4,10,b,c,2,+,5V,+,第18页,111 电压表内阻，S断开时，电压表读数12V；S闭合时，电压表读数11.8V，求,U,S,、,R,0,。,U,S,R,R,0,-,+,V,S,S断开时,U,S,U,开,12V,S闭合时,R,0,0.169,10,第19页,112,求S闭合和打开情况下A点电位。,A,+24V,6V,S,10k,10k,10k,10k,2V,I,=(24+6)/30=1mA,U,A,=2+1016=2V,解：,S打开:,S闭合:,I,=6/20=0.3mA,U,A,=2+36=5V,I,第20页,解：,+,10V,6,3A,4,I,1,I,2,I,1,I,2,+,3,=0,4,I,1,+6,I,2,10,=0,联解得,I,1,=0.8A，,I,2,=2.2A,P,R1,=40.8,2,W=2.56W,P,R2,=62.2,2,W=29.04W,P,Us,=10 0.8 W=8W,P,Is,=3(62.2)W=39.6W,21,用支路电流法求图中各支路电流，,及各电源及线性电阻功率。,第21页,+,6V,2,3A,1,1,2,22 用支路电流法求图中各支路电流。,I,4,I,2,I,1,I,3,a,b,解：,列KCL方程,a：,I,1,I,2,I,3,=0,b：,I,3,I,S,I,4,=0,联解得,I,1,=2.5A，,I,2,=0.5A,I,3,=,2A，,I,4,=1A,列KVL方程,：,R,1,I,1,R,2,I,2,U,S,=0,：,R,2,I,2,+,R,3,I,3,+,R,4,I,4,=0,第22页,电压源单独作用时,R,0,=6（4+2)k=3k,U,=12(36)(4 6)V,=4V,U,=6V+4V=10V,解：,+,12V,6k,3mA,23,用叠加原理求图中电压U。,4k,2k,3k,电流源单独作用时,R,0,=36+2 k=4k,U,=3(48)4V=6V,+,U,12V,+,第23页,+,-,35V,24,用叠加原理求图中电流,I,。,4,7A,2,3,1,I,电压源单独作用时,I,=35（34）A,=5A,I,5A3A8A,电流源单独作用时,I,=73(34)A=3A,第24页,+,125V,25 用叠加原理求图示电路中,I,。,解:,125V,单独作用时,120V,单独作用时,40,36,60,+,-,120V,60,I,I,=,I,+,I,=0.75A,第25页,26 已知,U,ab,=10V，去掉,E,后，,U,ab,=7V，求,E,=？,R,+,-,E,R,R,R,I,S1,I,S2,解：,依叠加原理，,U,ab,=10V是,E,，,I,S,1,，,I,S,2,共同作用结果。,U,ab,=7V是,I,S,1,，,I,S,2,共同作用结果。,a,b,设,U,ab,为,E,单独作用电压。,则,U,ab,=10V7V,3V,U,ab,=,E,R,4,R,=3V,E,12V,第26页,27,将以下各电路化为等效电压源,-,+,4,5,3A,5,15V,-,+,3A,4,4,4V,1A,4,2,4A,-,+,2,8V,第27页,28,将以下各电路化为等效电流源,-,+,2,2,8V,2,2A,2,2,4A,1,-,+,2,4V,2,第28页,1A,29,用等效变换法求图示电路中电流,I,。,2A,10,10,5,40,10V,+,I,20V,10,4A,+,10V,5,+,5V,I,=(105)V/(5+5+40),=0.1A,第29页,6V,-,+,8V,-,+,12V,-,+,210 用等效变换法求图示电路中电流,I,。,3,6,1,2A,2,1,I,2A,2A,4A,2,+,-,2V,I,(82)A/(2+1+3),1A,第30页,-,+,6V,4,3,211 用等效变换法求图示电路中电流,I,。,2,6A,I,1.5A,8,I,=（6+1.5）4V(48),=2.5A,第31页,+,-,150V,+,-,20V,212,用戴维宁定理求图中电流,I,。,+,-,120V,10,10,10,4,I,解：,令,R,开路,U,ab,=(20150+120)V,=10V,R,0,0,U,ab,I,U,ab,/(,R,0,+10)1A,第32页,+,-,2V,213,用戴维宁定理求图中电流,I,。