Node Voltage Method - University Of California, Berkeley
Node Voltage MethodThis simplest among many circuit analysis methods isapplicable only for connected circuits N made of linear 2terminal resistors and current sources. The only variables in thelinear equations are the n-1 node voltages e1, e2 , , en-1 for ann-node circuit.Step 1. Choose an arbitrary datum node and label theremaining nodes consecutively 1 , 2 , , n-1 , andlet e1, e2 , , en-1 be node-to-datum voltages.Step 2. Express the current of each resistor Rj via Ohm’slaw in terms of 2 node-to-datum voltages:ij e jRj Ωe jij e j e jRj(1)Step 3. Apply KCL to each node 1 , 2 , , n-1 with each resistor current ij expressed in terms of e j and e j .Step 4. Solve the (n-1) independent linear equations for e1, e2 , , en-1.Step 5. Solve for the resistor currents via Eq.(1).
i1 EXAMPLE 1.2v6 e25Ai6 e1 e22v1 2Ωe2i4 8v48Ω-1e1e1 e3i2 4v2v34Ω- 2Ai3v5 -6Ω3e3e3i5 6(e1 e2 ) (e1 e3 )(2) 524 (e1 e2 ) e22(3) 2KCL at:28 (e1 e3 ) e33(4) 2KCL at:4611 Recast Eqs. (2), (3), 3and (4) in 4 2 4 matrix form : e1 5 1 5(5)Node0 e2 2 2 8 voltage e3 2 5Equations 1 0 412 KCL at1:
S olving eq. (5) by C ram er's R ule (or any other m ethod) :11 3 424 2015 det 0 (6 )28384 5 1 0 41 2 (a) T o solve for e1 , replace colum n 1 of m atrix from (5) and calculate11 5 24 1155 1 det 20 (7 )968 5 2 01 2 C ram er's R ule e1 1 115 96 20 23 V384 (8 )(b) T o solve for e 2 , replace colum n 2 of m atrix from (5) and calculate1 35 44 1 76(9 ) 2 det 0 296 12 5 2 41 2 C ram er's R ule e2 2 76 96 20 15.2 V384 (c) T o solve for e3 , replace colum n 3 of1 35 42 15 3 det 228 2 10 412C ram er's R ule e3 3 93 96 m atrix from (5) and calculate 93 (1 0 )96 20 18.6 V 384 (1 1 )
O nce the node-to-datum voltages {e1 , e 2 , e 3 } are found,all resistor currents and voltages are found trivially viaK V L, K C L, and O hm 's law :7.8 3.9 A24.4 1.1 A4 2 A15.2 1.9 A818.6 3.1 A6 5Av1 e1 e 2 7.8 V, i1 v 2 e1 e 3 4.4 V, i2v 3 e 3 e 2 3.4 V, i3v 4 e 2 15.2 V, i4v 5 e 3 18.6 V, i5v 6 e1 23 V, i6V erification of S olu tionA pp ly T ellegen's T he o rem :6 vj 1ji j (v1 i1 ) (v2 i2 ) (v3 i3 ) (v4 i4 ) (v5 i5 ) (v6 i6 ) (7.8)(3.9) (4.4)(1.1) (3.4)( 2) (15.2)(1.9) (18.6)(3.1) ( 23)(5)? 0
Modified Node Voltage Methodv1 ee2Circuit N1 i1e e1i 24Ω4121 2Av4v2 i3i23Ω3v3- 6Vi2 e13i4Step 1.When the circuit contains "α " voltages vs1 , vs2 , " , vsα , usetheir associated currents is1 , is2 , " , isα when applying KCL.KCL at 1 :KCL at 2 :e1 (e1 e2 ) 234(e e ) 1 2 i3 04(1)(2)Step 2.For each voltage source v s j , add an equation e j e j v s j .e2 6(3)Step 3.Solve the ( n -1) α equations for e1 , e 2 , " e n 1 , is1 , is2 , " isα .S u b s titu tin g (3 ) in to (1 ), w e o b ta in :e1 (e1 6) 2 e1 6 V34S u b s titu tin g (4 ) in to (2 ), w e o b ta in :i3 0(4)(5)
Explicit matrix form of Node Voltage EquationsAssumption: Circuit N contains only linear resistors andindependent current sources which do not form cut sets.Step 1. Delete all “β” current sources from N and draw thereduced digraph G of the remaining pure resistor circuit.Assume G has “n” nodes and “b” branches.Step 2. Pick a datum node and label the node-to-datum voltages{ e1, e2 , , en-1 }, and derive the reduced-incidence matrix A.Define the “branch admittance matrix” Yb and independentcurrent source vector is as follow: is1 0 0 " 0 v1 i1 Y1 i 0 Y 0 " 0 v i2 2 2 (1) , i s2 (2)s # # ## ### # is β ib 0 0 0 " Yb vb iYvb1where Y j , R j resistance of branch j.Rjism algebraic sum of all current sources entering node m ,m 1, 2, , n -1.Step 3. Form the Node voltage equationYne i s(3)where Yn AYb AT is called the node - admittance matrix.
