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1、 第第 一一 章章 第第 二二 章章 第第 三三 章章 第第 四四 章章 第第 五五 章章 第第 七七 章章返回返回Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.第第 一一 章章v 11v
2、12v 13v 14v 15v 16v 17v 18v 19v 110v 111v 112Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.第第 二二 章章v 29v 210v 211v 212
3、v 213v 214v 215v 216v 21v 22v 23v 24v 25v 26v 27v 28v 217v 218v 219v 220Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.
4、第第 三三 章章v 31v 32v 33v 34v 35v 36v 37v 38v 39v 310v 311v 312v 313v 314v 315v 316Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose P
5、ty Ltd.第第 四四 章章v 49v 410v 411v 412v 413v 414v 415v 416v 41v 42v 43v 44v 45v 46v 47v 48v 417v 418v 419v 420v 421v 422v 423Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Pr
6、ofile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.第第 五五 章章v 51v 52v 53v 54v 55v 56v 57v 58v 59v 510v 511v 512v 513v 514v 515v 516Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Cl
7、ient Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.第第 七七 章章v 71v 72v 73v 74v 75v 76v 78v 79v 710v 711Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.C
8、opyright 2004-2011 Aspose Pty Ltd.11 确定元件电压电流的实际方向，计算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。功率、指出元件时发出功率还是吸收功率。N110V2.5AN260V1.5A实际方向：如实际方向：如依照参考方向依照参考方向PN1U I10V2.5A 25W 发出发出PN1U I60V1.5A 90W 吸收吸收Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose P
9、ty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.12 计算电流、电压。计算电流、电压。+ +60VU1020V+60u200U40VI40V/104A+1AU2I104AII4A1A3AU3A（210） 36VEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011
10、 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.13 试求图中线性电阻两端电压；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。各电源及线性电阻的功率，并判断其性质。（a） 解：解：+3Vu21Au=3VIR=u / R=3 / 2=1.5APR=1.53=4.5WPIs=13=3WPUs=(1.51) 3 =1.5W（b） 解：解：+ 3Vu21Au=12=2VPR
11、=(1)(2)=2WPUs=13=3WPIs=(3+2)=5WEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.+3Vu121A+ 10（c） 解：解：u1=3VIR=u / R=3 / 2=1
12、.5Au2=110=10Vu2PR1=1.53=4.5WPR2=(1)(10)=10WPUs=(1.51) 3 =1.5W PIs=1(3+10)=13WEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty
13、 Ltd.14 电压表、电流表都正偏，输出功率的是电压表、电流表都正偏，输出功率的是 ；电压表正偏，电流表反偏，；电压表正偏，电流表反偏，输出功率的是输出功率的是 。A1A2VA 电压表正偏：所测电压与电压表极性一致；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电流表正偏：所测电流由电流表正极流入负 极流出。极流出。A1A2Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluat
14、ion only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.15 计算各支路电流。计算各支路电流。 22111V3V2VI1I2I3I4I5I6I2221AI3313A 21I430 I41A 1I42I110 I11A对对a：I5I1I2 I41A对对b：I6I3I4 I15AabEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Co
15、pyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.16 已知已知US1=6V，US2=12V，IS=2A；R1=2，R2=1，试计算试计算各电压源及电流各电压源及电流源的功率，并判断其性质。源的功率，并判断其性质。+US1ISR1+US2R2解：解： UR1=ISR1= 4VUIS= UR1 +US1 =10V UR2= US2 US1 = 6V I2=
16、 UR2 / R2 =6A I1=IS+I2=8AI2I1PUS1=68W=48WPUS2=6 12W=72W PIs=102W=20W UISEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.
17、17 计算计算R0，R=时的电流时的电流I与与IE。IE7AR62A43I 12VR0电路被分为两部分电路被分为两部分6、3电阻两端电电阻两端电压为压为12VI12V62AI112V34AIE （ 247）A 1AI1R=I112V34AIE4A2A2AI7A2A4A5AEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NE
18、T 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.15A12A18 计算计算I、 US 、R。 US16A123RI6A9A18A3A123V9R151V0R73US 123V 183V0US90VEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .
19、NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.19 求图中求图中A点电位及恒流源功率。点电位及恒流源功率。对对A：I211mA UA=I11VA1k6V1k1k1mA 2mA IU1U2U2121I0U23VP22(3)6mW6U111UA0U16VP1166mWEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation
20、only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.3V110 求图中求图中a、b、c各点电位。各点电位。电路中存在两个独立电电路中存在两个独立电流，如图流，如图10a1A UC110 10V410bc2 +5V+ Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.E
21、valuation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.111 电压表内阻电压表内阻，S断开时，电压表读数断开时，电压表读数12V；S闭合时，电压表读数闭合时，电压表读数11.8V，求，求US、R0。USRR0- -+VSS断开时断开时USU开开12VS闭合时闭合时R00.16910Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.