,-,+,12V,I,解：,令,R,开路,a,b,U,ab,(23+8,+2)V,4V,R,0,6/3+3,5,I,U,ab,/(,R,0,+5),4V/10,0.4A,6,3,5,3,2A,U,ab,U,1,第33页,214,用戴维宁定理求图中电流,I,。,+,-,16V,1A,4,8,3,I,3,20,U,ab,解：,U,ab,1689813,1693,4V,R,0,(4+20)/83,9,168,I,14,I,1,20,I,2,0,I,1,I,2,1,I,1,98A,I,1,I,2,I,U,ab,/(,R,0,+3)1/3A,第34页,8V,-,+,12V,-,+,215 用戴维宁定理求图示电路中电流,I,。,4,4,2,2,I,2A,3A,5A,2,R,0,2/4/4,1,U,ab,U,ab,=（5/2）2,=5V,I,U,ab,/（,R,0,+2）,1.67A,第35页,2A,+,19V,3,2,I,X,216 求图示电路中电流,I,X,、,U,X,、,P,X,。,3A,1,9A,U,X,由KCL推广定律可得,I,X,9A,求电路戴维宁等效电路,U,OC,23+1923,7V,R,0,3+1+2)6,电路可等效为,+,-,7V,6,I,X,U,X,7V69V47V,P,X,U,X,I,X,947W423W （电源）,第36页,217 求图中,I,。,+,-,6V,+,-,4V,3,1,I,2A,6,4,2,2A,1A,应用等效变换法,2,4A,+,-,8V,4,I,=3 2/（12）A=2A,3A,2,第37页,-,+,5V,218,求S闭合和打开情况下电流,I,。,-,+,10V,5,I,2A,5,S,S断开，,105,I,1,5,I,50,I,I,1,+2,得,I,1.5A,I,1,S闭合，,105,I,1,5,I,0,I,I,1,+2,得,I,2A,第38页,-,+U,1,219 无源二端网络N,0,，,U,1,1V，,I,2,1A，时,U,3,0；,U,1,10V，,I,2,0A，时,U,3,1V；求,U,1,0，I,2,10A，时,U,3,？,N,0,I,2,U,3,由叠加原理可知,U,3,有,U,1,、,I,2,共同作用,设,U,3,K,1,U,1,K,2,I,2,有 0,K,1,K,2,110,K,1,得,K,1,0.1，,K,2,0.1,U,1,0，,I,2,10A，时,U,3,10,K,2,1V,第39页,220 无源二端网络N,0,，,I,S,1A，求,I,？,N,0,10V,I,S,N,0,2V,I,S,N,0,I,S,5,I,则其戴维宁等效电路为,U,开,10V,将N,0,、,I,S,看作有源二端网络,R,0,(21)2,I,(102+5)A1.43A,第40页,31,t,0,时，,S闭合后,u,C,及各电流初始值及稳态值。,u,C,(0,)1.5 4V 6V,在S闭合瞬间，依据换路定律有：,u,C,(0,)=,u,C,(0,+,)=6V,i,1,(0,+,),u,C,(0,+,)/4 =1.5A,i,2,(0,+,)=,u,C,(0,+,)/2 =3A,i,C,(0,+,)1.51.533A,S,C,i,2,i,C,u,C,i,1,1.5A,2,4,u,C,(,)1.5(4/2)V 2V,i,C,(,)0,i,1,(,)2/4A0.5A,i,2,(,)2/2A 1A,第41页,32 求开关闭合后初始值,及稳态值。,i,L,(0,)=124,3A,在S闭合瞬间，依据换路定律有：,i,L,(0,+,),i,L,(0,)=3A,4,i,1,(0,+,)+,u,L,(0,+,)=12,u,L,(0,+,)=4.8V,+,S,i,2,u,L,解：,i,1,4,6,i,L,12V,i,1,(0,+,)=3 6/（46）=1.8A,i,2,(0,+,)=3 4/（46）=1.2A,i,L,(,)12(4/6)5A,u,L,(,)0,i,1,(,)12/4 3A,i,2,(,)12/6 2A,第42页,33,电路中,已知,R1,=5k,R,2,=10k,C,=4,F,，,U,S,=20V,t,=0，打开S，求,u,C,(,t,)、,i,C,(t)、并画出改变曲线,。