Nodal Mesh
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LEARNING EXAMPLE
CONVERT THIS Y INTO A DELTA?
SHOULD KEEP THESE TWO NODES!
IF WE CONVERT TO Y INTO A DELTA THERE ARE SERIES PARALLEL REDUCTIONS!
++=
RRRRRRRR
b
accbba1
3*12 *12 3612k k k
kΩ Ω
= = ΩΩ
++
++=
RRRRRRR
RRRRRRR
R
c
accbba
b
2
12kΩ
36k12k
THE RESULTINGCIRCUIT IS A CURRENT DIVIDER
Δ−
++=
YR
RRRRRRRa
accbba3 4mA 36k
36k12k OV
+
−
CIRCUIT AFTER PARALLEL RESISTORREDUCTION
4mA 36k
36 ||12 9k k k=
+
REDUCTION
OI36 8k
OV−
9k 36 8436 18 3O
kI mA mAk k
= × =+
89 9 243O OV k I k mA V= Ω× = Ω× = NOTICE THAT BY KEEPING
THE FRACTION WE PRESERVEFULL NUMERICAL ACCURACY
NODAL AND LOOP ANALYSIS TECHNIQUES
NODAL ANALYSISLOOP ANALYSIS
Develop systematic techniques to determine all the voltagesp y q gand currents in a circuit
NODE ANALYSIS
• One of the systematic ways to determine every voltage and y gcurrent in a circuit
The variables used to describe the circuit will be “Node Voltages”-- The voltages of each node with respect to a pre-selected
reference nodereference node
A PROBLEM SOLVED BEFORE USING SERIES/PARALLEL RESISTOR COMBINATIONS
COMPUTE ALL THE VOLTAGES AND CURRENTS IN THIS CIRCUITCOMPUTE ALL THE VOLTAGES AND CURRENTS IN THIS CIRCUIT
k12kk 12||4
SECOND: “BACKTRACK” USING KVL, KCL OHM’S
V
k63I
kVI a62 = :SOHM' 0321 =−− III :KCL
3*3 IkVb = :SOHM' …OTHER OPTIONS...
12
3
4
34
*4124
12
IkV
II
b =+
=
345 0III =−+ :KCL
kk 6||6FIRST REDUCE TO A SINGLE LOOP CIRCUIT
5*3 IkVC = :SOHM'
V12kVI
1212
1 =)12(
933+
=aV
DEFINING THE REFERENCE NODE IS VITAL
−+ 12V
+V4
−V2
− +
SMEANINGLESIS4V VSTATEMENT THE 1 =
UNTIL THE REFERENCE POINT IS DEFINEDUNTIL THE REFERENCE POINT IS DEFINED
BY CONVENTION THE GROUND SYMBOLSPECIFIES THE REFERENCE POINT.
ALL NODE VOLTAGES ARE MEASURED WITHRESPECT TO THAT REFERENCE POINT
THE STRATEGY FOR NODE ANALYSIS1. IDENTIFY ALL NODES AND SELECT
A REFERENCE NODESV aV bV cV
2. IDENTIFY KNOWN NODE VOLTAGES
3. AT EACH NODE WITH UNKNOWN VOLTAGE WRITE A KCL EQUATION
( 0)(e.g.,SUM OF CURRENT LEAVING =0)
0:@ 321 =++− IIIVa
4. REPLACE CURRENTS IN TERMS OF NODE VOLTAGES
VVVVV
REFERENCE
0369
=−
++−
kVV
kV
kVV baasa AND GET ALGEBRAIC EQUATIONS IN
THE NODE VOLTAGES ...
0:@ 543 =++− IIIVb
0:@ =+− IIV
0943
=−
++−
kVV
kV
kVV cbbab
SHORTCUT:SHORTCUT: SKIP WRITING THESE EQUATIONS...