22、2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.112 求求S闭合和打开情况下的闭合和打开情况下的A点电位。点电位。A+24V6VS10k10k10k10k2V I=(24+6)/30=1mAUA=2+1016=2V解：解：S打开打开: S闭合闭合: I=6/20=0.3mAUA=2+36=5VIEvaluation only.Crea
23、ted with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.解：解：+10V63A4I1I2 I1I2+3 = 04I1+6I210 = 0联解得联解得 I1=0.8A，I2=2.2APR1=40.82W=2.56WPR2=62.22 W=29
24、.04WPUs=10 0.8 W=8W PIs=3(62.2)W=39.6W21 用支路电流法求图中各支路电流，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。及各电源及线性电阻的功率。Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright
25、2004-2011 Aspose Pty Ltd.+6V23A11222 用支路电流法求图中各支路电流。用支路电流法求图中各支路电流。I4I2I1I3ab解：解：列列KCL方程方程a：I1I2I3=0b：I3ISI4=0联解得联解得I1=2.5A，I2=0.5AI3=2A，I4=1A列列KVL方程方程：R1I1R2I2US=0：R2I2+R3I3+R4I4 =0Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Eval
26、uation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.电压源单独作用时电压源单独作用时 R0 = 6（4+2) k = 3k U = 12(36) (4 6)V = 4VU = 6V+4V=10V解：解：+12V6k3mA23 用叠加原理求图中电压用叠加原理求图中电压U。4k2k3k电流源单独作用时电流源单独作用时 R0 = 36+2 k= 4k U=3(48) 4V = 6V+U12V+Evaluation only.Creat
27、ed with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.+ +- -35V24 用叠加原理求图中电流用叠加原理求图中电流I。47A231I电压源单独作用时电压源单独作用时 I = 35（34）A = 5AI 5A3A8A电流源单独作用时电流源
28、单独作用时I= 73(34)A = 3AEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.+125V 25 用叠加原理求图示电路中的用叠加原理求图示电路中的I。解解: : 125V单独作用时单独
29、作用时120V单独作用时单独作用时403660+ +- -120V60II=I+I=0.75AEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.26 已知已知 Uab=10V，去掉，去掉E后，后
30、，Uab=7V，求，求E=？R+ +- -ERRRIS1IS2解：解： 依叠加原理，依叠加原理，Uab=10V是是E，IS1，IS2共同作用的结果。共同作用的结果。 Uab=7V是是IS1，IS2共同作用共同作用的结果。的结果。ab设设Uab为为E单独作用的电压。单独作用的电压。 则则Uab=10V7V3VUab = ER4R=3V E12VEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation onl
31、y.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.27 将以下各电路化为等效电压源将以下各电路化为等效电压源- -+ +453A515V- -+ +3A444V1A424A- -+ +28VEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.
32、Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.28 将以下各电路化为等效电流源将以下各电路化为等效电流源- -+ +228V22A2 24A1- -+ +24V2Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with A
33、spose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.1A29 用等效变换法求图示电路中的电流用等效变换法求图示电路中的电流I。2A101054010V+I20V104A+10V5+5VI=(105)V/(5+5+40) =0.1AEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation on
34、ly.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.6V- -+8V- -+12V- -+210 用等效变换法求图示电路中的电流用等效变换法求图示电路中的电流I。3612A21I2A2A4A2+ +- -2VI(82)A/(2+1+3) 1AEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspos
35、e Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.- -+ +6V43211 用等效变换法求图示电路中的电流用等效变换法求图示电路中的电流I。26AI1.5A8I = （6+1.5）4V(48) = 2.5AEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-201
36、1 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.+ +- -150V+ +- -20V212 用戴维宁定理求图中电流用戴维宁定理求图中电流I。+ +- -120V1010104I解：解：令令R开路开路Uab=(20150+ 120)V =10VR00UabIUab/(R0+10)1AEvaluation only.Created with Aspose.Slides for .NET
37、 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.+ +- -2V213 用戴维宁定理求图中电流用戴维宁定理求图中电流I。- -+ +12VI解：解：令令R开路开路abUab(23+8 +2)V 4VR06/3+3 5IUab/(R0+5) 4V/10 0.4A63532AUabU1Evalua
38、tion only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.214 用戴维宁定理求图中电流用戴维宁定理求图中电流I。+ +- -16V1A483I320Uab解：解：Uab1689813 1693 4VR0(4+20
39、)/83 9168I14I120I20 I1I21 I198AI1I2IUab/(R0+3)1/3AEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.8V- -+12V- -+215 用戴维宁定
40、理求图示电路中的电流用戴维宁定理求图示电路中的电流I。4422I2A3A5A2R02/4/4 1UabUab=（5/2）2 = 5VIUab/ （ R0 +2） 1.67AEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 A
