Построение потенциальной диаграммы
Для построения потенциальной диаграммы, возьмем контур 4-3-2-4 содержащий два источника ЭДС
Рис 6. Контур 4-3-2-4
-
2
2a
4
4a
3
2
R, Ом
0
90
90
100
100
150
φ, В
0
-415,485
-215,485
-201,992
-51,992
0,008
Построим график по полученным данным:
|
J11 |
J22 |
J33 |
|
J44=-JK1 |
U24 |
U24*JK1 |
|
-5,6565 |
-5,9658 |
-4,6165 |
|
-9 |
215,487 |
1939,383 |
|
|
|
|
|
|
|
|
|
k |
Jk |
Rk |
Uk |
Pk=Jk^2*Rk |
Ek |
Pk=EkIk |
|
1 |
-3,0342 |
60 |
-182,052 |
552,3821784 |
0 |
0 |
|
2 |
-1,3493 |
10 |
-13,493 |
18,2061049 |
150 |
-202,395 |
|
3 |
-0,3093 |
60 |
-18,558 |
5,7399894 |
0 |
0 |
|
4 |
-3,3435 |
10 |
-33,435 |
111,7899225 |
0 |
0 |
|
5 |
-1,04 |
50 |
-52 |
54,08 |
0 |
0 |
|
6 |
4,6165 |
90 |
415,485 |
1918,086503 |
200 |
923,3 |
|
|
|
|
|
2660,284698 |
|
2660,288 |
|
|
|
|
|
|
|
|
|
|
SUM(Jk) |
|
|
|
|
|
|
node1 |
0 |
|
|
|
|
|
|
node2 |
0 |
|
|
|
|
|
|
node3 |
0 |
|
|
|
|
|
|
node4 |
0 |
|
|
|
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|
|
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|
|
|
|
|
SUM(Uk) |
SUM(Ek) |
Δ |
|
|
|
|
circuit1 |
-0,007 |
0 |
-0,007 |
|
|
|
|
circuit2 |
150,001 |
150 |
0,001 |
|
|
|
|
circuit3 |
-349,992 |
-350 |
0,008 |
|
|
|
|
|
|
|
|
|
|
|
|
Jr |
R3+R4+R5 |
R11 |
R22 |
R33 |
Rr |
|
|
-9 |
120 |
5 |
4,166667 |
25 |
30,51613 |
|
|
J1 |
|
|
|
|
|
|
|
-3,03421 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
point |
φ |
R |
|
|
|
|
|
2 |
0 |
0 |
|
|
|
|
|
2a |
-415,485 |
90 |
|
|
|
|
|
4 |
-215,485 |
90 |
|
|
|
|
|
4a |
-201,992 |
100 |
|
|
|
|
|
3 |
-51,992 |
100 |
|
|
|
|