,S,u,C,R,2,R,1,U,S,i,C,C,+,电路为零输入响应,解:,u,C,(,0,+,)=,u,C,(,0,),=,U,S,R,2,/（,R,1,+,R,2,）,=2010/15=13.3V,=,R,2,C,=410,2,s,u,C,()=0,i,C,()=0,u,C,(,t,),=13.3e,25,t,V,i,C,(,t,),=1.33e,25,t,mA,u,C,(,t,),t,i,C,(,t,),-,1.33mA,13.3V,第43页S,C,u,C,i,1mA,10K,34,S打开0.2s后，,u,C,8V，试求,C,、,i,(0,+,)、,u,C,(,t,)。,解:,u,C,(,0,+,)=,u,C,(,0,)=,0,=1010,3,C,电路为零状态响应,u,C,()=201010V,u,C,(,t,),=10（1e,t,/,10000,C,）V,t,0.2，,u,C,=8V,u,C,(,t,),=10（1e,1/50000,C,）8V,e,1/50000,C,0.2，,C,12.5F,S闭合瞬间，,C,相当于短路,i,(0,+,)0,u,C,(,t,),=10（1e,8,t,）,第44页,35,电路中,已知,R,1,3,R,2,=10,R,3,=6,C,=10,F,，,U,S,=100V,t,=0，闭合S，,求,i,1,(,t,)、,i,2,(,t,)、,i,3,(,t,),。,S,i,3,R,2,R,3,R,1,U,S,解:,u,C,(,0,+,)=,u,C,(,0,)=,0,=(,R,1,/,R,3,+,R,2,),C,=1210,5,s,i,2,i,1,C,+,电路为零状态响应,i,1,()=,i,3,()=,U,S,/,(,R,1,+,R,3,)=11.1A,i,2,()=0,i,3,(t),=11.11.85e,8333,t,A,i,1,(t),=11.1+3.7e,8333,t,A,i,2,t),=5.55e,8333,t,A,第45页,C,u,C,I,S,36,已知,R,1,=6k,R,2,=2k,，,C,=20,F,，,I,S,=4mA，,U,S,=12V,t,=0，闭合S,，试求,u,C,(,t,)。,解:,u,C,(,0,+,)=,u,C,(,0,),=,I,S,R,2,=,8V,=(,R,1,/,R,2,),C,=310,2,s,9V,u,C,(,t,),=9e,33.3,t,V,R,2,S,U,S,+,R,1,第46页,37,电路中,已知,R,1,=4,R,2,=20,R,3,=6,C,=4,F,，,U,S,=50V,t,=0，闭合S，求,u,C,(,t,)、,i,t,)、并画出改变曲线,。,S,R,2,R,3,R,1,U,S,i,C,+,u,C,解:,=(,R,1,/,R,3,+,R,2,),C,=89.610,6,s,u,C,(,0,+,)=,u,C,(,0,)=,50V,u,C,()=,U,S,R,3,/,(,R,1,+,R,3,),=30V,-,0.896,i,(,t,),50,u,C,30,t,u,C,(V).,i,(A),i,(,0,+,),=(,U,S,/,R,1,),(,R,3,/,R,1,),u,C,(,0,+,),/(,R,2,+,R,3,/,R,1,),=0.896A,i,()=0,第47页,38,电路中,已知,R,1,=,R,2,=4k,R,3,=2k,C,=100,F,，,E,1,=10V,E,2,=5V,，,t,0，S由a打向b，求,u,C,(,t,)、,i,0,(,t,),。,解:,u,C,(,0,+,)=,u,C,(,0,),=,E,1,R,2,/,(,R,1,+,R,2,),=,5V,=(,R,1,/,R,2,+,R,3,),C,=410,1,s,S,R,3,R,2,R,1,E,2,i,0,C,E,1,u,C,(,)=,E,2,R,2,/,(,R,1,+,R,2,),=,2.5V,=1.56mA,i,0,(,)=,E,2,/,(,R,1,+,R,2,)=,0.625mA,a,b,第48页,S,C,u,C,i,2,2A,3,39,C,=0.5F，试求,i,1,(,t,)、,i,2,(,t,)。,解:,u,C,(,0,+,)=,u,C,(,0,),=,23V 6V,=(6/3),C,=1s,电路为零输入响应,6,i,1,i,1,(0,+,)=6/6 A=1A,i,2,(0,+,)=6/3A=2A,i,2,()=0,i,1,()=0,i,1,(,t,),=e,t,A,i,2,(,t,),=2e,t,A,第49页,310,电路中,已知,R,1,=6,R,2,=4,L,=10mH，,U,S,=10V,t,=0，闭合S，求,u,L,(t)、,i,L,(,t,),。