41、spose Pty Ltd.2A +19V 32IX216 求图示电路中的电流求图示电路中的电流IX、UX、PX。3A1 9AUX由由KCL推广定律可得推广定律可得 IX9A求电路的戴维宁等效电路求电路的戴维宁等效电路UOC23+1923 7VR0(3+1+2)6电路可等效为电路可等效为+ +- -7V6I IX XUX7V69V47VPXUXIX947W423W （电源）（电源）Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pt
42、y Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.217 求图中的求图中的I。+- -6V+- -4V31I2A642 22A1A应用等效变换法应用等效变换法24A+- -8V4I = 3 2/（12）A = 2A3A2Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2
43、011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.- -+ + 5V 218 求求S闭合和打开情况下的电流闭合和打开情况下的电流I。- -+ +10V5I2A5SS断开，断开，105I15I50II1 + 2得得 I1.5AI1S闭合，闭合，105I15I0II1 + 2得得 I2AEvaluation only.Created with Aspose.Slides for .NE
44、T 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.- -+ +U1219 无源二端网络无源二端网络N0，U11V，I21A，时，时 U30；U110V，I20A，时，时U31V；求；求U10，I210A，时，时U3？N0I2U3由叠加原理可知由叠加原理可知U3有有U1、I2共同作用共同作用设设
45、U3K1U1K2 I2有有 0K1K2 110K1得得K10.1，K20.1 U10，I210A，时，时U310K21VEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.220 无源二端网络无源
46、二端网络N0，IS1A，求，求I？N010VISN02VISN0IS5I则其戴维宁等效电路为则其戴维宁等效电路为 U开开10V将将N0、IS看作有源二看作有源二端网络端网络R0(21) 2I(102+5)A1.43AEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0
47、.Copyright 2004-2011 Aspose Pty Ltd.31 t0时时，S闭闭合合后后uC及及各各电电流流的的初初始始值值及及稳态值。稳态值。uC(0 ) 1.5 4V 6V在在S闭合的瞬间，根据换路闭合的瞬间，根据换路定律有：定律有： uC(0 )= uC(0+ )= 6V, i1(0+ ) uC(0+ ) /4 =1.5A i2(0+ ) = uC(0+ ) /2 =3AiC(0+ )1.51.533ASC Ci2iCu uCi11.5A24uC( ) 1.5 (4/2)V 2ViC( ) 0i1( ) 2/4A0.5A , i2( ) 2/2A 1A Evaluation
48、 only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.32 求开关闭合后的初始值求开关闭合后的初始值及稳态值。及稳态值。iL(0 ) =1243A在在S S闭合的瞬间，根据换闭合的瞬间，根据换路定律有：路定律有： iL
49、(0+)iL(0) = 3A4 i1(0+) + uL(0+) =12 uL(0+)= 4.8V +S Si2uL解：解：i146iL12Vi1 (0+) = 3 6/（46） = 1.8Ai2 (0+) = 3 4/（46） = 1.2AiL( ) 12(4/6) 5AuL( ) 0i1( ) 12/4 3A i2( ) 12/6 2A Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only
50、.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.33 电路中电路中, ,已知已知R1=5k, R2=10k, C=4F，US= 20V, t =0，打开，打开S，求，求uC(t) 、iC(t) 、并画出变化曲线、并画出变化曲线 。SuCR2R1USiCC+电路为零输入响应电路为零输入响应解解: : uC(0+)=uC(0)=USR2/（ R1 +R2）=2010/15=13.3V= R2 C=4102s uC() =0 iC() =0 uC(t)
51、 =13.3e25t V iC(t) =1.33e25t mAuC(t)tiC(t)- -1.33mA13.3VEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.SC Cu uCi1mA10K3
52、4 S打开打开0.2s后，后，uC8V，试求，试求C、 i(0+ )、 uC(t)。 解解: :uC(0+)=uC(0)=0 = 10103 C电路为零状态响应电路为零状态响应 uC() =201010V uC(t) =10（1et/10000C ）Vt0.2，uC= 8VuC(t) =10（1e1/50000C ）8Ve1/50000C0.2， C12.5FS闭合的瞬间，闭合的瞬间，C相当于短路相当于短路 i(0+ )0uC(t) =10（1e8t ）Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile
53、5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.35 电路中电路中, ,已知已知R1=3, R2=10, R3=6,C=10F，US= 100V, t=0，闭合，闭合S， 求求i1(t) 、i2(t) 、i3 (t) 。Si3R2R3R1US解解: :uC(0+)=uC(0)=0 = (R1/R3+R2) C =12105si2i1
54、C+电路为零状态响应电路为零状态响应 i1() =i3() =US / /(R1+R3)=11.1A i2() =0 i3(t) =11.11.85e8333t A i1(t) =11.1+3.7e8333t A i2(t) = 5.55e8333t AEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Cli
55、ent Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.C Cu uCIS36 已知已知R1=6k, R2=2k，C=20F， IS= 4mA，US= 12V, t =0，闭合，闭合S ，试求，试求uC(t)。 解解: :uC(0+)=uC(0)=ISR2= 8V= (R1/R2) C = 3102s9V uC(t) =9e33.3t VR2SUS+R1Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2
56、011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.37 电路中电路中, ,已知已知R1=4, R2=20, R3=6, C=4F，US= 50V, t =0，闭合，闭合S，求，求uC(t) 、i(t) 、并画出变化曲线、并画出变化曲线 。SR2R3R1USiC+uC解解: := (R1/R3+R2) C =89.6106suC(0+)=uC(0)=50V uC() =US R3 /
57、 /(R1+R3) =30V- -0.896 i(t)50uC30tuC(V).i(A) i(0+)=(US /R1)(R3/R1)uC(0+)/(R2+R3/R1) =0.896A i() =0Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright
58、 2004-2011 Aspose Pty Ltd.38 电路中电路中, ,已知已知R1=R2=4k, R3=2k, C=100F，E1= 10V, E2= 5V，t0，S由由a打向打向b，求求uC(t) 、i0(t) 。解解: :uC(0+)=uC(0) = E1R2 / /(R1+R2) = 5V= (R1/R2+R3) C =4101sSR3R2R1E2i0CE1uC()= E2R2/ /(R1+R2) = 2.5V=1.56mAi0()= E2/ /(R1+R2) = 0.625mA abEvaluation only.Created with Aspose.Slides for .N
