§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:
3. If the circuit contains only current sources, apply Kirchhoff 's current law at each non-reference node. to obtain the conductance matrix if a circuit has only independent current sources, equate the total current leaving each node through all conductance to the total source current entering that node, and order the terms from v, to VN-1. For each dependent current source present, relate the source current and the controlling quantity to the variables vi, v2,..., VN-1, if they are not already in that form
3. If the circuit contains only current sources, apply Kirchhoff 's current law at each non-reference node. To obtain the conductance matrix if a circuit has only independent current sources, equate the total current leaving each node through all conductance to the total source current entering that node, and order the terms from v1 to vN-1. For each dependent current source present, relate the source current and the controlling quantity to the variables v1 , v2 ,…,vN -1 , if they are not already in that form
4.If the circuit contains voltage sources, temporaril modify the given circuit by replacing each such source by a short circuit, thus reducing the number of nodes by one for each voltage source that is present. The assigned node voltages should not be changed Using these assigned node-to-reference voltage, apply Kirchhoff s current law at each of the nodes or super- nodes in this modified circuit. Relate each source voltage to the variables vi, v2,..., VN-I, if it is not already in that form
4.If the circuit contains voltage sources, temporarily modify the given circuit by replacing each such source by a short circuit, thus reducing the number of nodes by one for each voltage source that is present. The assigned node voltages should not be changed. Using these assigned node-to-reference voltage, apply Kirchhoff ' s current law at each of the nodes or supernodes in this modified circuit. Relate each source voltage to the variables v1 , v2 ,…,vN-1 , if it is not already in that form
Example 2: (A)IfX=2S. Find the power supplied by the 10A source (B)lfX10k, Find the power supplied by the 10A source; (C=v 3. Find vi; DD)IfX=4v 3. Find vr 3v1-2v,-v2=10 Solution (4){-2v1+9v2-3v3=0 1-3v2+6v3=20 2S 0. 3S v1=25/3 3 20A Po4=v1×10=83.3W 10A (B)v3=10V,v1=144 4S 4v,/3 P,=144W (C)v3=v1/3,v1=4.3 (D)1(v3-v1)+3(v3-v2)=20+4v1/3,v1=23.8V
Example 2 : (A) If X=2S.Find the power supplied by the 10A source; (B) If X=10V, Find the power supplied by the 10A source; (C) If X=v1 /3. Find v1 ; (D)If X= 4v1 /3. Find v1 . Solution: v1 = 25/ 3 P10A = v1 10 = 83.3W − − + = − + − = − − = 3 6 20 2 9 3 0 3 2 10 1 2 3 1 2 3 1 2 3 v v v v v v v v v (A) v3 = 10V,v1 = 14.4 P10A = 144W (B) / 3, 4.3 (C) v3 = v1 v1 = X 2 (D) 1(v3 − v1 )+ 3(v3 − v2 ) = 20+ 4v1 / 3,v1 = 23.8V − 10V ++ − / 3 1 v 4 / 3 1 v
DP: use nodal analysis to find Ux if element (A) X)2A (B)X→89 (C)X→10V(-+) 10 D D 2 94 6 2 174 Re∫ (4)→>42 (B)→>24 (C)→-19.61
DP: use nodal analysis to find if element X (C ) X V( ) ( B ) X ( A) X A → − + → → 10 8 2 (A) → 42V (B) → 24V (C) → −19.6V 1 9A 17A 6 4 2 2 − + x Re f . X 2A 8 10V − +
DP-2: Find: V1,v2, V3 2 02-D1⊥b2⊥b3-0 +2 1A+ A 2 5 1,U2=3,U3=-2 (与电流源串联的电阻不介入节点方程) R Find: P2v =lA P=2w Find: p U2A=0P4=0
1 2 3 DP − 2 :Find : v ,v ,v 5 2 − 3 = 1, 3, 2 1 = 2 = 3 = − (与电流源串联的电阻不介入节点方程) Find:P2V . Find: P2A. i 1 = 1A P2V = 2W 2A = 0 P2A = 0 2 1 1 1 1 1 2 1 3 = − + − + 2 1 1 1 2 1 2 3 1 = − + + − − + 2V 2A 1 1 1 1 1 1 i − + 2A + 5V − 1 2 3 Re f