Page 60 - Schaum's Outline of Theory and Problems of Electric Circuits
P. 60

ANALYSIS METHODS
               CHAP. 4]
               4.3   Solve the network of Problems 4.1 and 4.2 by the node voltage method.  See Fig. 4-19.  49

                         With two principal nodes, only one equation is necessary.
                                                    V 1   60  V 1  V 1  V 1
                                                          þ   þ   þ    ¼ 0
                                                      7     12   6  12
                     from which V 1 ¼ 18 V.  Then,
                                                           60   V 1
                                                       I 1 ¼     ¼ 6A
                                                              7
















                                                        Fig. 4-19


               4.4   In Problem 4.2, obtain R input;1 and use it to calculate I 1 .
                                                             2880    2880
                                                        R
                                              R input;1 ¼  ¼         ¼    ¼ 10

                                                        11     18  6     288
                                                              6  18
                                                           60    60
                     Then                            I 1 ¼     ¼   ¼ 6A
                                                         R input;1  10




               4.5   Obtain R transfer;12 and R transfer;13 for the network of Problem 4.2 and use them to calculate I 2 and
                     I 3 .
                         The cofactor of the 1,2-element in   R must include a negative sign:


                                           1þ2  12  6                       R  2880

                                    12 ¼ð 1Þ     0  18     ¼ 216  R transfer;12 ¼    12  ¼  216  ¼ 13:33
                     Then, I 2 ¼ 60=13:33 ¼ 4:50 A:


                                            1þ3  12  18                      R  2880

                                      13 ¼ð 1Þ     0  6     ¼ 72  R transfer;13 ¼    13  ¼  72  ¼ 40
                     Then, I 3 ¼ 60=40 ¼ 1:50 A:



               4.6   Solve Problem 4.1 by use of the loop currents indicated in Fig. 4-20.

                         The elements in the matrix form of the equations are obtained by inspection, following the rules of
                     Section 4.2.
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