,U,S,L,R,1,R,2,i,L,S,u,L,解:,=,L,/,R,2,=2.510,3,s,i,L,(,0,+,)=,i,L,(,0,),=,U,S,/(,R,1,+,R,2,)=1A,u,L,(,0,+,)=,-,R,2,i,L,(,0,+,),=,-,4V,i,L,()=0,u,L,()=0,第50页,311,电路中,已知,R,1,=3k,R,2,=6k,C,1,=40,F,，,C,2,=,C,3,20,F,，,U,S,=12V,t,0，闭合S，求,u,C,(,t,)，画出曲线,。,S,R,2,R,1,U,S,解:,u,C,(,0,+,)=,u,C,(,0,)=,0,=(,R,1,/,R,2,)C,=410,2,s,C,1,电路为零状态响应,C,2,C,3,u,C,C,=,C,2,/,C,3,+,C,1,u,C,(,)=,U,S,R,2,/,(,R,1,+,R,2,),=,8V,u,C,(,t,)=88e,25,t,V,t,u,C,(,t,),8V,第51页,I,S,312,已知,R,1,=2,R,2,=,R,3,=3,，,R,4,=6,L,=10mH,I,S,=1mA,U,S,=8V,t,0，断开S,，试求,u,L,(,t,)、,i,(,t,)。,解:,R,2,S,U,S,+,R,1,R,3,R,4,i,u,L,=0.001/（6+6/2)=1.3310,3,s,第52页,313,已知,R,1,=200,R2,=400,C,1,=0.1,F,，,C,2,0.05,F,，,U,=25V,t,=0，断开S，求,u,C1,(,t,)、,u,C2,(,t,)、,i,(,t,),。,S,R,1,R,2,U,解:,u,C1,(,0,+,)=,u,C1,(,0,),=,25400/(200+400),=50/3V,u,C2,(,0,+,)=,u,C2,(,0,),=,25200/(200+400),=25/3V,1,=,R,1,C,1,=210,5,s,2,=,R,2,C,2,=210,5,s,C,1,+,C,2,u,C2,u,C2,u,C1,(,)=,u,C2,(,)=25V,u,C1,(,t,)=2583e,50000,t,V,u,C2,(,t,)=2516.7e,50000,t,V,第53页,314,t,=0，闭合S，求,i,L,，画出曲线。,12V,1H,3,i,L,解,:,i,L,(0,+,)=,i,L,(0,),=12/(3+3)=2A,依等效变换,i,(0,+,)=(33/62),612/(3+,6),=1A,=,1/,3+(63),=0.2s,33V,S,a,b,i,L,(,t,),=3.81.8e,5,t,A，,i,(,t,),=0.21.2e,5,t,A,6,3,i,L,(,A,),t,2,3.8,1,0.2,第54页,315 图示电路在开关S闭合前电路已处于稳态，在,t,0时刻开关闭合。试求开关闭合后,u,C,(,t,)及,i,L,(,t,)。,+,t,=0,100,3k,3k,1.5k,1,F,i,L,u,C,90V,1H,第55页,S,u,C,i,2mA,3k,316,试求,u,C,(,t,)、,i,(,t,)。,解:,u,C,(,0,+,)=,u,C,(,0,),=,236V,1k,1,F,1k,+,6k,12V,=,1+(63)110,3,=310,3,s,u,C,(,t,)=82e,333,t,V,i,(,t,)=0.670.22e,333,t,mA,第56页,41,已知,i,=5.6sin(314,t,37)A,u,=311sin314,t,V。解：,U,m,=311V,I,m,=5.6A,u,=,i,=314 rad/s,f,u,=f,i,=,/2 =50Hz,T,u,=,T,i,=1/,f =,0.02s,u,=0,i,=37,u,.,i,(V.A),5.6,311,37,(2),=,u,i,=37,u,超前,i,37,(3),37,(4),第57页,42,已知,u,1,=100sin(,t,+,/,6)V,u,2,=60sin(,t,+,)V,试求,u=u,1,+,u,2,在,U,为,最大和最小时，,=？