Deriving the Node Voltage Equation( Yne i s )The “β ” current sources can bedeleted since they can be trivially accountedfor by representing their net contribution ateach node mby the algebraic sum of allcurrent sources entering node m , m 1, 2, , n-1. The KCL equations therefore takesthe “augmented” formA i is(4)Substituting (1) for i in (4), we obtainA Yb v i s(5)Substituting KVLv A eT(6)for v in (5), we obtain()A Yb AT e i sYn(7)
Writing Node-Admittance MatrixYn By InspectionNode Votage Equation :Y12 " Y1,n 1 e1 is1 Y11 Y i "YY222, n 1 e2 s 21 2 ##"# # # Yn 1,1 Yn 1,2 " Yn 1,n 1 en 1 isn 1 eYn(8)isDiagonal Elements of YnY m m su m o f ad m ittan ces Y j connected to node m1o f all resisto rsRj, m 1, 2, , n - 1Off-Diagonal Elements of YnY jk - ( sumo f ad m ittances Y j connected across node j1of all resisto rsRjand node k )Symmetry Property:Y n is a s ym m e tric m a tri x , i.e .,Y jk Y k jP ro o f :S in c e Y b in (1 ) is a d ia g o n a l m a trix , Y b Y b T(Y nT A Y b A T)T A Y bT A T A Y b A T Y n
i1 EXAMPLE4A-2v65Ae2i6 3Ae1 e22v1 2Ωe2i4 8v48Ω-1e1e1 e3i2 4v2v34Ω- 2Ai3v5 -6Ω3e3ei5 361A(e1 e2 ) (e1 e3 )(2) 5 424 (e1 e2 ) e22(3) 4 2 3KCL at:28 (e1 e3 ) e33(4) 2 1KCL at:46Matrix Form : 311 4 2 4 e1 1 Node 1 5(5)0 e2 1Equations 2 8 e3 3 5 1 0 412 KCL at1:
Mesh Current MethodThe next simplest among many circuitanalysis methods is applicable only forconnected planar circuit N (with a planardigraph) made of 2-terminal linear resistorsand voltage sources. The only variables in theequations are “ l ” conceptual mesh currentslˆm1 , lˆm 2 , " , lˆml circulating in the “ l ” meshes ina clockwise direction (by convention) 4R3i13i8i5i11R11iˆ5R6i6R13i9iˆ6R9
iˆcijiˆaRjiˆbiˆdi j iˆa iˆb
Mesh Current MethodThe next simplest among many circuitanalysis methods is applicable only for connectedplanar circuits made of 2-terminal linear resistors andvoltage sources. The only variables in the associated“mesh-current equations” are “ l ” mesh current lˆm1 ,lˆm 2 , " , lˆml which we define to be circulating in the“ l ” meshes in a clockwise direction (by convention).Unlike node-to-datum voltages in the node voltagemethod which are physical in the sense they can bemeasured by a volt meter, the “mesh” currents areabstract variables introduced mathematically forwriting a set of equations whose solution can be usedto find each resistor current ij trivially viai j iˆa iˆbWhere iˆa (resp., iˆb ) is the circulating current flowingthrough Rj in the same (resp., opposite) direction as thereference current ij.