59、ET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.SC Cu uCi22A 339 C= 0.5F，试求试求i1 (t )、 i2(t)。 解解: :uC(0+)=uC(0) =23V 6V = (6/3) C=1s电路为零输入响应电路为零输入响应 6i1i1 (0+) = 6/6 A= 1
60、Ai2 (0+) = 6/3A = 2A i2() =0 i1() =0 i1(t) =etA i2(t) =2etAEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.310 电路中电路中, ,
61、已知已知R1=6, R2=4, L=10mH，US= 10V, t =0，闭合，闭合S，求，求uL(t) 、iL(t) 。US LR1R2iLSuL解解: := L/R2 = 2.5103siL(0+)= iL(0) =US/(R1+R2)=1A uL(0+) = - -R2 iL(0+) = - -4V iL() =0 uL() =0Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.C
62、reated with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.311 电路中电路中, ,已知已知R1=3k, R2=6k, C1=40F， C2=C320F，US= 12V, t0，闭合，闭合S，求，求uC(t) ，画出曲线，画出曲线。SR2R1US解解: :uC(0+)=uC(0)=0 = (R1/R2) C =4102sC1+电路为零状态响应电路为零状态响应C2C3uCC = C2/C3+C1uC()= USR2/ /(R1+R2) = 8VuC(t)= 88
63、e25tVtuC(t)8VEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.IS312 已知已知R1=2, R2=R3=3， R4=6, L=10mH, IS= 1mA, US=8V, t0，断
64、开，断开S ，试求，试求uL(t)、 i(t) 。 解解: :R2SUS+R1R3R4iuL= 0.001/（6+6/2) = 1.33103sEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.
65、313 已知已知R1=200, R2=400, C1=0.1F， C20.05F，U= 25V, t=0，断开，断开S，求，求uC1(t) 、 uC2(t) 、 i(t) 。SR1R2U解解: :uC1(0+)=uC1(0) =25400/(200+400) =50/3VuC2(0+)=uC2(0) =25200/(200+400) =25/3V 1= R1 C1=2105s2= R2 C2=2105sC1+C2uC2uC2uC1()= uC2()= 25VuC1(t)= 258.3e50000tVuC2(t)= 2516.7e50000tVEvaluation only.Created wi
66、th Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.314 t = 0，闭合，闭合S，求，求iL，画出曲线。，画出曲线。12V1H3iL解解: iL(0+)= iL(0) =12/(3+3)=2A依等效变换依等效变换i(0+)=(33/62)612
67、/(3+6) =1A= = 1/1/3+(63) =0.2s33VSab iL(t)=3.81.8e5tA， i(t)=0.21.2e5tA63iL (A )t 2 3.81 0.2Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2
68、011 Aspose Pty Ltd.315 图示电路在开关图示电路在开关S闭合前电路已处于稳态，闭合前电路已处于稳态，在在t =0时刻开关闭合。试求开关闭合后的时刻开关闭合。试求开关闭合后的uC(t)及及iL(t)。 +t=01003k3k1.5k1FiLuC90V1HEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NE
69、T 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Su uCi2mA 3k316 试求试求uC (t )、 i(t)。 解解: :uC(0+)=uC(0) =236V 1k1F 1k+6k12V= = 1+(63)1103 =3103suC(t)= 82e333tVi(t)= 0.670.22e333tmAEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose P
70、ty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.41 已知已知 i = 5.6sin(314t37 )A, u = 311sin314tV。 解：解：(1) Um=311V, Im=5.6A u = i = 314 rad/sfu = fi = /2 = 50HzTu = Ti = 1 / f = 0.02su=0 i=37 u. i ( V.A )5.631137 (2) = u i = 37 u 超
71、前超前 i 37 (3)37 (4)Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.42 已知已知 u1 = 100sin(t+ / 6)V, u2 = 60sin (t+) V, 试求试求u
72、=u1+u2 在在U为为 最大和最小时，最大和最小时， =？Um=？解：解：当当 = / 6时，时，Um最大，最大， Um= 100+60 =160V当当 = - - 5 / 6时，时，Um最小最小 ， Um= 10060 = 40V3030Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Pr
73、ofile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.43 已知电路中某元件已知电路中某元件u和和i，试问是什么元件、，试问是什么元件、元件的参数、储能的最大值、平均功率。元件的参数、储能的最大值、平均功率。(1) u = 100sin(314t)V i = 10cos (314t) A= 100sin(314t+180)V= 10sin(314t90)Au超前超前i 90，元件为电感，元件为电感XL1001010Q100102500Var P0(2) u = 282sin(314t+60)V i=56.6cos(314t30)A = 56.6sin
74、(314t60)Au、i 同相，元件为电阻同相，元件为电阻R28256.65Q0 P28256.627980.6W(3) u = 70.7cos(314t30)V i = 5sin (314t30)A= 70.7sin(314t+60)V= 5sin(314t150)AU滞后滞后i 90，元件为电容，元件为电容XC70.7514.14Q70.752176.75Var P0Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd
75、.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.解：解： XL 2f L8 Z6+j81053.13 IU/ z 22A44 将电感将电感L25.5mH， R=6的线圈接到的线圈接到 f = 50Hz，U=220V的电源上，的电源上，求求XL、Z、 ，画相量图，画相量图 。 RU LU I U53.13 设设Evaluation only.Created with Aspose.Slides for .NET 3.