AND PRACTICE WRITING 0:@ 65 =+− IIVc
039
=+−
kV
kVV cbc
THESE DIRECTLY
CIRCUITS WITH ONLY INDEPENDENT SOURCES
@ NODE 1
02
21
1
1 =−
++−R
vvRviA SRESISTANCE USING
@ NODE 2@ NODE 2
THE MODEL FOR THE CIRCUIT IS A SYSTEMOF ALGEBRAIC EQUATIONS
EXAMPLE
WRITE THE KCL EQUATIONS
@ NODE 1 WE VISUALIZE THE CURRENTSLEAVING AND WRITE THE KCL EQUATION
REPEAT THE PROCESS AT NODE 2REPEAT THE PROCESS AT NODE 2
03
12
4
122 =
−+
−+−
Rvv
Rvvi
LEARNING EXTENSION: FIND NODE VOLTAGES V1 and V2
REARRANGE AND MULTIPLY BY 10k
][402 21 =− VVV eqs.addand2/*02
][402
21
21
=+ VVVVV eqs.addand 2/
VVVV 16805 11 =⇒=
04:@ 2111 =
−+−
VVmAVV
NODE EQUATIONSVVVV 8
2 21
2 −=⇒−=
010
410
:@ 1 +k
mAk
V
010
210
:@ 2122 =++
−k
VIkVVV O
CONTROLLING VARIABLE (IN TERMS ON NODEVOLTAGES)
kVIO 10
1=k10
REPLACE
010
410
211 =−
+−kVVmA
kV
01010
210
10102112 =++
−k
Vk
VkVV
kk
3 nodes plus the reference. In principle one needs 3 equations
CIRCUITS WITH INDEPENDENT VOLTAGE SOURCES Find V1, V2 and V3
principle one needs 3 equations...
…but two nodes are connected tothe reference through voltage g gsources. Hence those node voltages are known!!!
O l C i …Only one KCL is necessary
012126
12322 =−
+−
+kVV
kVV
kV
Q O SSO GHi t E h lt
][6][12
3
1
VVVV
−==
THESE ARE THE REMAININGTWO NODE EQUATIONS
0)()(2 12322 VVVVV =−+−+EQUATIONSTHESOLVINGHint: Each voltage source
connected to the referencenode saves one node equation
][5.1][64 22 VVVV =⇒=
We will use this example to introduce the concept of a SUPERNODE
THE SUPERNODE TECHNIQUE
SUPERNODE
SI
Efficient solutionEfficient solution: enclose the source, and all elements in parallel inside a surface
Conventional node analysis requires all currents at a node
@V 1 06 1 =++ IVmA
parallel, inside a surface.
Apply KCL to the surface!!!
046 21 =+++− mAVVmA@V_1 06
6 =++− SIk
mA
@V_2 012
4 2 =++−k
VmAIS
04126
6 +++ mAkk
mA
The source current is interiorto the surface and is not required
2 eqs, 3 unknownsThe current through the source is notrelated to the voltage of the source
h l i dd i
We STILL need one more equation
][621 VVV =−Math solution: add one equation
][621 VVV =−Only 2 eqs in two unknowns!!!
Equations The
ALGEBRAIC DETAILS
][6
046126
21
VVV
mAmAk
Vk
V=+−+
(2)
(1)
][621 VVV =− (2)
...by Multiply Eq(1). in rsdenominato Eliminate 1.Solution
][6][242
21
21
VVVVVV
=−=+
][21
][10][303 11 VVVV =⇒=2Veliminateto equations Add2.
][][ 11
][4][612 VVVV =−=2Vcompute to Eq(2) Use 3.
][][612 VVVV
SUPERNODE
LEARNING EXTENSION
VV 6 SOURCES CONNECTED TO THE
VVVV4
6
4
1
−== SOURCES CONNECTED TO THE
REFERENCE
CONSTRAINT EQUATION VVV 1223 =−
KCL @ SUPERNODE
02
)4(212
6 3322 =−−
+++−
kV
kV
kV
kV
k2/*
VVV 223 32 =+VVV 1232 =+− addand 3/*VV 385 3 =
V
OIV FORNEEDED NOT IS 2
mAk
VIO 8.32
3 == LAW SOHM'
CIRCUITS WITH DEPENDENT SOURCESPRESENT NO SIGNIFICANT ADDITIONAL COMPLEXITY THE DEPENDENT SOURCESCOMPLEXITY. THE DEPENDENT SOURCESARE TREATED AS REGULAR SOURCES
WE MUST ADD ONE EQUATION FOR EACHWE MUST ADD ONE EQUATION FOR EACHCONTROLLING VARIABLE
SUPER NODE WITH DEPENDENT SOURCE Find Io
VOLTAGE SOURCE CONNECTED TO REFERENCE
VV 63 =
SUPERNODE CONSTRAINT xVVV 221 =−
KCL AT SUPERNODE
CONTROLLING VARIABLE IN TERMS OF NODES
2VVx = 21 3VV =⇒
k12/*
062)6(2 2211 =−+++− VVVV1833 VV 184V1833 21 =+ VV 184 1 =⇒ V