MeshCurrentEquations 6 0 2 iˆ1 2 0 14 8 iˆ 2 2 2 8 10 iˆ3 5 (1)Then calculate :13819iˆ1 , iˆ2 , iˆ3 2020203ˆˆi1 i3 i1 1013i2 iˆ1 205ˆˆi3 i1 i2 2011i4 iˆ3 iˆ2 20,,,,(2)6v1 2i1 1013v2 4i2 5v3 222v4 8i4 5(3)128,ˆv5 6i5 i5 i2 52019,v6 5i6 iˆ3 20Verification by Tellegen's Theorem :6 v ij jj 113 6 3 13 ( 2 ) 10 10 5 20 ? 05 22 11 12 8 19 ( 5 ) 20 5 20 5 20 20
5V -v6iˆ3i6-v1 2Ωi1 iˆ3 iˆ1i2 iˆ1i3 iˆ1 iˆ2i4 iˆ3 iˆ2i5 iˆ2i6 iˆ3 v2ˆ1 4 Ωiv-3 i4 i2i12V8Ωv4,,,,,, i3v5 iˆ2 (v1 2 iˆ3 iˆ1v2 4iˆ16Ω)v3 2v4 8 iˆ3 iˆ2(i5)v5 6iˆ2v6 5Loop equation around mesh 1: v1 v2 v3 0 2(iˆ3 iˆ1 ) 4iˆ1 2 0 6iˆ1 2iˆ3 2(1)Loop equation around mesh 2: v3 v5 v4 0 ( 2) 6iˆ2 8(iˆ3 iˆ2 ) 0 14iˆ2 8iˆ3 2(2)Loop equation around mesh 3:v6 v1 v4 0 5 2(iˆ3 iˆ1 ) 8(iˆ3 iˆ2 ) 0 2iˆ1 8iˆ2 10iˆ2 5(3)
5V -iˆ3v6-v1 2Ωv4ˆ1 4 Ωiv 2V v2-3i4i6i2i18Ω- i3v5 iˆ2 -6Ω6 iˆ1 2 iˆ3 2M esh 2 : 1 4 iˆ2 8 iˆ3 2M esh 1 :M esh 3 : 2 iˆ1 8 iˆ2 1 0 iˆ3 51ˆ2ˆ i1 i3 3341S o lvin g iˆ2 fro m (2 ) iˆ2 iˆ3 77S u b stitu tin g (4 ) an d (5 ) in to (3 ) ˆi3 1 9 A128iˆ2 A(5 ) an d (6 ) 2013ˆi1 A(4 ) an d (6 ) 20S o lvin g iˆ1 fro m (1 )i5(1 )(2 )(3 )(4 )(5 )(6 )(7 )(8 )
i213i1i5i3i624i4We can redraw this circuit so that there areno intersecting branches.1i2i1i63i5i32i44Hence the above circuit is planar and it is possibleto formulate mesh current equations.
4Ωi21i13i3 2Ωi55V2V6Ω i68Ω24i4We can redraw this circuit so that there areno intersecting branches.1i24Ωi1- 5Vi6i5i3 2Ω32V6Ω8Ω2i44Hence the above circuit is planar and it is possibleto formulate mesh current equations.
