《電路與電子技術(shù)：第二章 電阻電路分析》由會員分享，可在線閱讀，更多相關(guān)《電路與電子技術(shù)：第二章 電阻電路分析（53頁珍藏版）》請在裝配圖網(wǎng)上搜索。
1、+_R1Rn+_uki+_u1+_u1uRk等效等效u+_ReqiinR1R2RkRni+ui1i2ik_等效等效inR1R2RkRni+ui1i2ik_+u_iReq令令G=1/R2 4 3 6 R例例601005010ba4080206010060ba12020100100ba204010060ba20例例4V+-10AU =?2+-3AA7A) 310(I024IUV1042 IU解解I-10V10V+-1AI =?10I1解解A3A) 12(11 IIA21I0)10(10101IR1R2R3R4R5R6+i2i3i4i1i5i6uS1234R1R2R3R4R5R6+i2i3i4i1i
2、5i6uS1234R1R2R3R4R5R6+i2i3i4i1i5i6uS1234u2+_u4+_u3+_u5+_u1+_u6+_u1 =R1i1， u2 =R2i2， u3 =R3i3，u4 =R4i4， u5 =R5i5， u6 =uS+R6i6節(jié)點節(jié)點 1：i1 + i2 i6 =0節(jié)點節(jié)點 2： i2 + i3 + i4 =0節(jié)點節(jié)點 3： i4 i5 + i6 =0節(jié)點節(jié)點 4： i1 i3 + i5 =03R1R2R3R4R5R6+i2i3i4i1i5i6uS123412回路回路1：u1 + u2 + u3 = 0回路回路2：u3 + u4 u5 = 0回路回路3： u1 + u5
3、+ u6 = 0R1 i1 + R2 i2 + R3 i3 = 0R3 i3 + R4 i4 R5 i5 = 0 R1 i1 + R5 i5 + R6 i6 uS = 03R1R2R3R4R5R6+i2i3i4i1i5i6uS123412 i1 + i2 i6 =0 i2 + i3 + i4 =0 i4 i5 + i6 =0R1 i1 + R2 i2 + R3 i3 = 0R3 i3 + R4 i4 R5 i5 = 0 R1 i1 + R5 i5 + R6 i6 uS = 0US1=130V, US2=117V, R1=1 , R2=0.6 , R3=24 .US1US2R1R2R3ba+I1
4、I3I212節(jié)點節(jié)點a：I1I2+I3=0(1) n1=1個個KCL方程：方程：(2) b( n1)=2個個KVL方程：方程：R2I2+R3I3 US2=0 U= USR1I1R2I2+US2US1=00.6I2+24I3 117=0I10.6I2+117130=0(3) 聯(lián)立求解聯(lián)立求解I1I2+I3=00.6I2+24I3 117=0I10.6I2+117130=0解得解得I1=10 AI3= 5 AI2= 5 A(4) 功率分析功率分析PU S1=US1I1=130 10=1300 WPU S2=US2I2=130 (10)=585 W驗證功率守恒：驗證功率守恒：PR 1吸吸=R1I12
5、=100 WPR 2吸吸=R2I22=15 WPR 3吸吸=R3I32=600 WP吸吸=715 WP發(fā)發(fā)= P吸吸P電源電源= 715 W(P發(fā)發(fā)=715 W)US1US2R1R2R3ba+I1I3I2123i1i3i2i5i4+ubacuSiSR1R2R3+R4- - i1- - i2 + i3 = 0 (1)- - i3+ + i4 - - i5 = 0 (2)R1 i1- -R2i2 = uS (3)R2 i2+ +R3i3 + + R4 i4 = 0 (4)- - R4 i4+ +u = 0 (5)i5 = iS (6)KVL方程：方程：KCL方程：方程：* 理想電流源的處理：由于理
6、想電流源的處理：由于i5 = iS，所以在選擇獨立回路時，可，所以在選擇獨立回路時，可不選含此支路的回路。不選含此支路的回路。對此例，可不選回路對此例，可不選回路3，即去掉，即去掉方程方程(5)，而只列，而只列(1)(4)及及(6)。 (uA- -uB)+uB- -uA=0KVL自動滿足自動滿足uA- - uBuAuBun1un2iS1iS2iS3R1i1i2i3i4i5R2R5R3R4012i1+i2+i3+i4 = iS1- -iS2+iS3- -i3 - - i4 + i5 = - - iS3 n1n2n1n2n1n2S1S2S31234uuuuuuiiiRRRRn1n2n1n2n2S3
7、345uuuuuiRRR n1n2S1S2S3123434111111() ()uuiiiRRRRRRn1n2S33434511111()() uuiRRRRR G11un1+G12un2 = iSn1G21un1+G22un2 = iSn2G11un1+G12un2 = iSn1G21un1+G22un2 = iSn2un1un2iS1iS2iS3R1i1i2i3i4i5R2R5R3R4012n111uiRn222uiRn1n233uuiRn1n244uuiRn255uiRG11un1+G12un2+G1,n- -1un,n- -1=iSn1G21un1+G22un2+G2,n-1un,n-
