chap2電路的分析方法

1、第第2章章 電路的分析方法電路的分析方法URRRU2111 URRRU2122 IRRRI2121 IRRRI2112 21111 RRR ROY- ROCBADCADBbcRaRcRbabCRaRcRbaababcababcbcabcaaccaabbc/()/()/()RRRRRRRRRRRRRRRCbCRaRcRbabaccbbacaaaccbbabccccbbaabRRRRRRRRRRRRRRRRRRRRRRRRa cabcabbccaccabcababbcbcabcabcaabaRRRRRRRRRRRRRRRRRR Y Y a等效變換等效變換bcRaRcRbbcRaRcRba例例1：1

2、2由圖：由圖：R12=2.684 12CD12 284484cabcabcaaba RRRRRR 184444b R 284448c R 52) 1(5)24() 1(5)24( RA A 2 .I OSREI 電壓源電壓源0SRUII 電流源電流源設(shè)設(shè) IS = 10 A，接上，接上RL 后，恒流源對(duì)外輸出電流。后，恒流源對(duì)外輸出電流。RL IUISOIISU+_0SREI 0SRUII 注意事項(xiàng)：注意事項(xiàng)：解解:+abU2 5V(a)+ +abU5V(c)+ a+-2V5VU+-b2 (c)+ (b)aU 5A2 3 b+ (a)a+5V3 2 U+ a5AbU3

3、 (b)+ A1A22228 I例例3： 解：解：2 +-+-6V4VI2A 3 4 6 1 2A3 6 2AI4 2 1 1AI4 2 1 1A2 4AA2A3122 II4 2 1 1A2 4A1 I4 2 1A2 8V+-I4 1 1A4 2AI2 1 3AA10A110111 RUIA6A22102S1 IIIaIRISbI1R1(c)IR1IR1RISR3+_IU1+_UISUR2+_U1ab(a)aIR1RIS+_U1b(b)(2)由圖由圖(a)可得：可得：R1S2A6A4AIIIA2A51031R3 RUI理想電壓源中的電流理想電壓源中的電流A6A)4(A2R1R3U1 III理

4、想電流源兩端的電壓理想電流源兩端的電壓V10V22V61S2S2IS IRRIIRUUIR1IR1RISR3+_IU1+_UISUR2+_U1ab(a)各個(gè)電阻所消耗的功率分別是：各個(gè)電阻所消耗的功率分別是：W36=61=22RIPRW16=41=22111）（RRIRPW8=22=22S22IRPRW20=25=22333RRIRP兩者平衡：兩者平衡：(60+20)W=(36+16+8+20)W80W=80W(3)由計(jì)算可知，本例中理想電壓源與理想電流源由計(jì)算可知，本例中理想電壓源與理想電流源 都是電源，發(fā)出的功率分別是：都是電源，發(fā)出的功率分別是：W60=610=111UUIUPW20=2

5、10=SSSIUPII 當(dāng)不需求當(dāng)不需求a、c和和b、d間的電流時(shí)，間的電流時(shí)，(a、c)( b、d)可分別看成一個(gè)結(jié)點(diǎn)。可分別看成一個(gè)結(jié)點(diǎn)。 支路數(shù)支路數(shù)b =4，且恒流，且恒流源支路的電流已知。源支路的電流已知。I1= 2A， I2= 3A， I3=6A 例例3：試求各支路電流。試求各支路電流。對(duì)結(jié)點(diǎn)對(duì)結(jié)點(diǎn) a： I1 + I2 I3 = 7對(duì)回路對(duì)回路1：12I1 6I2 = 42對(duì)回路對(duì)回路2：6I2 + UX = 0baI2I342V+I112 6 7A3 cd123+UX對(duì)回路對(duì)回路3：UX + 3I3 = 0結(jié)點(diǎn)電壓方程的推導(dǎo)：結(jié)點(diǎn)電壓方程的推導(dǎo)：設(shè)：設(shè)：Vb = 0 V 結(jié)點(diǎn)電

6、壓為結(jié)點(diǎn)電壓為 U，參，參考方向從考方向從 a 指向指向 b。111RIEU 因因?yàn)闉?11RUEI 所所以以2. 應(yīng)用歐姆定律求各支路電流應(yīng)用歐姆定律求各支路電流 ：111RUEI 222RUEI 33RUI 1. 用用KCL對(duì)結(jié)點(diǎn)對(duì)結(jié)點(diǎn) a 列方程：列方程： I1 I2 + IS I3 = 0E1+I1R1U+baE2+I2ISI3E1+I1R1R2R3+U3211RUIRUERUE S23212211111RRRIREREUS RIREUS1 RIREU1Sab V18V316112171242 A2 A1218421242ab1 UIA3 A618 6 ab2 UIA6 3183ab3

7、 UI212S1S2211ab11RRIIREREU V24V312127330250 1ab11RUEI A13A22450 A18A324302ab22 RUEI515A1VI 5A2VI 10AB3VVI 10B4VI 1565B5VI 2121RREII SIRRRI2121 S21221111IRRRRREIII SIRRRI2112 S21121IRRRRRE 22112S1RIRIEIII S212211 IRRRRREI S211212 IRRRRREI 12112112111211 ) (RIRIRIIRIP A1A5510322 RREIV5V5122S RIUA5 . 0A5 . 0A1 222 III所所以以A5 . 01555S3232 IRRRIV5 . 2V55 . 022S RIUV5 . 72.5V5VSSS UUUA5 . 2A4420402121 RREEI 221210RRRRR，所所以以A2A13230303 RREIA2 . 1A5512211 RREIA8 . 0A51012432 RREIR0abR3R4R1R2 8 . 5434321210RRRRRRRRR，所以所以A126. 0A1