Lecture 18

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					PHY301 – Circuit Theory


                                         Virtual University
                                              PHY 301
                                            LECTURE 18

Example:      Calculate the voltage V
                                        o.




Solution:   We want to calculate the voltage V        The circuit can be redrawn as
                                                 o.




                 Now
                          Vx = (I - I ) 6k -------------(A)
                                 2 3
                           I = Vx/2k ---------------(B)
                            1
                          Put Vx from A into B we have
                                     I = (I – I )6k/2k
                                      1    2 3
                              = 3(I – I )
                                   2 3
                 Now
                          I = 2mA
                           2
                        I = 6mA - 3I
                         1           3
                   Now KVL equation for mesh 3
                                   6kI +6k(I – I ) +2k(I -I ) =0
                                      3     3 2          3 1
                          6kI + 6kI – 12 +2kI -2k(6mA -3I )=0
                             3     3         3            3
                              14kI – 12 -12 + 6kI = 0
                                  3               3
                                             20kI = 24
                                                  3
                                                 I = 1.2mA
                                                  3
                                                    V = 6kI
                                                     0      3
                                              = 6k(1.2mA)
                                          =7.2 volts




Virtual University of Pakistan                                                        Page 75
PHY301 – Circuit Theory

Example:     Calculate the voltage V
                                         a.




Solution:   We want to calculate the voltage V The circuit can be redrawn as
                                              a.




        Va can also be found by super node technique but it will take a lengthy
        calculation lets see how it becomes very easy with loop analysis.




                 Here
                         I =1A
                          4
                 KVL for mesh 1
                         1(I – I ) +3(I -3I )-5 =0
                            1 2        1 3
                    I – I +3I -3I -5 = 0
                     1 2       1 3
                           4I – 3I – I = 5 --------(A)
                             1    3 2


        KVL for mesh 2
                                    2I2 +3I2 -2Va+I2-I1 = 0
                                              6I2 –I1 = 2Va------(B)
                                                    Va = 3I2
                 Put this value in (a)
                                                  6I2 –I1 = 3I2
                                                       I1 = 0
        KVL for mesh 3




Virtual University of Pakistan                                                    Page 76
PHY301 – Circuit Theory

                  2Va + 5(I – I ) +3(I –I ) =0
                            3 4        3 1
                  Putting the values of I ,I and Va
                                         1 4
                            6I + 3I -0 +5I – 5 = 0
                              2    3        3
                                       8I + 6I = 5
                                         3    2
                  Solving equation for mesh 1 and mesh 3
                                     - 18I - 6I = 30
                                          3    2
                               8I + 6I = 5
                                 3    2

                                       -10I     = 35
                                         3
                                     I = - 3.5A
                                      3
                  Putting values of I and I in eq of mesh 1 i.e in (A) we have
                                     3      1
                                    10.5– I =5
                                            2
                                                  I = 5.5 A
                                                   2
                  As                                  Va =3I
                                                             2
                  Therefore                Va =3 * 5.5
                                       Va = 16.5 V
    Example:       Find Current through all meshes.




Solution:




        Incorrect approach
                           I =-5A
                            2
                           I =+5A
                            3
        We drive all these
                 KVL equation for mesh 1
                  -5+I -I +2I =0 -------------------- (1)
                      1 3 1




Virtual University of Pakistan                                                   Page 77
PHY301 – Circuit Theory


                KVL equation for super mesh
                         1(I -I )+4I +3I +2+4I - 4I +6I =0 ------------ (2)           KVL
                             3 1    2 2         3 4 3
       equation for Mesh 4
                                             4I – 4I +2I -3=0 ------------ (3)
                                               4    3 4
                Simplifying
                         Equation (1)                           3I - I = 5
                                                                  1 3
                         Equation (2)                -I +7I +11I -4I =-2
                                                       1 2      3 4
                         Equation (3)                        - 4I + 6I =3
                                                                 3      4
                Also                       -I +I =5
                                             2 3
                            Therefore          I =2.481A
                                                1
                                               I =-2.556A
                                                2
                                         I =2.444A
                                          3
                                          I =2.130A
                                           4
Example:    Calculate current I .
                                0




Solution: We want to calculate the current I The circuit can be redrawn as
                                            0.




                  Here
                           I = -1mA
                            1
                                    I = -2mA
                                     2
                           KVL equation for loop 3
                           1k( I + I ) + 2kI + 1k(I + I ) = 2 + 4
                                3 1         3       3 2
                            1kI + 1kI + 2kI + 1kI + 1kI = 6
                                3     1      3      3     2
                                              4kI – 1 – 2 = 6
                                                 3
                                                          4kI – 3 = 6
                                                              3
                                                                I = 9/4 mA
                                                                 3
                                                                   I = 2.25 mA
                                                                    3
                                                   I = I = 2.25 mA
                                                    3 0




Virtual University of Pakistan                                                   Page 78
PHY301 – Circuit Theory


Example:     Calculate current I .
                                0




Solution: We want to calculate the current I The circuit can be redrawn as
                                            0.




                 Here
                          I = 2mA
                           1
                           I = - 4mA
                            2
                 KVL for loop 3
                                    2kI + 2k(I + I ) = 12
                                       3      3 2
                                    2kI + 2kI + 2kI = 12
                                       3     3     2
                                      4kI – 8 =12
                                         3
                                                     4kI = 20
                                                         3
                                                        I = 5mA
                                                         3
                 Now
                                             I =I +I
                                              0 3 1
                                  I = 5mA + 2mA
                                   0
                                     = 7mA




Virtual University of Pakistan                                               Page 79

				
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