76、5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Cu1u2R45 R100，u1的的 f 500Hz，要求输出信号，要求输出信号u2比输入信号比输入信号u1超前超前60，求求C？RXCz60XC=|Rtan | =173.2C12f XC=1.84F设设 为参考正弦量为参考正弦量则则 的初相角为的初
77、相角为0则则 的初相角为的初相角为60Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.46 已知已知 i = 5.6sin（314t17 )A， u = 314sin（314t+20 ) V，
78、 试求串联等效电路。试求串联等效电路。 i+uN解：解：串联等效电路串联等效电路R=zcos37 =44.7X= zsin37 =33.7R=70.2 X=93.2 并联等效电路并联等效电路Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 20
79、04-2011 Aspose Pty Ltd.1Z2Z3Z47 已知已知İS20A, Z1=j5 ，Z2=(2+j)，Z3=(3+j4)， 求求各个各个电流及各个流及各个电压。按分流公式：按分流公式：Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyrig
80、ht 2004-2011 Aspose Pty Ltd.R1XLXCİİ1İ245作相量图：作相量图：İ1=1090Aİ= İ1+İ2100AUab= USU150Vab48 图中图中,= 314rad/s, US=100V，R1 =5 ，I1 =10A，I210 A，RL=XL ，求，求I、RL、L、C 。 RLİ2=10 45A RL=XLzL=Uab/I2=3.5, RL= XL=2.5L=2.5/314 = 7.96mHC=1/5314=0.637mF设设 为参考正弦量为参考正弦量Evaluation only.Created with Aspose.Slides for .NET 3
81、.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.49 电路如图，标出各表读数。电路如图，标出各表读数。A1LCA2A03A5Ai1 、i2 反相反相I0I1I2532AV1RCV2V010V10VEvaluation only.Created with Aspose.Slides for .NET
82、 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.LCA1A010ARV1100V5510İ1=1090İR=10 45İ0 = İ1+İR= 100V0Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile
83、5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.RLCV1V2V3V5V4V6410 电路如图，电路如图， R=XL5，XC10 ，V1读数数为10V，求各表读数。求各表读数。I =V1/R= 2AV2IXL10VV3IXC20VV4V3V210V+CLUU RU CU LU IEvaluation only.Created wit
84、h Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.411 已知已知U=220V, U1=120V, U2=130V,f = 50Hz, R1=50, 求求RL和和L。LUU2R1U1RL解：解：I = U1/R1 = 2.4Az = U/I = 2
85、20/2.4 =91.67I由已知可列方程组：由已知可列方程组：z2=(R1+RL)2+XL2解得解得: RL=29.69 XL=45.3 L = XL/2f = 0.144HEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011
86、 Aspose Pty Ltd.1ZZ2Z412 已知已知 120, Z =（ +j） ， Z1=(3+j4) ，Z2=(44j)， 求求各个各个电流及各个流及各个电压。Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 As
87、pose Pty Ltd.413 已知已知 u = 2202sin314t V, R1=3， R2=8，L=12.7mH, C=531F, 求求i1、 i2、 i，cos、P、Q、S,画相量图。画相量图。 R1LuiCi2i1R2解：解：Z1=jXL+R1=j4+3 =553.13Z2= R2 jXC+=8j6 =10 36.86XC=6XL=4oooZ1UI113.534413.5350220 = = = = = oooZ2UI286.3622 36.18100220 = = = = = Evaluation only.Created with Aspose.Slides for .NET
88、3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.İ = İ1+İ2= 44j22 = 49.226.56cos = cos26.56= 0.9 P = UIcos = 9741.6W Q =UIsin = 4840.6Var S = UI = 10824VAEvaluation only.Cre
89、ated with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.414 已知已知U=220V，I=4A，P160W,f=50Hz，试求试求R、L。RLuWA0.18 =79.5UR= Ucos = 40VRPI210XL=Rtan=54LXL2f
90、 =0.172HEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.1ZR2Z415 已知已知İ530, R =2 ， Z1=j10 ，Z2=(40+30j)， 求求İ1、 İ2、U。Evalua
91、tion only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.416 图示正弦交流电路中图示正弦交流电路中, U=220V，U与与I同同相，相，R=XC， 求求UL、 UC。 RXLXCİ45作相量图：作相量图：UL=
92、U=220VEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.R1LuiRR2 i2i1设İ= 100按分流公式：按分流公式：5824= 453.13A= 8.2522.83A= 81.429V
93、P=UIcos9=81.42100.988=804.2WEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.R1LuCi2i1R2418 电路如图，电路如图， R1= 5 , R2= 3, XL2
