Lecture 26

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					PHY301 – Circuit Theory


                                        Virtual University
                                             PHY 301
                                           LECTURE 26
Example: Calculate the voltage V by using Thevenin’s theorem .
                                o




Solution:
        We want to calculate V by using Thevenin’s theorem .we will follow the
                              0
        following steps .

        First step:   Removing R
                                    L




               Here R is 4k resistor at which we want to calculate the voltage V
                     L                                                           o.
        Second step: Calculating V
                                   th




                 We want to calculate V        . Apply KVL in two loops to calculate the individual loop
                                          th
                 currents.




                             Here
                                               I = 2mA
                                                1
                 KVL for the super mesh
                          -6 +6kI +12k(I +I ) + 12k(I +I ) =0
                                 2      1 2          1 2
                             -6 +6kI +12kI + 12kI +24kI =0
                                    2     2       2      1
                                        30kI – 6 +24k(2m)=0
                                            2
                                              30KI -6 +48 =0
                                                   2

                                            I = -42/30mA
                                             2
                                             I = -1.4mA
                                              2
                 Now we will calculate the voltage across 6k and 12 k to find the value of V th
                 Therefore,
                                            V = 6kI
                                              6k      2
                                                  = 6k(-1.4m)
                                             V = -8.4volts
                                               6k
                                            V      = 12k(I1 +I2)
                                              12k




Virtual University of Pakistan                                                                             Page 107
PHY301 – Circuit Theory

                                                   = 12k( 2m -1.4m)
                                              V    = 7.2volts
                                               12k
                 So,
                                              V = V6k + V12k
                                               th
                                                   = -8.4 +7.2
                                              V = -1.2volts
                                                th

        Third step: Calculating R
                                    th




                6k is in series with 12k .The resultant of these two is in parallel with 12k.
                                    (6k+12k)||12k = 18k x12k/(12k+18k)
                                             R = 7.2k
                                               th
        Fourth step: Calculating the unknown quantity.
                After calculating V and R , re-inserting the load resistance R in the circuit in series with
                                    th      th                                      L
                R and considering the V as a battery in series with these two resistances.
                  th                        th




                                              V = -1.2 x4k/(7.2k + 4k)
                                               0
                                              V = - 0 .4285 volts
                                                0

Example:     Calculate the value of RL and the maximum power dissipation across it by Thevenin’s
             Theorem.




Solution:
        We want to calculate R and the maximum power across it by using Thevenin’s theorem .we will follow
                                 L
        the steps given earlier.

                 First step: Removing R
                                          L




                           To remove R to calculate V .In this case our R is our R .
                                      L              th                  th         L
                 Second step: Calculating V
                                             th
                 We want to calculate V . Apply KVL in two loops to calculate the individual loop
                                       th
                 currents.




Virtual University of Pakistan                                                                      Page 108
PHY301 – Circuit Theory

        Here
                                              I = 2mA
                                               1
                 Applying KVL to loop 2
                                             6kI + 3k(I – I ) +3 =0
                                                2       2 1
                                                   9kI -3kI + 3 =0
                                                      2    1
                                                      9kI – 6 +3 =0
                                                         2
                                                             I = 1/3 mA
                                                              2
                                                             I = 0.33mA
                                                              2
                 Now we will calculate the voltage across 4k and 6k to find the value of V th
                 Therefore,
                                   V     = 4k(I )
                                    AB         1
                                        = 4k( 2m)
                                   V     = 8 volts
                                    AB
                                   V     = 6k(I )
                                    BC         2
                                        = 6k x 0.33m
                                   V     = 2 volts
                                    BC
                 So
                                   V =V         +V
                                     th     AB     BC
                                        = 8 +2
                                   V = 10 volts
                                     th
        Third step: Calculating R
                                  th




                    To calculate R       Short circuiting the voltage source and open circuiting the current source.
                                    th .
                          3k is in parallel with 6k and 4k is series with these two so
                                     3k||6k + 4k = 3k x 6k/(3k + 6k) +4k
                                                  = 2k +4k
                                                  = 6k = R =R
                                                            th   L
                                         R =R = 6k
                                           th    L
        Fourth step: Calculating unknown quantity.
                After calculating V and R , re-inserting the load resistance R in the circuit in series with
                                     th       th                                    L
                R and considering the V as a battery in series with these two resistances.
                  th                          th




                 For maximum power dissipation
                                R =R = 6k
                                 L    th
                                      2
                                P = I R ------------- (A)
                                 L
                         From Ohm’s law V=IR and I = V/R
                         so              I =10/12 put in A we have
                                                       2
                                        P = (10/12k) (6k)
                                         L
                                                      2
                                           = (0.83) X 6
                                        P = 4.1 mW
                                         L




Virtual University of Pakistan                                                                          Page 109
PHY301 – Circuit Theory

Example:     Calculate the value of RL and the maximum power dissipation across it by Thevenin’s
             Theorem.




