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Lecture 20

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					PHY301 – Circuit Theory


                                         Virtual University
                                              PHY 301
                                            LECTURE 20

Example:      Calculate the voltage V
                                        o.




Solution:   we want to calculate the voltage V        The circuit can be redrawn as
                                                 0.




                 Here
                          I = - 2mA
                           1
                          KVL equation for loop 2
                          2kI + 4kI + 2k(I + I ) = 12
                             2     2       2 1
                                        6kI + 2kI + 2kI = 12
                                            2      2     1
                                                       8kI – 4 = 12
                                                          2
                                                  8kI = 16
                                                     2
                                                               I = 2mA
                                                                2
                 Now
                                                       V = 2k(I )
                                                         o     2
                                                 = 2k (2m)
                                             V = 4 volts
                                              o
Example:      Calculate the voltage V
                                        o.




Solution:   We want to find the voltage V The circuit can be redrawn as
                                         0.




                 Here
                           I =3mA
                            1
                          I = 1mA
                           2




Virtual University of Pakistan                                                        Page 83
PHY301 – Circuit Theory

                  KVL for loop 3
                           4kI + 2k(I +I ) + 4k(I –I ) +2k(I – I ) =6
                              3      3 2         3 1        3 1
                           4kI +2kI +2kI +4kI – 4kI +2kI -2kI =6
                              3    3      2    3      1     3     1
                                                         12kI +2 -12 – 6 =12
                                                             3
                                                                        12kI =22
                                                                            3
                                                                   I =11/6mA
                                                                    3
                                                                           I =1.833mA
                                                                            3
                                                    V = 4kI
                                                      0      3
                                                       = 4k x 1.833m
                                                    V = 7.33 volts
                                                      0
Example:       Calculate the current I
                                       o.




Solution:    We want to find the current I        The circuit can be redrawn as
                                             o.




                           Here
                                    I =-2mA
                                     1
                                    I = 1mA
                                     3
                  Now KVL for mesh 2
                          4k(I +I ) +2k(I – I ) +2k(I – I ) =0
                              2 4          2 3       2 1
                                       8kI +4kI – 2 +4 =0
                                          2    4
                                                       4kI +4kI = -1
                                                          2     4
                  Now KVL for loop 4
                                   4k(I +I ) +2k(I +I ) +6kI -6 =0
                                        4 2       4 3        4
                                            12kI +4kI +2 -6 =0
                                                4    2
                                               12kI +4kI -4 =0
                                                   4    2
                                                             3kI +1kI =1
                                                                4     2
        Multiplying equation of mesh2 by 3 and equation of loop 4 by 2 and        subtracting

                                        12kI +6kI = -3
                                             2     4
                                        -2kI -6kI = -2
                                            2    4

                                             10kI =-5
                                                 2
                                        I = -o.5mA
                                         2
        Now
                                                      I =I –I
                                                       0 1 2
                                                        I = -2 +0.5
                                                         0
        so
                                                      I       = -1.5mA
                                                          0




Virtual University of Pakistan                                                                  Page 84
PHY301 – Circuit Theory


Example:         Find the currents I ,I and I .
                                    1 2      3




Solution:   We want to find the currents I ,I and I . The circuit can be redrawn as
                                          1 2      3.




                 For mesh 1
                        4I -4I +1 = 0 ...............(1)
                          1 2




        For Super mesh (mesh 2 and mesh 3)
                         3I + 2I + 4I - 4I = 0 ...............(2)
                           2    3    2    1
        Coupling Equation
                                  I -I =2V
                                   3 2       X
        Here
                                    V = 3I
                                     X      2

        Solving equation 1
                                             4I - 4I = -1
                                               1    2
        Solving equation 2
                                     - 4I + 7I + 2I = -1
                                         1    2    3

        For Mesh 3
                                               -7I + I = 0 --------------(3)
                                                  2 3
        Solving equation 1,2 and 3
                                             I = - 308.8 mA
                                              1
                                              I = - 58.82 mA
                                               2
                                             I = - 411.8 mA
                                              3




Virtual University of Pakistan                                                        Page 85

				
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