94、 ，XC3，V读数数为45V，求求A表读数。表读数。AVI2 = UR2 = 15A设İ= 150İ = İ1+İ2= 22.4628.9 I22.46AEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty
95、 Ltd.R1LuCi2i1R2419 R1=R2= 30, XLXC30 ，a biİ = İ1+İ2= 20Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.1Z2Zi2i1iu420 u=
96、220 sin314t V，总的，总的S=5KVA，Z1的的 cos1= =1，Z2的的 cos2= =0.5，电路总的，电路总的cos= =0.8，试，试求求P1、P2。PScos=4KW = P1 +P2QSsin=3KVar = Q2P2UI2 cos2=1731.4WP1=PP2=2268.6WEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspos
97、e.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.421 iR =2sin2000tA, R=200，L=0.1H, C=2.5F, 求求i1、 iC、uR、UL、US、 cos、P、Q、S,画相量图。画相量图。 RXLXCİ1XC=200XL=200İR= 10İC= 190cos=cos45=0.707 P = UIcos = 200W Q =UIsin = 200Var S = UI = 282.84VAEvaluation only.Created with Aspose.Sl
98、ides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.R1X1uXCii1R422 图示正弦交流电路中图示正弦交流电路中, U=120V, İ1超前于超前于İ相位相位90,R+jX=11+j8, R1+jX1=30+j50, 求求XC。 X1=(90 36)= 54 XC=
99、50（41.3）91.3Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.423 已知已知R=10，L=0.12mH, C=600PF, 外加电压外加电压10mV，试求：，试求：f0、Q、Z、I
100、、UR、UL、 UC、P.iuRLC解：解： z=R=10 , I=U/R=1mAUR=U=10mV UL=UC=QU=447.1mV P = UI =10WEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose P
101、ty Ltd.51 对称三相负载，对称三相负载，Z=17. 32+j10，额额定电压定电压UN=220V， 三相四线制电源，线电压三相四线制电源，线电压uUV = 3802sin(314t+30)(V) ，负载如何，负载如何接？求线电流，画相量图。接？求线电流，画相量图。解：解：应采用星形联接，相电压应采用星形联接，相电压UU=220V，满足满足额定电压需要。额定电压需要。3030Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pt
102、y Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.52 有日光灯有日光灯120只，每只功率只，每只功率P40W，额定，额定电压电压UN220V，cosN=0.5，电源是三相四线制，电源是三相四线制，电压是电压是380220V，问日光灯应如何接？当日光灯，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？全部点亮时，相电流、线电流是多少？解：解：电源线电压是电源线电压是380V相电压是相电压是220
103、V将将120只日光灯均匀接在三相上，每相只日光灯均匀接在三相上，每相40只只（并联）（并联）IL = IP =P40UPcos1101600=14.54 AEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pt
104、y Ltd.53 Z=26.87+j26.87，UN=380V， 相电相电压压uU = 2202sin(314t30)(V) ，负载如何，负载如何接？求相电流和线电流，画相量图。接？求相电流和线电流，画相量图。解：解：应采用角形联接，线电压应采用角形联接，线电压UUV=380V，满足满足额定电压需要。额定电压需要。Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with
105、 Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.54 三相绕组角形联接的对称三相电路三相绕组角形联接的对称三相电路 。已。已知知UL =380V, IL=84.2A。三相总功率。三相总功率P=48.75KW, 求绕组复阻抗求绕组复阻抗Z 。zUL /IP =7.82=28.36 Z7.82 28.36 Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyrig
106、ht 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.ZZZZUVİUİVİW55 电源电源UL =380V，对称负载，对称负载Z38.1+j22, ZUV9.8+j36.7 。求。求İU、 İV、 İW 。İUVEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0
107、.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.ZZZİUİVİW56 电路如图，电源电路如图，电源 ，负载，负载ZU10, ZV 6j8, ZW8.66j5 。求。求İU、 İV、 İW、 İN 。İNİN = İU +İV + İW 14.838.9Evaluation only.Created with Aspose.Slides for
108、.NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.57 对称三相负载对称三相负载ZL50+j28.9星形联接，已星形联接，已知知UL =380V, 端线阻抗端线阻抗ZL 2+j1, 中线阻抗中线阻抗ZN1j 。负载端的电流和线电压，并画相量图。负载端的电流和线电压，并画相量图 。Eval
109、uation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.UVWZ58 S闭合时，电流表读数闭合时，电流表读数7.6A，求求S打开时各电流表读数（设三相电压不变）。打开时各电流表读数（设三相电压不变）。AAAS
110、S打开打开IL7.6AEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.59 电压表读数电压表读数220V，负载，负载Z=8+j6，求电流表读数，求电流表读数，cos，P。解解：电源是星形联接，