8、1=iSn2 Gn- -1,1un1+Gn- -1,2un2+Gn-1,nun,n- -1=iSn,n- -1un1un2uS1iS2iS3R1i1i2i3i4i5R2R5R3R4012+- -Us1/R1(G1+G2+G3+G4)un1- -(G3+G4) un2 = G1 uS1 - -iS2+iS3- -(G3+G4) un1 + (G3+G4 + G5)un2= - -iS3等效電流源等效電流源20k 10k 40k 20k 40k +120V- -240VUAUBI4I2I1I3I5(1) 列節(jié)點電壓方程：列節(jié)點電壓方程：UA=21.8V， UB=- -21.82VI1=(120-
9、-UA)/20k= 4.91mAI2= (UA- - UB)/10k= 4.36mAI3=(UB +240)/40k= 5.45mAI4= UB /40=0.546mAI5= UB /20=- -1.09mA(0.05+0.025+0.1)UA- -0.1UB= 0.006- -0.1UA+(0.1+0.05+0.025)UB=- -0.006(2) 解方程，得：解方程，得：(3) 各支路電流：各支路電流：* 可先進(jìn)行電源變換可先進(jìn)行電源變換20k 10k 40k 20k 40k +120V- -240VUAUBI4I2I1I3I5(0.05+0.025+0.1)UA- -0.1UB- UC
10、1/(20k)=0- -0.1UA+(0.1+0.05+0.025)UB-UD1/(40k)=0UCUDUC =120UD =-240S2211S2uRRi iiii 111S212S211RRRRR1 iiiui S211RR1ui僅與電壓源僅與電壓源u uS S有關(guān)有關(guān)S2121RRRii 僅與電流源僅與電流源i is s有關(guān)有關(guān)為什么把電流為什么把電流分為兩項？分為兩項？i2uSS211RR1uiS2121RRRii i2uS+10V4A6 +4 uu=4Vu= - -4 2.4= - -9.6Vu=u+u= 4+(- - 9.6)= - - 5.6V4A6 +4 u+10V6 +4 u
11、AabiiabRiUoc+- -(a)(b)abAi+uNiUoc+uNab+Ri=+電流源電流源i為零為零網(wǎng)絡(luò)網(wǎng)絡(luò)A中獨立源全部置零中獨立源全部置零abAi+uabA+uabPi +uRi根據(jù)疊加定理，可得根據(jù)疊加定理，可得u= Uoc (外電路開路時外電路開路時a 、b間開路電壓間開路電壓) u= Ri i則則u = u + u = Uoc - - Ri i此關(guān)系式恰與圖此關(guān)系式恰與圖(b)電路相同。證畢！電路相同。證畢！IRxab+10V4 6 6 4 ab+10V4 6 6 +U24 +U1IRxIabUoc+RxRiUoc = U1 + U2 = - -10 4/(4+6)+10 6
12、/(4+6) = - -4+6=2Vab+10V4 6 6 +U24 +U1+- -UocRiab4 6 6 4 Ri=4/6+6/4=4.8 I= Uoc /(Ri + Rx) =2/6=0.333ARx =5.2 時，時，I= Uoc /(Ri + Rx) =2/10=0.2AAababRiIsc12V2 10 +24Vab4 I+4 Iab RiIsc(1)求求IscI1 =12/2=6A I2=(24+12)/10=3.6AIsc=- -I1- -I2=- - 3.6- -6=- -9.6A2 10 +24VabIsc+I1I212V(2) 求求Ri：串并聯(lián)：串并聯(lián)Ri =10 2/(
13、10+2)=1.67 (3) 諾頓等效電路諾頓等效電路:I = - - Isc 1.67/(4+1.67) =9.6 1.67/5.67 =2.83ARi2 10 abb4 Ia1.67 - -9.6A8513UIII13abURIab+一UI+-5I83 3 6 I+9V+U0ab+6IabUoc+Ri3 U0- -+3 6 I+9V+Uocab+6IUoc=6I+3II=9/9=1AUoc=9V3 6 I+U0ab+6II03 6 I+9VIscab+6II1U0=6I+3I=9II=I0 6/(6+3)=(2/3)I0U0 =9 (2/3)I0=6I0Ri = U0 /I0=6 (Uoc=9V)3 電阻中電流電阻中電流I為：為：I=- -6I/3=- -2I I=0Isc=9/6=1.5ARi = Uoc / Isc =9/1.5=6 abUoc+Ri3 U0- -+6 9V0393V63U R12R31R23i3 i2 i1 123R1R2R3i1Yi2Yi3Y123R12R31R23i3 i2 i1 123R1R2R3i1Yi2Yi3Y123R12R31R23i3 i2 i1 123R1R2R3i1Yi2Yi3Y123
電路與電子技術(shù)：第二章 電阻電路分析
