3.5 RLC串联正弦交流电路

3.5RLC串并联正弦交流电路1.阻抗和导纳下页上页返回对单个元件时：URIICU+-IRU+-ILU+-jjLULIXIZI1()(j)jcUIXIZILUZI正弦稳态情况下IZU+-无源线性网络IU+-zφZIUZ||defiuzIUZ阻抗模阻抗角欧姆定律的相量形式下页上页返回导纳S||yφYUIY定义导纳uiyUIY导纳模导纳角下页上页IYU+-返回2.RL串联电路.....jRLUUURILI[j()][j()]LRLIRXIZIjzUZRLZI下页上页LRuuLi+-+-+-uRR+-+-+-.IjLULURU返回Z—复阻抗；|Z|—复阻抗的模；z—阻抗角；R—电阻(阻抗的实部)；X—电抗(阻抗的虚部)。转换关系：22||arctanLLzZRXXφR或R=|Z|coszX=|Z|sinz阻抗三角形|Z|RXLziuzIUZ下页上页返回XL0，z0，电路为感性，电压超前电流。下页上页相量图：选电流为参考向量电压三角形22RLUUU返回IRUUzUL...RLUUU3.RC串联电路.....jcRIUUURIC[j()]CRXIZI下页上页RR+-+-+-.I1/jCUcURU返回CuuLi+-+-+-uRzZCjRIUZZ—复阻抗；|Z|—复阻抗的模；z—阻抗角；R—电阻(阻抗的实部)；X—电抗(阻抗的虚部)。转换关系：22||1arctanCzZRXφCR或R=|Z|coszX=|Z|sinz阻抗三角形|Z|RXCziuzIUZ下页上页返回XL0，z＜0，电路为容性，电压滞后电流。下页上页相量图：选电流为参考向量电压三角形22RCUUU返回...CRUUUIRUUzcU|Z|RXCz4.RLC串联电路KVL：.......1jjICILIRUUUUCLRIXXRICLRCL)](j[)]1(j[IXR)j(zZXRCLRIUZj1jj下页上页LCRuuLuCi+-+-+-+-uRR+-+-+-+-.IjLULUCU.Cj1RU返回Z—复阻抗；|Z|—复阻抗的模；z—阻抗角；R—电阻(阻抗的实部)；X—电抗(阻抗的虚部)。转换关系：arctan||22RXφXRZz或R=|Z|coszX=|Z|sinz阻抗三角形|Z|RXziuzIUZ下页上页返回I分析R、L、C串联电路得出：（1）Z=R+j(L-1/C)=|Z|∠z为复数，称复阻抗（2）L1/C，X0，z0，电路为感性，电压超前电流。0i下页上页相量图：一般选电流为参考向量，CURULUUzUX电压三角形2CL222)(UUUUUURXRjLeqXUR+-+-+-RU等效电路返回I（3）L1/C，X0，z0，电路为容性，电压落后电流。CURULUUzUX等效电路下页上页XUeqj1CR+-+-+-RU.UI（4）L=1/C，X=0，z=0，电路为电阻性，电压与电流同相。IRULUCUR+-+-IRUU等效电路2222)(LCRXRUUUUUU返回例已知：R=15,L=0.3mH,C=0.2F,.Hz103),60(cos254ftu求i,uR,uL,uC.解画出相量模型V605UCLRZ1jjΩ5.56j103.0103π2jj34LΩ5.26j102.0103π21j1j64C5.26j5.56j15Ω4.6354.33o下页上页LCRuuLuCi+-+-+-+-uRR+-+-+-+-.IjLULUCU.Cj1RU返回A4.3149.04.6354.33605oooZUI则A)4.3ω(cos2149.0otiV4.3235.24.3149.015ooIRURV4.8642.84.3149.0905.56joooILULV4.9395.34.3149.0905.26C1joooIUCV)4.3(cos2235.2otωuRV)6.86(cos242.8otωuLV)4.93(cos295.3otωuC下页上页返回下页上页UL=8.42U=5，分电压大于总电压。相量图注意ULUCUIRU-3.4°返回例已知：R=15,L=0.3mH,C=0.2F,.Hz103),60(cos254ftu求i,uR,uL,uC.