§2-1 Nodal analysis 5Q 34 2 24 Ref Re∫ We select one node as a reference node and then define a voltage between each remaining node and the reference node We must apply kcl to nodes 1 and 2 65 a×2 0.7U1-0.2U2=3 U1=5)bsg=0.54 -0.201+.22=2U2=25P3A=5×3=15W 十
§2-1 Nodal analysis We must apply KCLto nodes 1 and 2 − + = − = 0 2 1 2 2 0 7 0 2 3 1 2 1 2 . . . . or We select one node as a reference node, and then define a voltage between each remaining node and the reference node. = − − − + = − + ( 2 ) 1 5 3 2 5 2 2 1 1 1 2 = = . V V 2 5 5 2 1 P W i . A A 5 3 15 0 5 3 5 = = = Re f 1 2 3A − 2A 2 1 5 1 Re f 2 3A − 2A 2 1 i 5 5
Now let us increase the number of nodes by one 4S nodel:3(u1-U2)+4U1-u3)=-8-3 7U1-3U2-4U3=-11 2S nde2:3(u2-u1)+2(2-U3)+12=3 8A -3U,+6U,-2U2=3 IS ode 5U3+2{U3-U2)+4(U3-01)=25 254 r-4U,-20,+11U,=25 Ref By cramer s rule and determinants, we have 11-3-4 25 211191 2 L 3-4191 36 211
2 : 3( ) 2( ) 1 3 node 2 −1 + 2 − 3 + 2 = 3 : 5 2( ) 4( ) 25 node 3 + 3 −2 + 3 −1 = 7 3 4 11 or 1 − 2 − 3 = − 3 6 2 3 or − 1 + 2 − 3 = 4 2 11 25 or − 1 − 2 + 3 = By Cramer' s rule and determinants, we have 1 191 191 4 2 11 3 6 2 7 3 4 25 2 11 3 6 2 11 3 4 1 = = − − − − − − − − − − − = 2 = 2 and 3 = 3 Now let us increase the number of nodes by one. 1 3 4 8 3 node : (1 − 2 )+ (1 − 3 ) = − − 1 2 3 − 25A −8A − 3A 1S 5S 4S 3S 2S Re f
R4 000V 3000V 2 0.25 ) 000V R1-3A R5 0.3333 0.5 个 R2 R3 3 8A 0.2 25A M 0
I 1 -8A R 4 0.25 R 2 1 3.000V 2.000V I 2 -3A R 5 0.5 R 1 0.3333 1.000V R 3 0.2 0V 0 I 3 -25A
nodl:3(U1-U2)+4(b1-U3)=-8-3 D,-5N+ orTu1-3U2-4U2=-11(1 -34 3S node2:3(U2-b1)+1U2+i,-3=0 or-3U1+4U2+i=3 (2) IS node3:5U3+4(U3-U1)-i,-25=0 +9U2-i=25 3 254 2)+(3)→-+42+9=28(4) Ref D3-D2=22 (5 Equations(1),(4),(5)may be solved 11-3-4 849 22 189 =-4.5 155U3=65 3 749
22 ( 5 ) 3 − 2 = ( 2 ) ( 3 ) 7 4 9 28 ( 4 ) + − 1 + 2 + 3 = node2 : 3(2 −1 )+12 + i s − 3 = 0 node3 : 53 + 4(3 −1 )− i s − 25 = 0 7 3 4 11 (1) or 1 − 2 − 3 = − 3 4 3 (2) or − 1 + 2 + i s = 4 9 25 (3) or − 1 + 3 − i s = Equations (1), (4), (5) may be solved 4 5 42 189 0 1 1 7 4 9 7 3 4 22 1 1 28 4 9 11 3 4 1 = − . − = − − − − − − − − = 1 3 4 8 3 node : (1 − 2 )+ (1 − 3 ) = − − 2 = −15.5 3 = 6.5 − 22V + s i 1 2 3 − 25A −8A − 3A 1S 5S 4S 3S Re f
Let us summarize the method by which we obtain a set of nodal equations for any resistive circuit: 1 Make a neat, simple, circuit diagram. Indicate all element and source values. Each source should have its reference symbol 2. Assuming that the circuit has N nodes, choose one of these nodes as a reference node. Then write node voltages v1, V2,.WN-I at their respective nodes, remembering that each node voltage is under-stood to be measured with respect to the chosen reference
1 .Make a neat, simple, circuit diagram. Indicate all element and source values. Each source should have its reference symbol. 2. Assuming that the circuit has N nodes, choose one of these nodes as a reference node. Then write node voltages v1 , v2 ,…vN-1 at their respective nodes, remembering that each node voltage is under-stood to be measured with respect to the chosen reference. Let us summarize the method by which we obtain a set of nodal equations for any resistive circuit: