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Lecture 15

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					PHY301 – Circuit Theory


                                         Virtual University
                                              PHY 301
                                            LECTURE 15

Example:      Calculate the voltages Vad , Vac and Vbd.




Solution:         We want to calculate V        and V        . The circuit can be redrawn as
                                           ad           ac




        As seen form the figure there is only a voltage source between points a and d
        so the voltage V will be 12 volts. There is no physical closed path between             point a and c we can
                        ad
        consider two closed paths acba and acda.

        KVL equation for the path acba
                                   V -4–6=0
                                    ac
                                   V = 10 Volts
                                    ac
        Second path in which only unknown value is V                is acda. KVL equation for
                                                               ac
        this will be
                                    V + 2 – 12 =0
                                     ac
                                    V = 10 Volts
                                     ac
                 Note:                    Calculate Vbd yourself.
Example:      Calculate the voltages Vac and Vdb.




Solution:   We want to calculate Vac and Vdb. The circuit can be redrawn as




Virtual University of Pakistan                                                                            Page 65
PHY301 – Circuit Theory


                 To calculate V  , the two paths will be acda and acba.
                              ac
                          KVL equation for the path acda
                                   V + 8 – 12 =0
                                     ac
                                   V = 4 volts
                                     ac

                 KVL equation for the path acba will be
                                   V + 6 – 10 = 0
                                    ac
                                     V = 4 volts
                                      ac
                 Two paths for V    will be bdcb and bdab.
                                bd
                 KVL equation for bdcb
                                   V -8 +6 = 0
                                    bd
                               V = 2 volts
                                 bd
                                   V     = - 2 volts
                                    db
                 KVL equation for the path bdab
                                   V – 12 +10 =0
                                    bd
                                    V = 2 volts
                                      bd
                                               V    = - 2 volts
                                                db


Example:     Calculate the voltage V .
                                    0




Solution:  We want to calculate V0 .
                         Here
                                    I = 2 mA
                                     2
                 KVL for mesh 1
                                    3kI + 6k (I – I ) -12 = 0
                                        1         1 2
                                           9kI – 6kI = 12
                                               1      2
                                    9kI – 12 =12
                                       1
                                           I = 8/3 mA
                                            1
        Voltage across 6k resistor
                                    V = 6k( I – I )
                                              1 2
                                      = 6k(8/3 – 2
                             = 4 volts
                 As current through 2k resistor is
                                                 I = 2mA
                                                  2
                         Voltage across 2k = 2k x 2mA
                                       = 4 volts
                                   V = V–V
                                     0             2k
                                          =4–4
                                      V = 0 volts
                                       0




Virtual University of Pakistan                                            Page 66

				
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