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Class 3 Thevenin_ Norton

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Class 3  Thevenin_ Norton Powered By Docstoc
					    EC 101

Electrical Sciences




                      1
Welcome to Lecture 3

     CIRCUIT
    ANALYSIS
   TECHNIQUES

                       2
            Thevenin’s Theorem

 M.L.Thevenin published this theorem in 1883.
Theorem : Any two-node linear network may be replaced by a voltage
source equal to the open circuit voltage between the nodes in series with
the resistance as seen by a load at this port


 Ex Find voltage Vab across terminal ab

                  5Ω                       4Ω
                                                      a
      25V                                 3A
                          20Ω                          24Ω

                                                      b
                                                                       3
             5Ω                      4Ω
                                                 a
 25V                                3A
                     20Ω                          24Ω

                                                 b

                                          Replace by
              Req                         Thevenin’s Equivalent Circuit
                            a

VTh                           24Ω

                             b
  Thevenin’s Equivalent Circuit
                                                                   4
  VTh – Open circuit voltage across ab

  Req - Equivalent Resistance as seen by the load after deactivating
        all independent sources

1. Open Circuit ab

                5Ω                        4Ω
                                                     a
     25V                                 3A
                         20Ω

                                                     b



                                                                       5
VTh         VTh = v1
              5Ω   v1         4Ω
                                         a
                                    +
      25V                    3A     VTh
                       20Ω
                                     -
                                         b

  Applying nodal analysis,
      v1  25 v1
                 3
         5     20             v Th  32V
       v1  32V
                                             6
Req

Deactivate all independent sources

Voltage sources – short circuit

Current sources – Open circuit


       5Ω                 4Ω
                                   a

             20Ω                       Req

                                   b

      R eq  4  20 5  4  4  8Ω
                                             7
            Req = 8Ω
                       a
                                +


 VTh                    24Ω    Vab
=32V
                        b       -


                 24
     Vab  32          24V
                24  8




                                     8
             Norton’s Theorem

    E.L.Norton published this theorem

Theorem : Any two -node linear network may be replaced by a current
source equal to the short circuited current between the nodes in parallel
with the resistance as seen by a load at this port

 Ex Find voltage Vab across terminal ab

              5Ω                       4Ω
                                                   a
  25V                                3A
                      20Ω                          24Ω

                                                   b
                                                                        9
  Norton’s Equivalent Circuit

        
                                      a
                  ISC           Req   24Ω

                                      b
1. Short Circuit ab

                  5Ω                      4Ω
                                               a
      25V                             3A       ISC
                          20Ω

                                               b
                                                     10
            5Ω   v1         4Ω
                                 a
  25V                      3A    ISC
                  20Ω

                                 b
Applying nodal analysis,

  v1  25 v1 v1
               3
     5      20 4
  v1  16V
           v1 16
  I SC          4A
           4   4                       11
Req

Deactivate all independent sources

Voltage sources – short circuit

Current sources – Open circuit


       5Ω                 4Ω
                                   a

             20Ω                       Req

                                   b

      R eq  4  20 5  4  4  8Ω
                                             12

                                a       +
                   Req
    ISC = 4A                           Vab
                   =8Ω           24Ω

                                b       -

         Norton’s Equivalent Circuit

                    8
       Vab  4          24  24V
                 8  24



                                             13
   Thevenin’s Equivalent Circuit

                     By Source Transformation

         Norton’s Equivalent Circuit

               Req = 8Ω              
                              a                                        a
                                                          Req
                                          ISC = 4A        =8Ω           24Ω
 VTh                           24Ω
=32V
                                       Source                          b
                              b        Transformation
   Thevenin’s Equivalent Circuit                Norton’s Equivalent Circuit

                 VTh                        VTh
        I SC                      R eq   
                 R eq                       I SC
                                                                        14
  Req     When the Circuit contains dependent sources also

Deactivate all independent sources only

Apply a test voltage source or a test current source to the Thevenin
 terminals a,b

            VTest
   R eq   
            I Test




                                                                   15
Ex Find I applying Thevenin’s Theorem

              2kΩ               3kΩ
                                                  +
                                        I
    4V                        VX            4kΩ   VX
                             4000
                                                  -




                                                       16
Solution    VTh

1. Open Circuit the 4kΩ resistance
               2kΩ        VTh    3kΩ
                                       +
    4V
                                 VTh   VTh
                                4000
                                        -
   Applying nodal analysis,

     VTh  4 VTh
            
      2000    4000
    or, 2VTh  8  VTh
         VTh  8V                            17
Req   1. Deactivate the independent sources

           2kΩ               3kΩ
                       VX
                                       +
                             VX         VX
                            4000
                                        -




                                              18
1. Apply a test voltage VTest


                  2kΩ                3kΩ
                                             I Test
                                  VTest      VTest
                                  4000


   Applying nodal analysis,

             VTest VTest  I Test  3000
  I Test         
             4000          2000

                                                      19
4000I Test  VTest  2VTest  6000I Test

 VTest  10000I   Test


  VTest
          10000  10k 
  I Test

   Req = 10kΩ




                                           20
Thevenin’s Equivalent Circuit
               Req = 10kΩ


                       I
  VTh                           4kΩ
  =8V



             8
        I         0.571mA
           10  4




                                      21
Alternative Way

              2kΩ     3kΩ
                                      +
                            I
   4V                VX         4kΩ   VX
                    4000
                                      -

   VTh  8V




                                           22
          2kΩ             3kΩ
                                   +

4V                        VX
                              0   ISC VX  0
                         4000
                                    -

            2kΩ            3kΩ

     4V                            ISC



                    4
          I SC         0.8mA
                   23
                                                23
         VTh      8V
R eq                 10k Ω
         I SC   0.8mA

                Req = 10kΩ


                        I                    8
                                        I         0.571mA
  VTh                             4kΩ
                                           10  4
  =8V


       Thevenin’s Equivalent Circuit




                                                         24

				
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