,U,m,=？,解：,当,=,/,6时，,U,m,最大，,U,m,=100+60=160V,当,=,-,5,/,6时，,U,m,最小，,U,m,=10060=40V,30,30,第58页,43,已知电路中某元件,u,和,i,，试问是什么元件、元件参数、储能最大值、平均功率。,(1),u,=100sin(,314,t,)V,i,=10cos(,314,t,)A,=100sin(314,t,+180)V,=10sin(314,t,90)A,u,超前,i,90，元件为电感,X,L,1001010,Q,100102500Var,P,0,(2),u,=282sin(,314,t,+60)V,i,=56.6cos(,314,t,30)A,=56.6sin(314,t,60)A,u,、,i,同相，元件为电阻,R,28256.65Q,0,P,28256.627980.6W,(3),u,=70.7cos(,314,t30)V,i,=5sin(,314,t30)A,=70.7sin(314,t,+60)V,=5sin(314,t,150)A,U,滞后,i,90，元件为电容,X,C,70.7514.14,Q,70.752176.75Var,P,0,第59页,解：,X,L,2,f,L,8,Z,6+j81053.13,I,U,/,z,22A,44 将电感,L,25.5mH，,R,=6线圈接到,f,=50Hz，,U,=220V电源上，求,X,L,、,Z,、,，画相量图,。,R,U,L,U,I,U,53.13,设,第60页,C,u1,u,2,R,45,R,100，,u,1,f,500Hz，要求输出信号,u,2,比输入信号,u,1,超前60，求,C,？,R,X,C,z,60,X,C,=,|,R,tan,|,=,173.2,C,12,f X,C,=1.84F,设 为参考正弦量,则 初相角为0,则 初相角为60,第61页,46,已知,i,=5.6sin（314,t,17)A，,u,=314sin（31,4t,+20),V，试求串联等效电路。,i,+,u,N,解：,串联等效电路,R,=,z,cos37=44.7,X,=,z,sin37=33.7,R,=70.2,X,=93.2,并联等效电路,第62页,1,Z,2,Z,3,Z,47 已知,İ,S,20A,Z,1,=j5,，,Z,2,=(2+j),，,Z,3,=(3+j4),，,求,各个电流及各个电压,。,按分流公式：,第63页,R,1,X,L,X,C,İ,İ,1,İ,2,45,作相量图：,İ,1,=1090A,İ,=,İ,1,+,İ,2,100A,U,ab,=,U,S,U,1,50V,a,b,48 图中,=314rad/s,U,S,=100V，,R,1,=5,，,I,1,=10A，,I,2,10 A，,R,L,=,X,L,，求,I,、,R,L,、,L,、,C,。,R,L,İ,2,=10 45A,R,L,=,X,L,z,L,=,U,ab,/,I,2,=3.5,R,L,X,L,=2.5,L,=2.5/314,=7.96mH,C,=1/5314=0.637mF,设 为参考正弦量,第64页,49 电路如图，标出各表读数。,A,1,L,C,A,2,A,0,3A,5A,i,1,、,i,2,反相,I,0,I,1,I,2,532A,V,1,R,C,V,2,V,0,10V,10V,第65页,L,C,A,1,A,0,10A,R,V,1,100V,5,5,10,İ,1,=1090,İ,R,=10 45,İ,0,=,İ,1,+,İ,R,=100,V,0,第66页,R,L,C,V,1,V,2,V,3,V,5,V,4,V,6,410 电路如图，R=X,L,5,，X,C,10,，V,1,读数为10V，,求各表读数。,I,=V,1,/R=2A,V,2,IX,L,10V,V,3,IX,C,20V,V,4,V,3,V,2,10V,+,C,L,U,U,R,U,C,U,L,U,I,第67页,411,已知U=220V,U,1,=120V,U,2,=130V,f,=50Hz,R,1,=50,求R,L,和L。