Расчёты по заданных схемах с применением законов Кирхгофа, методом контурных токов и методом узловых потенциалов

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Вариант 23   

R1

R2

R3

R4

R5

R6

E1

E2

E3

Ik1

Ik2

Ik3

Ом

В

А

16

20

12

30

42

52

50

-

34

0

-

0,5

Схема №1

                                                                                     c

                                                                                 

                                                                      

                                                             R5                                                                             

                                                                                             R4      

                                                                            Jk1       

                                                                                                                                                   

                                               b                                                        E1 

          Ik3                  R3                                                     d                                d                                 

                                                                       m       R1     

                              n                                                             I1                                                          

                                                                                            

                                                                          R6                              R2

                    E3                                              

                                                                                  a                                    

Изменим схему 1, заменяя в ней источники тока  на ЭДС, учтя что E3/=Jk3*R3, Е\1=Jк1*R1= 0, т.е. Jk1=0 ;

Схема №2.

                                            I3                             c           

                                                                                 

                                                                  I5                                                                                                                                     

                                                                      

        E\3                                                                                                                                         

                                                    R5                             R4 

                                                                          I                        I4       

              3                                                                                                                                   

                                                                         R1          1    E1

           R3             III     b                                                                         d                                                  

                                                             I1  

               2                                                        II                             

                                                                                                  R2      

         E3                                                        

                                                            R6                        I2          

                                                                 I6       

                                                                            a 

Произведем анализ схемы 2.

Число узлов n=4; Число ветвей m=6.

I.  Непосредственное применение законов Кирхгофа.

Расставим произвольно направления токов в ветвях. Составим по I закону Кирхгофа (n-1)=3 уравнение.

Для узла c: I3 + I- I4= 0;  Для узла b: I1-I5+I6=0; Для узла d: -I1-I2+I4=0.

Составим по 2 закону Кирхгофа (m-mj)-(n-1)=6-3=3 уравнения для схемы 2. Для этого произвольно выберем направление обхода  в каждом контуре.

1 контур: I1R1+I5R5+I4R4=E1 ;

2 контур:  -I1R1+I2R2+I6R6=-E1;

3 контур: I3R3-I5R5-I6R6=E3+E\3;

Получим систему уравнений c шестью неизвестными:

                                                    I1R1+I5R5+I4R41;

                                                     -I1R1+I2R2+I6R6=-E1;    

                                                    I3R3-I5R5-I6R6=E3+ Е\3;                                      (1)

                                                    I3+I5-I4=0;

                                                    I1-I5+I6=0;

                                                     -I1-I2+I4=0.           

II. Метод контурных токов (МКТ)                                                                                      

                                                     I1=I11-I22;

                                                     I2=I22;                                             

                                                     I3=I33;                                                                  (2)

                                                     I4=I11 ;

                                                     I5=I11 – I33 ;

                                                     I6=I22-I33;

            Решаем систему  относительно контурных токов

                                              I11(R1+ R4+ R5)-I22R1-I33R5=E1;

                                              -I11R1 +I22(R1+ R2+ R6)-I33R6=-E1;                                                                     

                                              -I11R5 –I22 R6 + I33(R3+ R5+R6)=E3 + Е\3.

Подставляя в получившуюся систему все численные значения известных величин, решаем её методом  Крамера-Копелли

- матрица неизвестных токов; - матрица ЭДС.

 -матрица коэффициентов при неизвестных;

 Найдём определитель матрицы R,  ∆ :   

                                                              

найдём определитель матрицы   ∆1

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