第二章-电阻电路分析-1

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 RR1R2R3R4R5R6+i2i3i4i1i5i6uS1234u1 =R1i1， u2 =R2i2， u3 =R3i3，u4 =R4i4， u5 =R5i5， u6 =uS+R6i6节点节点 1：i1 + i2 i6 =0节点节点 2： i2 + i3 + i4 =0节点节点 3： i4 i5 + i6 =0节点节点 4： i1 i3 + i5 =0R1R2R3R4R5R6+i2i3i4i1i5i

2、6uS1234u2+_u4+_u3+_u5+_u1+_u6+_3R1R2R3R4R5R6+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 + u6 = 0 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 = 0KCLKVLUS1=130V, US2=117V, R1=1 , R2=0.6 , R3=24 .

3、I1I3US1US2R1R2R3ba+I212节点节点a：I1I2+I3=0(1) n1=1个个KCL方程：方程：(2) b( n1)=2个个KVL方程：方程：R2I2+R3I3= US2 U= USR1I1R2I2=US1US20.6I2+24I3= 117I10.6I2=130117=13(3) 联立求解联立求解I1I2+I3=00.6I2+24I3= 117I10.6I2=130117=13解之得解之得I1=10 AI3= 5 AI2= 5 A(4) 功率分析功率分析PU S1发发=US1I1=130 10=1300 WPU S2发发=US2I2=130 (10)=585 W验证功率守恒

4、：验证功率守恒：PR 1吸吸=R1I12=100 WPR 2吸吸=R2I22=15 WPR 3吸吸=R3I32=600 WP发发=715 WP吸吸=715 WP发发= P吸吸123i1i3uSiSR1R2R3ba+i2i5i4ucR4- - 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：* 理想电流源的处理：由于理想电流源的处理：由于i5 = i

5、S，所以在选择独立回路时，可，所以在选择独立回路时，可不选含此支路的回路。不选含此支路的回路。对此例，可不选回路对此例，可不选回路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自动满足自动满足uA- - uBuAuBun1un2iS1iS2iS3R1i1i2i3i4i5R2R5R3R4012i1+i2+i3+i4 = iS1- -iS2+iS3- -i3 - - i4 + i5 = - - iS3 n1n2n1n2n1n2S1S2S31234uuuuuuiiiRRRRn1n2n1n2n2S3345uuuuuiRRR n1n

6、2S1S2S3123434111111() ()uuiiiRRRRRRn1n2S33434511111()() uuiRRRRR G11un1+G12un2 = iSn1G21un1+G22un2 = iSn2G11un1+G12un2 = iSn1G21un1+G22un2 = iSn2un1un2iS1iS2iS3R1i1i2i3i4i5R2R5R3R4012n111uiRn222uiRn1n233uuiRn1n244uuiRn255uiRG11un1+G12un2+G1,n- -1un,n- -1=iSn1G21un1+G22un2+G2,n-1un,n-1=iSn2 Gn- -1,1u

7、n1+Gn- -1,2un2+Gn-1,nun,n- -1=iSn,n- -1un1un2uS1iS2iS3R1i1i2i3i4i5R2R5R3R4012+- -Us1/R5(G1+G2+G3+G4)un1- -(G3+G4) un2 = G1 uS1 - -iS2+iS3- -(G3+G4) un1 + (G3+G4 + G5)un2= - -iS3等效电流源等效电流源20k 10k 40k 20k 40k +120V- -240VUAUBI4I2I1I3I5(1) 列节点电压方程：列节点电压方程：UA=21.8V， UB=- -21.82VI1=(120- -UA)/20k= 4.91mA

8、I2= (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) 各支路电流：各支路电流：* 可先进行电源变换可先进行电源变换20k 10k 40k 20k 40k +120V- -240VUAUBI4I2I1I3I5(0.05+0.025+0.1)UA- -0.1UB- UC 1/(20k)=0- -0.1U

