Lecture 08

							PHY301 – Circuit Theory

                                              Virtual University
                                                  PHY 301
                                                LECTURE 8

Example:         Write KCL equations for all nodes.




Solution:

       The KCL equations for nodes 1 through 4 follow
       For node 1
                        I +I –I =0
                         1 2 5
       For node 2
                       -I +I -50I = 0
                         2 3       2
       For node 3
                        -I + 50I + I = 0
                          1      2 4
       For node 4
                  I –I –I =0
                   5 3 4
Example         Write KCL equations for all nodes.




Solution:

        For node A 10mA is entering the node and the source current I is leaving and   60mA is also entering
                                                                     t
so,
        For node A
                         I -60mA – 10mA= 0
                          t
        For node B
                         60mA – 40mA – 20mA = 0


Example           Write KCL equation for node A.




Virtual University of Pakistan                                                                   Page 44
PHY301 – Circuit Theory

Solution:

                By KCL the equation for node A
                                 -12mA + 4mA + I =0


Example        Write KCL equation for node A and node B.




Solution:
              The equation of currents at node A and B by KCL
       For node A
                  -I +I + 3mA = 0
                    1 2
       For node B
                      -12mA + 4mA +I =0
                                    1
Example      Write KCL equation for node A .




               Writing equation for node A .
                For node A
                         -10 I + I +44 mA – 12 mA = 0
                              x x
Example         Write KCL equation for node A .




Solution:
                         For node A equation will be by KCL
                                 I + 10I – 44mA = 0
                                  x     x

Example         Calculate the values of I and I
                                         1      2.




Solution:




Virtual University of Pakistan                                  Page 45
PHY301 – Circuit Theory

                 We want to calculate the values of I and I
                                                     1      2
                 For node A
                                  4mA + 8mA- I = 0
                                                1
                                                     I = 12mA
                                                      1
                 Now for I
                           2
                 For node B
                                  -8mA + 2 mA + I =0
                                                   2
                                            I = 6mA
                                             2
Example          Calculate the values of I , I , and I
                                          1 2          3.




Solution:
        We want to calculate the values of I , I , and I so we will use node analysis.
                                            1 2         3,

                 For node A
                           -I – I +8mA = 0
                             1 2
                                 -I – I = -8mA
                                   1 2
                 For node B
                        I + I +4mA = 0
                         2 3
                               I + I = - 4mA
                                2 3

                 For node C
                                   -I + 2mA – 8mA = 0
                                     3
                                                       I = - 6mA
                                                        3
       Putting the value of I in equation of node B
                             3
                                         I - 6mA = -4mA
                                          2
                                      I = 2mA
                                       2
       Putting the value of I in equation of node A
                             2
                                    -I – 2mA = -8mA
                                      1
                                                      I = 6mA
                                                       1
Example:         Find the KCL equtations for node A,node B, node C and node D.




Solution:




Virtual University of Pakistan                                                           Page 46
PHY301 – Circuit Theory

        We want to write the equations for nodes A, B, C and D.
        For node A
                                    -5mA + 8mA + 4mA =0
        For node B
                                   I – I + 5mA =0
                                    1 2
        For node C
                              -I – 2mA + 3mA – 8mA =0
                                1
        For node D
                                       -4mA – 3mA + I = 00
                                                       3

Example         Calculate the current I .
                                       o




Solution:
                At node 1
                                   (V1/12k) + ((V1 – V2)/10k) = 6mA
                                                5V1 +6V1 – 6V2 = (60k)(6mA)
                                                       11V1– 6V2 = 360
                At node 2
                     (V2/3k) +(V2/6k) + ((V2- V1)/10k) =0
                              10V2 +5V2+3V2-3V1=0
                                                      18V2 -3V1 =0

                     Equating equation of node1 and node 2

                                    33V1 -18V2 =1080
                                   -3V1 + 18V2 = 0

                                            30V1 =1080
                                                V1 = 36 volts
                                             18V2 -3V1 =0
                                                      6V2 – V1=0
                                                       6V2 -36 =0

                                  V2 = 36/6
                                  V2 = 6volts
                                   I =V2/6k
                                    o
                                      = 6/6k
                                      = 1mA




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