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    1

    By:

    DR SITI NOORAYA MOHD TAWIL

    Department of Electronic Engineering

    Faculty of Electrical and Electronic Engineering

    Universiti Tun Hussein Onn Malaysia

    ELECTRICAL AND ELECTRONIC

    TECHNOLOGY

    (BDU10803)

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    2

    Topic 3:

    Direct Current CircuitAnalysis (III)

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    3

    Lecture Contents

    Thevenins and Nortons Theorems

    Superposition Theorem

    Maximum Power Transfer Theorem

    Circuit Theorems

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    SuperpositionTheorem

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    Superposition Theorem

    The superposition theoremstates that

    the voltage across (or current through) an

    element in a circuit is the algebraic sumofthe voltages across (or currents through)

    that elements due to each independent

    source acting alone.

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    How to apply the superposition

    theorem?1. Consider one independent sourceat a

    time while other independent sources are

    turned-off. [short-circuit the voltagesource and open-circuit the current

    source]

    2. Dependent source are left intact becausethey are controlled by circuit variables.

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    Example 1

    Use the superposition theorem to find V in thefollowing circuit.

    8

    6V

    4 V 2A

    +

    -

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    SolutionStep 1:

    0.5A48

    6'I

    circuited)-opensource(2Aoperationinsource6VOnly

    4 +

    TR

    V

    6V

    8

    4 V 2A

    +

    -6V

    8

    4 V

    +

    -I4I4

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    9

    SolutionStep 2:

    V7.321.83(4)RIV 44

    ++

    +

    A83.133.15.0"' 444

    4

    III,currentTotal

    1.33A(2)48

    8"I

    circuited)-shortsourceV6(operationinsource2AOnly

    6V

    8

    4 V 2A

    +

    -

    8

    4 V

    +

    -b

    I4I4

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    10

    Given the following circuit, calculate ixand the power

    dissipated by the 10-resistor using superposition.

    Example 2

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    Solution

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    Cont

    0.6A10

    6

    10

    VaI

    625

    150a

    10Va15Va150

    3Va12Va10Va150:x120

    40

    0Va

    10

    0Va

    12

    Va-1540

    0VaI

    12

    Va-15

    :a)(nodKCL

    (4A)sourcecurrentcircuit-Open(a)FigureConsider

    x1

    x1

    +

    +-

    -+

    -

    -+

    V

    a

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    13

    Cont

    17.43Watt(10)1.32)(RIP

    1.32A1.92)(0.6III

    1.92A10

    19.2

    10

    VaI

    19.2V25

    480-Va

    048025Va:(2)()

    (2)-----120I4803Va:x120

    440

    0-VaI

    VbVa

    440

    0-VbI

    4II

    :bnod

    2

    x

    2

    xx

    x2x1x

    x2

    2

    2

    2

    32

    -

    --++

    --

    -

    ++

    +

    +

    +

    +

    (1)120I22Va

    120I10Va12Va:x120

    I10

    Va

    12

    Va

    10

    0Va

    I12

    Va-0

    III

    :a)(nodKCL

    (15V)sourcevoltagecircuit-Short

    (b)FigureConsider

    2

    2

    2

    2

    x221

    -----

    -+

    -+

    -

    +

    +

    a bI1 I2

    I3

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    Caution!

    Superposition Theorem is much easier to

    use in a circuit that has independent source.

    If the circuit has dependent source, it is

    recommended not to use this theorem.

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    Equivalent circuit forms

    R1 + R2

    R1

    R2

    R2R1

    V2

    V1 - V2

    V1

    I1 I2 I1 - I2

    21

    21

    RR

    RR

    +

    (a)

    (b)

    (c)

    (d)

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    SOURCE TRANSFORMATION

    A source transformationis the process ofreplacing a voltage Vin series with resistor

    Rby a current sourceIin parallel with a

    resistor R, or vice versa.

    # Source transformation is another tool for simplifyingcircuits

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    Cont..

    V

    R

    I R

    a

    b

    V

    R

    I R

    a

    b

    Transformation of dependent sources

    Transformation of independent sources

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    Thevenins

    Theorem

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    Thevenins Theorem

    Thevenins Theorem states that a lineartwo-terminal circuit can be replaced by an

    equivalent circuit consisting of:

    A voltage source VTHin SERIES withA resistor RTH

    Complexcircuit VTH

    RTH

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    Thevenins Theorem

    Where VTHis the open-circuit voltage at theterminals and,

    RTHis the input or equivalent resistance at theterminals when the independent sources areturned off.

    VTH

    RTH

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    Example 3

    For the following circuit, find the Theveninsequivalent circuit seen from RB.

    24

    20V 3 5ARB

    6

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    Solution 31. Remove R

    B

    .

    2. Find RTHfrom terminal a-b with turned-off

    all independent sources

    24

    20V 3 5A

    6

    a

    b

    24

    3

    6

    a

    b

    ++

    9.71(4//3)62RTH RTH

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    Cont3. Find VTHfrom terminal a-b.

    (2)30VyVx

    056

    VyVx05I

    ynode

    -(1)---602Vy-9Vx

    2Vy2Vx4Vx3Vx60:x12

    6

    VyVx

    3

    Vx

    4

    Vx20

    xnode

    leavingIenteringInodes,forKCL

    analysis,nodeusingExample

    3

    --------

    +-

    +

    -+-

    -+

    -

    +

    321 III

    24

    20V 3 5A

    6

    a

    bVTH

    +

    -

    xy

    I1 I2

    I3

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    24

    47.14V0VyV

    47.14V7

    330

    Vy

    330270)(602

    1

    60

    30

    792

    2-9

    1-1

    60

    30

    Vy

    Vx

    2-9

    1-1

    :rulesCramer'Using

    TH

    y

    y

    -

    ---

    --

    +-

    -

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    4. Draw the Thevenins

    equivalent circuit.

