Nodal Mesh

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    11-Apr-2016

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Nodal analysis electrical science

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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!

    ++=R

    RRRRRRRb

    accbba1

    3*12 *12 3612k k kk = =

    ++

    ++=RRRRRR

    RRRRRRRR

    R

    c

    accbba

    b

    2

    12k36k

    12k

    THE RESULTINGCIRCUIT IS A CURRENT DIVIDER

    ++=Y

    RRRRRRRR

    a

    accbba3 4mA 36k 36k

    12k OV+

  • CIRCUIT AFTER PARALLEL RESISTORREDUCTION

    4mA 36k

    36 ||12 9k k k=+

    REDUCTION

    OI36 8k

    OV

    9k 36 8436 18 3OkI mA mA

    k k= =+

    89 9 243O O

    V k I k mA V= = = NOTICE THAT BY KEEPINGTHE 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 OHMS

    V

    k63I

    kVI a62

    = :SOHM' 0321 = III :KCL3*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:@ =+ IIV0

    943=++

    kVV

    kV

    kVV cbbab

    SHORTCUT:SHORTCUT: SKIP WRITING THESE EQUATIONS...

    AND PRACTICE WRITING 0:@ 65 =+ IIVc0

    39=+

    kV

    kVV cbc

    THESE DIRECTLY

  • CIRCUITS WITH ONLY INDEPENDENT SOURCES

    @ NODE 1

    02

    21

    1

    1 =++Rvv

    RviA 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 =++ R

    vvRvvi

  • 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 =+ VVmAVVNODE EQUATIONS

    VVVV 82 21

    2 ==0

    104

    10:@ 1 + kmAkV

    010

    210

    :@ 2122 =++ kVI

    kVVV O

    CONTROLLING VARIABLE (IN TERMS ON NODEVOLTAGES)

    kVIO 101=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 =++ SIkmA@V_2 0

    124 2 =++

    kVmAIS

    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

    VkV =++

    (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 THEVV

    VV4

    6

    4

    1

    == SOURCES CONNECTED TO THE

    REFERENCE

    CONSTRAINT EQUATION VVV 1223 =KCL @ SUPERNODE

    02

    )4(212

    6 3322 =+++k

    VkV

    kV

    kV

    k2/*

    VVV 223 32 =+VVV 1232 =+ addand 3/*VV 385 3 =

    V

    OIV FORNEEDED NOT IS 2

    mAkVIO 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

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