111、电源是星形联接，负载是三角形联接负载是三角形联接 A是负载的线电流，是负载的线电流， V是电源的相电压。是电源的相电压。10IAB = IP =ULz=380=38 AAUP = 220 V UL = 3 UP = 380V= IL = IU = 3 IP = 38 3 =65.8AZZZVAcos = cos(tan16/8) = 0.8P =3ULIP cos =34.656KWEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose P
112、ty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.510 负载负载Z1=19.1+j11， Z2=32.9+j19，UL=380V,求电流表求电流表İ、 İ1、 İ2 。ABCİZ1Z2İ2İ1解：解：设设UAB=3800为参考正弦量为参考正弦量Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Co
113、pyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.511 负载额定电压为负载额定电压为220V，现有两种电源：，现有两种电源：UL =380V、UL220V，应如何连接，对称负载，应如何连接，对称负载R24, XL18 。分别求相电流、线电流。画。分别求相电流、线电流。画出出UL =380V时的全部相量图。时的全部相量图。UL =380V，采用，采用Y接
114、接UL =220V，采用，采用接接3036.87Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.512 一台三相交流电动机，定子绕组星形联结，一台三相交流电动机，定子绕组星形联结，额定电压额定
115、电压380V，额定电流，额定电流2.2Acos=0.8，求每相绕，求每相绕组的阻抗。组的阻抗。UL380V，ILIP2.2A UP220VzUP /IP =100=36.87 Z100 36.87 Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyrigh
116、t 2004-2011 Aspose Pty Ltd.513 电流表读数电流表读数32.9A，负载，负载Z=12+j16，求电压表读数求电压表读数。解解：电源是星形联接，电源是星形联接，负载是三角形联接负载是三角形联接 A是负载的线电流，是负载的线电流， V是电源的相电压。是电源的相电压。ZZZVAA= IL 32.9AVEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created
117、with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.514 星接负载星接负载Z=30.8+j23.1，电源线电压电源线电压380V ，求，求cos，P、Q、S。38.5IU = IP =UPz=220=5.7 AUL = 380V UP = 220 Vcos = cos(tan123.1/30.8) = 0.8P =3UPIP cos =3KWQ =3UPIP sin =2.3KVarS =3UPIP =3.77KVA Evaluation only.Created
118、with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.515 电源线电压为电源线电压为220V，对称负载，对称负载R4, XL3 。求线电流、。求线电流、cos，P、Q、S 。画出电压。画出电压相量图。相量图。UWVUL = 220V UP =
119、127Vcos = cos(tan13/4) = 0.8P =3UPIP cos =7.74KWQ =3UPIP sin =5.8KVarS =3UPIP =9.68KVA 3036.87Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 200
120、4-2011 Aspose Pty Ltd.516 星接电源，角接负载，线电流星接电源，角接负载，线电流25.4A，负，负载载P7750W，cos0.8，求求UL、UP、S 、R、XL。UWVS =3ULIP =9.68KVA zUL /IP =15电源电源UP127V R =zcos = 12 XL=zsin = 9Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created wi
121、th Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.71 一台三相异步电动机额定转速，电源的一台三相异步电动机额定转速，电源的f 1= 50Hz，nN=1450r/min，空载转差率，空载转差率s00.26。试。试求同步转速求同步转速n1、磁极对数、磁极对数P、空载转速、空载转速n0、额定转、额定转差率差率sN。 解：解： nN=1450r/min nn1 P2 n1=1500r/min n0=1496r/minEvaluation only.Created with A
122、spose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.72 绕线转子异步电动机起动时，是不是转子绕线转子异步电动机起动时，是不是转子串电阻越大，起动转矩就越大？为什么？当转子串电阻越大，起动转矩就越大？为什么？当转子开路时，又怎样？开路时，又怎样？1.否定否
123、定 转子串入适当的电阻，起动转矩会增大，但转子串入适当的电阻，起动转矩会增大，但增大到增大到TMAX，就不会再增大了。，就不会再增大了。2. 当转子开路时，转子中电流为当转子开路时，转子中电流为0，不能转动。，不能转动。Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0
124、.0.Copyright 2004-2011 Aspose Pty Ltd.73 在电源电压不变时，若将在电源电压不变时，若将接电动机误接成接电动机误接成Y，或将，或将Y误接成误接成，会产生什么样的后果？，会产生什么样的后果？1.若将若将连接电动机误接成连接电动机误接成Y 绕组的相电压下降绕组的相电压下降 转矩减少转矩减少3倍，只能空载运行倍，只能空载运行 若带负载，将导致电流增加。若带负载，将导致电流增加。2.或将或将Y误接成误接成 绕组的相电压增加，电流增大，绕组发热甚至绕组的相电压增加，电流增大，绕组发热甚至烧坏。烧坏。Evaluation only.Created with Aspos
125、e.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.74 Y132S-6型三相异步电动机额定值如下型三相异步电动机额定值如下,试求试求线电压线电压380V,如何连接如何连接,p, sN , TN , Tst , Tmax ,P1N, S 。 PN nNUN N I