    BTotal

    TH

    R71.9

    14.47

    R

    VI

    +

    VTH=47.14V

    RTH=9.71

    a

    b

    RB

    A

    A

    58.081.71

    47.14I,27Rwhenii)

    17.2

    21.71

    47.14I,12Rwheni)

    B

    B

    I

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    Example 4

    Determine RThand VThat terminals 1-2 for

    the following circuits.

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    Solution

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    Nortons

    Theorem

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    Nortons Theorem

    Nortons Theorem states that a linear two-terminal circuit can be replaced by

    equivalent circuit consisting of:A current source I

    N

    in PARALLEL with

    A resistor RN

    Complexcircuit

    IN RN

    a

    b

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    Example 5

    Find the Nortons equivalent circuit for the

    following circuit seen from RB.24

    20V 3 5ARB

    6

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    3. Find INat a-b.

    (1)602Vy9Vx

    2Vy2Vx4Vx3Vx60:x12 3

    Vx

    6

    VyVx

    4

    Vx20

    III

    :xnodeleaving

    Ientering

    Inodes,forKCL

    analysis,nodeusingExample

    321

    ------

    -+-

    +-

    -

    +

    24

    20V3

    5A

    6

    a

    b

    IN

    x y

    (2)304VyVx

    3Vy3VyVx:x6

    2

    Vy5

    6

    VyVx

    I5I:ynodeN3

    --------+-

    +-

    +

    I2

    I1 I3

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    4.85A2

    9.71

    2

    0VyI

    9.71V34

    330-

    Vy

    -3306027030-1

    609

    342)(364-1

    29

    30-

    60

    Vy

    Vx

    4-1

    29

    rule;sCramer'Using

    N

    y

    y

    -

    -

    --

    -----

    -

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    4. Draw the equivalent

    circuit.

    IN=4.85A RN=9.71 RB

    a

    b

    B

    B

    N

    Total

    N

    R71.9

    09.47

    )85.4(

    R71.9

    71.9

    )I(R

    RI

    +

    +

    A

    A

    58.0

    81.71

    47.09I,27Rwhenii)

    17.221.71

    47.09I,12Rwheni)

    B

    B

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    Example 6

    Find the Norton equivalent with respect to

    terminals a-bin the following circuits.

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    Solution

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    Cont

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    Cont

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    Example 7

    Find the Norton equivalent in the following

    circuit.

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    Solution

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    Cont

    Current divider rules

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    RELATION BETWEEN THEVENIN

    THEOREM AND NORTON THEOREM

    VTH

    RTH

    IN RN

    a

    b

    TH

    THN

    NNTH

    NTH

    R

    VI

    RIV

    RR

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    IN=4.85A RN=9.71 RB

    a

    b

    VTH=47.14V

    RTH=9.71

    a

    b

    RB

    I

    Thevenins equivalent circuitNortons equivalent circuit

    A85.471.9

    14.47

    R

    VI

    47.14V1)(4.85)(9.7RIV9.71RR

    TH

    THN

    NNTH

    NTH

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    Maximum Power

    Transfer

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    Maximum Power Transfer

    Practicala circuit is designed to provide

    power to a load.

    Applicationcommunicationdesirableto maximize the power delivered to the

    load.

    Thevenin equivalentuseful in finding themaximum power a linear circuit can deliver

    to a load.

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    Cont

    Consider the following circuit for maximum

    power transfer.

    ThR

    ThV

    a

    b

    i

    LR

    L

    2

    LTh

    Th2 RRR

    VRip

    +

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    Cont

    Maximum power is transferred to the load

    when the load resistance equals the

    Thevenin resistance as seen from the load(RL=RTh).

    C t

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    Cont

    For the given circuit, VTh

    and RTh

    are fixed.

    Varying the RL, the power delivered to the

    load as the graph below.

    Pmax

    RThRL

    P

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    To prove the maximum power transfer differentiate p with

    respect to RLand set the results =0

    (watt)4R

    V

    4R

    VP

    ,RRWith

    .RRwhenplacespower takemaximumthethat,Showing]R[R),R(R)2RR(R0

    thatimpliesThis

    0)R(R

    )2RR(RV

    )R(R

    )R(R2R)R(RV

    dR

    dP

    0dR

    dPwhen,maximumispowerThe

    L

    2

    Th

    L

    2

    Thmax

    ThL

    ThL

    LThLThLLTh

    3LTh

    LLTh2

    Th

    2LTh

    LThL2

    LTh2

    Th

    L

    L

    --+

    +

    -+

    +

    +-+

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    FORMULA!!

    TH

    2

    TH

    L

    2

    THL

    2

    LTH

    THLMAX

    THL

    4RV

    4RVR

    RRVI2RP

    RR

    +

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    Example 8Find the value of RLfor maximum power

    transfer in the following circuit and find themaximum power.

    VTH=47.14V

    RTH=9.71

    a

    b

    RL

    I

    Thevenins equivalent circuit

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    Solution

    Thevenins equivalent circuit

    57.17

    4(9.71)

    (47.14)

    4R

    VP

    9.71RR

    2

    TH

    2

    THMAX

    THL

    VTH=47.14V

    RTH=9.71

    a

    b

    RL

    I