9、A+(0.1+0.05+0.025)UB-UD1/(40k)=0UCUDUC =120UD =-240S2211S2uRRi iiii 111S212S211RRRRR1 iiiui S211RR1ui仅与电压源仅与电压源u uS S有关有关S2121RRRii 仅与电流源仅与电流源i is s有关有关为什么把电流为什么把电流分为两项？分为两项？i2S211RR1uiS2121RRRii +10V4A6 +4 uu=4Vu= - -4 2.4= - -9.6Vu=u+u= 4+(- - 9.6)= - - 5.6V4A6 +4 u+10V6 +4 uAabiiabRiUoc+- -(a)(b)

10、abAi+uNiUoc+uNab+Ri=+电流源电流源i为零为零网络网络A中独立源全部置零中独立源全部置零abAi+uabA+uabPi +uRi根据叠加定理，可得根据叠加定理，可得u= Uoc (外电路开路时外电路开路时a 、b间开路电压间开路电压) u= Ri i则则u = u + u = Uoc - - Ri i此关系式恰与图此关系式恰与图(b)电路相同。证毕！电路相同。证毕！IRxab+10V4 6 6 4 ab+10V4 6 6 +U24 +U1IRxIabUoc+RxRiUoc = U1 + U2 = - -10 4/(4+6)+10 6/(4+6) = - -4+6=2Vab+1

11、0V4 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 时，时，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：串并联：串并联Ri =10 2/(10+2)=1.67 (3) 诺顿等效电路诺

12、顿等效电路:I = - - Isc 1.67/(4+1.67) =9.6 1.67/5.67 =2.83ARi2 10 abb4 Ia1.67 - -9.6Aab+一UI+-5I88513UIII13abURI+10V6 I14A+Us+10 I14 +10V6 I1+Us+10 I14 6 I14A+Us+10 I14 Us= - -10 I1+4 = - -10 1+4= - -6VUs= - -10I1+2.4 4 = - -10 (- -1.6)+9.6=25.6VUs= Us +Us“= - -6+25.6=19.6V3 3 6 I+9V+U0ab+6IabUoc+Ri3 U0- -

13、+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 电阻中电流电阻中电流I为：为：I=- -6I/3=- -2I I=0Isc=9/6=1.5ARi = Uoc / Isc =9/1.5=6 abUoc+Ri3 U0- -+6 9V0393V63U abUoc+U R0.5k Ri+10V1k 1k 0.5Iab R0.5k +UI1k 1k 0.5Iab+

14、U0II0U0 =(I0- -0.5 I0) 103+ I0 103 =1500I0Ri = U0 / I0 =1.5k +10V1k 1k 0.5IabIIscIsc = - -I，(I- -0.5I) 103 +I 103+10=01500I= - -10I= - -1/150 A即即 Isc=1/150 A Ri = Uoc / Isc =10 150=1500 Uoc =10V（已求出）（已求出）求短路电流求短路电流Isc (将将a、b短路短路)：1k 1k 0.5Iab+U0II0I= I0U0 =0.5I0 103 +I0 103 =1500I0 Ri = U0 /I0=1500

15、ab10V+U R0.5k 1.5k U=Uoc 500/(1500+500)=2.5V静态电阻 :iuR diduRd动态电阻:据据KVLKVL方程得方程得: : ( ( 该方程确定了直该方程确定了直流负载线流负载线.).)uiRuUisS00uIRUiS对整个电路据对整个电路据KVLKVL方程得方程得 : :该直线与伏安特性曲该直线与伏安特性曲(I(I0 0=g(U=g(U0 0)相交得静态工作点相交得静态工作点. .u u1 1,i,i1 1 在在(U(U0 0 ,I ,I0 0) ) 附近工作附近工作, ,可用可用(U(U0 0 ,I ,I0 0) ) 附近的切线来近似附近的切线来近似10iIi10uUu00010001( )(