电子电工学习题答案1教学提纲.ppt

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 一 章,11,12,13,14,15,16,17,18,19,110,111,112,第 二 章,29,210,211,212,213,214,215,216,21,22,23,24,25,26,27,28,217,218,219,220,第 三 章,31,32,33,34,35,36,37,38,39,310,311,312,313,314,315,316,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 五 章,51,52,53,54,55,56,57,58,59,510,511,512,513,514,515,516,第 七 章,71,72,73,74,75,76,78,79,710,711,11 确定元件电压电流的实际方向，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,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,13,试求图中线性电阻两端电压；,各电源及线性电阻的功率，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,+,+,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,14 电压表、电流表都正偏，输出功率的是,；电压表正偏，电流表反偏，输出功率的是,。,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,2,15 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,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,源的功率，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,17 计算,R,0，,R,=时的电流,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,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,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,1,11,U,A,0,U,1,6V,P,1,166mW,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,+,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,0,0.169,10,112,求S闭合和打开情况下的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,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的功率。,+,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,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,+,+,-,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,+,125V,25 用叠加原理求图示电路中的,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,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,共同作用的结果。,U,ab,=7V是,I,S,1,，,I,S,2,共同作用的结果。,a,b,设,U,ab,为,E,单独作用的电压。,则,U,ab,=10V7V,3V,U,ab,=,E,R,4,R,=3V,E,12V,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,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,2,1A,29,用等效变换法求图示电路中的电流,I,。,2A,10,10,5,40,10V,+,I,20V,10,4A,+,10V,5,+,5V,I,=(105)V/(5+5+40),=0.1A,6V,-,+,8V,-,+,12V,-,+,210 用等效变换法求图示电路中的电流,I,。,3,6,1,2A,2,1,I,2A,2A,4A,2,+,-,2V,I,(82)A/(2+1+3),1A,-,+,6V,4,3,211 用等效变换法求图示电路中的电流,I,。,2,6A,I,1.5A,8,I,=（6+1.5）4V(48),=2.5A,+,-,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,+,-,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,1,214,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,1,4,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,8V,-,+,12V,-,+,215 用戴维宁定理求图示电路中的电流,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,2A,+,19V,3,2,I,X,216 求图示电路中的电流,I,X,、,U,X,、,P,X,。,3A,1,9A,U,X,由KCL推广定律可得,I,X,9A,求电路的戴维宁等效电路,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）,217 求图中的,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,-,+,5V,218,求S闭合和打开情况下的电流,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,-,+,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,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,31,t,0,时，,S闭合后,u,C,及各电流的初始值及稳态值。,u,C,(0,)1.5 4V 6V,在S闭合的瞬间，根据换路定律有：,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,32 求开关闭合后的初始值,及稳态值。,i,L,(0,)=124,3A,在S闭合的瞬间，根据换路定律有：,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,33,电路中,已知,R,1,=5k,R,2,=10k,C,=4,F,，,U,S,=20V,t,=0，打开S，求,u,C,(,t,)、,i,C,(t)、并画出变化曲线,。,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,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闭合的瞬间，,C,相当于短路,i,(0,+,)0,u,C,(,t,),=10（1e,8,t,）,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,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,1,37,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,)、并画出变化曲线,。,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,()=0,38,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,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,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,()=0,311,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,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,313,已知,R,1,=200,R,2,=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,)=258.3e,50000,t,V,u,C2,(,t,)=2516.7e,50000,t,V,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,2,3.8,1,0.2,315 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。试求开关闭合后的,u,C,(,t,)及,i,L,(,t,)。,+,t,=0,100,3k,3k,1.5k,1,F,i,L,u,C,90V,1H,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,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),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,30,30,43,已知电路中某元件,u,和,i,，试问是什么元件、元件的参数、储能的最大值、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,28256.65,Q,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,解：,X,L,2,f,L,8,Z,6+j81053.13,I,U,/,z,22A,44 将电感,L,25.5mH，,R,=6的线圈接到,f,=50Hz，,U,=220V的电源上，求,X,L,、,Z,、,，画相量图,。,R,U,L,U,I,U,53.13,设,C,u,1,u,2,R,45,R,100，,u,1,的,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 的初相角为0,则 的初相角为60,46,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,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：,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,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,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,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,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,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,。,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,1,13,.,53,44,13,.,53,5,0,220,-,=,=,=,o,o,o,Z,2,U,I,2,86,.,36,22,36.18,10,0,220,=,=,=,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,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,1,Z,R,2,Z,415 已知,İ530,R=2,，Z,1,=j10,，Z,2,=(40+30j),，,求,İ,1,、İ,2,、U。,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,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,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,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,=20,1,Z,2,Z,i,2,i,1,i,u,420 u=220 sin314t V，总的S=5KVA，Z,1,的 cos,1,=,1,，,Z,2,的 cos,2,=,0.5,，电路总的,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,421,i,R,=2sin2000tA,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,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）91.3,423,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,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)，负载如何,接？求线电流，画相量图。,解：,应采用星形联接，相电压,U,U,=220V,，,满足,额定电压需要。,30,30,52 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，问日光灯应如何接？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,53,Z=26.87+j26.87,，,U,N,=380V，相电压u,U,=2202sin(314t30)(V)，负载如何,接？求相电流和线电流，画相量图。,解：,应采用角形联接，线电压,U,UV,=380V,，,满足,额定电压需要。,54 三相绕组角形联接的对称三相电路。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,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,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,14.838.9,57 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。负载端的电流和线电压，并画相量图。,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,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是负载的线电流，V是电源的相电压。,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,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,511 负载额定电压为220V，现有两种电源：U,L,=380V、U,L,220V，应如何连接，对称负载R24,X,L,18。分别求相电流、线电流。画出U,L,=380V时的全部相量图。,U,L,=380V，采用Y接,U,L,=220V，采用接,30,36.87,512 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,，求每相绕组的阻抗。,U,L,380V，I,L,I,P,2.2A,U,P,220V,z,U,P,/I,P,=100,=36.87,Z100 36.87,513,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是负载的线电流，V是电源的相电压。,Z,Z,Z,V,A,A,=I,L,32.9A,V,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=3U,P,I,P,cos,=3KW,Q=3U,P,I,P,sin,=2.3KVar,S=3U,P,I,P,=3.77KVA,515 电源线电压为220V，对称负载R4,X,L,3。求线电流、cos,，P、Q、S,。画出电压相量图。,U,WV,U,L,=220V U,P,=127V,cos,=cos(tan,1,3/4)=0.8,P=3U,P,I,P,cos,=7.74KW,Q=3U,P,I,P,sin,=5.8KVar,S=3U,P,I,P,=9.68KVA,30,36.87,516 星接电源，角接负载，线电流25.4A，负载P7750W，cos,0.8，,求U,L,、,U,P,、S,、R、X,L,。,U,WV,S=3U,L,I,P,=9.68KVA,z,U,L,/I,P,=15,电源U,P,127V,R=,z,cos,=,12,X,L,=,z,sin,=,9,71 一台三相异步电动机额定转速，电