126、N NIst /INTm /TNTst /TN 3 KW960 r/min2203800.8312.8 7.2 0.75 6.5 2.2 2.0解：解：磁极对数磁极对数 p= 3, n1=1000r/min电源线相电压电源线相电压380V，若满足电子绕组相电，若满足电子绕组相电压压220V，应接成，应接成Y形连接。形连接。 Tst=2TN=59.68Nm Tmax=2.2TN=65.65NmNPNP1N P1N30.833.6KWIN = 7.2A S32207.24.74KVAEvaluation only.Created with Aspose.Slides for .NET 3.5 Cl
127、ient Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.75 三相异步电动机额定值为三相异步电动机额定值为PN=7.5KW, , UN=380V, nN=1450r/min, IN14.9A， Ist / /IN=7, Tst / /TN = 1.7。试求。试求(1) TN 。(2) 电动机若要采用电动机若要采用换
128、接起动，则换接起动，则Ist，Tst各为多少；（各为多少；（3）在）在额定负载下直接起动，最小电源电压额定负载下直接起动，最小电源电压Umin=? Ist = 7 IN =104.3A Tst=1.4TN=69.2Nm解：解：采用采用换接起动换接起动 Ist = = Ist/3=34.77A Tst = Tst/3 = 23Nm Tst(U)2 Tst=(X)2Tst=49.4Nm X=0.845 Umin=321.1V Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyrigh
129、t 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.76 三相异步电动机额定值为三相异步电动机额定值为PN=7KW, p=3, sN=0.03, Tst /TN = 1.4, f1 =50Hz。试求。试求(1) TN ，(2) 电动机若要采用电动机若要采用换接起动，换接起动，T20.3TN；能；能否起动？否起动？磁极对数磁极对数 p= 3, n1=1000r/minT
130、st=1.4TN=96.46Nm 解：解：采用采用换接起动换接起动 Tst = Tst/3 = 32.15Nm Tst TL= 68.90.3=20.67Nm 能起动能起动Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 A
131、spose Pty Ltd.78 三相异步电动机，三相异步电动机， UN=380V, Y Y接法接法, , 额定状额定状态下，态下，PN=10KW, N=0.9, IN=20A，轻载状态下，轻载状态下，P=2KW, =0.6, I=10.5A，分别计算两种情况，分别计算两种情况的功率因数，并比较。的功率因数，并比较。解：解：额定状态下额定状态下轻载状态下轻载状态下 显然显然N ，电动机工作在额定状态下，电动机工作在额定状态下最高。最高。Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.C
132、opyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.79 两台三相异步电动机两台三相异步电动机Y Y系列，系列， 额定额定PN=4KW, nN1=2860r/min， nN2=720r/min，比较二者的额定，比较二者的额定转矩。转矩。同样的额定功率下，转速越高，转矩越小。同样的额定功率下，转速越高，转矩越小。Evaluation only.Create
133、d with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.710 三相异步电动机三相异步电动机p=3，PN=30KW, 接法接法, , UN=380V, sN=0.02, N=0.9, IN=57.5A, Ist / /IN=7, Tst / /
134、TN = 1.2，f =50Hz。试求。试求(1) TN , N。(2) 电电动机若要采用动机若要采用换接起动，则换接起动，则Ist，Tst各为各为多少；当负载转矩为额定转矩的多少；当负载转矩为额定转矩的60%和和25%时，时，能否起动？能否起动？TN= 9550301470 = 194.9Nm解：解：磁极对数磁极对数 p= 2, n1=1500r/min Ist = 7 IN = 757.5A =402.5A换接起动时换接起动时 Ist = Ist/3 = 402.53 = 134.17ATst = Tst/3 = 1.2194.93 =77.96Nm TL=0.6TN=0.6194.9=1
135、16.94Nm TL Tst 不可以不可以 TL=0.25TN=0.25194.9=48.725Nm TL Tst 可以可以Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.711 Y160M-
136、6型三相异步电动机型三相异步电动机PN=7.5KW, UN=380V, nN=970r/min, N=0.86, N=0.78, Ist / /IN=6.5, S = 2, , M=2，起动电流不允许超过，起动电流不允许超过80A。若负载转矩若负载转矩TL=100Nm,试问试问能否带此负载能否带此负载(1)长长期运行期运行,(2)短时运行短时运行,(3)直接起动？直接起动？Tst=2TN=147.6Nm Tmax=2TN=147.6Nm Ist = 6.5 IN =110.5A 解：解： TN TL 能短时运行能短时运行 TST TL 但但IST100A 不能直接起动不能直接起动Evaluat
137、ion only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.九九七七级级计计计计算算算算题题题题86 试用按钮、交流接触器、试用按钮、交流接触器、指示灯设计三人抢答竞赛电路。指示灯设计三人抢答竞赛电路。要求：要求：SB
138、1、SB2、SB3、SB分别是三个参赛人和主持人的分别是三个参赛人和主持人的按钮，抢答时按一下；按钮，抢答时按一下；HL1、HL2、HL3分别是三个参赛人分别是三个参赛人的抢答指示灯，谁抢上了，谁的抢答指示灯，谁抢上了，谁的灯就亮，否则不亮；的灯就亮，否则不亮；主持主持人按一下按钮，使指示灯全部人按一下按钮，使指示灯全部熄灭，准备再次抢答；熄灭，准备再次抢答； 交交流接触器、指示灯的额定电压流接触器、指示灯的额定电压均为均为220V；电路要有短路电路要有短路保护。保护。 220VHL1KM1KM1KM 1SB1KM2KM3FUFUFUFUSBKM2KM 2SB2KM1KM3KM3KM 3SB2KM1KM2HL2KM2HL3KM3FUFUEvaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2004-2011 Aspose Pty Ltd.