Solution:
        We want to calculate R and the maximum power across it by using Thevenin’s theorem .we will follow
                               L
        the following steps .
        First step: Removing R
                                 L
                  To remove R to calculate V .In this case our R is our R .
                              L             th                  th       L




        Second step: Calculating V
                                       th




        Apply KVL to the circuit
                          For loop1
                                  -12 +6I + 6(I - I ) = 0
                                         1      1 2
                                           -12 +12I – 6I = 0
                                                   1    2
                                                   2I – I = 2 --------------- (A)
                                                     1 2


                          For loop 2

                                   12I – 6I +3 =0
                                       2    1
                                    4I – 2I +1 =0
                                      2    1
                                             2I = 4I
                                               1     2
                 put in equation (A) we have
                                   4kI +1 –1kI =2
                                      2         2
                                            3kI =1
                                                2
                                               I = 0.33mA
                                                2
                 from equation (A)
                                               I = (I2 +2)/2
                                                1
                                                  = 0.33+2/2
                                               I = 1.166mA
                                                1
                 Now voltage across 6k resistor
                                            V      = 6kI
                                              R6k        1
                                                   = 6k(1.166m)
                                            V      = 7volts
                                              R6k
                                            V = 7V +3V
                                              th
                                            V =10 volts
                                              th




Virtual University of Pakistan                                                                  Page 110
PHY301 – Circuit Theory

                  Third step: Calculating R
                                              th




         We want to calculate R 6k is in parallel with 6k .The resultant is again parallel with third 6k resistor.
                                th
                           6k||6k = (6k x 6k)/6k + 6k
                                  = 3k
                           3k||6k = 6k x 3k/(6k + 3k)
                                  = 2k
         So
                                 R =R = 2k
                                   th    L
         Fourth step: Calculating unknown quantity.
                 After calculating V and R , re-inserting the load resistance R in the circuit in series with
                                     th       th                                   L
                 R and considering the V as a battery in series with these resistances.
                   th                         th




                                   V      = 10 x2k/4k
                                     RL
                                          = 5 volts
         To calculate the power dissipation we have
                                             2
                                     P = V /R = 25/2k
                                       L
                                      P = 12.5mW
                                        L

THEVENIN’S THEOREM AND DEPENDENT SOURCES:
   Working with dependent sources is different from working with independent sources while applying
   Thevenin’s theorem .
   While calculating Rth we can simply open circuit current sources and short circuit voltage sources.
   Because the voltage or current of the dependent sources is dependent on the voltage or current of these
   independent spources.
   While calculating R we will short circuit the open terminals of the Thevenin circuit and will calculate the I
                       th                                                                                        sc
   and then divide V with I to calculate R .
                     th     sc                th
Example: Calculate the voltage V by using Thevenin’s theorem .
                                   o




Solution:
        We want to calculate V by using Thevenin’s theorem .we will follow the following steps .
                              0
        First step:  Removing R
                                  L




                  Here R is 6k resistor at which we want to calculate the voltage V
                        L                                                          o.




Virtual University of Pakistan                                                                           Page 111
PHY301 – Circuit Theory

        Second step: Calculating V
                                      th
                           Voltage across 4k resistor
                                    V = (4k/6k) 12
                                      4k
                                     V = 8volts
                                       4k
                           Voltage across 2k resistor
                                     V = (2k/6k) 12
                                       2k
                                    V = V = 4volts
                                      A      2k
                                    V = 8 – 4V
                                     th            A
                                          = 8 – 4(4)
                                          = 8 – 16
                                     V = - 8 volts
                                       th
        Third step:
                 When we use dependent sources we will use the following technique .
                 In the above circuit output which is opened after removing the load resistance now we will
                 replace this open circuit with short circuit. As in the circuit below.




                 We will first find I        to calculate R .
                                        sc                 th
                           Here
                                        V = 2kI
                                         A     1
        Applying KVL to loop 1
                           -12 +2kI +(I – I )4k= 0
                                   1 1 sc
                          -12 +2kI +4kI – 4kI = 0
                                  1      1        sc
                                      6kI -4kI = 12
                                          1      sc
        For other loop
                                              (I –I )4k +4V = 0
                                                sc 1         A
                                              4kI -4I +4(2kI )=0
                                                  sc 1         1
                                                  4kI -4I +8I =0
                                                     sc 1      1
                                                             I = -I
                                                              1     sc
                Putting in equation for loop1
                                               3I -2I = 6
                                                 1 sc
                                          3(-I ) -2I = 6
                                               sc     sc
                                                      I = - 6/5 mA
                                                       sc
                So
                                    Rth = Vth/Isc
                                        = -8/(-6/5 m)
                                   Rth = 6.67k
        Fourth step:
                After calculating V and R , re-inserting the load resistance R in the circuit in series with
                                   th        th                               L
                R and considering the V as a battery in series with these resistances.
                  th                        th




                 So
                                        V = (6k/6k +6.67k) x8
                                         0
                                           = 6k/12.67k x8
                                        V = 3.78 volts
                                         0




Virtual University of Pakistan                                                                       Page 112

				
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