Исследование цепей постоянного тока

МИНИСТЕРСТВО ОБРАЗОВАНИЯ  РОССИЙСКОЙ ФЕДЕРАЦИИ

 

УФИМСКИЙ ГОСУДАРСТВЕННЫЙ АВИАЦИОННЫЙ  ТЕХНИЧЕСКИЙ УНИВЕРСИТЕТ

 

 

 

 

Кафедра теоретических  основ электротехники

 

 

 

 

 

 

РАСЧЕТНАЯ ГРАФИЧЕСКАЯ  РАБОТА

 

ИССЛЕДОВАНИЕ  ЦЕПЕЙ  ПОСТОЯННОГО  ТОКА

 

 

 

 

 

 

 

               Выполнил:

                 студент группы Т28-219

                 Плюхин В.В.

 

 

 

                 Принял:

                

 

 

 

 

 

 

 

 

 

 

УФА – 2002

Вариант №8.

 

Задание: рассчитать токи в схеме  следующим методом контурных  токов;

       проверить законы  Кирхгофа;

               проверить баланс мощности;

               построить потенциальную диаграмму  для контура, содержащего два   

               источника Э.Д.С.           

 

                                             3 

                                   E₄              E₅                                         

                                                                                                    

                         R₄                                  R₅                                 

                I₄                      I₃                             I₅             R₆         

                               E₃               R₃                                              

   J_k1           1                                                     2                         

                                                                                         I₆        

                                   R₁              R₂ 

              J_k2                         I₁ I₂                          J_k3         E₆        

                                                                                                   

                                   E₁                E₂                                         

                                                 4                                                                                                                     

                                                                                                  

              R₁ = 90 Ом                        E₁ = 0 В                       

              R₂ = 40 Ом                        E₂ = 0 В                    J_k1 = 0 A                                                                                                       

                      R₃ = 90 Ом                        E₃ = 0 В                    J_k2 = 5 A                                     

                      R₄ = 40 Ом                        E₄ = -200 В               J_k3 = 0 A 

                      R₅ = 90 Ом                                    E₅ = 450 В

                      R₆ = 30 Ом                        E₆ = 0 В

 

Расчет  методом  контурных  токов.

 

                                                                                  Рассчитаем собственные токи

                            I₄                E₅                                          I₁ = I_a

                                                                  I_c                        I₂ = - I_a + I_c – J_k3 

                R₄                    I₅             R₅                                 I₃ = - I_a + I_b  

                                     I_b                                    R₆             I₄ = I_b

                             I₃            R₃                                             I₅ = -I_b + I_c   

                                                                                             I₆ = I_c 

                        R₁         I_a         I₂                       

                                                            E_u           E₆  

                                  I₁                             

                                                   R₂                    I₆

 

 

Рассчитаем собственные сопротивления  контуров и сопротивления, находящиеся на границе контуров:

 

                          R_a = R₁ + R₂ + R₃ = 40 + 80 + 50 = 170 Ом                         

                          R_b = R₃ + R₄ + R₅ = 50 + 40 + 50 = 140 Ом                         

                          R_c = R₂ + R₅ + R₆ = 80 + 50 + 90 = 220 Ом                         

 

                                 R_ab = R_ba = R₃ = 50 Ом                     

                                 R_ac = R_ca = R₂ = 80 Ом               

                                 R_bc = R_cb = R₅ = 50 Ом                          

 

Рассчитаем сумму Э.Д.С., действующих  в контуре:

 

                               E_a = -E_u = -160 B

                               E_b = -E₅ = -250 B

                               E_c = E_u + E₅ + E₆ = -160 + 250 – 150 = -60 B

 

Cоставим матрицы сопротивлений и Э.Д.С.:

 

                     R_a    R_ab   R_ac             170  -50  -80                    E_a           160

              R = R_ba   R_b    R_bc      =     -50  140   -50         E =    E_b     =   -250

                     R_ca   R_cb   R_c              -80  -50   220                   E_c           -60   

 

Рассчитав определители данных ниже матриц, находим контурные  токи I_a, I_b, I_c по приведенным ниже формулам.

                    R_a    R_ab   R_ac       170  -50  -80    

           D =   R_ba   R_b    R_bc  =   -50   140 -50   = 2965000 

                    R_ca   R_cb   R_c        -80   -50  220                        

   

                    E_a    R_ab   R_ac           160 -50  -80

          D₂ =   E_b   R_b    R_bc    =    -250 140 -50   = -44000

                    E_c    R_cb   R_c           -60  -50  220

                                                      

                                                                                                

                    R_a    E_a   R_ac          170 160  -80

          D₂ =   R_ba  E_b   R_bc    =   -50 -250 -50   = -6100000

                    R_ca  E_c   R_c           -80 -60  220

              

 

 

 

 

                    R_a    R_ab   E_a         170 -50  160

          D₂ =   R_ba  R_b    E_b    =   -50 140 -250  = -2211000

                    R_ca  R_cb   E_c         -80  -50  -60

 

               I_a = D₁/D = -44000/2965000 = -0.01484 А

               I_b = D₂/D = -6100000/2965000 = -2.0573 А

               I_c = D₃/D = -2211000/2965000 = -0.7457 А

 

       I₁ = - 0.01484 А                                                 I₄ = - 2,0573 А    

       I₂ = 1,26914 А                                                   I₅ = 1,3116 А                                                                                                                                                                                                

       I₃ = -2.04246 А                                                 I₆ = -0.7457 A              

 

Проверим I закон Кирхгофа для узла 2:

                   I₂ – I₃ – I₅ + J_k3 = 0 

                   1.26914 + 2.04246 – 1.3166 – 2 = 0

I закон Кирхгофа для узла 2 выполняется.

 

Проверим II закон Кирхгофа для контура 1,2,3.

                   -R₃*I₃ + R₅*I₅ – R₄*I₄ = E₅

                   50*2.04246 + 50*1.3116 + 40*2.0573 = 249.25 ~ E₅ = 250

II закон Кирхгофа для контура 1,2,3 выполняется. 

 

Баланс мощностей.

 

P_ист. = I₁²*R₁ + I₂²*R₂ + I₃²*R₃ + I₄²*R₄ + I₅²*R₅ + I₆²*R₆ = 642.808 Вт

 

P_потр. = E₅*I₅ + E₆*I₆ + J_k3*j₂ = 642.815 Вт

 

Р_ист. ~ Р_потр.

 

Баланс мощностей сходится.

 

Потенциальная диаграмма  контура 4-1-2-3-4.

 

 

 

 

 

 

 

 

 

 

 

 

φ₄=0

φ_4.1=E₁=-300

φ₁= φ_4.1-I₁R₁=-367.08

φ_1.2= φ₁-I₄R₄=-381.6

φ₂= φ_1.2+E₄=-31.6

φ₃= φ₂- I₅R₅=-122.7

φ₄= φ₃- I₂R₂=-0.02