,L,U,U,2,R,1,U,1,R,L,解：,I=U,1,/R,1,=2.4A,z,=U/I=220/2.4,=91.67,I,由已知可列方程组：,z,2,=(R,1,+R,L,),2,+X,L,2,解得:R,L,=29.69,X,L,=45.3,L=X,L,/2,f=,0.144H,第68页,1,Z,Z,2,Z,412 已知,120,Z=（+j）,，Z,1,=(3+j4),，Z,2,=(44j),，,求,各个电流及各个电压,。,第69页,413,已知,u,=2202sin314,t,V,R,1,=3，,R,2,=8，,L,=12.7mH,C,=531F,求,i,1,、,i,2,、,i,，cos,、,P,、,Q,、,S,画相量图。,R,1,L,u,i,C,i,2,i,1,R,2,解：,Z,1,=jX,L,+R,1,=j4+3,=553.13,Z,2,=R,2,jX,C,+=8j6,=10 36.86,X,C,=,6,X,L,=4,o,o,o,Z,1,U,I,1,1353,44,13,.,53,5,0,220,-,=,=,=,o,o,o,Z,2,U,I,2,86,.,36,22,36.18,10,0,220,=,=,=,第70页,İ,=,İ,1,+,İ,2,=44j22=49.226.56,cos,=,cos26.56,=0.9,P=UIcos,=,9741.6,W,Q,=UIsin,=,4840.6,Var,S=UI=10824,VA,第71页,414 已知U=220V，I=4A，P160W,f,=50Hz，试求R、L。,R,L,u,W,A,0.18,=79.5,U,R,=Ucos,=,40V,RPI,2,10,X,L,=Rtan,=,54,LX,L,2,f,=0.172H,第72页,1,Z,R,2,Z,415 已知,İ530,R=2,，Z,1,=j10,，Z,2,=(40+30j),，,求,İ,1,、İ,2,、U。,第73页,416 图示正弦交流电路中,U=220V，U与I同相，R=X,C,，求U,L,、U,C,。,R,X,L,X,C,İ,45,作相量图：,U,L,=U=220V,第74页,R,1,L,u,i,R,R,2,i,2,i,1,设İ,=100,按分流公式：,5,8,2,4,=453.13,A,=8.2522.83,A,=81.429,V,P=UIcos,9,=81.42100.988=804.2W,第75页,R,1,L,u,C,i,2,i,1,R,2,418 电路如图，R,1,=5,R,2,=3,X,L,2,，X,C,3,，V读数为45V，,求A表读数。,A,V,I,2,=U,R,2,=15A,设İ,=150,İ,=,İ,1,+,İ,2,=22.4628.9,I22.46A,第76页,R,1,L,u,C,i,2,i,1,R,2,419 R,1,=R,2,=30,X,L,X,C,30,，,a b,i,İ,=,İ,1,+,İ,2,=20,第77页,1,Z,2,Z,i,2,i,1,i,u,420 u=220 sin314t V，总S=5KVA，Z,1,cos,1,=,1,，,Z,2,cos,2,=,0.5,，电路总,cos,=,0.8，试求P,1,、P,2,。,PScos,=,4KW=P,1,+P,2,QSsin,=,3KVar=Q,2,P,2,UI,2,cos,2,=,1731.4W,P,1,=PP,2,=2268.6W,第78页,421,i,R,=2sintA,R=200，L=0.1H,C=2.5F,求,i,1,、,i,C,、,u,R,、U,L,、U,S、,cos,、P、Q、S,画相量图。,R,X,L,X,C,İ,1,X,C,=,200,X,L,=200,İ,R,=10,İ,C,=190,cos,=cos45,=0.707,P=UIcos,=,200,W,Q,=UIsin,=,200,Var,S=UI=282.84,VA,第79页,R,1,X,1,u,X,C,i,i,1,R,422 图示正弦交流电路中,U=120V,İ,1,超前于İ相位90,R+jX=11+j8,R,1,+jX,1,=30+j50,求X,C,。,X,1,=(90 36)=54,X,C,=50（41.3）91.3,第80页,423,已知R=10，L=0.12mH,C=600PF,外加电压10mV，试求：,f,0,、Q、Z、I、U,R,、,U,L,、U,C,、P.,i,u,R,L,C,解：,z,=R=10,I=U/R=1mA,U,R,=U=10mV U,L,=U,C,=QU=447.1mV,P =UI=10,W,第81页,51,对称三相负载，Z=17.32+j10,，,额,定电压U,N,=220V，三相四线制电源，线电压,u,UV,=380,2sin(314t+30)(V)，负载怎样,接？