解画出相量模型V605UCLRZ1jjΩ5.56j103.0103π2jj34LΩ5.26j102.0103π21j1j64C5.26j5.56j15Ω4.6354.33o下页上页LCRuuLuCi+-+-+-+-uRR+-+-+-+-.IjLULUCU.Cj1RU返回A4.3149.04.6354.33605oooZUI则A)4.3ω(cos2149.0otiV4.3235.24.3149.015ooIRURV4.8642.84.3149.0905.56joooILULV4.9395.34.3149.0905.26C1joooIUCV)4.3(cos2235.2otωuRV)6.86(cos242.8otωuLV)4.93(cos295.3otωuC下页上页返回5.RLC并联电路由KCL：CLRIIII1/jj1[1/()]URUCULRCjUL下页上页iLCRuiLiC+-iRR+-IjLULICICj1RI返回分析方法：设电压为参考方向，电流随着频率的变化而变化，频率不同，分流不同（2）C1/L，ImZ0，y0，电路为容性，电流超前（1）相量图：选电压为参考向量，2222)(LCGBGIIIIII0u分析R、L、C并联电路得出：RLC并联电路会出现分电流大于总电流的现象UGI.CI.IyLI.IB下页上页注意返回（3）C1/L，y0电路为感性，电流落后；2222)(CLGBGIIIIIIUGI.LI.IyCI.等效电路下页上页IUBIeqj1CRIR+-返回（4）C=1/L，y=0，电路为电阻性，电流与电压同相。等效电路等效电路下页上页IjLegUBIRIR+-R+-+-IRUUUGIICILI返回例1求图示电路的等效阻抗，＝105rad/s。解感抗和容抗为：Ω100j130100)100j100(100j30jj)j(j221CLCLXRXXRXRZ1001011035LXLΩ100101.0101165CXC下页上页1mH301000.1FR1R2返回例2图示电路对外呈现感性还是容性？解1等效阻抗为：75.4j5.54j81.53256j3)4j3(5)4j3(56j30Z下页上页33－j6j45电路对外呈现容性返回1、据原电路图画出相量模型图（电路结构不变）LCRRLjXCjXuUiIeE、、、、2、根据相量模型列出相量方程式或画相量图一般正弦交流电路的解题步骤3、用复数符号法或相量图求解4、将结果变换成要求的形式复杂电路的相量分析法例1画出电路的相量模型7.175.1049901047.31847.318j1000)47.318j(1000)(3111CjRCjRZ,rad/s314,V100,μF10,mH500,10,100021UCLRR求:各支路电流。已知：解下页上页R2+_Li1i2i3R1CuZ1Z2U1I2I3IC1jLjR2+_R1返回157j10j22LRZ3.5299.16613.132j11.102157j1013.289j11.9221ZZZ下页上页13.28911.923.7245.0331jZZ1Z2U1I2I3IC1jLjR2+_R1返回A3.526.03.5299.16601001ZUIA20181.03.526.07.175.104947.318j1j1j112ICRCIA7057.03.526.07.175.104910001j1113ICRRI下页上页Z1Z2U1I2I3IC1jLjR2+_R1返回瞬时表达略SILjC1jSU+_R1R2R3R4列写电路的结点电压方程例2解下页上页+_susiLR1R2R3R4C返回1nU2nU3nU结点方程SnUU11232123311111()0jnnnUUURRLRRR123334111j(j)nnnSCUUCUIRRR下页上页SILjC1jSU+_R1R2R3R4返回.Ω,45,Ω30Ω,30j,A904321oSIZZZZI求电流已知：方法1：电源变换15j1530j30)30j(30//31ZZ解例3ZZZZZZII23131S//)//(4530j15j15)15j15(4joo36.9-5455.657A9.8113.1o下页上页S31)//(IZZZ2Z1Z3ZI+-Z2SIZ1ZZ3I返回方法2：戴维宁等效变换V4586.84)//(o310ZZIUS求开路电压：求等效阻抗：Ω45j15//231ZZZZeqA9.8113.14545j154586.84o00ZZUI下页上页ZeqZ0U+-I+-0UZ2SIZ1Z3返回例4用叠加定理计算电流2I解:)()1(SS短路单独作用UI323S2ZZZIIoooo30503050305004A3031.235030200ooV45100:oSU已知oooS13240A,5030Ω,5030Ω.IZZZ下页上页Z2SIZ1Z32ISU+-Z2SIZ1Z32I返回32S2ZZUIA135155.13031.2oo222III35045100o:)()2(SS开路单独作用IU下页上页A135155.1oZ2Z1Z32ISU+-Z2SIZ1Z32ISU+-返回在图示电路中，VU00100，3R，1LX，2CX。求总电流I，并画相量图课堂练习UI1I2IRLjXCjX10161213)3)(62(3622jjjjj01.5326.12.1jAI001.53501.532100I01.5304.18033.218.4RLZj1I2IU()2(3)23CLCLjXRjXjjZjXRjXjj相量图课堂练习已知V010SU，阻抗Z为多大时取得最大功率，最大功率为多少？（