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

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      

                                                                            J_k1       

                                                                                                                                                   

                                               b                                                        E₁ 

          I_k3                  R₃                                                     d                                d                                 

                                                                       m       R₁     

                              n                                                             I₁                                                          

                                                                                            

                                                                          R₆                              R₂

                    E₃                                              

                                                                                  a                                    

Изменим схему 1, заменяя в ней источники тока  на ЭДС, учтя что E₃^/=J_k3*R₃, Е^\₁=J_к1*R₁= 0, т.е. J_k1=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 + I₅  - I₄= 0;  Для узла b: I₁-I₅+I₆=0; Для узла d: -I₁-I₂+I₄=0.

Составим по 2 закону Кирхгофа (m-m_j)-(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+I₄R₄=Е₁;

                                                     -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 :