All branch voltages and currents can be triviallycalculated from e1 and i3.v1 e1 e 2 0 Vv 2 e1 6 Vv3 e2 6 Vv 4 e1 6 Vv1 0A4v2, i2 2A3, i3 0 A, i1 , i4 2 AV erification of S olution by T ellegen's T heor em :4 vj 1ji j (v1 i1 ) (v2 i2 ) (v3 i3 ) (v4 i4 ) (0)(0) (6)(2) (6)(0) ( 6)(2)? 0
Note:The unknown variables in themodified node voltage method consistoftheusualn-1node-to-datumvoltages, plus the unknown currentsassociated with the voltage sources.Hence, if there are “α” voltagesources, the modified node voltagemethod would consist of (n-1) αindependent linear equations involving(n-1) α unknown variables{e , e ," , e12n 1(n-1) node-to-datumvariables}, is1 , is2 , " isα .α currentvariables
Conservation of ElectricalEnergyThealgebraicsumofelectrical energy flowing into alldevices in a connected circuit iszero for all times t .Proof .Tellegen's Theorem b j 1t v j (t ) i j (t ) dt 0for all t.
000 0 0 000010 0 0 0 11 0 0 0 00 0 0 1 0 00 1 0 0 1 0det A1 0 0 0 0 0 1100010 10010000000000000060001 0 0 0 4 0 00 1 0 0 0 3 00 0 1 0 0 0 0000200 0 0 01A1000 1
000 0 0 00 1010 0 0 0 11 0 0 0 0 1 0 0 0 0 00 1 0 0 1 0det A 4 1 0 0 0 0 1100010 10010000000000000000001 0 0 0 4 0 00 0 0 0 0 3 00 6 1 0 0 0 0000000 2 0 01A4000 4
000 0 0 00 1010 0 0 0 11 0 0 0 0 1 0 0 1 0 00 1 0 0 1 0det A 9 1 0 0 0 0 1100010 10000000000000000000001 0 0 0 4 0 00 1 0 0 0 3 00 0 1 0 0 0 6000000 0 0 01A9002 9
1e1 1 a11 A11 a21 A21 a31 A31 a41 A41 a51 A51 a61 A61 a71 A71 a81 A81 a91 A91 a10,1 A10,1 1 ( a91 A91 a10,1 A10,1 ) , because a11 a21 " a81 0 A10,1A91 a10,1 )( a91 ) ( k11k12 k11 6 k12 2 k11 vs1 k12 is1
Instantaneous Powerof a 2-terminal devicei (t )1 v(t ) D2p(t ) v(t ) i (t )Under Associated Reference Convention,p (t ) 0 at t T1means p(T1 ) Watts of power enters(flows into) D at t T1.p (t ) 0 at t T2means p(T2 ) Watts of power leaves(flows out of) D at t T2.Energy entering D from time T1 to T2 :T2WT1 T2 v(t ) i (t ) dtT1
Instantaneous Powerof an n-terminal devicek2vkv2 i2v1 i1ikD1in 1vn 1n-1nn 1p(t ) v j (t ) i j (t )j 1Energy entering D from time T1 to T2 :n 1WT1 T2 v j (t ) i j (t ) dtj 1T2T1
Tellegen’s Theorem has many deepapplications. For this course, it can be used tocheck whether your answers in homeworkproblems, midterm and final exams arecorrect.
Node Voltage Method This simplest among many circuit analysis methods is applicable only for connected circuits N made of linear 2-terminal resistors and current sources. The only variables in the linear equations are the n-1 node voltages e1, e2, , en-1 for an n-node circuit. Step 1. Choo
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Sep 19, 2014 · The node-voltage is the first (and maybe most used) of our three formal methods. The node-voltage method allows for the calculation of the voltages at each node of the circuit, relative to a reference node. Once the node voltages are kno
potential of node a or b with respect to the reference node, c. To solve for the unknown node voltages in this circuit, begin by applying Kirchhoff's current law at node a. Using Ohm’s Law, the current through R 1 and R 2 can be expressed in terms of the unknown node voltage at node
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A RPL node may attach to a DODAG as a leaf node only. One example of such a case is when a node does not understand or does not support (policy) the RPL Instance's OF or advertised metric/constraint,the node may either join the DODAG as a leaf node or may not join the DODAG. A node operating as a leaf node must obey the following rules: 1.