求线电流，画相量图。,解：,应采取星形联接，相电压,U,U,=220V,，,满足,额定电压需要。,30,30,第82页,52 有日光灯120只，每只功率P40W，额定电压U,N,220V，cos,N,=0.5,，电源是三相四线制，电压是380220V，问日光灯应怎样接？当日光灯全部点亮时，相电流、线电流是多少？,解：,电源线电压是380V相电压是220V,将120只日光灯均匀接在三相上，每相40只（并联）,I,L,=,I,P,=,P40,U,P,cos,110,1600,=,=,14.54 A,第83页,53,Z=26.87+j26.87,，,U,N,=380V，相电压u,U,=2202sin(314t30)(V)，负载怎样,接？求相电流和线电流，画相量图。,解：,应采取角形联接，线电压,U,UV,=380V,，,满足,额定电压需要。,第84页,54 三相绕组角形联接对称三相电路。已知U,L,=380V,I,L,=84.2A。三相总功率P=48.75KW,求绕组复阻抗Z。,z,U,L,/I,P,=7.82,=28.36,Z7.82 28.36,第85页,Z,Z,Z,Z,UV,İ,U,İ,V,İ,W,55 电源U,L,=380V，对称负载Z38.1+j22,Z,UV,9.8+j36.7。求İ,U,、İ,V,、İ,W,。,İ,UV,第86页,Z,Z,Z,İ,U,İ,V,İ,W,56 电路如图，电源 ，负载Z,U,10,Z,V,6j8,Z,W,8.66j5。求İ,U,、İ,V,、İ,W,、İ,N,。,İ,N,İ,N,=İ,U,+İ,V,+İ,W,14.838.9,第87页,57 对称三相负载Z,L,50+j28.9星形联接，已知U,L,=380V,端线阻抗Z,L,2+j1,中线阻抗Z,N,1j。负载端电流和线电压，并画相量图。,第88页,U,V,W,Z,58,S闭合时，电流表读数7.6A，,求S打开时各电流表读数（设三相电压不变）。,A,A,A,S,S打开,I,L,7.6A,第89页,59,电压表读数220V，负载Z=8+j6,，,求电流表读数，cos,，P。,解,：,电源是星形联接，负载是三角形联接,A是负载线电流，V是电源相电压。,10,I,AB,=,I,P,=,U,L,z,=,380,=,38 A,A,U,P,=220 V U,L,=,3 U,P,=380V,=I,L,=I,U,=,3,I,P,=,38,3,=,65.8A,Z,Z,Z,V,A,cos,=cos(tan,1,6/8)=0.8,P=3U,L,I,P,cos,=34.656KW,第90页,510,负载Z,1,=19.1+j11,，,Z,2,=32.9+j19,，,U,L,=380V,求电流表,İ、İ,1,、İ,2,。,A,B,C,İ,Z,1,Z,2,İ,2,İ,1,解：,设,U,AB,=380,0,为参考正弦量,第91页,511 负载额定电压为220V，现有两种电源：U,L,=380V、U,L,220V，应怎样连接，对称负载R24,X,L,18。分别求相电流、线电流。画出U,L,=380V时全部相量图。,U,L,=380V，采取Y接,U,L,=220V，采取接,30,36.87,第92页,512 一台三相交流电动机，定子绕组星形联结，额定电压380V，额定电流2.2Acos,=0.8,，求每相绕组阻抗。,U,L,380V，IL,I,P,2.2A,U,P,220V,z,U,P,/I,P,=100,=36.87,Z100 36.87,第93页,513,电流表读数32.9A，负载Z=12+j16,，,求电压表读数,。,解,：,电源是星形联接，负载是三角形联接,A是负载线电流，V是电源相电压。,Z,Z,Z,V,A,A,=I,L,32.9A,V,第94页,514,星接负载Z=30.8+j23.1,，,电源线电压380V，求cos,，P、Q、S。,38.5,I,U,=,I,P,=,U,P,z,=,220,=,5.7 A,U,L,=380V U,P,=220 V,cos,=cos(tan,1,23.1/30.8)=0.8,P