高二物理競賽課件支路電流分析法

支路電流分析法

支路電流分析法BRbIb=–B(Usb–RbISb)矩陣形式方程的分析視察法列寫方程的規(guī)則解題主要步驟第j個方程的左邊

bjkRkik第j個方程的右邊

–bjkuSk

–B(Usb–RbISb)=–BUsb

支路電流分析法以支路電流為求解對象，直接根據(jù)KCL和KVL建立方程支路電流方程（1）（2）（3）（4）-R1R4R2R3R5R6+usisi1i6i2i3i4i5i1+i2+i6=0-i2+i3+i4=0-i4+i5-i6=0AIb=0–R3i3+R4i4+R5i5=–uS–R2i2–R4i4+R6i6=0–u1+u2+u3=0–(R1i1+R1iS)+R2i2+R3i3=0–R1i1+R2i2+R3i3=R1iSl1l3l2BUb=0Ub=RbIb+Usb-RbISbBRbIb=–B(Usb–RbISb)5-2-2含受控電源電路的支路電流方程處理方法(1)(2)(3)312-i1+i2+i3=0-i3-i4+i5=0-i2+i4+i6=0R1i1+R2i2+R6i6=R1iS-R2i2+R3i3-R4i4-rmi5+gmR4R6i6=0R4i4+R5i5

–(R6+gmR4R6)i6=05-2支路電流分析法–R2i2+R3i3–R4i4=rmi5–gmR4u6–R2i2+R3i3–R4i4=rmi5–gmR4R6i6gmu6+-i2isrmi5R3R1R2R6R4i1i3i4i5R5i6例1例2-usu1R1R3R2+-++-

u1i1i2i3-i1+i2+i3=0R1i1+R3i3=us-μR1i1+R2i2-R3i3=0支路電流分析法評述5-2-3網(wǎng)絡(luò)含無伴電流源的處理問題兩種方法解決(1)虛設(shè)電壓變量(2)選合適獨立回路

(與合適樹對應(yīng)）R3R4R6R5++--us4us6is2R1i1i3i4i5i65-2支路電流分析法支路電流分析法AIb=0BRbIb=–B(Usb–RbISb)=–BUsb

+例寫圖示電路的支路法方程+++++------1A2V2V3V1VU34U3I16

4

2

2

8

I2I3I4I512I1–I2–I3–1=0I3+I4–I5+1=0l1l2l3–6I1–4I2=–2–34I2

–2I3+2I4=1+3–2–2I4–8I5=–1–4U3=–1–4(–2I3)–8I3–2I4–8I5=–15-3節(jié)點電壓分析法以n-1個節(jié)點到參考節(jié)點的電壓為求解對象，根據(jù)KCL建立方程（1）（2）（3）（4）-R1R4R2R3R5R6+usisi1i6i2i3i4i55-3-1節(jié)點電壓方程AIb=0AGbATΦ=AGbUSb-AISbIb=GbUb+Isb-GbUSbUb=ATΦGnΦ=InS（1）（2）（3）（4）-R1R4R2R3R5R6+usisi1i6i2i3i4i5例1100010-11100000-11-1A=i1i2i3i4i5i6G1000000G2000000G3000000G4000000G5000000G6-is00000+=u1u2u3u4u5u6-G1000000G2000000G3000000G4000000G5000000G60000uS0GbISbUSbAGbATΦ=AGbUSb-AISbGnΦ=InS5-3-2節(jié)點電壓方程的分析1、節(jié)點電導(dǎo)矩陣GnGn=AGbAT1）對稱性2）gii—節(jié)點i的自電導(dǎo)3）gij—節(jié)點i與j的互電導(dǎo)2、節(jié)點電流源電流列向量

InS=AGbUSb-AISb=-AI

bSG1+G2+G6-G2-G6Φ1iS-G2G2+G3+G4-G4Φ2=0

-G6-G4G4+G5+G6Φ3G5uS(1)(2)(3)(4)-R1R4R2R3R5R6+usisi1i6i2i3i4i5gii=

aikGka’kik=1bgij=

aikGka’kjk=1biiS=–

aikikSk=1b-404030204030100v35v++-例112140130140130+++140130140120+++140130+–()140130+–()100403540

1

2=-us（1）（2）（3）（4）+isi1i2i3i5i6i4R2R1R6R3R5R4例25-3-3含受控電源網(wǎng)絡(luò)的節(jié)點電壓方程問題AGbATΦ=AGbUSb-AISb=-AI'bSGnΦ=InS處理辦法u1u2u3u4u5u610000002000000205000001010000050000005i1i2i3i4i5